Logical proof of SRT invalidity [Archive] - Page 3

Ziggurat

26th October 2008, 09:48 PM

It is true that invariance needs to be handled and it is also true that I beleieve it is handled incorrectly. Simultaneity can be maintained with invariance of 'c' but you got to throw most of SRT out the window to do it.

No, it cannot. You will notice that my diagrams make no reference to length contraction or time dilation, but only the invariance in c (that is, the slope of the red lines is always the same). The invariance of c requires non-absolute simultaneity. There's simply no way to get those blue lines to intersect the red line at the same t when the blue lines are tilted (meaning we're in a reference frame where they move). It's simple geometry, and there's no getting around it. Whether or not special relativity is the correct description of reality is a question for experiments (which is irrelevant to the question of internal consistency), but simultaneity MUST break down if you enforce constancy of c. If you don't understand why, then you don't understand relativity, Galilean or special.

Both Galilean and special relativity are mathematical theories. They can be expressed in terms of equations which describe coordinate transformations. We know from experiments that Galilean relativity is inconsistent with reality. Every experimental test indicates that special relativity is a correct description of reality. But both theories are internally consistent. One can form other internally consistent theories, but not in which simultaneity and c are both absolute. You are claiming that it is possible. The only way to prove you're correct is to write down a set of transforms (meaning equations representing how coordinates change when switching from one reference frame to another) which do what you claim is possible. But such transforms are not, in fact, even possible. It is exactly the sort of theory which you propose which would be internally inconsistent, not the existing Galilean or special relativity theories.

So here's a little exercise for you: try to figure out what the equations would be to describe Galilean transforms. If this assignment seems difficult, then you don't have the math skills to understand either theory, and no basis on which to make any claims about inconsistencies in a theory you don't understand. Physics is mathematical. If you can't do the math, you can't speak the language, and you can't form theories which mean anything.

ozziemate

27th October 2008, 01:41 AM

No, it cannot. You will notice that my diagrams make no reference to length contraction or time dilation, but only the invariance in c (that is, the slope of the red lines is always the same). The invariance of c requires non-absolute simultaneity. There's simply no way to get those blue lines to intersect the red line at the same t when the blue lines are tilted (meaning we're in a reference frame where they move). It's simple geometry, and there's no getting around it. Whether or not special relativity is the correct description of reality is a question for experiments (which is irrelevant to the question of internal consistency), but simultaneity MUST break down if you enforce constancy of c. If you don't understand why, then you don't understand relativity, Galilean or special.

Both Galilean and special relativity are mathematical theories. They can be expressed in terms of equations which describe coordinate transformations. We know from experiments that Galilean relativity is inconsistent with reality. Every experimental test indicates that special relativity is a correct description of reality. But both theories are internally consistent. One can form other internally consistent theories, but not in which simultaneity and c are both absolute. You are claiming that it is possible. The only way to prove you're correct is to write down a set of transforms (meaning equations representing how coordinates change when switching from one reference frame to another) which do what you claim is possible. But such transforms are not, in fact, even possible. It is exactly the sort of theory which you propose which would be internally inconsistent, not the existing Galilean or special relativity theories.

So here's a little exercise for you: try to figure out what the equations would be to describe Galilean transforms. If this assignment seems difficult, then you don't have the math skills to understand either theory, and no basis on which to make any claims about inconsistencies in a theory you don't understand. Physics is mathematical. If you can't do the math, you can't speak the language, and you can't form theories which mean anything.

The invariance of light can be handled that allows for simultaneity but it means that our understanding of the universe has to shift dramatically and that is another thread... If I felt that invariance was as you say I would not have even started this thread.
Put it this way if someone was to show how invariance could be handled with out breaking simultaneity you would say what? That it is impossible or would you seriously consider it if presented in a professional scientific standard?

you are correct though I must admit not having the language skills is increedibly frustrating but we shall see if we can get some of the way.
the rest of the actual math work will be done not by me but others once there is enough reason to pursue the concepts and abstractions.

Hence the logical proof issue has come up , because this issue requires very little math but very clean reasoning and logic.

ozziemate

27th October 2008, 01:59 AM

Let's shorten that a bit: "Every member of a subset of the HSP are simultaneous". That hasn't changed the meaning from what you said, has it?

Did you forget that the HSP is defined as a set of simultaneous events? If we use that definition, the statement says that "Every member of a subset of this particular set of simultaneous events are simultaneous".

It took almost 500 posts to find a way to state your starting assumption in an unambiguous way. If it's starting to look like it's going to take 500 more to find a way to make you see that it's an empty statement, then I won't be around much longer either.

OK so we have defined simultaneity of observer/light in the HSP with a zero duration.

next agreement needed before proceeding.

1] Do we also agree that regardless of what zero moment of time in all of eternity we take our zero duration snap shot simultaneity of observer/photon is maintained in that zero duration moment [hsp] ?
2] If agreed how many hsp's do you think could exist in all of eternity given they are zero duration?
-the answer must be infinity but I shall ask you to confirm your agreement again.

I will present a number of logical arguments to refute non-simultaneity once this is agreed to.
note : if you all could allow me to just present this case with your agreements along the way we shall get this over with more quickly. Avoid making assumptions and speculations about what I am going to do next and let it unfold.

Tubbythin

27th October 2008, 02:43 AM

The invariance of light can be handled that allows for simultaneity but it means that our understanding of the universe has to shift dramatically and that is another thread... If I felt that invariance was as you say I would not have even started this thread.
Put it this way if someone was to show how invariance could be handled with out breaking simultaneity you would say what? That it is impossible or would you seriously consider it if presented in a professional scientific standard?

you are correct though I must admit not having the language skills is increedibly frustrating but we shall see if we can get some of the way.
the rest of the actual math work will be done not by me but others once there is enough reason to pursue the concepts and abstractions.

Hence the logical proof issue has come up , because this issue requires very little math but very clean reasoning and logic.

You don't seem to have really read Ziggurat's post. It isn't that we don't know how to devise a universe in which invariant c and absolute simultaneity are possible, it is actually mathematically impossible.
The only way you could achieve it is if you either:
a)Have a universe in which speed isn't equal to the rate of change of distance. But this internally inconsistent and nonsensensical, given that c is defined as the rate of change of distance of light.
b)Reinvent maths. Completely. Such that statements like:
1+1=3-1 are false.

ozziemate

27th October 2008, 02:55 AM

You don't seem to have really read Ziggurat's post. It isn't that we don't know how to devise a universe in which invariant c and absolute simultaneity are possible, it is actually mathematically impossible.
The only way you could achieve it is if you either:
a)Have a universe in which speed isn't equal to the rate of change of distance. But this internally inconsistent and nonsensensical, given that c is defined as the rate of change of distance of light.
b)Reinvent maths. Completely. Such that statements like:
1+1=3-1 are false.
I shall wait until after the logical proofs are presented before talking to much about this issue but I can assure you you are incorrect in saying it is impossible yet maintain 'c' and invariance and mathematical integrity.
It only a question of what you are applying invariance to that provides the clue.
Re: the proofs:
It should be pretty obvious what One of those proofs is going to be... the other is not so obvious.

ozziemate

27th October 2008, 03:37 AM

hmmmm all quiet in here:eek:

Reality Check

27th October 2008, 04:30 AM

It's starting to look more and more like "if x=x then SR is wrong". :)
Have a look at the Logical proof of ZPT invalidity thread (http://forums.randi.org/showthread.php?t=127275) to see the absurdity of ozziemate's 'logical proof'.

Reality Check

27th October 2008, 04:35 AM

I shall wait until after the logical proofs are presented before talking to much about this issue but I can assure you you are incorrect in saying it is impossible yet maintain 'c' and invariance and mathematical integrity.
It only a question of what you are applying invariance to that provides the clue.
Re: the proofs:
It should be pretty obvious what One of those proofs is going to be... the other is not so obvious.
So you start a thread called "Logical proof of SRT invalidity" and now you admit that you don't have one?

If you still do not then get you will be demoted from crackpot to troll.
If you do then present it.

ozziemate

27th October 2008, 04:50 AM

So you start a thread called "Logical proof of SRT invalidity" and now you admit that you don't have one?

If you still do not then get you will be demoted from crackpot to troll.
If you do then present it.
Dear readers,
It is because of posts like the one above that this thread has run for as long as it has. I have been attempting to explain the proofs needed but due to constant harassment and flaming by certain members of the board this thread has taken over 500 pages to get to a point where it may very well conclude successfully either one way or the other.
Reality Check has started a new thread that claims no proofs have yet been presented yet he has failed to read post #509, which is awaiting agreement before the proofs are finally presented.
In fact the proof #1 is already easily anticipated the 2nd proof is not quite so obvious.
Invariance issues can be accommodated and simultaneity can be maintained just by changing what it is exactly we apply invariance to.

Unfortunately some members of the board are so convinced that SRT is impossible to be wrong that they are unable to open their eyes to the possibility that the application of reality based data is being incorrectly applied.

We shall await an outcome from post number #509
and then proceed if we can...

As to the other thread that Reality check has created due to his paranoia, I shall post the proofs there as well when agreement is finally achieved and no doubt at many forums around the globe to have them tested again and again until they no longer need testing.
If it can be explained reasonably and rationally why the proofs are not valid after they are presented and not before as poster Reality Check wishes to do, then and only then will I admit defeat and apologies for testing your patience.

ozziemate

27th October 2008, 05:13 AM

Dancing David

27th October 2008, 05:14 AM

That's as good as it gets. It is an experimental fact that the laws of physics are boost and rotation invariant. Those two symmetries (really there are six, since one can rotate or boost around any of the three axes) form the Lorentz group and imply many things, including time dilation, relativity of simultaneity, and Lorentz contraction.

Why the laws of physics are Lorentz invariant is another question, and one better not discussed in this thread (because it's off-topic). The topic was supposed to be a logical proof that SRT is invalid, not an argument over why it is valid.

What! Reality?

Who needs reality!

Dancing David

27th October 2008, 05:16 AM

Dancing David

27th October 2008, 05:19 AM

Great, then we know what your starting assumption is. The next step for you is to realize that it implies nothing.

If you're asking how to draw it in a spacetime diagram, this is the answer:

1. Draw a point. This is the event where the photons are emitted.
2. Draw a a straight vertical line through the point. This is the world line of the light source.
3. Draw a cone with the tip at the emission event, and the rest of the cone located above that point. The cone must be pointing straight down, and the angle between the surface of the cone and a horizontal plane should be 45 degrees. The cone represents the motion of photons away from the source.
4. Draw a horizontal plane that intersects the cone somewhere (anywhere) above the emission event. This is your HSP.
5. Draw the circle where the plane intersects the cone. This circle represents the photon sphere at a specific time.

Now consider your statement (as it stands after our discussion): "If the points where the photon world lines intersect the HSP are simultaneous, then all photon detection events in the HSP are simultaneous".

The set of points where photon world lines intersect the HSP is the circle you drew.

You can only detect a photon at a location in space where there actually is a photon. Those locations in the HSP are represented by the circle. So the photon detection events in the HSP must be a subset of the circle.

This means that what the statement is really saying is this: "If the points in the circle are all simultaneous, then the points in the circle that represent detection events are all simultaneous".

This statement is true, but it's useless since it's of the form "if P then Q" where P and Q are both true by definition.

Thanks Fredrick, I appreciate your efforts.

ozziemate

27th October 2008, 05:21 AM

It's useful for us to who are trying to discuss this with you to know what your starting assumption is. I just meant that your starting assumption doesn't really say anything. (I explained why).

I described how to draw this in a spacetime diagram, and I showed how to express the statement using references to the diagram. Isn't that sufficient? What more do you want? We could choose to let the time coordinate be 0 in the HSP. Then we have:

\mbox{HSP=}\{(t,x,y,z)\in\mathbb R^4|t=0\}

\mbox{Photon sphere at t=0\ \ \ \ }=\{(0,x,y,z)\in\mbox{HSP}|x^2+y^2+z^2=c^2T^2\}

where T is the time that has passed since the emission event.

Is that the sort of thing you'd like to see? I don't think this sort of mathematical notation is very helpful in a discussion like this. The spacetime diagram that I described represents the same things in a way that's much easier to understand (especially for you), so maybe you should draw it and post it.
I just noticed your constructions and thank you very sincerely for your effort...
They will be tested I can assure you if the proofs get a chance to get up.

because possibly they will be used eventually in showing part of the math needed to show the correct way invariance needs to be handled.

ozziemate

27th October 2008, 05:23 AM

Thanks Fredrick, I appreciate your efforts.
DD when you finally catch up with post #509 let us all know ok....

ozziemate

27th October 2008, 05:24 AM

deleted

Dancing David

27th October 2008, 05:24 AM

Dear readers,
It is because of posts like the one above that this thread has run for as long as it has. I have been attempting to explain the proofs needed but due to constant harassment and flaming by certain members of the board this thread has taken over 500 pages to get to a point where it may very well conclude successfully either one way or the other.
Reality Check has started a new thread that claims no proofs have yet been presented yet he has failed to read post #509, which is awaiting agreement before the proofs are finally presented.
In fact the proof #1 is already easily anticipated the 2nd proof is not quite so obvious.
Invariance issues can be accommodated and simultaneity can be maintained just by changing what it is exactly we apply invariance to.

Unfortunately some members of the board are so convinced that SRT is impossible to be wrong that they are unable to open their eyes to the possibility that the application of reality based data is being incorrectly applied.

We shall await an outcome from post number #509
and then proceed if we can...

As to the other thread that Reality check has created due to his paranoia, I shall post the proofs there as well when agreement is finally achieved and no doubt at many forums around the globe to have them tested again and again until they no longer need testing.
If it can be explained reasonably and rationally why the proofs are not valid after they are presented and not before as poster Reality Check wishes to do, then and only then will I admit defeat and apologies for testing your patience.

So in other words you can't make critical arguments, can't think critically, can't discuss critiques of your statements, and you much less can't make a cohernet theory out of your ramblings...

Gosh, that some suprise!

Now it is time to say "If you won't play the way i want, I will take my ball and go home."

You need help Ozziemate, this onsession of your is unhealthy, you have othet choices that would probably be better for you.

So please keep blaming other people for your inability to articulate a coherent argument.

Check out some of Zeuzzz's PC threads in you want to see what it looks like when this forum gets a little rude, you hav been babied and pampered, and still you whine!

ozziemate

27th October 2008, 05:26 AM

[moved to later in the thead to keep in on topic]

Dancing David

27th October 2008, 05:33 AM

DD when you finally catch up with post #509 let us all know ok....

Let me know when you decide to come join us in reality, you are stuck in your head.

I pointed out earlier, in fact at the beggining of the original thread, the inadequate nature of your theory, it is all in mental space, you have no experimental verification of a single statement you have made, it is all NeoPlatonic twaddle of the first order.

The universe does not give a dead rodent's posterior for logic and reason, we can never know what mechanisms underlie the workings of the universe.

But you like to live in your convinient little head space and it is farther removed from reality everyday.

I will spell it out for you but i know you are too wedded to your own ego to understand:

The universe does not care about human logic and reason, it does as it will.

here it is
http://forums.randi.org/showpost.php?p=4087250&postcount=30
http://forums.randi.org/showpost.php?p=4087255&postcount=31

ozziemate

27th October 2008, 05:33 AM

Summary of process in debate:

Step 1,

Confirm agreement or disagreement with the tautology that all observer/photon events are simultaneous in a zero duration moment.

Step 2,
Ensure that this is clear and agreed to by testing, using extended logic to include eternity.

Step 3,
Present proofs in clear simple terms with out need for mathematics.

Step 4,
Present mathematics if available to confirm proofs in a way that can be recognised by mathematicians and physicists. [ commonly ]

ozziemate

27th October 2008, 05:36 AM

Let me know when you decide to come join us in reality, you are stuck in your head.

I pointed out earlier, in fact at the beggining of the original thread, the inadequate nature of your theory, it is all in mental space, you have no experimental verification of a single statement you have made, it is all NeoPlatonic twaddle of the first order.

The universe does not give a dead rodent's posterior for logic and reason, we can never know what mechanisms underlie the workings of the universe.

But you like to live in your convinient little head space and it is farther removed from reality everyday.

I will spell it out for you but i know you are too wedded to your own ego to understand:

The universe does not care about human logic and reason, it does as it will.

here it is
http://forums.randi.org/showpost.php?p=4087250&postcount=30
http://forums.randi.org/showpost.php?p=4087255&postcount=31
DD ...relax man, have a beer , you are starting to rant.

we can never know what mechanisms underlie the workings of the universe.
absolutely correct if you stick with SRT. and believe what you just said.
absolutely wrong if you can think past your own little box.

ozziemate

27th October 2008, 05:39 AM

Copy of post #509 to keep this thread on topic:

OK so we have defined simultaneity of observer/light in the HSP with a zero duration.

\mbox{HSP=}\{(t,x,y,z)\in\mathbb R^4|t=0\}

\mbox{Photon sphere at t=0\ \ \ \ }=\{(0,x,y,z)\in\mbox{HSP}|x^2+y^2+z^2=c^2T^2\}
contruction provided by poster Fredrik with many thanks

Next agreement needed before proceeding.

1] Do we also agree that regardless of what zero moment of time in all of eternity we take our zero duration snap shot simultaneity of observer/photon is maintained in that zero duration moment [hsp] ?
2] If agreed how many hsp's do you think could exist in all of eternity given they are zero duration?
-the answer must be infinity but I shall ask you to confirm your agreement again.

I will present a number of logical arguments to refute non-simultaneity once this is agreed to.
note : if you all could allow me to just present this case with your agreements along the way we shall get this over with more quickly. Avoid making assumptions and speculations about what I am going to do next and let it unfold.

Thabiguy

27th October 2008, 05:51 AM

1] Do we also agree that regardless of what zero moment of time in all of eternity we take our zero duration snap shot simultaneity of observer/photon is maintained in that zero duration moment [hsp] ?
In any single chosen reference frame, yes.
2] If agreed how many hsp's do you think could exist in all of eternity given they are zero duration?
-the answer must be infinity but I shall ask you to confirm your agreement again.
Yes.
note : if you all could allow me to just present this case with your agreements along the way we shall get this over with more quickly. Avoid making assumptions and speculations about what I am going to do next and let it unfold.
Go ahead then.

ozziemate

27th October 2008, 05:54 AM

In any single chosen reference frame, yes.

Given that we are talking about a HSP that has zero duration is there really a need to qualify it down to a single reference frame, after all we have just defined the entire hsp [ all reference frames ~aka observer/light events] as simutaneous not just one reference frame?
qualifying to a single reference frame means that your "yes" is in fact a "no" ..you do not agree to:
1] Do we also agree that regardless of what zero moment of time in all of eternity we take our zero duration snap shot simultaneity of observer/photon is maintained in that zero duration moment [hsp] ?

by including photon/observer simultaneity [ a tautology ] the HSP has been redefined.

drkitten

27th October 2008, 06:00 AM

The Man

27th October 2008, 06:05 AM

Light as time (or even just a clock).

On a normal “dial” clock face the second hand completes one cycle (a minute) every 60 seconds. We can relate the change in position of the second hand as an angular velocity which is not dependent on the physical size or scale of the clock. Angular velocity w is measured in radians per second. With 2p radians per cycle the angular velocity of the second hand becomes p/30 or 0.104 Radians per Second. Frequency f is also related to angular velocity w as f = w/2p, this give us a frequency of 1 cycle every 60 seconds or 0.01667 cycles per second. Using the speed of light c and the relationship of wavelength to velocity over frequency we can come up with a light (or EM radiation) wavelength for a standard minute as c/0.01667 or about 1,800,0000,000 meters per cycle, far above the visual spectrum that we refer to as light. The real problem comes when we consider the rate at which light or EM radiation represents time as dependent on frequency and wavelength. Even without evoking SRT, the Doppler Effect, or the dependence of frequency on relative motion, is readily apparent. So we have both that the different frequencies of any given EM radiation represents clocks moving at different angular velocities (different measures of time) as well as the dependency of that frequency on relative motion, without evoking SRT. Light (or EM radiation) is clearly not an absolute but a relative temporal reference just as any standard clock is not an absolute temporal reference, even without evoking SRT.

ozziemate

27th October 2008, 06:06 AM

Yes, you need to qualify it down to a single reference frame.

... and that seems to be the point you don't get. Your understanding of "simultaneous" is neither what theory predicts, nor what experiment reveals.
ahh but it has taken 500 + posts to show what simultaneity in the hsp is...

Observer/Photon simultaneity has been shown to be a tautology, in fact I have been ridiculed heavilly for asking such innane questions.
tautology: you canot have an observer ebvent wiyth out a photon.
all I am doing is formalising this relationship as being simultaneous across trhe entire HSP at zero duration. another tautology, and ridiculed earlier as innane and worthlless.

Now you are saying that it isn't worthless but extremely important which is why this thread has happened to begin with.

Thabiguy

27th October 2008, 06:08 AM

Given that we are talking about a HSP that has zero duration is there really a need to qualify it down to a single reference frame, after all we have just defined the entire hsp [ all reference frames ~aka observer/light events] as simutaneous not just one reference frame?
We have done no such thing.

The HSP (hypersphere of the present), as defined in the equation you have just used, is defined within a single reference frame.

The HSP does in no way include "all reference frames".

Your statement "reference frames ~aka observer/light events" makes no sense. Reference frames are in no way the same as observer/light events. In fact, they could hardly be more different.

A reference frame is a coordinate system in which you are providing your measurements of events, such as observer/light events.

I'll try to explain in simpler terms:

A: "This observer/light event has coordinates x=1, y=5, z=3, t=8."
B: "What the hell do those numbers mean?"
A: "Well, t is the number of seconds since I snapped my fingers, and x,y,z are the distances along respective axes from the Earth."

In the first line, A has described an observer/light event, in the third line, she described a reference frame. A reference frame defines how coordinates are interpreted.

ozziemate

27th October 2008, 06:09 AM

ozziemate

27th October 2008, 06:11 AM

We have done no such thing.

The HSP (hypersphere of the present), as defined in the equation you have just used, is defined within a single reference frame.

The HSP does in no way include "all reference frames".

Your statement "reference frames ~aka observer/light events" makes no sense. Reference frames are in no way the same as observer/light events. In fact, they could hardly be more different.

A reference frame is a coordinate system in which you are providing your measurements of events, such as observer/light events.

I'll try to explain in simpler terms:

A: "This observer/light event has coordinates x=1, y=5, z=3, t=8."
B: "What the hell do those numbers mean?"
A: "Well, t is the number of seconds since I snapped my fingers, and x,y,z are the distances along respective axes from the Earth."

In the first line, A has described an observer/light event, in the third line, she described a reference frame. A reference frame defines how coordinates are interpreted.
well then we shall just have to call it PHS instead of HSP....to get past the SRT's dependancy on reference frame use.
every observer in this PHS is effectively inertial as the entire shebang is zero duration and all observers are static or stationary. so in a sense HSP is still valid.
Does that help?

drkitten

27th October 2008, 06:11 AM

Light as time (or even just a clock).

What????

On a normal “dial” clock face the second hand completes one cycle (a minute) every 60 seconds. We can relate the change in position of the second hand as an angular velocity which is not dependent on the physical size or scale of the clock. Angular velocity w is measured in radians per second. With 2p radians per cycle the angular velocity of the second hand becomes p/30 or 0.104 Radians per Second. Frequency f is also related to angular velocity w as f = w/2p, this give us a frequency of 1 cycle every 60 seconds or 0.01667 cycles per second. Using the speed of light c and the relationship of wavelength to velocity over frequency we can come up with a light (or EM radiation) wavelength for a standard minute as c/0.01667 or about 1,800,0000,000 meters per cycle, far above the visual spectrum that we refer to as light.

This is mostly irrelevant, and the part that isn't irrelevant is gibberish.

The real problem comes when we consider the rate at which light or EM radiation represents time as dependent on frequency and wavelength.

Since light or EM radiation doesn't represent time as dependent on frequency and wavelength, this "real problem" is trivial to solve.

So we have both that the different frequencies of any given EM radiation represents clocks moving at different angular velocities (different measures of time) as well as the dependency of that frequency on relative motion, without evoking SRT. Light (or EM radiation) is clearly not an absolute but a relative temporal reference just as any standard clock is not an absolute temporal reference, even without evoking SRT.

Gibberish. If I have a 24-hour clock instead of a 12-hour clock, the hands of the clock move exactly half the speed and therefore cover half the distance per unit time. Either way, the measurement of time is absolute regardless of the units. The length of a song is a constant when measured in seconds, regardless of whether the song has 140 bpm or 90 bpm.

You're making the same argument here that I would if I said that "length" is not absolute because I can measure something as being 12 inches long instead of 1 foot, and therefore the same object has two different lengths. And for exactly this reason, you're wrong.

Time is relative, but that has nothing to do with different measurements; if Newtonian mechanics were actually correct, then time would be absolute despite your being able to measure it with any units you choose.

drkitten

27th October 2008, 06:14 AM

ahh but it has taken 500 + posts to show what simultaneity in the hsp is...

No. I knew what it was before post #1 -- and you still don't know what it is.

Observer/Photon simultaneity has been shown to be a tautology,

No, it has not.

in fact I have been ridiculed heavilly for asking such innane questions.

And rightly so.

all I am doing is formalising this relationship as being simultaneous across trhe entire HSP at zero duration.

And doing it incorrectly.

Now you are saying that it isn't worthless but extremely important

No, I'm saying that it's worthless.

More specifically, your incorrect understanding of both SR theory and experiment is worthless.

drkitten

27th October 2008, 06:16 AM

every observer in this PHS is effectively inertial as the entire shebang is zero duration and all observers are static or stationary.

No.

Does that help?

Not at all. It simply makes another incorrect assumption; that observers are "stationary" in the limiting case of zero duration in different inertial frames.

(It also makes another incorrect assumption that all frames are inertial in the same limiting case. Basically, you've never heard of calculus or Zeno's paradox, have you?)

ozziemate

27th October 2008, 06:16 AM

The HSP (hypersphere of the present), as defined in the equation you have just used, is defined within a single reference frame.

just to clarify my point.
If every possible observer in the HSP is stationary due to the HSP being zero in duration then the observers must all be inertial or sharing the same t=0 thus d=0?

Therefore the hsp being described is a single reference frame

ozziemate

27th October 2008, 06:18 AM

No.

Not at all. It simply makes another incorrect assumption; that observers are "stationary" in the limiting case of zero duration in different inertial frames.

(It also makes another incorrect assumption that all frames are inertial in the same limiting case. Basically, you've never heard of calculus or Zeno's paradox, have you?)

so if t= 0 what does d=?

drkitten

27th October 2008, 06:19 AM

so if t= 0 what does d=?

An excuse to take calculus.

ozziemate

27th October 2008, 06:19 AM

Basically, you've never heard of calculus or Zeno's paradox, have you?)
Actually I am a little familiar with it and your point being?

ozziemate

27th October 2008, 06:21 AM

An excuse to take calculus.
rather useful post that....hmmmmm.....
t=0 and d= (excuse to take calculus):)

drkitten

27th October 2008, 06:22 AM

Actually I am a little familiar with it and your point being?

My point being that there is a difference between zero and an infinitesimal, and that the whole point of calculus was to provide a method of analysis of simple (classical) problems of motion that didn't result in division by zero errors when trying to calculate instantaneous rates of change.

ozziemate

27th October 2008, 06:24 AM

My point being that there is a difference between zero and an infinitesimal, and that the whole point of calculus was to provide a method of analysis of simple (classical) problems of motion that didn't result in division by zero errors when trying to calculate instantaneous rates of change.

ahh now you are talking, I thought the use of infinitesimals would finally show up...

So you would say that t= infinitesiimal and d= infiniteseimal....hmmmm..this is going to be good because you will actually be able to prove the existance of absolute zero [ inverse sphere mentioned in another thread.]

ozziemate

27th October 2008, 06:25 AM

so now we are saying that the hsp has infinitesimal duration yes?

Tubbythin

27th October 2008, 06:29 AM

just to clarify my point.
If every possible observer in the HSP is stationary due to the HSP being zero in duration then the observers must all be inertial or sharing the same t=0 thus d=0?

Therefore the hsp being described is a single reference frame

Just because you have a "snapshot" does not mean the objects were all stationary. If you take a picture of a moving car, does it affect the speed at which the car is moving?
The fact that the HSP is "zero in duration" has no effect on the relative velocities of the observers.

Thabiguy

27th October 2008, 06:31 AM

just to clarify my point.
If every possible observer in the HSP is stationary due to the HSP being zero in duration then the observers must all be inertial or sharing the same t=0 thus d=0?

The observers are not stationary due to the HSP being zero in duration.

Within a zero-duration snapshot, it becomes meaningless to talk about stationary and moving things. But this does not mean that all objects become stationary, i.e. that they will not change their position as time changes.

Stationary means: not changing its position as time changes.

The word inertial means something completely different from stationary, and is not relevant right now.

Reality Check

27th October 2008, 06:32 AM

well then we shall just have to call it PHS instead of HSP....to get past the SRT's dependancy on reference frame use.
every observer in this PHS is effectively inertial as the entire shebang is zero duration and all observers are static or stationary. so in a sense HSP is still valid.
Does that help?

Actually it does.
You are making yet another assumption: that all observers are static or stationary (I assume in space). It is not an assumption that is made in SRT. So now you are describing a theory that is not SRT. Please feel free to logically prove that that theory (where observers cannot move) is invaild.

Oddly enough I have just observed my typing (I am an observer!), I have just moved (the Earth has rotated!) and time has passed (my computer clock is ticking away!). Are you an observer, do you move and does time pass?

FYI: Reference frames (http://en.wikipedia.org/wiki/Frame_of_reference) are related to the observers and their state of motion. SRT uses them because it uses observers. Reference frames are also used in classical mechanics.

ozziemate

27th October 2008, 06:32 AM

Just because you have a "snapshot" does not mean the objects were all stationary. If you take a picture of a moving car, does it affect the speed at which the car is moving?
The fact that the HSP is "zero in duration" has no effect on the relative velocities of the observers.

I suppose you are going to answer the following the same way:

t=0 then d= ( an excuse to learn calculus)

tubby normally what is 'd' if t=0?

ozziemate

27th October 2008, 06:34 AM

Actually it does.
You are making yet another assumption: that all observers are static or stationary (I assume in space). It is not an assumption that is made in SRT. So now you are describing a theory that is not SRT. Please feel free to logically prove that that theory (where observers cannot move) is invaild.

Oddly enough I have just observed my typing (I am an observer!), I have just moved (the Earth has rotated!) and time has passed (my computer clock is ticking away!). Are you an observer, do you move and does time pass?

FYI: Reference frames (http://en.wikipedia.org/wiki/Frame_of_reference) are related to the observers and their state of motion. SRT uses them because it uses observers. Reference frames are also used in classical mechanics.
hey your the math guy you tell me?

what is the value of d (distance) if t (time ) is zero?

Tubbythin

27th October 2008, 06:35 AM

I suppose you are going to answer the following the same way:

t=0 then d= ( an excuse to learn calculus)

tubby normally what is 'd' if t=0?

d is a distance? Between what? (Apologies, I've forgotten)

Thabiguy

27th October 2008, 06:36 AM

what is the value of d (distance) if t (time ) is zero?

If change in time is zero, then change in distance is zero.
(ETA: Within a single reference frame. ;))

ozziemate

27th October 2008, 06:39 AM

If change in time is zero, then change in distance is zero.
of course it is and thanks.....
t=0 therefore d must = 0 another tautology.
And if the HSP has "infinitesimal" duration SRT is stuffed as well if I am not mistaken. [I may actually be incorrect.:confused:]

Tubbythin

27th October 2008, 06:39 AM

Tubbythin

27th October 2008, 06:41 AM

ozziemate

27th October 2008, 06:42 AM

Steady-on. We haven't made it quite that far... yet.
you are kidding me... it's post 558zillion and it's time for bed and we aint there yet....awww:(:D

Thabiguy

27th October 2008, 06:42 AM

And if the HSP has "infinitesimal" duration SRT is stuffed as well if I am not mistaken. [I may actually be incorrect.:confused:]
Let's not talk about infinitesimal stuff right now.

If it needs be, then I will explain calculus to you. But if I do that, I will choose the epsilon-delta approach. No infinitesimals.

ozziemate

27th October 2008, 06:45 AM

Huh? What is it you think you've proved?

Copy of post #509 to keep this thread on topic:

OK so we have defined simultaneity of observer/light in the HSP with a zero duration.

\mbox{HSP=}\{(t,x,y,z)\in\mathbb R^4|t=0\}

\mbox{Photon sphere at t=0\ \ \ \ }=\{(0,x,y,z)\in\mbox{HSP}|x^2+y^2+z^2=c^2T^2\}
contruction provided by poster Fredrik with many thanks

Next agreement needed before proceeding.

1] Do we also agree that regardless of what zero moment of time in all of eternity we take our zero duration snap shot simultaneity of observer/photon is maintained in that zero duration moment [hsp] ?
2] If agreed how many hsp's do you think could exist in all of eternity given they are zero duration?
-the answer must be infinity but I shall ask you to confirm your agreement again.

I will present a number of logical arguments to refute non-simultaneity once this is agreed to.
note : if you all could allow me to just present this case with your agreements along the way we shall get this over with more quickly. Avoid making assumptions and speculations about what I am going to do next and let it unfold.

<>

So now we have to rule out infinitesimal duration of the HSP...maybe drKitten would like to respond.

Tubbythin

27th October 2008, 06:47 AM

I suppose you are going to answer the following the same way:

t=0 then d= ( an excuse to learn calculus)

tubby normally what is 'd' if t=0?

Note that distance travelled in time t = 0 according to a single observer in a single inertial reference frame is 0.

Since speed = distance/time
speed=0/0.
Which is mathematically undefined. And does not support the following:

"If every possible observer in the HSP is stationary due to the HSP being zero in duration"

ozziemate

27th October 2008, 06:48 AM

Let's not talk about infinitesimal stuff right now.

If it needs be, then I will explain calculus to you. But if I do that, I will choose the epsilon-delta approach. No infinitesimals.
can I ask you then that if the HSP has zero duration then all locations on that hsp must be simultaneous?

drkitten

27th October 2008, 06:50 AM

of course it is and thanks.....
t=0 therefore d must = 0 another tautology.
And if the HSP has "infinitesimal" duration SRT is stuffed as well if I am not mistaken. [I may actually be incorrect.:confused:]

No. Because infinitesimals (such as dx) are non-zero (as they are, by definition), the value 2*dx is different than dx itself (and if dx is positive, as by standard convention, 3dx > 2dx > dx > 0.5(dx).

So if t is infinitesimal, then a fast-moving object covers a longer infinitesimal distance than a slow-moving object, exactly as common sense and observation suggest.

Tubbythin

27th October 2008, 06:50 AM

can I ask you then that if the HSP has zero duration then all locations on that hsp must be simultaneous?

Simultaneous according to who?

ETA: I'm not really sure how a position can have simultaniety with another position for that matter.

ozziemate

27th October 2008, 06:51 AM

Note that distance travelled in time t = 0 according to a single observer in a single inertial reference frame is 0.

Since speed = distance/time
speed=0/0.
Which is mathematically undefined. And does not support the following:

"If every possible observer in the HSP is stationary due to the HSP being zero in duration"
so are you saying that the HSP is undefinable or that the HSP's duration is >0?

drkitten

27th October 2008, 06:51 AM

can I ask you then that if the HSP has zero duration then all locations on that hsp must be simultaneous?

... for the observer for whom the HSP has zero duration.

But a differently moving observer will not observe the HSP as having zero duration.

ozziemate

27th October 2008, 06:52 AM

Simultaneous according to who?
to all locations...sheesh!

Tubbythin

27th October 2008, 06:53 AM

to all locations...sheesh!

Locations are not observers. Simultaneity must be with respect to one or more observers.

ozziemate

27th October 2008, 06:54 AM

... for the observer for whom the HSP has zero duration.

But a differently moving observer will not observe the HSP as having zero duration.

thanks for posting drkitten you obviously have a really good working knowledge....can you explain what you mean in more detail by the above....but a differently moving observer will not.....

you may actually be able to provide a targeted proof killer....

ozziemate

27th October 2008, 06:56 AM

Locations are not observers. Simultaneity must be with respect to one or more observers.
Copy of post #509 to keep this thread on topic:

OK so we have defined simultaneity of observer/light in the HSP with a zero duration.

\mbox{HSP=}\{(t,x,y,z)\in\mathbb R^4|t=0\}

\mbox{Photon sphere at t=0\ \ \ \ }=\{(0,x,y,z)\in\mbox{HSP}|x^2+y^2+z^2=c^2T^2\}
contruction provided by poster Fredrik with many thanks

Next agreement needed before proceeding.

1] Do we also agree that regardless of what zero moment of time in all of eternity we take our zero duration snap shot simultaneity of observer/photon is maintained in that zero duration moment [hsp] ?
2] If agreed how many hsp's do you think could exist in all of eternity given they are zero duration?
-the answer must be infinity but I shall ask you to confirm your agreement again.

I will present a number of logical arguments to refute non-simultaneity once this is agreed to.
note : if you all could allow me to just present this case with your agreements along the way we shall get this over with more quickly. Avoid making assumptions and speculations about what I am going to do next and let it unfold.

<>

this exactly what we had confirmed earlier observer/photon simultaneity...
I was using locations as a way of abreviating the issue

drkitten

27th October 2008, 06:56 AM

OK so we have defined simultaneity of observer/light in the HSP with a zero duration.

\mbox{HSP=}\{(t,x,y,z)\in\mathbb R^4|t=0\}

\mbox{Photon sphere at t=0\ \ \ \ }=\{(0,x,y,z)\in\mbox{HSP}|x^2+y^2+z^2=c^2T^2\}
contruction provided by poster Fredrik with many thanks

Okay. The problem is that unless two observers share the same reference frame, their perceptions of x, y, z --- and t, for that matter -- will differ, and therefore their solutions to the "photon sphere" will differ.

1] Do we also agree that regardless of what zero moment of time in all of eternity we take our zero duration snap shot simultaneity of observer/photon is maintained in that zero duration moment [hsp] ?

No.

2] If agreed how many hsp's do you think could exist in all of eternity given they are zero duration?

Not relevant as it's a badly posed question.

So now we have to rule out infinitesimal duration of the HSP...maybe drKitten would like to respond.

Yes. You won't be able to do that without running into Zeno's paradox.

But beyond that, you're making the invalid assumption that length is frame-invariant, which (of course) we know from experiment is not true.

Tubbythin

27th October 2008, 06:56 AM

so are you saying that the HSP is undefinable or that the HSP's duration is >0?

Neither. I'm saying that you cannot determine an observers velocity over a time span of 0 (this should be obvious). Therefore you're assertion that every possible observer in the HSP must be stationary is false.

Ziggurat

27th October 2008, 06:56 AM

Put it this way if someone was to show how invariance could be handled with out breaking simultaneity you would say what? That it is impossible or would you seriously consider it if presented in a professional scientific standard?

I'd consider it if it was framed as mathematical expressions describing the transforms so that it could be tested. Your ideas (or what I've seen of them) cannot be tested in any meaningful way, and they aren't even really coherent.

you are correct though I must admit not having the language skills is increedibly frustrating but we shall see if we can get some of the way.

Apparently we cannot. The problem is not that you can't express your ideas, but that you cannot understand the ideas that already exist. I had hoped that the simple geometry of my figure would be enough to show you why simultaneity MUST break (as in, it's a geometrical proof akin to demonstrating that the hypoteneus of a right triangle is longer than either side), but apparently it is not.

the rest of the actual math work will be done not by me but others once there is enough reason to pursue the concepts and abstractions.

No, it will not, because it cannot be done. And anybody with the math skills is going to figure that out soon enough.

Hence the logical proof issue has come up , because this issue requires very little math but very clean reasoning and logic.

Look at my diagram again: those tilted observer lines CANNOT intersect the light cone at the same time if the light cone doesn't tilt. That is the ONLY thing you need to demonstrate that simultaneity cannot be absolute if c is invariant. It's really quite simple, and you still don't get it.

The fault is not with us. It's with you.

Fredrik

27th October 2008, 07:01 AM

OK so we have defined simultaneity of observer/light in the HSP with a zero duration.

No, what we have done is to define the HSP as a hypersurface of events that are simultaneous in an inertial frame in which the light source is stationary.

We have not agreed on a definition of simultaneity.

I posted the standard definition in #436 and #446, but your only comment was that there's "math" in #436. (I used the fact that the number at the center of the interval from A to B is A+(B-A)/2, i.e. the smaller number plus half the distance between them).

1] Do we also agree that regardless of what zero moment of time in all of eternity we take our zero duration snap shot simultaneity of observer/photon is maintained in that zero duration moment [hsp] ?
2] If agreed how many hsp's do you think could exist in all of eternity given they are zero duration?
-the answer must be infinity but I shall ask you to confirm your agreement again.

I agree.

Tubbythin

27th October 2008, 07:02 AM

this exactly what we had confirmed earlier observer/photon simultaneity...
I was using locations as a way of abreviating the issue

Which is a nice illustration of why you should learn about a theory before tyring to dispute it. SR includes a 4D construct we call spacetime. The time direction is mutually orthogonal to those of space. Simultaneity implies two or more events that occur at the same height on the time axis. Suggesting that points in space are simultaneous makes no sense.

drkitten

27th October 2008, 07:03 AM

thanks for posting drkitten you obviously have a really good working knowledge....can you explain what you mean in more detail by the above....but a differently moving observer will not.....

you may actually be able to provide a targeted proof killer....

See http://www.spacetimetravel.org/fussball/fussball.html

Basically, what a static observer sees as a sphere, a moving observer will see in various distorted shapes, depending upon her direction and speed of motion. In particular, a moving observer will see the "sphere" of photons as compressed in one direction or another, and will therefore consider the points that are "near" the center of the sphere as being "closer" to the emission point in time, and therefore earlier. So the shape that the stationary observer sees as being spherical and stationary, the moving observer sees as being nonspherical and dispersed in time.

ozziemate

27th October 2008, 07:03 AM

Okay. The problem is that unless two observers share the same reference frame, their perceptions of x, y, z --- and t, for that matter -- will differ, and therefore their solutions to the "photon sphere" will differ.
you are of course assuming that SRT non-simultaneity is valid and true. But Of course it is the issue of non-simultaneity that is in question so to apply this as a counter arguement would be invalid.
If the use of HSP is causing concern I could always just use PHS instead

in fact just assume any reference to HSP as PHS [present hyper surface ] and it make it easier.

Also if all observer/photon events are simultaneious due to zero PHS duration all observer/photon events share the same reference frame do they not? As their relative velocity is zero.

Thabiguy

27th October 2008, 07:04 AM

can I ask you then that if the HSP has zero duration then all locations on that hsp must be simultaneous?

Once more: it's events that get to be simultaneous, not objects or locations.

But, I will let it go in this case, because in the HSP, time is defined, so we could say that every location defines a single unique event taking place at that location and the present time.

So the answer to your question is yes. The time of the HSP applies all over the HSP, so it will always be shared between any two locations.

Within the reference frame in which the HSP was defined.

ozziemate

27th October 2008, 07:06 AM

Once more: it's events that get to be simultaneous, not objects or locations.

But, I will let it go in this case, because in the HSP, time is defined, so we could say that every location defines a single unique event taking place at that location and the present time.

So the answer to your question is yes. The time of the HSP applies all over the HSP, so it will always be shared between any two locations.

Within the reference frame in which the HSP was defined.

and I repeat what I just posted to drkitten "if all observer/photon events are simultaneious due to zero HSP/PHS duration all observer/photon events share the same reference frame do they not? As their relative velocity is zero. ?
after all all events are simultaneous with t=0 and d=0....thus one reference frame

Tubbythin

27th October 2008, 07:08 AM

Also if all observer/photon events are simultaneious due to zero PHS duration all observer/photon events share the same reference frame do they not? As their relative velocity is zero.

No no no. I just explained this!

"Note that distance travelled in time t = 0 according to a single observer in a single inertial reference frame is 0.

Since speed = distance/time
speed=0/0.
Which is mathematically undefined. And does not support the following:

"If every possible observer in the HSP is stationary due to the HSP being zero in duration"

drkitten

27th October 2008, 07:08 AM

If the use of HSP is causing concern I could always just use PHS instead

in fact just assume any reference to HSP as PHS [present hyper surface ] and it make it easier.

The problem is that the "present" varies from observer to observer, and therefore one person's PHS will not necessarily be the same as any other person's PHS.

Also if all observer/photon events are simultaneious due to zero PHS duration all observer/photon events share the same reference frame do they not?

No.

Thabiguy

27th October 2008, 07:08 AM

If the use of HSP is causing concern I could always just use PHS instead

in fact just assume any reference to HSP as PHS [present hyper surface ] and it make it easier.

Whoa, whoa. What is your definition of PHS?

Also if all observer/photon events are simultaneious due to zero PHS duration all observer/photon events share the same reference frame do they not? As their relative velocity is zero.

If you are talking about HPS, then yes, all objects share the same reference frame (of course, because they're described in it), but their relative velocity definitely doesn't have to be zero.

If you're talking about PHS, I have no idea, because I don't know what you mean by PHS.

ozziemate

27th October 2008, 07:22 AM

The problem is that the "present" varies from observer to observer, and therefore one person's PHS will not necessarily be the same as any other person's PHS.

Again I shall repeat the PHS [ which is a non SRT version of the HSP to avoid confusion] has been defined as entirely universaly and absolutely simulatneous in all possible observer/photon events with a zero duration.
this is part of the proof against non-simultaneity which is derived by the use of transforms.
To saythat the simutaneity of the PHS is borken by the very thing in question and yet to be re-confirmed as valid by this debate is an invalid arguemtn and I don't mean to use this as a way of copping out it is in fact an invalid argument.
non-simultaneity is being intially invalidated as a step towards final inavidation by the confirming theat all observer/photon events are simultaneous in a zero duration moment PHS

The first logic proof:
If the PHS is held as universally simultaneous in the zero duartion moment then every PHS possible thoughout eternity is also simultaneous there fore non-simultaneity is impossible. Regardless of issues regarding invariance and so called impossibilites of handling such.
If all phs's eternally are simultaneous universally then it is impossible for non-simultaneity to exist.
An infinite number of phs's can exist in eternity....as they are zero duration and even if infinitesimal duration [ which invalidates SRT any way if I am not mistaken] the same applies an infinite number of phs's can occur in eternity past present and future.

Fredrik

27th October 2008, 07:23 AM

I just noticed your constructions and thank you very sincerely for your effort...
They will be tested I can assure you if the proofs get a chance to get up.

I just want to mention once more that those definitions (HSP={(t,x,y,z)|t=0} and "photon sphere at t=0" =...) contain no information that wouldn't also be present in the spacetime diagram I described. I think you should draw that diagram and post it. I would do it myself if I didn't suck so much at drawing stuff.

can I ask you then that if the HSP has zero duration then all locations on that hsp must be simultaneous?
The question was not for me, but I will answer it. All points in the HSP are simultaneous (in the frame where the light source is stationary), but this is not a consequence of its "zero duration". They are simultaneous because we defined the HSP as a hypersurface of simultaneous events.

All spacelike hypersurfaces have zero "duration" (thickness), even the ones that consist of events that are not simultaneous.

ozziemate

27th October 2008, 07:25 AM

Whoa, whoa. What is your definition of PHS?

If you are talking about HPS, then yes, all objects share the same reference frame (of course, because they're described in it), but their relative velocity definitely doesn't have to be zero.

If you're talking about PHS, I have no idea, because I don't know what you mean by PHS.

PHS is just a HSP and used to avoid this frame dependancy issue that appears to be entrenched in it's SRT definition of HSP

drkitten

27th October 2008, 07:27 AM

Again I shall repeat the PHS [ which is a non SRT version of the HSP to avoid confusion] has been defined as entirely universaly and absolutely simulatneous in all possible observer/photon events with a zero duration.

Repeating an incorrect statement doesn't make it right.

The problem is that if you define the PHS as a zero-duration event, then nothing has any velocity at all, due to division-by-zero errors, and so it can't be a counterexample to anything at all.

If you define it as a infinitesimal-duration event, then things have different relative velocities and you can't assume that it is universal (and in fact, observationally, it isn't.)

The first logic proof:
If the PHS is held as universally simultaneous in the zero duartion moment

But that's exactly what cannot hold. You've been told this over and over again.

ozziemate

27th October 2008, 07:29 AM

I just want to mention once more that those definitions (HSP={(t,x,y,z)|t=0} and "photon sphere at t=0" =...) contain no information that wouldn't also be present in the spacetime diagram I described. I think you should draw that diagram and post it. I would do it myself if I didn't suck so much at drawing stuff.

The question was not for me, but I will answer it. All points in the HSP are simultaneous (in the frame where the light source is stationary), but this is not a consequence of its "zero duration". They are simultaneous because we defined the HSP as a hypersurface of simultaneous events.

All spacelike hypersurfaces have zero "duration" (thickness), even the ones that consist of events that are not simultaneous.

do you mean light sourse as in photon/observer event type light source or light source as in star etc etc...I would assume you mean the prior

Thabiguy

27th October 2008, 07:30 AM

PHS is just a HSP and used to avoid this frame dependancy issue that appears to be entrenched in it's SRT definition of HSP

How does PHS avoid frame dependency?

Mashuna

27th October 2008, 07:32 AM

How does PHS avoid frame dependency?

Because Ozziemate said it does. Obviously.

ozziemate

27th October 2008, 07:33 AM

The problem is that if you define the PHS as a zero-duration event, then nothing has any velocity at all, due to division-by-zero errors, and so it can't be a counterexample to anything at all.

therefore the SRT HSP is invalidf??
after all it is the HSP that has the zero duration and thus no velocities are able to be determined even if frame dependant.

the same rules apply to both forms of hyper surface surely...

ozziemate

27th October 2008, 07:40 AM

How does PHS avoid frame dependency?

the HSP of SRT is frame dependant as in a single frame...ok in the normal way that imlpies non-simultaneity is occurring.
the PHS is essentially the same but recognises that the entire Hypersphere is one frame due to the fact that no relative v can be determined, due to the fact that is of zero ration.

The two are actually more or less the same except that Reality Check was offended when I used HSP in a way that he feels SRT does not like.

So to avoid uneccessary conflict with Reality Check I coined the term PHS which is just a partial reversal of the letters HSP

but in the end the HSP has proved so far to be one simulatneous o/p [observer/photon frame any way as there is no relative velocity present due to it's zero duration nature.

Fredrik

27th October 2008, 07:41 AM

well then we shall just have to call it PHS instead of HSP....to get past the SRT's dependancy on reference frame use.

No no no no no. You can't solve your problems by renaming something and pretending that this gives it magical properties.

The HSP is a set of events that are simultaneous in the inertial frame in which the light source is stationary. If you think you can prove that events in the HSP are also simultaneous in other frames, then please go ahead, but do not try to replace well-defined mathematical objects with magical nonsense!

ozziemate

27th October 2008, 07:47 AM

No no no no no. You can't solve your problems by renaming something and pretending that this gives it magical properties.

The HSP is a set of events that are simultaneous in the inertial frame in which the light source is stationary. If you think you can prove that events in the HSP are also simultaneous in other frames, then please go ahead, but do not try to replace well-defined mathematical objects with magical nonsense!
thanks and advice well taken, so HSP it is...{phs is now a defunct and redundent strategy ]

Fredrik are you still happy with your definition of the hyperpshere. drKitten has issues with infinitesimals?

Thabiguy

27th October 2008, 07:49 AM

the HSP of SRT is frame dependant as in a single frame...ok in the normal way that imlpies non-simultaneity is occurring.

Not really. In SRT, the relativity of simultaneity definitely does not manifest within a single reference frame. It manifests between two different reference frames only.

the PHS is essentially the same but recognises that the entire Hypersphere is one frame due to the fact that no relative v can be determined, due to the fact that is of zero ration.

It is one frame, but not due to the fact that no relative velocity can be determined. It is one frame because it was defined in that frame.

The two are actually more or less the same except that Reality Check was offended when I used HSP in a way that he feels SRT does not like.

So to avoid uneccessary conflict with Reality Check I coined the term PHS which is just a partial reversal of the letters HSP

Well I'm not offended by permutation of letters, so we can just keep talking about HSP.

but in the end the HSP has proved so far to be one simulatneous o/p [observer/photon frame any way as there is no relative velocity present due to it's zero duration nature.

You are dangerously close to confusing "no relative velocity present" with "zero relative velocity".

ozziemate

27th October 2008, 07:50 AM

No no no no no. You can't solve your problems by renaming something and pretending that this gives it magical properties.

The HSP is a set of events that are simultaneous in the inertial frame in which the light source is stationary. If you think you can prove that events in the HSP are also simultaneous in other frames, then please go ahead, but do not try to replace well-defined mathematical objects with magical nonsense!
this is why I chose to use PHS instead of HSP because HSP assumes non-simultaneity as self eviedent. Where as PHS makes no such presumption.

Fredrik

27th October 2008, 07:52 AM

do you mean light sourse as in photon/observer event type light source or light source as in star etc etc...I would assume you mean the prior
I mean a star, a light bulb or whatever. I don't understand what you mean by the first option.

ozziemate

27th October 2008, 07:52 AM

In the end it doesn't matter so we may as call our RF BOB and every other observer/photon event is simultaneous with BOB's observer/photon event.
the end result will be the same.

ozziemate

27th October 2008, 07:56 AM

I mean a star, a light bulb or whatever. I don't understand what you mean by the first option.
OK but I don't think it matters given that t=0 and d=0 for the entire frame. The star is stationary like everything else in the hypersphere. duration zero.

and on that note I gotta retire 'tis 2 am and I goota work tomorrow oops today.

back in 6

Fredrik

27th October 2008, 08:02 AM

Edit: I deleted a few comments that I would have known were inappropriate if I had seen this post before I started writing:

thanks and advice well taken, so HSP it is...{phs is now a defunct and redundent strategy ]

Fredrik are you still happy with your definition of the hyperpshere. drKitten has issues with infinitesimals?
I am, but the HSP is a hyperplane, not a hypersphere.

I don't think he has issues with it. He just doesn't want to get into a discussion about the proper definition of an infinitesimal. (After reading #571, I also think he has issues with your language. It's harder to decode than you think).

this is why I chose to use PHS instead of HSP because HSP assumes non-simultaneity as self eviedent.

No it doesn't. It assumes a few other things though, e.g. that a position in space and time can be represented by four numbers, and those assumptions combined with the invariance of the speed of light will eventually force us to conclude that inertial observers disagree about simultaneity.

The star is stationary like everything else in the hypersphere. duration zero.

What you should be saying is that its world line intersects the HSP at a single point.

drkitten

27th October 2008, 09:13 AM

therefore the SRT HSP is invalidf??
after all it is the HSP that has the zero duration and thus no velocities are able to be determined even if frame dependant.

Yup. Although you can at least define the HSP (although you can't determine relative velocities of anything in the HSP) because it's defined in terms of a static property (distance, relative to a particular point in space-time in a particular reference frame).

Without a static distance, you can't even define the PHS.

Fredrik

27th October 2008, 10:49 AM

I wouldn't answer "yup" to that. The HSP is just a subset of spacetime, so it's definitely something "valid" that we can talk about. The HSP is defined as the set of all events that are simultaneous, in the inertial frame in which the light source is stationary, with some point on the world line of the light source. What Ozzie has to do to complete his proof is to show us that any two members of the HSP are simultaneous in all inertial frames.

This will of course fail, but I see no harm in letting him try.

Unfortunately Ozzie's proof starts with the observation that since all the events in a particular subset of the HSP are simultaneous, all the events in a subset of that subset must be simultaneous. This is obviously true, but the statement contains no information since the HSP by definition is a set of simultaneous events.

Reality Check

27th October 2008, 12:10 PM

The Man

27th October 2008, 12:18 PM

and in summation? [ facinating btw ]

It was at the end of the post in case you missed it. If you need it put more succinctly..

Your assertion of light (or a photon) as time or as an absolute temporal reference is invalid even without considering SRT.

The Man

27th October 2008, 12:37 PM

What????

Light as time (or as an absolute temporal reference) is one of ozziemate’s assertions on this thread.

This is mostly irrelevant, and the part that isn't irrelevant is gibberish.

It is not irrelevant to ozziemate’s assertion of light as time, which I agree is gibberish, but the relationships as given certainly are not.

Gibberish. If I have a 24-hour clock instead of a 12-hour clock, the hands of the clock move exactly half the speed and therefore cover half the distance per unit time. Either way, the measurement of time is absolute regardless of the units. The length of a song is a constant when measured in seconds, regardless of whether the song has 140 bpm or 90 bpm.

Unless of course you’re moving towards or away from the source then the bpm you hear is different from what is being played and thus so is the length of the song or the time each of you ascribe to any given number of beats. If your only measure of time is the number of beats you hear for a given unit of time (sound as time or an absolute temporal reference) then what you measure as time is dependent on your motion relative to the source (the same can be said for light) without considering SRT.

You're making the same argument here that I would if I said that "length" is not absolute because I can measure something as being 12 inches long instead of 1 foot, and therefore the same object has two different lengths. And for exactly this reason, you're wrong.

Only if 12 inches becomes 11 due relative motion, which it can under SRT but the intent here was to demonstrate to ozziemate that light is not time or an absolute temporal reference without even evoking SRT .

Time is relative, but that has nothing to do with different measurements; if Newtonian mechanics were actually correct, then time would be absolute despite your being able to measure it with any units you choose.

Well, actually the fact that “Time is relative” has every thing to do with, well, measuring time differently. Sure “if Newtonian mechanics were actually correct, then time would be absolute…” but the units you choose to measure with are not absolute, the count of 20 beats per minute at the source could become 20 beats in .75 minutes or in 1.12 minutes when heard, based just on the Doppler effect and “Newtonian mechanics”.

I appreciate your fervor drkitten but you can put away your claws, as ozziemate opposes arguments based on the validity of SRT, this was just a demonstration of the fallacy of light as time (or an absolute temporal reference) without evoking SRT.

I must apologize though, I should have included the quote from ozziemate to make it clear what I was talking about and prevent you from wasting your time. Oh wait, what am I saying, we’re on an ozziemate thread, so wasting our time, with him, is pretty much a given, sorry again.

you may recall that earlier in this thread I said that the photon IS time and not a product of time.
Therefore invariance of light speed is also invariance of fundamental time flow as well.

ozziemate

27th October 2008, 02:34 PM

Light as time (or as an absolute temporal reference) is one of ozziemate’s assertions on this thread.

It is not irrelevant to ozziemate’s assertion of light as time, which I agree is gibberish, but the relationships as given certainly are not.

Unless of course you’re moving towards or away from the source then the bpm you hear is different from what is being played and thus so is the length of the song or the time each of you ascribe to any given number of beats. If your only measure of time is the number of beats you hear for a given unit of time (sound as time or an absolute temporal reference) then what you measure as time is dependent on your motion relative to the source (the same can be said for light) without considering SRT.

Only if 12 inches becomes 11 due relative motion, which it can under SRT but the intent here was to demonstrate to ozziemate that light is not time or an absolute temporal reference without even evoking SRT .

Well, actually the fact that “Time is relative” has every thing to do with, well, measuring time differently. Sure “if Newtonian mechanics were actually correct, then time would be absolute…” but the units you choose to measure with are not absolute, the count of 20 beats per minute at the source could become 20 beats in .75 minutes or in 1.12 minutes when heard, based just on the Doppler effect and “Newtonian mechanics”.

I appreciate your fervor drkitten but you can put away your claws, as ozziemate opposes arguments based on the validity of SRT, this was just a demonstration of the fallacy of light as time (or an absolute temporal reference) without evoking SRT.

I must apologize though, I should have included the quote from ozziemate to make it clear what I was talking about and prevent you from wasting your time. Oh wait, what am I saying, we’re on an ozziemate thread, so wasting our time, with him, is pretty much a given, sorry again.
so lets waste some more time shall we....

DRkitten mentioned something which needs further clarification.
There is a very real difference between our imaginary construction of time using a metric we have determined to be a second, minute hour and that which provokes change in the universe in a very real way.
Light speed invariance means that the universe is changing universally at it's own rate using it's own metric which we have qualified as a metre per second and have continued as a way of accommodating invariance to allow that metric to be relative and thus variable.

So Light speed is actually the uivserses rate of change therefore it could be argued that Light IS time and not a product of time.

The issue is not so much this, but how we commonly mix up and loose the distinction between what is imaginary [ abstraction or constructs and what is proven to be actual and real provided by reality and not some product of someones furtile imagination.

The time being referred to is commonly in the form of a human mind construct and taken from an equally human mind construct called a clock, neither of which is actually time according to the universe as they are merely conveniances of measurement that we use to quantify change.

At a universal level though time as change is very constant and invariant and is occuring at the rate of 'c'. simultaneously across the entire universe.

Light as time (or as an absolute temporal reference) is one of ozziemate’s assertions on this thread.
totally false as it is a very real phenonema premsied on the invariance of light. It is the use of clock time that is temporal and currently being misused by yourself.
Please provide a link to show where I have made such an assertion. if not apologise for your insinuations.

It is not irrelevant to ozziemate’s assertion of light as time, which I agree is gibberish, but the relationships as given certainly are not.
gibberish only because it is based on your obviously limited understanding of the logic of the topic. [ make the distinction between abstract and reality and your situation may improve]

Unless of course you’re moving to wards or away from the source then the bpm you hear is different from what is being played and thus so is the length of the song or the time each of you ascribe to any given number of beats. If your only measure of time is the number of beats you hear for a given unit of time (sound as time or an absolute temporal reference) then what you measure as time is dependent on your motion relative to the source (the same can be said for light) without considering SRT.

I must agree entirely with drkitten when he/she states:

Either way, the measurement of time is absolute regardless of the units
there must be an underpinning time or change constant that is invariant for any theoretical models to function. And that invariance is light speed.

Well, actually the fact that “Time is relative” has every thing to do with, well, measuring time differently. Sure “if Newtonian mechanics were actually correct, then time would be absolute…” but the units you choose to measure with are not absolute, the count of 20 beats per minute at the source could become 20 beats in .75 minutes or in 1.12 minutes when heard, based just on the Doppler effect and “Newtonian mechanics”.
As time is only variant because of the metric used time is actually absolute and subject to the relativity of the metric...however the change rate or underpinning speed is invariant.

[gibberish I hear you cry...and you know why]

I must apologize though, I should have included the quote from ozziemate to make it clear what I was talking about and prevent you from wasting your time. Oh wait, what am I saying, we’re on an ozziemate thread, so wasting our time, with him, is pretty much a given, sorry again.
all you are doing is making me look good so keep on wasting time...

btw I stil think you post about time was facinatiing even though you attempt to use it as a way of offensive commment by your own admission. There is potential to develop it further.

[ as using an invariant light clock is actually possible once the issue of non-simultaneity is resolved.]

Maybe if you concentrated on the issue at hand rather than you hatred for me you will achieve more benefit for yourself. "eye on the ball and not the man"

ozziemate

27th October 2008, 02:58 PM

It was at the end of the post in case you missed it. If you need it put more succinctly..

Your assertion of light (or a photon) as time or as an absolute temporal reference is invalid even without considering SRT.
according to your limited understanding of the subject this is an understandable outcome.

ozziemate

27th October 2008, 03:09 PM

ozziemate seems to think that 'non-simultaneity' is an assumption of the hypersphere of the present. This is not true - the light cone diagram where the HSP is defined has only 1 observer. Non-simultaneity requires at least 2 observers so that their times can be compared.
Of course he does tend to redefine terms so perhaps he should tell us exactly what he means by 'non-simultaneity'.

SRT does not make any assumptions about whether 2 observers have the same or different times. The only assumption that it makes is that its postulates are correct. Relativity of simultaneity (http://en.wikipedia.org/wiki/Relativity_of_simultaneity) is a result of deductions from the postulates.
by the time this issue is close to resolution you can have as many observers as you like all at relative velocity and you will still find non-simultaneity is impossible due to our mishandling of the light invariance pheno.
The first logic proof:
If the HPS/PHS is held as universally simultaneous in the zero duration moment then every HPS/PHS possible thoughout eternity is also simultaneous there fore non-simultaneity is impossible. Regardless of issues regarding invariance and so called impossibilites of handling such.
If all hps/phs's eternally are simultaneous universally then it is impossible for non-simultaneity to exist.
An infinite number of hps/phs's can exist in eternity....as they are zero duration and even if infinitesimal duration [ which invalidates SRT any way if I am not mistaken] the same applies an infinite number of hps/phs's can occur in eternity past present and future.

This of course relies entirely on SRT's definition of the HSP when refined to include the tautological proof of observer/photon simultaneity.
Including two observers may placate your need to relate it to SRT in a way you can handle. when in fact the use of two observers is non necessary to prove the point. As the proof avoids their participation any way.
"non-simultaneity cannot be valid as every HSP is zero duration and therefore all frames regardless of velocity must share the observers simultaneity of observer/photon locations. and as every HSP through out eternity past, present and future has this factor non-simultaneity can not be valid.

UncaYimmy

27th October 2008, 03:09 PM

Maybe if you concentrated on the issue at hand rather than you hatred for me you will achieve more benefit for yourself. "eye on the ball and not the man"

If you think anybody has shown you hatred in this thread, you should take this to an unmoderated forum like a newsgroup. You'd be in tears. The only reason I keep reading is because I don't think you're a jerk. That said, if you start making it personal when clearly it's not, then I think you will find yourself alone.

From where I sit I'd say that everyone has been quite indulgent. You have not demonstrated an understanding of SRT. Therefore, you have pretty much zero credibility when it comes to poking a hole in it. You refuse to even attempt to understand it, which is also quite frustrating.

Einstein didn't need an interactive forum to write up this theories. Why do you think you need one? Why do you feel the need to drag out this process and force people to accept what are poorly stated points before moving on?

Just write it up, start to finish. There's no need to wait for a "yes, go on" from anyone. Just spit it out.

If, however, you insist on continuing in this direction, expect people to get a little frustrated with the process and take it out on you. Such is the price of being able to overturn a century of thinking and testing by the greatest minds in the world. I'd say it's a small price. Just think of all the groupies you'll have when you're done!

Ziggurat

27th October 2008, 03:10 PM

according to your limited understanding of the subject this is an understandable outcome.

Given your inability to express any of your ideas in a mathematical form which can be examined objectively, you're not really in a position to call anyone else's understanding "limited". I was formerly willing to give you the benefit of the doubt, but it's become quite clear that you're a troll. I have plenty of patience for people who are ignorant of relativity and want to learn, but I have no more patience for someone like you who is not only ignorant, but unwilling to learn.

The Man

27th October 2008, 03:11 PM

so lets waste some more time shall we....

DRkitten mentioned something which needs further clarification.
There si a very real difference between our imaginary construction of time using a metric we have determined to be a second, minute hour and that which provokes change in the universe in a very real way.

Light speed invariance means that the universe is changing universally at it's own rate using it's own metric which we have qualified as a metre per second and have continued as a way of accommodating invariance to allow that metric to be variable.
So Light speed is actually the uivserses rate of change therefore it could be argued that Light IS time and not a product of time.

As the speed of light is meters per second it is just a measure of change in position over change in time or the product of distance and inverse time DD*DT-1. As it is not a measure of time itself you would need some invariant distance to divide by c in order to obtain your invariant measure of time. Do you have such a distance? How are you sure it is frame invariant?

Your whole premise is based on the invariance of the speed of light which is one of the basic postulates of SRT. The others, invariance of physical laws in inertial reference frames, length contraction and time dilation in some sense all stem from that invariance of the speed of light. What you are basically saying is that given the validity of a fundamental postulate of SRT you can logically invalidate SRT. That is not a tautology but the negation of a tautology or a contradiction, to put it in Fredrik’s form “given P then not P”.

The issue is not so much this but how we commonly mix yup theand loose the distinction between what is imaginary [ abstraction or constructs and wha is proven to be actual an dreal provided by reality and not some product of someones furtile imagination.

Which is why we use experimentation and as you have been well informed, SRT is consistent with the experimental data, however your statement does give me some hope that we might not be wasting our time with you.

The time being referred to is commonly in the form of a human mind construct and taken from an equally human mind construct called a clock, neither of which is actually time according to the universe as they are merely conveniances of measurement that we use to quantify change.

Quantifying change is in a sense time and it is not just a human mind construct but a physical construct, at this point down to the atomic level.

http://en.wikipedia.org/wiki/Second

Under the International System of Units, the second is currently defined as
“ the duration of 9 192 631 770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium 133 atom.[1]
”
This definition refers to a caesium atom at rest at a temperature of 0 K (absolute zero). The ground state is defined at zero magnetic field. The second thus defined is equivalent to the ephemeris second, which was based on astronomical measurements. (See History below.)
The international standard symbol for a second is s[2] (see ISO 31-1)
The realization of the standard second is described briefly in NIST Special Publication 330; Appendix 2, pp. 53 ff, and in detail by National Research Council of Canada.

At a universal level though time as change is very constant and invariant and is occuring at the rate of 'c'. simultneaouly across the entire universe.

So what makes you think that your “human mind construct” above represents anything “At a universal level”? You’re going to make me take back my statement about having hope that we are not wasting are time with you.

Fredrik

27th October 2008, 03:12 PM

There si a very real difference between our imaginary construction of time using a metric we have determined to be a second, minute hour and that which provokes change in the universe in a very real way.

As usual it's very hard to decode what you're saying. I didn't quite pull it off this time. In classical physics, "that which provokes change" is a force. So you seem to be talking about the difference between a force and...something else..

Light speed invariance means that the universe is changing universally at it's own rate using it's own metric...

No it doesn't. It just means that all light cones have the same shape in every spacetime diagram that represent the coordinates of an inertial observer.

I can also inform you that "metric" is a mathematical term with a very specific meaning that's used a lot in SR and even more in GR.

...which we have qualified as a metre per second and have continued as a way of accommodating invariance to allow that metric to be variable.

This doesn't make sense.

So Light speed is actually the uivserses rate of change therefore it could be argued that Light IS time and not a product of time.

This makes even less sense.

The issue is not so much this but how we commonly mix yup theand loose the distinction between what is imaginary [ abstraction or constructs and wha is proven to be actual an dreal provided by reality and not some product of someones furtile imagination.

All theories, and all concepts they define, are "imaginary". The only difference between a good theory and a bad theory is that a good theory is able to correctly predict the results of experiments.

The connection between the real world and the concepts defined by the theory is always a bunch of postulates. For example, in special relativity we have to postulate what "thing" in the mathematical model of spacetime that corresponds to what a clock measures.

The time being referred to is commonly in the form of a human mind construct and taken from an equally human mind construct called a clock, neither of which is actually time according to the universe as they are merely conveniances of measurement that we use to quantify change.

So time is not what we measure with a clock? Then how do you propose that we measure time?

At a universal level though time as change is very constant and invariant and is occuring at the rate of 'c'. simultneaouly across the entire universe.
Doesn't make sense.

The Man

27th October 2008, 03:15 PM

according to your limited understanding of the subject this is an understandable outcome.

Well, fortunately some of us are not burdened by the ominous foreboding and infinite responsibility of thinking we have unlimited knowledge about anything.

ozziemate

27th October 2008, 03:26 PM

We have again unfortunately drifted away for the topic of logical proofs that invalidate non-simultaneity thus SRT.

I shall copy again post #509
OK so we have defined simultaneity of observer/light in the HSP with a zero duration.
http://bizzymate.com/uploads/images/physics/contruct01.jpg

next agreement needed before proceeding.

1] Do we also agree that regardless of what zero moment of time in all of eternity we take our zero duration snap shot simultaneity of observer/photon is maintained in that zero duration moment [hsp] ?
2] If agreed how many hsp's do you think could exist in all of eternity given they are zero duration?
-the answer must be infinity but I shall ask you to confirm your agreement again.

I will present a number of logical arguments to refute non-simultaneity once this is agreed to.
note : if you all could allow me to just present this case with your agreements along the way we shall get this over with more quickly. Avoid making assumptions and speculations about what I am going to do next and let it unfold.

Now obviously we still have no agreement about the HSPs zero duration which automatically extends the logic to every HSP in eternity...

Is it possible to determine the HSP thoroughly or is it just another SRT fudge...?

Fredrik

27th October 2008, 03:28 PM

[I]The first logic proof:
If the HPS/PHS is held as universally simultaneous in the zero duration moment then every HPS/PHS possible thoughout eternity is also simultaneous there fore non-simultaneity is impossible.

Hold it right there. All you know is that the events in the HSP are simultaneous in one inertial frame. If you can prove that this implies that they are "universally simultaneous" (I assume that means "simultaneous in all frames"), you have succeeded. But you can't start by assuming the very thing you're trying to prove.

I didn't try to understand the rest of your post. There's no point moving on when there's such an enormous flaw in the first sentence.

ozziemate

27th October 2008, 03:44 PM

Hold it right there. All you know is that the events in the HSP are simultaneous in one inertial frame. If you can prove that this implies that they are "universally simultaneous" (I assume that means "simultaneous in all frames"), you have succeeded. But you can't start by assuming the very thing you're trying to prove.

I didn't try to understand the rest of your post. There's no point moving on when there's such an enormous flaw in the first sentence.

ok, Fredrik you have pointed out more or less exactly what the real issue is and why I have been so stubborn in getting a definition of the HSP refined to include zero duration.
try this
logic:
If all observer/photon events are simultaneous as all have zero duration in a given moment then all events within the HSP can be considered as inertial frames as relative velocity in the HSP is a nul concept.
as t=0, d must also = 0 [ a tautology. ]

have a close look at the above logic and if you can clearly state why it is invalid. As if it is true non-simultaneity is non existant by logical extension into eternity.
It is not about assuming it is about logic and reasoning. I am not making an assumption I am merely applying logic. Sure that logic is having difficulty in being understood but it is not an assumption. [ except that I am testing that logic by rpresenting it here]

also the second proof also relies on the refined definition of the HSP
and that is how you must use a simultaneous HSP to generate a nonsimultaneous universe wihch is circular and flawed logic. "use absolute time to generate relative time..sort of reaonsing"

Fredrik

27th October 2008, 03:47 PM

I shall copy again post #509

There are some serious language problems in your #509. Look at it:

1] Do we also agree that regardless of what zero moment of time in all of eternity we take our zero duration snap shot simultaneity of observer/photon is maintained in that zero duration moment [hsp] ?
2] If agreed how many hsp's do you think could exist in all of eternity given they are zero duration?
-the answer must be infinity but I shall ask you to confirm your agreement again.

When you say things like "simultaneity of observer/photon", most people won't know what you're talking about. I understand that you're saying that "in any hypersurface of constant time, all events are simultaneous", but others don't. They might disagree if they think you meant something else. (It's actually pretty natural to think that you meant something else, because what you said is so obvious that it's pointless to say it). They might also disagree because they interpret the word "simultaneous" as "simultaneous in all frames" rather than "simultaneous in one frame".

Is it possible to determine the HSP thoroughly or is it just another SRT fudge...?

The HSP is well-defined.

The Man

27th October 2008, 03:48 PM

so lets waste some more time shall we....

DRkitten mentioned something which needs further clarification.
There is a very real difference between our imaginary construction of time using a metric we have determined to be a second, minute hour and that which provokes change in the universe in a very real way.
Light speed invariance means that the universe is changing universally at it's own rate using it's own metric which we have qualified as a metre per second and have continued as a way of accommodating invariance to allow that metric to be relative and thus variable.

So Light speed is actually the uivserses rate of change therefore it could be argued that Light IS time and not a product of time.

The issue is not so much this, but how we commonly mix up and loose the distinction between what is imaginary [ abstraction or constructs and what is proven to be actual and real provided by reality and not some product of someones furtile imagination.

The time being referred to is commonly in the form of a human mind construct and taken from an equally human mind construct called a clock, neither of which is actually time according to the universe as they are merely conveniances of measurement that we use to quantify change.

At a universal level though time as change is very constant and invariant and is occuring at the rate of 'c'. simultaneously across the entire universe. "

Since I already replied to the above I now reply to the edited additions.

totally false as it is a very real phenonema premsied on the invariance of light. It is the use of clock time that is temporal and currently being misused by yourself.
Please provide a link to show where I have made such an assertion. if not apologise for your insinuations.

you may recall that earlier in this thread I said that the photon IS time and not a product of time.
Therefore invariance of light speed is also invariance of fundamental time flow as well.

Happy now?

gibberish only because it is based on your obviously limited understanding of the logic of the topic. [ make the distinction between abstract and reality and your situation may improve]

Can you make that distinction (you have not made that ability evedent)? Do you want your “situation” to improve?

I must agree entirely with drkitten when he/she states:

Oh I’m sure you think you do, but if time is not absolute (as all evidence indicates) then neither is it’s measurement “regardless of the units”.

there must be an underpinning time or change constant that is invariant for any theoretical models to function. And that invariance is light speed.

Why? Because your “human mind construct” requires it, regardless of experimental evidence? I now have to take back my statement about having hope about not wasting time with you.

As time is only variant because of the metric used time is actually absolute and subject to the relativity of the metric...however the change rate or underpinning speed is invariant.

[gibberish I hear you cry...and you know why]

I do not think you understand the meaning of the term “metric” or that when constructed in 4d space time it is frame invariant.

all you are doing is making me look good so keep on wasting time...

Well then you must have been very ugly to start with and I am always glad to help.

btw I stil think you post about time was facinatiing even though you attempt to use it as a way of offensive commment by your own admission. There is potential to develop it further.

[ as using an invariant light clock is actually possible once the issue of non-simultaneity is resolved.]
Well I hope you at least try, liner and rotation motion relationships would be a good place to start. If my being offensive gets you to actual look at thing then I have no problem with that. We have to utilize those talents we have.

Maybe if you concentrated on the issue at hand rather than you hatred for me you will achieve more benefit for yourself. "eye on the ball and not the man"

Aw, and just when we were getting along so good, do not confuse my indifference for you as hatred, much like you confuse everything else.

ozziemate

27th October 2008, 03:57 PM

There are some serious language problems in your #509. Look at it:

When you say things like "simultaneity of observer/photon", most people won't know what you're talking about. I understand that you're saying that "in any hypersurface of constant time, all events are simultaneous", but others don't. They might disagree if they think you meant something else. (It's actually pretty natural to think that you meant something else, because what you said is so obvious that it's pointless to say it). They might also disagree because they interpret the word "simultaneous" as "simultaneous in all frames" rather than "simultaneous in one frame".

The HSP is well-defined.
would most people conversant iwth physics agree that an event requires at least two aspect for it to be defined as an event.
to say observer event on it's own of course imlpies that the observer has had something impact on it other wise no event.
to use the "long hand" observer /photon event is only attempting to keep it clear what events we are discussing.
eg
"all [observer/photon] events must occur simutaneousy if the moment is of zero duration..."

how would you word it in a way that refines the use of the word "event" in this context of HSP definition?

you may feel that it is unecessary for example to use t=0 in defining the HSP but as proven it is essential.
you could I guess simply state "all events" on a zero duiration HSP are simultaneous" but alas because SRT is so intrenched this will achieve nothing but confusion.

clearly showing that obsrever and light colide to geerate an event is obviously important to bring the light aspect into the HSP more clearly. IMO

Fredrik

27th October 2008, 03:59 PM

...a definition of the HSP refined to include zero duration.

The requirement that it's a hypersurface of constant time coordinate takes care of that.

try this
logic:
If all observer/photon events are simultaneous as all have zero duration in a given moment then all events within the HSP can be considered as inertial frames as relative velocity in the HSP is a nul concept.
as t=0, d must also = 0 [ a tautology. ]

The language is ambiguous and confusing. I think this is what you're trying to say:

"If all photon observation events in the HSP are simultaneous in the inertial frame used to define the HSP, then all events in the HSP can be considered stationary in that inertial frame".

The problem is that it doesn't make sense to say that an event is stationary. (It wouldn't make any more sense to say that it isn't stationary. You just can't assign a velocity to it).

This is what you need to learn about motion: The motion of a particle is represented by a curve called its world line. The velocity of the particle at the event where it intersects the HSP is defined by the angle its world line makes with the HSP at that event.

sol invictus

27th October 2008, 04:02 PM

Now obviously we still have no agreement about the HSPs zero duration which automatically extends the logic to every HSP in eternity...

There is no problem with the HSP having zero duration. Fredrik's definition of it was fine. The problem is that your "logic" is completely wrong.

Let me present a rather exact analogy. Consider 3D space in Cartesian coordinates x,y,z. Think of z as oriented vertically. z will be our proxy for time.

The "HSP" in this analogy is the plane z=0, or (at a later time z_0) the plane z=z_0. That plane is infinitely thin. "Simultaneous" simply means every point on the plane has the same value of z.

Your claim was:

"If the HPS/PHS is held as universally simultaneous in the zero duration moment then every HPS/PHS possible thoughout eternity is also simultaneous there fore non-simultaneity is impossible."

Let me translate it into this language:

If all the points in the plane z=0 have the same value of z, then every point in every plane at fixed z has the same value of z, and therefore all points have the same value of z.

Of course this is wrong, for obvious reasons. The error is much more elementary than the relativity of simultaneity (which by the way corresponds to tilting the planes) - there is no logic to the sentence at all, and the conclusion is patently false.

ozziemate

27th October 2008, 04:04 PM

The velocity of the particle at the event where it intersects the HSP is defined by the angle its world line makes with the HSP at that event.
however the intersection is of zero duration....ahh ! ok I see, however how does that impinge upon the fact that we have a "snapshot" simultaneous events. the world line, t=0 ?

sol invictus

27th October 2008, 04:06 PM

ozziemate

27th October 2008, 04:07 PM

There is no problem with the HSP having zero duration. Fredrik's definition of it was fine. The problem is that your "logic" is completely wrong.

Let me present a rather exact analogy. Consider 3D space in Cartesian coordinates x,y,z. Think of z as oriented vertically. z will be our proxy for time.

The "HSP" in this analogy is the plane z=0, or (at a later time z_0) the plane z=z_0. That plane is infinitely thin. "Simultaneous" simply means every point on the plane has the same value of z.

Your claim was:

"If the HPS/PHS is held as universally simultaneous in the zero duration moment then every HPS/PHS possible thoughout eternity is also simultaneous there fore non-simultaneity is impossible."

Let me translate it into this language:

If all the points in the plane z=0 have the same value of z, then every point in every plane at fixed z has the same value of z, and therefore all points have the same value of z.

Of course this is wrong, for obvious reasons. The error is much more elementary than the relativity of simultaneity (which by the way corresponds to tilting the planes) - there is no logic to the sentence at all, and the conclusion is patently false.
ok I think I see your point,
by stacking planes throughout eternity assuming z=0 on all planes then no movement is possible and the universe would be in an eternal state of sus'pended animation....

ahhhh....ok....
am I correct in saying thats what you want me to realise?

drkitten

27th October 2008, 04:12 PM

sol invictus

27th October 2008, 04:12 PM

ok I think I see your point,
by stacking planes throughout eternity assuming z=0 on all planes then no movement is possible and the universe would be in an eternal state of sus'pended animation....

ahhhh....ok....
am I correct in saying thats what you want me to realise?

No, not quite. Forget about time for a moment. In this analogy, height (the z coordinate) is the proxy for time. HSPs are planes at constant height. Objects and photons are lines, either vertical (a stationary observer) or at most at 45 degrees down from the vertical (that would be a photon).

Now try to make your argument.

Fredrik

27th October 2008, 04:14 PM

would most people conversant iwth physics agree that an event requires at least two aspect for it to be defined as an event.

No. An event is by definition a point in spacetime, and you can specify what point you're talking about by stating its coordinates (t,x,y,z) in one coordinate system.

I suspect that you misunderstood this comment:
Non-simultaneity requires at least 2 observers so that their times can be compared.
What he's saying is that in order to see that different frames disagree about which events are simultaneous, you have to consider at least two frames.

"all events" on a zero duiration HSP are simultaneous" but alas because SRT is so intrenched this will achieve nothing but confusion.

I think there's no chance that people would find that as confusing as what you said.

ozziemate

27th October 2008, 04:16 PM

... which is Zeno's paradox and the reason you need to approach this using infinitesimals and limits.
ha and of course once 'ole zeno is invloved the hare is chasing the tortise for ever....ha

hmmmmm....but to include infinitiesimals in the situation such a descibing the HSP to have infinitesimal duration would corrupt SRT [ I think but as yet do not know] ....hmmmm ..

again Zeno at work...

hmmm shall think on it some...
good point!:)

sol invictus

27th October 2008, 04:19 PM

... which is Zeno's paradox and the reason you need to approach this using infinitesimals and limits.

If ozzie will agree that lines can puncture the plane z=0 at an angle other than 90 degrees even though it's infinitely thin, he have agreed that objects passing through the HSP don't have to be at rest, even though it only lasts for an instant.

Do you agree that lines can puncture a plane at an angle other than 90 degrees, ozzie?

ozziemate

27th October 2008, 04:19 PM

What he's saying is that in order to see that different frames disagree about which events are simultaneous, you have to consider at least two frames.
but of course we have virtually and infinite number of observers thus frames in our HSP so how many do you want....

Fredrik

27th October 2008, 04:23 PM

however the intersection is of zero duration....ahh ! ok I see,

Yes, that was my point. You have to look at a region of spacetime that doesn't have zero duration to define a velocity.

however how does that impinge upon the fact that we have a "snapshot" simultaneous events. the world line, t=0 ?
I don't understand what you're asking. The equation t=0 defines a plane (actually the HSP), not a world line. The world line of a photon moving in the positive x direction would be defined by "t>-T, x=ct, y=0, z=0".

ozziemate

27th October 2008, 04:26 PM

If ozzie will agree that lines can puncture the plane z=0 at an angle other than 90 degrees even though it's infinitely thin, he have agreed that objects passing through the HSP don't have to be at rest, even though it only lasts for an instant.

Do you agree that lines can puncture a plane at an angle other than 90 degrees, ozzie?
actually to be honest when studying th nature of metastability with regards to gravity I had to show that lines can indeed puncture a two dimensional plane at an angle of infinite variations and allow that plane to acquire that information somehow.
So yes I do agree....
as to whether that defeats the 1st logical proof I shall think about it some...and post my assessment later.
In some ways this has been a good exercise in that it will help later when discussing the nature of metastability and gravity to show why we have movement in the universe to start with. It dos get into this sort of dimensional ism, that being the seemingly paradox ed position of zero distance having time value [ aka the use of infinitesimals is designed to deal with such a paradox c/of Zeno as drkitten has already pointed out.]

ozziemate

27th October 2008, 04:29 PM

Yes, that was my point. You have to look at a region of spacetime that doesn't have zero duration to define a velocity.

I don't understand what you're asking. The equation t=0 defines a plane (actually the HSP), not a world line. The world line of a photon moving in the positive x direction would be defined by "t>-T, x=ct, y=0, z=0".
meaning that even though the photon could be deemed stationary on the HSP it still has speed....hmmmm yes?
as an aside:
but isn't this the same as discussing "movement potential" rather than actual movment

Fredrik

27th October 2008, 04:29 PM

but of course we have virtually and infinite number of observers thus frames in our HSP so how many do you want....
That's wrong. The events in the HSP do not define inertial frames. There is however an infinite number of inertial frames associated with each straight line that's "more vertical in the spacetime diagram" than a photon's world line. Each such straight line can be interpreted as the time axis of some other inertial observer's coordinates.

I need to do something else for a couple of hours, but I'll check in later.

ozziemate

27th October 2008, 04:37 PM

That's wrong. The events in the HSP do not define inertial frames. There is however an infinite number of inertial frames associated with each straight line that's "more vertical in the spacetime diagram" than a photon's world line. Each such straight line can be interpreted as the time axis of some other inertial observer's coordinates.

I need to do something else for a couple of hours, but I'll check in later.
thanks for your patience and help

sol invictus

27th October 2008, 04:41 PM

actually to be honest when studying th nature of metastability with regards to gravity I had to show that lines can indeed puncture a two dimensional plane at an angle of infinite variations and allow that plane to acquire that information somehow.

Word salad - I have absolutely no idea what you're talking about. This was a trivial and completely obvious fact from basic Euclidean geometry (the kind most people learn as young children) - it has nothing to do with gravity, metastability, or infinite variations (the angle ranges from 90 degrees to 0).

So yes I do agree....

Well, that's good, I guess. Are we done?

ozziemate

27th October 2008, 04:43 PM

ozziemate

27th October 2008, 04:47 PM

Word salad - I have absolutely no idea what you're talking about. This was a trivial and completely obvious fact from basic Euclidean geometry (the kind most people learn as young children) - it has nothing to do with gravity, metastability, or infinite variations (the angle ranges from 90 degrees to 0).

Well, that's good, I guess. Are we done?

this use of dimensionalism can actually get quite complex.
If I am not mistaken the branch of mathematics - calculus was developed simply from the need to work with such concepts so pooh poohing me about it is not all that valid given that an 8 year old child doesn't get to study calculus in prmary school.
If you think physics is trivial at any level then may be you need to reconsider your interests

Tubbythin

27th October 2008, 04:58 PM

ok, Fredrik you have pointed out more or less exactly what the real issue is and why I have been so stubborn in getting a definition of the HSP refined to include zero duration.
try this
logic:
If all observer/photon events are simultaneous as all have zero duration in a given moment then all events within the HSP can be considered as inertial frames as relative velocity in the HSP is a nul concept.
as t=0, d must also = 0 [ a tautology. ]

have a close look at the above logic and if you can clearly state why it is invalid. As if it is true non-simultaneity is non existant by logical extension into eternity.

First of all its wrong because it doesn't make sense. Lets take it step by step:

If all observer/photon events are simultaneous
This makes no sense in SR, as I have explained multiple time, without stating who is observing. If you mean simultaneous to all observers then you are just saying "If SR is wrong then SR is wrong." Which, while logical, doesn't prove a thing.

If all observer/photon events are simultaneous as all have zero duration in a given moment
The second bit of the sentence doesn't follow from the first. Eg. two zero duration moments could occur 3 hours apart. They would not be simultaneous since simultaneous implies the time difference between the two events is 0.

then all events within the HSP can be considered as inertial frames
You really need to learn the difference between an event and a reference frame. They are not interchangeable. Secondly, even if all observers were in inertial reference frames that does not mean we can ignore their relative velocities.

as relative velocity in the HSP is a nul concept.
Just because we cannot define the relative velocity of observers does not mean we can ignore their relative motions.
Remember that moving car and the picture you were taking of it. Now imagine there's a man stood in the road a metre away from the car. The car is travelling at 60mph relative to you when you take the photo. By your logic we can look at the photo and go "the car doesn't move in the photo therefore the man is safe".

ozziemate

27th October 2008, 04:59 PM

http://bizzymate.com/uploads/images/physics/grav.gif.....http://bizzymate.com/uploads/images/physics/grav01.gif.....http://bizzymate.com/uploads/images/physics/grav02.gif
an example of how complex this issue can get...

move all intersecting points including the tips of the >'s into a zero dimensional point and you have a proximation of why meteastability in a field of attrraction exists...such as gravity and magnetism.
magnetism : on one side of the plane is north pole and the other is the south pole and they face inversely to the field so that movement is possible and actual at all times showing why nothing is at absolute rest...and I might add there is experimental evidence to support it...

so yeah it can get complex...and you have shown the path to that field of physics that may actually be involved in this sort of thing...

sol invictus

27th October 2008, 05:23 PM

does this image show what you mean, just to be clear....if we imagine the plane as shown it means that we have a single point that is also zero dimensional that has two intersecting lines at angle travaleing through it. That zero dimensional point must be able to show the potential of not only angle but in this case speed or velocity.
http://bizzymate.com/uploads/images/physics/intersecting%20lines.gif

There's no need to have two lines intersecting the plane at the same point - that shows that two objects at the same position at the same time can have different velocities, which is of course true, but a little more than was needed. But fine.

is there a name for this field of study in physics that deals with such dimensional concepts?

I'm not sure what you're asking. Things like that arise in all fields of physics, mathematics, computer science, engineering, and chemistry all the time.

this use of dimensionalism can actually get quite complex.
If I am not mistaken the branch of mathematics - calculus was developed simply from the need to work with such concepts so pooh poohing me about it is not all that valid given that an 8 year old child doesn't get to study calculus in prmary school.

Calculus is not necessary to understand that - that's why I used this analogy. The Greeks understood Euclidean (hint - Euclid) geometry very well, 2,000 years before calculus was invented.

If you want to define the slope of a curve locally, that's calculus. But the fact that straight lines can intersect a plane at any angle - that's trivial, and every child understands it at least intuitively. What maybe wasn't so obvious here was that velocity at an instant is almost exactly the same thing, mathematically - but know you know.

move all intersecting points including the tips of the >'s into a zero dimensional point and you have a proximation of why meteastability in a field of attrraction exists...such as gravity and magnetism.
magnetism : on one side of the plane is north pole and the other is the south pole and they face inversely to the field so that movement is possible and actual at all times showing why nothing is at absolute rest...and I might add there is experimental evidence to support it...

so yeah it can get complex...and you have shown the path to that field of physics that may actually be involved in this sort of thing...

Gibberish.

ozziemate

27th October 2008, 05:24 PM

Just because we cannot define the relative velocity of observers does not mean we can ignore their relative motions.
Remember that moving car and the picture you were taking of it. Now imagine there's a man stood in the road a metre away from the car. The car is travelling at 60mph relative to you when you take the photo. By your logic we can look at the photo and go "the car doesn't move in the photo therefore the man is safe".
so tell me which vehicles are parked and which vehicles are moving....

http://bizzymate.com/uploads/images/physics/sss.jpg

:D

ozziemate

27th October 2008, 05:32 PM

There's no need to have two lines intersecting the plane at the same point - that shows that two objects at the same position at the same time can have different velocities, which is of course true, but a little more than was needed. But fine.

I'm not sure what you're asking. Things like that arise in all fields of physics, mathematics, computer science, engineering, and chemistry all the time.

Calculus is not necessary to understand that - that's why I used this analogy. The Greeks understood Euclidean (hint - Euclid) geometry very well, 2,000 years before calculus was invented.

If you want to define the slope of a curve locally, that's calculus. But the fact that straight lines can intersect a plane at any angle - that's trivial, and every child understands it at least intuitively. What maybe wasn't so obvious here was that velocity at an instant is almost exactly the same thing, mathematically - but know you know.

Gibberish.
sole invictus imagine a plane of zero duration and you see absolutely nothing on it as you are looking at from it's edge. In fact the plane cannot even be seen but only imagined.
yet you know that something exists in that plane and that it has velocity....you see absolutely nothing becaue it has zero duration ...zero...and therefore invisible.

but yet look around you and what do you see? everything is there... yes?

if the moment is zero duration then what deos that tell you about what you see?

so trivialise it as much as you want but to me it is pretty amazing that what you see is purely temporal generated by a contantly changing moment of zero duration...
in this senes the universe we see is a 3 dimesnional rendition of a two dimensional reality

which means that 3 dimensional +time space is purey a temporal pheno

Tubbythin

27th October 2008, 05:41 PM

so tell me which vehicles are parked and which vehicles are moving....

Do you see why your statement was inherently wrong yet?

ozziemate

27th October 2008, 05:56 PM

Do you see why your statement was inherently wrong yet?
In a sense I can but have yet to decide whether the issue is enough to kill the logical proposition.

As the issue comes back to light and the behaviours or photons in it's way of impacting on observer and not the other way round, observer impacting on light. We agree that light is your bench mark and that is why this thead has been structured the way it has.

Observer/light event occur usually because photons arrive at the observer and not the observer arriving at the photon.
So the issue of zero duration rel v observers in the HSP does not disqualify the fact that simultaneity exists due to the nature of the photon and not the observer.
so if all photons in existance exist simultaneously within the HSP then it can be said that regardless of relative Velocity of observer, simultaneity is still maintained. Even if length is contracted or time dilation is present light still can only impact the observer simultaneously with all other observers in the HSP because it is the photons that determine the simultaneity issue and not the observer.
Do you see what I am getting at?

even if you have a universal HSP with relative time involved the photons events are still simultaneous even though percpetions of world views differ....any way I shall work through ti a little more before comenting further. The main pont to recognise I guess that the observer /photon event duration is also zero and not just the HSP duration.

Tubbythin

27th October 2008, 06:09 PM

In a sense I can but have yet to decide whether the issue is enough to kill the logical proposition.

As the issue comes back to light and the bahavius or photons in it;s way of inpacting on observer and nto the other way round, observer impacting on light. We agree that light is your bench mark and that is why this trhead has been structured the way it has.
This thread has sturcture?

Observer/light event occur ususally because photons arrive at the observer and not the observer arriving at the photon.
So the issue of zero duration of the HSP does not disqualify the fact that simultaneity exists due to the nature of the photon and not the observer.
so if all photons in existance exist simultaneously within the HSP then it can be said that regardless of relative Velocity of observer simultaneity is still maintained. even if length is contracted or time dilation is present light still can nonlt impact the observer simultaneously with all other observers in the HSP because it is the photons that determine the simultaneity issue and not the observer.
Do you see what I am getting at?
Not a clue. The point is that you cannot ignore the relative velocities of different observers even in a "zero duration moment". That is why your "logical proof of SRT invalidity" MUST include consideration of observers in different inertial reference frames.

even if you have a universal HSP with relative time involved the photons events are still simultaneous even though percpetions of world views differ....
One other thing to understand is that the results of SRT are not just a perception issue. Length contraction isn't just an optical illusion. The effects are real and measurable.

any way I shall work through ti a little more before comenting further. The main pont to recognise I guess that the event duration is also zero and not just the HSP duration.
The main point for who to realise?

ozziemate

27th October 2008, 06:09 PM

back in 6

Reality Check

27th October 2008, 06:37 PM

sol invictus

27th October 2008, 06:46 PM

sole invictus imagine a plane of zero duration and you see absolutely nothing on it as you are looking at from it's edge.

How do you look at a plane of "zero duration" from its "edge"? Quite a trick, that... if you mean an infinitely thin plane in space that would only make slightly more sense. Have you forgotten that the plane in question is imaginary? Make it thick, make it thin, make it green - you still can't see it, because it's not there.

In fact the plane cannot even be seen but only imagined.

That's because it's imaginary.

Reality Check

27th October 2008, 06:54 PM

but of course we have virtually and infinite number of observers thus frames in our HSP so how many do you want....
By definition the HSP is the present time of an (one) observer. This observer has a reference frame.
A different observer can have a different reference frame and a different HSP.

The nice thing about SRT is that it starts with a spacetime that allows any observer to have any inertial frame of reference and any HSP. Note there is no assumption of simultaneity or non-simultaneity.
Then 2 simple postulates are applied and all of the experimentally verified predictions of SRT emerge.

Fredrik

27th October 2008, 06:59 PM

What is d?
I'm pretty sure that it's the distance represented by the part of a particle's world line that's actually in the HSP.

This is what he said:

If all observer/photon events are simultaneous as all have zero duration in a given moment then all events within the HSP can be considered as inertial frames as relative velocity in the HSP is a nul concept.
as t=0, d must also = 0 [ a tautology. ]

This is what he meant:

If all photon observation events in the HSP are simultaneous in the inertial frame used to define the HSP, then all events in the HSP are stationary in that inertial frame.

(OK, I'm not 100% sure that's what he meant, but it seems very likely).

Reality Check

27th October 2008, 07:06 PM

I am also fairly sure that d = distance but with ozziemate you never know. It might be d = the dot that marks the position of something.

On the other hand he must have had some education or may have even read some of the posts in this thread. So he would know that distance is a difference between 2 points and that t=0 (I assume that this is the time coordinate) is part of the coordinates of a single point.

UncaYimmy

27th October 2008, 07:10 PM

How do you look at a plane of "zero duration" from its "edge"? Quite a trick, that... if you mean an infinitely thin plane in space that would only make slightly more sense. Have you forgotten that the plane in question is imaginary? Make it thick, make it thin, make it green - you still can't see it, because it's not there.

That's because it's imaginary.

Wait a sec...Bigfoot is a two-dimensional plane?

ozziemate

28th October 2008, 12:09 AM

Wait a sec...Bigfoot is a two-dimensional plane?
ha...
naah "big foot" is 1 dimensional and...wait for it....

"big feet" is 2 dim to realise he has only one footl...:D

ozziemate

28th October 2008, 12:11 AM

I'm pretty sure that it's the distance represented by the part of a particle's world line that's actually in the HSP.

This is what he said:

If all observer/photon events are simultaneous as all have zero duration in a given moment then all events within the HSP can be considered as inertial frames as relative velocity in the HSP is a null concept.
as t=0, d must also = 0 [ a tautology. ]

This is what he meant:

If all photon observation events in the HSP are simultaneous in the inertial frame used to define the HSP, then all events in the HSP are stationary in that inertial frame.

(OK, I'm not 100% sure that's what he meant, but it seems very likely).

yeah,,,just thinking about it what I meant was that time is zero therefore distance achieved with the use of time must also be zero [as in t/d ]
the length of the intersecting world line as in t/d is also zero...hard to make sense of it in those terms. But maybe the idea is easy to see...
In absolute terms the HSP can not presume a future exists as the future may not. It is impossible to be 100% sure that the next moment will eventuate.
hmmmm...but still to me this is fudging the issue, even if true....and certainly not good enough for a logical proof...[ possibly as reinforcing logic ]
the logical proof has to be bullet proof if it is to stand up, with out any doubt clear as a bell and as of this morning that proof is getting a little fuzzy around the edges and as it stands it is not going to go any where but end up inconclusive.

so I shall think on it some more.... as I am not interested specifically in trying to tear down a good theory just for the hell of it, I am actually interested in finding a better way if possible, so the proof has to be demonstrably clear and concise.

ozziemate

28th October 2008, 12:45 AM

The question that i am trying to work out is this:

A photon or waves speed is determined by t/d where according to SRT d is variable and dependant on relative velocity? correct?
So Iask myself:
even if we assume length conrtraction and time dilation manages to bring the invariance of the speed of light so that the rel v observers frame expereinces no obvious distortion to his vector, is that photon still in a simultaneous relationship with all other non spacaily and time distorted observers.

So if we have two observers at rel v and space is contracted by 50% for either of them but not both of them and like wise with time dilation by the appropriate amount then I ask:

Is the photons involved in their events still simultaneous with all other photons in a zero duration moment.

say the frozen HSP includes all spacial distortions possible [ big bloody mess of all possible time and length distortions] are the photons involved due to invariance still simultaneous to each other?

I tend to think that invariance would require this absolutely but this would be a hard case to prove.
we are running a two bit scenario
observer is one bit and the photon is the other.
test:
If one observer is chosen the entire universe HSP would be observer/photon simultaeous.
If two observers are involved one observer is relative in time yet is the photon relative in time also?

2 possible outcomes:

If the photon is relative in time also, then invariance is threatened.
If the photon is simultaneous then non-simultaneity is threatened.

Say I use the following cryptic statements:

option pair one |Observs. .....sim | Photons sim
_____________|Observs. nonsim | photons sim

option pair two |Observs. .....sim | Photons sim
_____________|Observs. nonsim | photons non sim
sim= simultaneous and nonsim = non-simultaneous
at this point I cannot see how photon invariance can be maintained with either option....
note the red line in the middle between observer and photons can be used to indicate the simultateous relationship between observer and photon event in all options because we have already agreed that both observer and photon are simultaneously necessary for an event to occur.
as proven in :
http://bizzymate.com/uploads/images/physics/contruct01.jpg

ozziemate

28th October 2008, 01:02 AM

possibly we can use the followng to make a better job of the question :

observer a and observer b
[a=hsp] and

now we have the following construction for both observers but they are not simultaneous due to relative velocity >0
a:hsp has:
http://bizzymate.com/uploads/images/physics/contruct01.jpg
b:hsp has exactly the same construction:
http://bizzymate.com/uploads/images/physics/contruct01.jpg

we know that the observers are non-simultaneous. according to SRT
however are the two [B]photon spheres simultaneous or are they non-simultaneous?

any ways thats the line of thought I am progressing with at the moment...

Reality Check

28th October 2008, 02:00 AM

The question that i am trying to work out is this:

A photon or waves speed is determined by t/d where according to SRT d is variable and dependant on relative velocity? correct?
So Iask myself:
even if we assume length conrtraction and time dilation manages to bring the invariance of the speed of light so that the rel v observers frame expereinces no obvious distortion to his vector, is that photon still in a simultaneous relationship with all other non spacaily and time distorted observers.

So if we have two observers at rel v and space is contracted by 50% for either of them but not both of them and like wise with time dilation by the appropriate amount then I ask:

Is the photons involved in their events still simultaneous with all other photons in a zero duration moment.

say the frozen HSP includes all spacial distortions possible [ big bloody mess of all possible time and length distortions] are the photons involved due to invariance still simultaneous to each other?

I tend to think that invariance would require this absolutely but this would be a hard case to prove.
we are running a two bit scenario
observer is one bit and the photon is the other.
test:
If one observer is chosen the entire universe HSP would be observer/photon simultaeous.
If two observers are involved one observer is relative in time yet is the photon relative in time also?

2 possible outcomes:
If the photon is relative in time also, then invariance is threatened.
If the photon is simultaneous then non-simultaneity is threatened.
Say I use the following cryptic statements:

option pair one |Observs. .....sim | Photons sim
_____________|Observs. nonsim | photons sim

option pair two |Observs. .....sim | Photons sim
_____________|Observs. nonsim | photons non sim
sim= simultaneous and nonsim = non-simultaneous
at this point I cannot see how photon invariance can be maintained with either option....
note the red line in the middle between observer and photons can be used to indicate the simultateous relationship between observer and photon event in all options because we have already agreed that both observer and photon are simultaneously necessary for an event to occur.
as proven in :
http://bizzymate.com/uploads/images/physics/contruct01.jpg

You really have gone off on a tangent fueled by ignorance.

What are the two postulates of the Theory of Special Relativity?

ozziemate

28th October 2008, 02:04 AM

You really have gone off on a tangent fueled by ignorance.

What are the two postulates of the Theory of Special Relativity?

1. The Principle of Relativity - The laws by which the states of physical systems undergo change are not affected, whether these changes of state be referred to the one or the other of two systems in uniform translatory motion relative to each other.
2. The Principle of Invariant Light Speed - Light in vacuum propagates with the speed c (a fixed constant) in terms of any system of inertial coordinates, regardless of the state of motion of the light source.
wiki

so...?

what do you think I am attempting to do?

Reality Check

28th October 2008, 02:04 AM

Reality Check

28th October 2008, 02:12 AM

1. The Principle of Relativity - The laws by which the states of physical systems undergo change are not affected, whether these changes of state be referred to the one or the other of two systems in uniform translatory motion relative to each other.
2. The Principle of Invariant Light Speed - Light in vacuum propagates with the speed c (a fixed constant) in terms of any system of inertial coordinates, regardless of the state of motion of the light source.
wiki

so...?
So...

ozziemate

28th October 2008, 02:14 AM

t=0 is only right for a single observer where we can assume that they count from a time of zero.

You need to put:
t=ta for observer a.
t=tb for observer b.
ta may be zero.
ta need not be the same as tb unless the observers have synchronized their clocks.
Maybe you can easilly answer this question .

we know that the observers are non-simultaneous. according to SRT
however are the two photon spheres simultaneous or are they non-simultaneous?

Reality Check

28th October 2008, 02:27 AM

Maybe you can easilly answer this question .
we know that the observers are non-simultaneous. according to SRT
however are the two photon spheres simultaneous or are they non-simultaneous?

We know that the two observers may be non-simultaneous according to SRT.
The 2 photon spheres are just sets of points in spacetime. They will be non-simultaneous or simultaneous according to whether the observers are non-simultaneous or simultaneous.

Tubbythin

28th October 2008, 02:30 AM

we know that the observers are non-simultaneous. according to SRT
however are the two photon spheres simultaneous or are they non-simultaneous?

This has been explained earlier by someone else. If a light source emits light isotropically then an observer that is stationary relative to the source will see the wavefronts as spherical. An observer with a constant and non-zero velocity relative to the source will not.

ozziemate

28th October 2008, 02:30 AM

ahh this reminds me of a discussion I had with a retired Russian Phsyicist who did a significant amount of math to prove just this issue, spent his whole life time supporting SRT and GR and when he got the crux of his math work on this issue he stopped and disappeared off the internet under his user name. Haven't heard from him since and that was about 5 years ago.

ozziemate

28th October 2008, 02:32 AM

This has been explained earlier by someone else. If a light source emits light isotropically then an observer that is stationary relative to the source will see the wavefronts as spherical. An observer with a constant and non-zero velocity relative to the source will not.
ahh but is that an observer issue or is that a light issue....

are the photon spheres simultaneous is the question currently under test.
and am still thinking on it I might add.

talking about what the obserer sees is irrelevant I feel.

as we are talking about simultaneity of the photon sphere universally

ozziemate

28th October 2008, 02:35 AM

We know that the two observers may be non-simultaneous according to SRT.
The 2 photon spheres are just sets of points in spacetime. They will be non-simultaneous or simultaneous according to whether the observers are non-simultaneous or simultaneous.
yeah that is the standard SRT position and have heard it many times before...hmmmm it means of course that the photons across all rfs must now be non-simultaneous.

and I just wanted to confirm that that was the case.

Tubbythin

28th October 2008, 02:37 AM

The question that i am trying to work out is this:

A photon or waves speed is determined by t/d where according to SRT d is variable and dependant on relative velocity? correct?
No. speed = distance/time. d is variable from frame to frame but so is t such that the speed of light is invariant.

So Iask myself:
even if we assume length conrtraction and time dilation manages to bring the invariance of the speed of light so that the rel v observers frame expereinces no obvious distortion to his vector, is that photon still in a simultaneous relationship with all other non spacaily and time distorted observers.

You cannot say this observer is distorted and this observer is not. This is assuming there is a 'special' reference frame. You can only describe how different observers agree or disagree on events and objects.

So if we have two observers at rel v and space is contracted by 50% for either of them but not both of them and like wise with time dilation by the appropriate amount then I ask:

Is the photons involved in their events still simultaneous with all other photons in a zero duration moment.
I don't know what you're trying to ask here.

say the frozen HSP includes all spacial distortions possible [ big bloody mess of all possible time and length distortions] are the photons involved due to invariance still simultaneous to each other?
It doesn't. Therefore this question is nonsensical.

ozziemate

28th October 2008, 02:42 AM

in the object /photon rfs the folowing is now what is suggeste to be the case:

observer ....A hsp : sim |photonsphere: sim
observer B hsp: nonsim | photonsphere: nonsim

which means at some point the photonsphere must have lost simultaneity as one frame accelerated from a position of rel v=0

...just thinking....

Tubbythin

28th October 2008, 02:43 AM

ahh but is that an observer issue or is that a light issue....
I'm not sure what you mean. If you mean "is this an optical illusion?" the answer is no.

are the photon spheres simultaneous is the question currently under test.
and am still thinking on it I might add.
Erm. Don't really understand what you're trying to ask. You need to define two events before you can can ask whether the events are simultaneous.

talking about what the obserer sees is irrelevant I feel.
Huh? If we're not talking about observers what are we talking about?

as we are talking about simultaneity of the photon sphere universally
Like I said, the wavefronts are only spherical in frames at rest with respect to the source.

Tubbythin

28th October 2008, 02:44 AM

Reality Check

28th October 2008, 02:45 AM

yeah that is the standard SRT position and have heard it many times before...hmmmm it means of course that the photons across all rfs must now be non-simultaneous.

and I just wanted to confirm that that was the case.
You did read the post correctly. What you meant was
"hmmmm it means of course that the photons across all rfs may or may not now be non-simultaneous."

ETA: But even that is not totally correct. It is the comparison between the observer's and their 'photon spheres' that tells whether the photons in those spheres can be called non-simultaneous or simultaneous.

Tubbythin

28th October 2008, 02:46 AM

ozziemate

28th October 2008, 03:13 AM

There is no rest frame for the photon.
read post #656 as to what I have been going on about

ozziemate

28th October 2008, 03:22 AM

You did read the post correctly. What you meant was
"hmmmm it means of course that the photons across all rfs may or may not now be non-simultaneous."

ETA: But even that is not totally correct. It is the comparison between the observer's and their 'photon spheres' that tells whether the photons in those spheres can be called non-simultaneous or simultaneous.
I am attempting to put together a coherant gedanken to show this issue so bear with me if you can...

basically
1] observer ship A and observer ship B are at rel v = 0
the hsp and photonsphere are simultaneous in both aspects - observer and photon.

http://bizzymate.com/uploads/images/physics/contruct01.jpg

2] Observer B accellerates away to .5c
The hsp and photonsphere are no longer simultaneuous in both aspects - observer ship and photon.
Observer A has
http://bizzymate.com/uploads/images/physics/contruct01.jpg
Observer B has
http://bizzymate.com/uploads/images/physics/contruct01.jpg

so how did we get from the initial frame of both obsrevers being simultaneous in all aspects to A and B's non-simultaneity position with out breaking invariance.?

again just thinking out loud....

Reality Check

28th October 2008, 03:28 AM

Reality Check

28th October 2008, 03:43 AM

I am attempting to put together a coherant gedanken to show this issue so bear with me if you can...

basically
1] observer ship A and observer ship B are at rel v = 0
the hsp and photonsphere are simultaneous in both aspects - observer and photon.

http://bizzymate.com/uploads/images/physics/contruct01.jpg

2] Observer B accellerates away to .5c
The hsp and photonsphere are no longer simultaneuous in both aspects - observer ship and photon.
Observer A has
http://bizzymate.com/uploads/images/physics/contruct01.jpg
Observer B has
http://bizzymate.com/uploads/images/physics/contruct01.jpg

so how did we get from the initial frame of both obsrevers being simultaneous in all aspects to A and B's non-simultaneity position with out breaking invariance.?

again just thinking out loud....
To be more exact: basically you have it wrong:
observer ship A has their own set of coordinates (use the image with a xB, yB, zB, t=tA) and their own photon sphere).
observer ship B has their own set of coordinates (use the image with A replaced by B) and their own photon sphere).

"Observer B accellerates away to .5c". At the end thay have own set of coordinates (use the image with B replaced by C) and their own photon sphere).

Now take the posulates of your logical proof and find the transformation between the A and C coordinates. This tells you how observer A sees observer B and how observer B sees observer A.
You can now state whether the 2 observer's are simultaneous or non-simultaneous.

Tubbythin

28th October 2008, 05:55 AM

Tubbythin

28th October 2008, 05:57 AM

read post #656 as to what I have been going on about

This doesn't stop the fact that photons don't have a rest frame.

Thabiguy

28th October 2008, 07:33 AM

This has been explained earlier by someone else. If a light source emits light isotropically then an observer that is stationary relative to the source will see the wavefronts as spherical. An observer with a constant and non-zero velocity relative to the source will not.

(Actually, all observers will see the light wavefront as spherical at any given moment. This is a direct consequence of the invariance of speed of light. But it doesn't mean that events that happen to these photons retain their simultaneity across reference frames.)

(But don't let this distract the current thread of conversation.)

Fredrik

28th October 2008, 07:53 AM

now we have the following construction for both observers but they are not simultaneous due to relative velocity >0
a:hsp has:
http://bizzymate.com/uploads/images/physics/contruct01.jpg
b:hsp has exactly the same construction:
http://bizzymate.com/uploads/images/physics/contruct01.jpg

...except that the t,x,y,z in B's construction aren't the same as the t,x,y,z in A's construction. You could e.g. put primes on B's coordinates to distinguish them from A's: (t',x',y',z').

we know that the observers are non-simultaneous. according to SRT
however are the two photon spheres simultaneous or are they non-simultaneous?

The first sentence makes no sense. Only events can be simultaneous or non-simultaneous. Hm, OK, I think I get it. You mean that the observers disagree about which events are simultaneous. At least I hope you do.

The photon spheres are subsets of spacetime, not events, so the only way to make sense of the question is to interpret it as "Are all the events in the union of A's or B's photon spheres simultaneous?". (The union of two sets X and Y is the set that consists of all the members of X and all the members of Y). The answer is no. The events in A's photon sphere are simultaneous in A's frame, and the events in B's photon sphere are simultaneous in B's frame, but you can definitely pick two points in the union that both observers will agree are not simultaneous.

Fredrik

28th October 2008, 08:18 AM

A few pages back, I described how to draw the spacetime diagram:

1. Draw a point. This is the event where the photons are emitted.
2. Draw a a straight vertical line through the point. This is the world line of the light source.
3. Draw a cone with the tip at the emission event, and the rest of the cone located above that point. The cone must be pointing straight down, and the angle between the surface of the cone and a horizontal plane should be 45 degrees. The cone represents the motion of photons away from the source.
4. Draw a horizontal plane that intersects the cone somewhere (anywhere) above the emission event. This is your HSP.
5. Draw the circle where the plane intersects the cone. This circle represents the photon sphere at a specific time.

Now that you're talking about a second observer, you should know how to include him in the diagram:

6. Draw a straight line through the point where the world line of the light source intersects the HSP. Make it almost vertical but slightly tilted to the right (clockwise). This is B's world line, and therefore it's also the time axis of his coordinate system.
7. Draw a second plane through that point. Make it tilted in the exact opposite direction, by an amount that makes the angle between this plane and the HSP the same as the angle between the two world lines. This is B's version of the HSP. It has to be tilted that way to ensure that the speed of light is the same in B's frame. (Note that this is the first time we have used any part of special relativity in the construction of this diagram).

It would be a good exercise for you to draw this diagram.

Fredrik

28th October 2008, 08:54 AM

You just introduced a non-inertial reference frame (see bolding). And thus went outside of SRT.
Ouside of the original 1905 version of SR, but not outside of what the experts today have in mind when they talk about special relativity. This includes a few things that were introduced by general relativity: The manifold structure of spacetime (which among other things includes non-inertial coordinate systems) and a more sophisticated set of postulates about identifications between the mathematical model and the real world.

sol invictus

28th October 2008, 10:47 AM

Ouside of the original 1905 version of SR, but not outside of what the experts today have in mind when they talk about special relativity. This includes a few things that were introduced by general relativity: The manifold structure of spacetime (which among other things includes non-inertial coordinate systems) and a more sophisticated set of postulates about identifications between the mathematical model and the real world.

That's right.

General relativity is really a theory of gravity (the response of the metric to the presence of mass/energy/pressure/stress). Non-inertial observers in flat space experience various bizarre and interesting phenomena, but one doesn't need the full machinery of GR to figure out what they are. In fact I think all they all follow from the postulates of SR, plus the (almost tautological) requirement that physics be invariant under changes of coordinates.

Tubbythin

28th October 2008, 11:38 AM

(Actually, all observers will see the light wavefront as spherical at any given moment. This is a direct consequence of the invariance of speed of light. But it doesn't mean that events that happen to these photons retain their simultaneity across reference frames.)

(But don't let this distract the current thread of conversation.)

(Yes. Thats a good point. I was thinking that if that the wavefronts define a sphere in the source's rest frame and a sphere looks ellipsoidal when you move relative to it then the wavefronts must be non-spherical. I should really think about these things more closely.
I'm right in thinking that if we had a stationary (relative to the source) physical sphere centred on the source, an observer in a different inertial reference frame would say the sphere is squashed (ie ellipsoidal) and the wavefront arrives at different parts of the "sphere" at different times?)

Tubbythin

28th October 2008, 11:40 AM

Ziggurat

28th October 2008, 11:51 AM

(Yes. Thats a good point. I was thinking that if that the wavefronts define a sphere in the source's rest frame and a sphere looks ellipsoidal when you move relative to it then the wavefronts must be non-spherical. I should really think about these things more closely.
I'm right in thinking that if we had a stationary (relative to the source) physical sphere centred on the source, an observer in a different inertial reference frame would say the sphere is squashed (ie ellipsoidal) and the wavefront arrives at different parts of the "sphere" at different times?)

Yes, that is correct. The physical sphere is also moving, so it will be off-center when the wave front first touches it on one side.

ozziemate

28th October 2008, 03:54 PM

Please read the entire post before posting to save time later

If Fedrik can offer critique as well to this commentary as he composed the construction I would appreciate it

Lets refine the gedanken a little and chase an answer to this question:

If A and B are separating at the rate of .5c and do so for 1 million years [according to observer A], how much separation in time exists between their respective zero duration moments?

Both observer A and observer B are sitting on a space platform [ plat ]. Observer A and B are at rest regarding each other and the platform.

Both observer A,B share the same HSP
http://bizzymate.com/uploads/images/physics/contruct01.jpg

Now Observer B leaves the platform and heads in a straight line away from the platform and observer A.

Eventually he achieves a steady v=.05c and holds this v and vector indefinitely.

At this stage we have :

Observer A and the platform with a HSP of
http://bizzymate.com/uploads/images/physics/contruct01.jpg [with a zero duration.]
and we have
Observer B has a HSP the same from his perspective:
http://bizzymate.com/uploads/images/physics/contruct01.jpg
[with zero duration ]

[the reason I am leaving the reference to the observer out of the construction image is to maintain what I called global or universal perspective.]
<>
Now we can confirm at this stage that both Observers A and B are no longer in a simultaneous zero duration moment HSP in fact they each have their own unique zero duration HSP and their zero duration moments are different in a universal time line sense. [ universal means [ all rel.v frames included ]

Point 1.
to declare two frames as non-simultaneous one has to know what simultaneous is. Thus the need even if imaginary to use a universal/world time line.

So from a universal/global point of view the mathematical construction:
http://bizzymate.com/uploads/images/physics/contruct01.jpg
is now invalid. To be valid it must include at least 2 zero duration moments with in the single formulation/construction. - no big deal...ok

SRT treats the non-simultaneity issue symmetrically, by this I mean that form one Observers perspective the other is non-simultaneous when viewed from his perspective using the original HSP as a way of determining the degree of non-simultaneity.
What this means is one observer takes his HSP has a look from the other observers HSP and compares HSP's therefore concluding and measuring the degree of non-sim between those HSP's.
Symmetrically means that observers can swap symmetrically from one perspective to the other. [ except when one considerer's acceleration I believe as this is handled differently]

However ignoring acceleration at the moment for the sake of clarity we can conclude how much time separates the two zero duration HSP"s

So if we include in our gedanken the over all time of relative v to say million light years of time at rel.v .05

How much time would separate the zero duration moments of the two observers Hsp's?

How non-simultaneous are they?

How would you create the contruction that would show how much non-simultaneity exists by describing the amount of time separation between the two zero duration moments of both HSP's?

note:
The Russian physicists I referred to earlier attempted to do this and unfortunately for reasons he only knows failed to complete the task.

The other issue is that originally we had only one HSP/Photon sphere.

and now we have two hsp/photon spheres that are non-simultaneous.

If say the non-simultaneity was calculated to say 1000 years separation in the universal/global time line that is to say that one RF can be seen to be 1000 years into the future or past of the other RF it means that the light information from one observer is either 1000 years into the future of the other or 1000 years into the past of the other.
which of course makes little sense because what tis simply implies i s that once a RF accelerates to rel.v it is now traveling into the future or past and thus time traveling Adan not just merely non-simultaneous.

so to sum up:

Does non-simultaneity outcome of SRT generate a state of actually time travel for the appropriate Ref frame or does that frame maintain the same zero duration moment as the other yet retain non-simultaneity of time and the relativity of distance along vector?

If you calculate the amount of time separating the two zero duration moments [ the degree of non-simultaneity ] you will find I think that SRT is a time traveling model, sending one Reference frame in to the future and because of symmetry issues the other one can also go in to the future..depending on which perspective you use.

So,
If A and B are separating at the rate of .5c and do so for 1 million years [according to observer A], how much separation in time exists between their respective zero duration moments?

Reality Check

28th October 2008, 04:43 PM

Please read the entire post before posting to save time later

If Fedrik can offer critique as well to this commentary as he composed the construction I would appreciate it

Lets refine the gedanken a little and chase an answer to this question:

If A and B are separating at the rate of .5c and do so for 1 million years how much separation in time exists between their respective zero duration moments?

Both observer A and observer B are sitting on a space platform [ plat ]. Observer A and B are at rest regarding each other and the platform.

Both observer A,B share the same HSP
http://bizzymate.com/uploads/images/physics/contruct01.jpg

Now Observer B leaves the platform and heads in a straight line away from the platform and observer A.

Eventually he achieves a steady v=.05c and holds this v and vector indefinitely.

At this stage we have :

Observer A and the platform with a HSP of
http://bizzymate.com/uploads/images/physics/contruct01.jpg [with a zero duration.]
and we have
Observer B has a HSP the same from his perspective:
http://bizzymate.com/uploads/images/physics/contruct01.jpg
[with zero duration ]

[the reason I am leaving the reference to the observer out of the construction image is to maintain what I called global or universal perspective.]
<>
Now we can confirm at this stage that both Observers A and B are no longer in a simultaneous zero duration moment HSP in fact they each have their own unique zero duration HSP and their zero duration moments are different in a universal time line sense. [ universal means [ all rel.v frames included ]

Point 1.
to declare two frames as non-simultaneous one has to know what simultaneous is. Thus the need even if imaginary to use a universal/world time line.
So from a universal/global point of view the mathematical construction:
http://bizzymate.com/uploads/images/physics/contruct01.jpg
is now invalid. To be valid it must include at least 2 zero duration moments with in the single formulation/construction. - no big deal...ok

SRT treats the non-simultaneity issue symmetrically, by this I mean that form one Observers perspective the other is non-simultaneous when viewed from his perspective using the original HSP as a way of determining the degree of non-simultaneity.
What this means is one observer takes his HSP has a look from the other observers HSP and compares HSP's therefore concluding and measuring the degree of non-sim between those HSP's.
Symmetrically means that observers can swap symmetrically from one perspective to the other. [ except when one considerer's acceleration I believe as this is handled differently]

However ignoring acceleration at the moment for the sake of clarity we can conclude how much time separates the two zero duration HSP"s

So if we include in our gedanken the over all time of relative v to say million light years of time at rel.v .05

How much time would separate the zero duration moments of the two observers Hsp's?

How non-simultaneous are they?

How would you create the contruction that would show how much non-simultaneity exists by describing the amount of time separation between the two zero duration moments of both HSP's?

note:
The Russian physicists I referred to earlier attempted to do this and unfortunately for reasons he only knows failed to complete the task.

The other issue is that originally we had only one HSP/Photon sphere.

and now we have two hsp/photon spheres that are non-simultaneous.

If say the non-simultaneity was calculated to say 1000 years separation in the universal/global time line that is to say that one RF can be seen to be 1000 years into the future or past of the other RF it means that the light information from one observer is either 1000 years into the future of the other or 1000 years into the past of the other.
which of course makes little sense because what tis simply implies i s that once a RF accelerates to rel.v it is now traveling into the future or past and thus time traveling Adan not just merely non-simultaneous.

so to sum up:

Does non-simultaneity outcome of SRT generate a state of actually time travel for the appropriate Ref frame or does that frame maintain the same zero duration moment as the other yet retain non-simultaneity of time and the relativity of distance along vector?

If you calculate the amount of time separating the two zero duration moments [ the degree of non-simultaneity ] you will find I think that SRT is a time traveling model, sending one Reference frame in to the future and because of symmetry issues the other one can also go in to the future..depending on which perspective you use.

So,
If A and B are separating at the rate of .5c and do so for 1 million years how much separation in time exists between their respective zero duration moments?
Can you give us the citations to "Russian physicists I referred to earlier" and what was their task that failed?

There is not enough information in the gedanken to determine the answers to your questions.

You last question is ambiguous. Whose time are you talking about?
IMHO: The time between between their respective HSPs will be zero. They started with the same HSP, i.e. their clocks are synchronized. General relativity will make the accelerating observer's HSP different from observer A's HSP. But we will ignore this since you don't like SRT and thus probably think that GR is invalid as well.

Observer A may measure that observer B has a different HSP from when they separated. Observer B may measure that observer A has a different HSP from when they separated. It all depends on the transformation between the observer's coordinates that I assume you are about to deduce.

ETA: About '[the reason I am leaving the reference to the observer out of the construction image is to maintain what I called global or universal perspective.]'
There is no such thing as a 'global or universal perspective' - you need an observer to have a perspective.

The Man

28th October 2008, 04:59 PM

Please read the entire post before posting to save time later

If Fedrik can offer critique as well to this commentary as he composed the construction I would appreciate it

Lets refine the gedanken a little and chase an answer to this question:

If A and B are separating at the rate of .5c and do so for 1 million years how much separation in time exists between their respective zero duration moments?

Both observer A and observer B are sitting on a space platform [ plat ]. Observer A and B are at rest regarding each other and the platform.

Both observer A,B share the same HSP
http://bizzymate.com/uploads/images/physics/contruct01.jpg

Now Observer B leaves the platform and heads in a straight line away from the platform and observer A.

Eventually he achieves a steady v=.05c and holds this v and vector indefinitely.

At this stage we have :

Observer A and the platform with a HSP of
http://bizzymate.com/uploads/images/physics/contruct01.jpg [with a zero duration.]
and we have
Observer B has a HSP the same from his perspective:
http://bizzymate.com/uploads/images/physics/contruct01.jpg
[with zero duration ]

[the reason I am leaving the reference to the observer out of the construction image is to maintain what I called global or universal perspective.]
<>
Now we can confirm at this stage that both Observers A and B are no longer in a simultaneous zero duration moment HSP in fact they each have their own unique zero duration HSP and their zero duration moments are different in a universal time line sense. [ universal means [ all rel.v frames included ]

Point 1.
to declare two frames as non-simultaneous one has to know what simultaneous is. Thus the need even if imaginary to use a universal/world time line.

So from a universal/global point of view the mathematical construction:
http://bizzymate.com/uploads/images/physics/contruct01.jpg
is now invalid. To be valid it must include at least 2 zero duration moments with in the single formulation/construction. - no big deal...ok

SRT treats the non-simultaneity issue symmetrically, by this I mean that form one Observers perspective the other is non-simultaneous when viewed from his perspective using the original HSP as a way of determining the degree of non-simultaneity.
What this means is one observer takes his HSP has a look from the other observers HSP and compares HSP's therefore concluding and measuring the degree of non-sim between those HSP's.
Symmetrically means that observers can swap symmetrically from one perspective to the other. [ except when one considerer's acceleration I believe as this is handled differently]

However ignoring acceleration at the moment for the sake of clarity we can conclude how much time separates the two zero duration HSP"s

So if we include in our gedanken the over all time of relative v to say million light years of time at rel.v .05

How much time would separate the zero duration moments of the two observers Hsp's?

How non-simultaneous are they?

How would you create the contruction that would show how much non-simultaneity exists by describing the amount of time separation between the two zero duration moments of both HSP's?

note:
The Russian physicists I referred to earlier attempted to do this and unfortunately for reasons he only knows failed to complete the task.

The other issue is that originally we had only one HSP/Photon sphere.

and now we have two hsp/photon spheres that are non-simultaneous.

If say the non-simultaneity was calculated to say 1000 years separation in the universal/global time line that is to say that one RF can be seen to be 1000 years into the future or past of the other RF it means that the light information from one observer is either 1000 years into the future of the other or 1000 years into the past of the other.
which of course makes little sense because what tis simply implies i s that once a RF accelerates to rel.v it is now traveling into the future or past and thus time traveling Adan not just merely non-simultaneous.

so to sum up:

Does non-simultaneity outcome of SRT generate a state of actually time travel for the appropriate Ref frame or does that frame maintain the same zero duration moment as the other yet retain non-simultaneity of time and the relativity of distance along vector?

If you calculate the amount of time separating the two zero duration moments [ the degree of non-simultaneity ] you will find I think that SRT is a time traveling model, sending one Reference frame in to the future and because of symmetry issues the other one can also go in to the future..depending on which perspective you use.

So,
If A and B are separating at the rate of .5c and do so for 1 million years how much separation in time exists between their respective zero duration moments?

Do you mean elapsed time, since time is not actually a “separation”? We can say how much time passed between two events from a certain reference frame but that does not mean those events are separated by that amount of time. Another observer in another reference frame might see the events as simultaneous or not separated in time (as has been explained to you). “Separation” generally refers to the space-time separation and takes the form s2 =(c2*t2)- x2 - y2 - z2

A space like separation refers to events where not enough time passes between there occurrences for there to be a causal influence (would require superluminal travel of transfer of information)

A light like separation refers to an to events where the spatial and temporal values are balanced (s2 = 0) or in other words along the world line of a photon.

A time like separation refers to events where enough time passes between events for a causal relationship.

It might be more understandable if you phased your “gedanken” and questions within those considerations and considered as well as expressed the precise reference frame for your “1 million years” time frame, since chances are in that frame the answer will be “1 million years” of elapsed time.

ozziemate

28th October 2008, 05:04 PM

Can you give us the citations to "Russian physicists I referred to earlier" and what was their task that failed?
no, as it was anonomous internet forum posting and also he faile dto complete his task thus any reference is merely by the by.

There is not enough information in the gedanken to determine the answers to your questions.
what information would you like to include?

You last question is ambiguous. Whose time are you talking about?
either observers time will do...as the situation can be treated symmetrically [ not including accelleration at this stage]
IMHO: The time between between their respective HSPs will be zero. They started with the same HSP, i.e. their clocks are synchronized. General relativity will make the accelerating observer's HSP different from observer A's HSP. But we will ignore this since you don't like SRT and thus probably think that GR is invalid as well.
and thus this poses a problem for SRT....If both hsps share the same zero moment we have simultaneity of the phton sphere. but still have relative distances due to length contraction.

General relativity will make the accelerating observer's HSP different from observer A's HSP. But we will ignore this since you don't like SRT and thus probably think that GR is invalid as well
what I do and what I don't like is not relevant. I happen to like SRT really very much. It is an amazing piece of early 1900's scientific work.

Observer A may measure that observer B has a different HSP from when they separated. Observer B may measure that observer A has a different HSP from when they separated. It all depends on the transformation between the observer's coordinates that I assume you are about to deduce.
given the complexity of the task I do not think doing a transform using the standrad Lorenz form is going to achieve the detail required.
And I will probably stuff the math up any way.

ETA: About '[the reason I am leaving the reference to the observer out of the construction image is to maintain what I called global or universal perspective.]'
There is no such thing as a 'global or universal perspective' - you need an observer to have a perspective.
However to calculate the degree of separation in time between zero duration HSPs we need to use a universal/global time line or at the very least derive one for our two observers at rel.v.

To say that non-simultaneity exist is well and good but this has to be quantified with more detail.
And even if it does not provide logical proof to invalidate SRT it will provide a great tool for clarifying the non-sim issue.

ozziemate

28th October 2008, 05:26 PM

Do you mean elapsed time, since time is not actually a “separation”? We can say how much time passed between two events from a certain reference frame but that does not mean those events are separated by that amount of time. Another observer in another reference frame might see the events as simultaneous or not separated in time (as has been explained to you). “Separation” generally refers to the space-time separation and takes the form s2 =(c2*t2)- x2 - y2 - z2

A space like separation refers to events where not enough time passes between there occurrences for there to be a causal influence (would require superluminal travel of transfer of information)

A light like separation refers to an to events where the spatial and temporal values are balanced (s2 = 0) or in other words along the world line of a photon.

A time like separation refers to events where enough time passes between events for a causal relationship.

It might be more understandable if you phased your “gedanken” and questions within those considerations and considered as well as expressed the precise reference frame for your “1 million years” time frame, since chances are in that frame the answer will be “1 million years” of elapsed time.
Thanks for taking the gedanken seriously.
I have ammended the posted gedanken to include that the 1 million years is according to observer A's clock.
and that should supply the necessary information to work something up.

we are discussing non-simultaneity of a very fundamental part of universal physics. it is not time duration between zero duration moments of each observer as according to SRT this time duration of separation is nonexistant. [ i think]
Meaning that the separation between zero duration hsp's at their repsective t=0 is only a separation in time only [ according to what ever observer you wish to utilise] and maybe we shall stick with observer A's clock all the way.
What I am trying to say is that
if we find for example that observer A and observer B are separated by 1000 years then
when observer A takes a snap shot of his HSP obsrever B will be at that instant 1000 years of A's time into the future or past [ I don't know which]
so in effect there is zero tiem duration between past and fture events with 1000 years of separation...shesh...tis tough to explain...
So,
A = t=0 hsp
B = t=0 hsp
yet 1000 years may separate those moments frmo A's perspective and as it is symmetrical B also woud find the same result.

So if we stick to A as a preference, the results can be swapped latter.

ozziemate

28th October 2008, 05:29 PM

example:
the duration of time between A and B may be zero
Yet the separation may be 1000 years according to either clocks as one of the observers is in the future but being observed in his HSP t=0
any ways something to think about...<:)>

Fredrik

28th October 2008, 05:38 PM

If A and B are separating at the rate of .5c and do so for 1 million years how much separation in time exists between their respective zero duration moments?

If you're asking what special relativity's answer is, the answer is that the question doesn't quite make sense. A's hypersurface of simultaneity is represented by a horizontal plane in a 2+1-dimensional spacetime diagram, B's hypersurface of simultaneity is represented by another plane. B's plane is tilted so that it makes a 25.6 degree angle with A's plane. If the second observer is moving to the right in the diagram, then it's the part of the plane that's to the right that's tilted "up" and the part to the left that's tilted "down".

You asked about the separation in time between these two planes. That's equivalent to asking about their separation in the vertical direction in the diagram. But it would only make sense to ask that if the planes were parallel, and they're not.

Both observer A and observer B are sitting on a space platform [ plat ]. Observer A and B are at rest regarding each other and the platform.

Both observer A,B share the same HSP
http://bizzymate.com/uploads/images/physics/contruct01.jpg

Now Observer B leaves the platform and heads in a straight line away from the platform and observer A.

Eventually he achieves a steady v=.05c and holds this v and vector indefinitely.

At this stage we have :

Observer A and the platform with a HSP of
http://bizzymate.com/uploads/images/physics/contruct01.jpg [with a zero duration.]
and we have
Observer B has a HSP the same from his perspective:
http://bizzymate.com/uploads/images/physics/contruct01.jpg
[with zero duration ]

As I said before, you have to put primes or something on B's coordinates to distinguish them from A's coordinates.

[the reason I am leaving the reference to the observer out of the construction image is to maintain what I called global or universal perspective.]
<>

You can't do that. We defined the HSP as the set of all events (t,x,y,z) with t=0 in A's frame, and B's version of the HSP as the set of all events (t',x',y',z') with t'=0 in B's frame. You can't assume that B's version of the HSP can also be expressed as (t,x,y,z) with t=0 (or whatever it is you're doing). That's what you're trying to prove, so you can't just assume it to be true. You're only confusing yourself by using the same symbols for different variables.

Now we can confirm at this stage that both Observers A and B are no longer in a simultaneous zero duration moment HSP...

This doesn't make sense to me. Observers are represented by curves (like the vertical line representing the light source). "Moments" (like the HSP) are represented by surfaces. A particular moment for a particular observer would be the event where the curve intersects the surface. To say that an observer isn't "in" a moment doesn't make sense.

...in fact they each have their own unique zero duration HSP and their zero duration moments are different in a universal time line sense. [ universal means [ all rel.v frames included ]

This is actually true. Everyone agrees that the two planes I described earlier are not the same. But your job is to prove that they are the same, or at least parallel.

to declare two frames as non-simultaneous one has to know what simultaneous is.

That's a good point, but you haven't defined simultaneity in any of your posts. I did in two of mine, but you haven't commented those.

ozziemate

28th October 2008, 06:04 PM

This is actually true. Everyone agrees that the two planes I described earlier are not the same. But your job is to prove that they are the same, or at least parallel.
fredrik,
With this gedanken I am not tryng to prove simulatenity in fact quite the opposite. I am asking SRT to prove non-simultaneity in a way that clarifies the degree of that non-simultaneiity.
And to determine whether of not non-simultaneity implies actual time travel.
In other words SRT shall show us the solutions to the issue of this thread.

which was summed up initially with:

"If all photons in existsance exist simultaneously , then it follows that all observation/ phton events must be simultaneous also."
In a Galalein universe I imagine there would be only one hsp for all time and observers. [ as the only thing that exists is a HSP as everything else is temporal ]
where as in SRT we can have an infinite number of HSPs due to non-simultaneity issues.
With this SRT gedanken I am attempting to define the time separation difference of their respective HSp's between at least, in this case, two rel. v obserers A and B. using A's clocks as a preferred reference and then swapping later to B's

Reality Check

28th October 2008, 06:12 PM

no, as it was anonomous internet forum posting and also he faile dto complete his task thus any reference is merely by the by.

what information would you like to include?

either observers time will do...as the situation can be treated symmetrically [ not including accelleration at this stage]

and thus this poses a problem for SRT....If both hsps share the same zero moment we have simultaneity of the phton sphere. but still have relative distances due to length contraction.

what I do and what I don't like is not relevant. I happen to like SRT really very much. It is an amazing piece of early 1900's scientific work.

given the complexity of the task I do not think doing a transform using the standrad Lorenz form is going to achieve the detail required.
And I will probably stuff the math up any way.

However to calculate the degree of separation in time between zero duration HSPs we need to use a universal/global time line or at the very least derive one for our two observers at rel.v.

To say that non-simultaneity exist is well and good but this has to be quantified with more detail.
And even if it does not provide logical proof to invalidate SRT it will provide a great tool for clarifying the non-sim issue.

From your latest post this is not a gedanken but a request for what SRT will state in this situation so:
Before the acceleration:
Observer A and Observer B were at rest with synchronized clocks.
After the acceleration:
Observer A is at rest.
Observer B is travelling at 0.5c.
Observer A's clock and Observer B's clock are still synchronized (we will ignore the acceleration and what GR might say about this).

Does this answer the question?

The Man

28th October 2008, 06:19 PM

Thanks for taking the gedanken seriously.

No problem as it seems you are now willing to include aspects of SRT in your attempt to refute it, a critical aspect of being taken seriously in that attempt.

I have ammended the posted gedanken to include that the 1 million years is according to observer A's clock.
and that should supply the necessary information to work something up.

Well that gives you your answer for time from A’s reference frame. If you are looking for the elapsed time from some other reference frame (like B’s) then you need to be more specific about what you are looking for. If you want the spacetime separation of A and B it would be a time like separation of the form I expressed before, but again you need to be more specific about what you are looking for.

we are discussing non-simultaneity of a very fundamental part of universal physics. it is not time duration between zero duration moments of each observer as according to SRT this time duration of separation is nonexistant. [ i think]

Well again separation in the common parlance has a somewhat different meaning then it does when discussing spacetime and SRT. Elapsed time would probably a more applicable term for most of your usage of separation. Since your are specifically referring to the elapsed time in A’s reference frame of “1 million years” then the discussion must include “time duration between zero duration moments of each observer” at least from A’s perspective

Meaning that the separation between zero duration hsp's at their repsective t=0 is only a separation in time only [ according to what ever observer you wish to utilise] and maybe we shall stick with observer A's clock all the way.

Well the spacetime separation would be time like (since you are not proposing superluminal travel) but it still has spatial elements as describe in the calculation of spacetime separation I gave in the previous post.

What I am trying to say is that
if we find for example that observer A and observer B are separated by 1000 years then
when observer A takes a snap shot of his HSP obsrever B will be at that instant 1000 years of A's time into the future or past [ I don't know which]
so in effect there is zero tiem duration between past and fture events with 1000 years of separation...shesh...tis tough to explain...
So,
A = t=0 hsp
B = t=0 hsp
yet 1000 years may separate those moments frmo A's perspective and as it is symmetrical B also woud find the same result.

So if we stick to A as a preference, the results can be swapped latter.

Again 1000 years would represent the elapsed time in some reference frame (A’s in this case) that A and B were traveling apart from being in close proximity. Also in A’s reference frame there would be a distance that they travel apart in that time. Both are required to calculate the spacetime separation of A and B, different observers might disagree about the temporal and spatial amounts but all will agree about the spacetime separation (given inertial reference frames).

ozziemate

28th October 2008, 06:20 PM

ozziemate

28th October 2008, 06:27 PM

From your latest post this is not a gedanken but a request for what SRT will state in this situation so:
Before the acceleration:
Observer A and Observer B were at rest with synchronized clocks.
After the acceleration:
Observer A is at rest.
Observer B is travelling at 0.5c.
Observer A's clock and Observer B's clock are still synchronized (we will ignore the acceleration and what GR might say about this).

Does this answer the question?
not realy as we are assuming synchronised clocks as both were on the platform and at rest to begin with.

no we are looking at something that is very difficult to describe in SRT terms let alon in mine...ha

ozziemate

28th October 2008, 06:45 PM

the only coloquial way I can describe it is that teh 300000 years is not horizonatal time on the HSP but vertcal time on the HSP as the eparation between hsps is vertical and hot horizontal.
if it were merely horizontal we would have dilation with out a loss of simultaneity and of course this is not SRT. We have a lossof simuotaneity so the separation of 300000 years in this case mustbe vertical in our light cones. thus at a given zero duration moment the other observer must be in A's future by 300000 years and is time travelling....according to A
I think that as you do the gedanken workings and if you constantly refer back to A's HSP yu will find this time travel business to be true.

Reality Check

28th October 2008, 06:52 PM

not realy as we are assuming synchronised clocks as both were on the platform and at rest to begin with.

no we are looking at something that is very difficult to describe in SRT terms let alon in mine...ha
You wanted the SRT answer and I gave it to you. It is extremely easy to describe in SRT.

In order to get your non-SRT answer you will have to work out the transformation between the coordinates of the 2 observers. For that you will have to state your postulates and deduce the transformation from them (as Einstein did from the SRT postulates).
After that you will have to compare your non-SRT predictions with actual observations and then make falsifiable predictions that can be tested.

Of course this has nothing to do with the OT since you are working on creating a non-SRT theory, not on a logical proof of SRT invalidity.

The Man

28th October 2008, 06:53 PM

If Iam not mistaken using an old excel spread sheet:
if according to observer 'A' 1,000,000 years passes, observer 'B' will experience those 1,000,000 years as 700,000 years {from A's perspective}

which means the time separation or our HSPs zero duration moment will be approximately 300,000 years into the future or past of either frame depending on observer used.
But because Observer A can see Observer B in his HSP [ say B flies past A with only inches of distance iseparation] which places Observer B as a simultaneious observer/photon in his hSP.
observer B is still part of A's HSP yet according to B his zero moment is 300000 years away into the future or past of observer A's zero duration HSP...

sheesh! in editing I just realised some thing....ha

Well you will have to show how you obtained that calculation and I do not know why you consider the difference in your calculation of A’s and B’s elapsed time as a “time separation” which really has no specific meaning in the parlance of SRT.

As you describe above A and B passing at 0.5 c and maintaining a constant relative velocity might be a better consideration as it maintains inertial reference frames which SRT is directly and easily applicable.

not realy as we are assuming synchronised clocks as both were on the platform and at rest to begin with.

no we are looking at something that is very difficult to describe in SRT terms let alon in mine...ha

No not at all SRT (and GR) describe it very well, but the problem is your terms which can result in confusion about what you are saying to, or asking of, us. As I motioned before your usage of separation, is it in the common parlance or do you refer to the spacetime application of separation?

Reality Check

28th October 2008, 06:55 PM

the only coloquial way I can describe it is that teh 300000 years is not horizonatal time on the HSP but vertcal time on the HSP as the eparation between hsps is vertical and hot horizontal.
if it were merely horizontal we would have dilation with out a loss of simultaneity and of course this is not SRT. We have a lossof simuotaneity so the separation of 300000 years in this case mustbe vertical in our light cones. thus at a given zero duration moment the other observer must be in A's future by 300000 years and is time travelling....according to A
You are forgetting that observer B experiences 1 million years but measures that observer A experiences 700,000 years.

ETA: HSP is neither horizontal or vertical - that is just a convention used in some diagrams.
The time between the HSP's of any observers who have synchronized their clocks is always zero in SRT.

ozziemate

28th October 2008, 07:02 PM

Reality Check

28th October 2008, 07:12 PM

If observer A did a light cone for his hsp he will according to SRT have to include in his list of simultaneous observer/photon events another HSP which is not simultaneous with his. but he would be forced to include it because he can observe B in his hsp/phton sphere.
which means that B must be in A's future by 300000 years and yet A can see him.
http://bizzymate.com/uploads/images/physics/hsp02.gif
B is receiving light information that is 300000 years in the furture of A yet A can see B in his zero duration moment.
That is wrong. Observer B is '300000 years in the furture of A' but still has the same HSP as when they departed A.
Observer A never includes any other HSP. Observers just have 1 HSP.

Observer A can include observer B in observer A's spacetime diagram. Any light from observer B will take 300,000 years (according to your figures) to reach observer A.

ozziemate

28th October 2008, 07:23 PM

That is wrong. Observer B is '300000 years in the furture of A' but still has the same HSP as when they departed A.
Observer A never includes any other HSP. Observers just have 1 HSP.

Observer A can include observer B in observer A's spacetime diagram. Any light from observer B will take 300,000 years (according to your figures) to reach observer A.

yeah even if both observers are 2kms away in a fly by, 'A' will get B's incomming light info 300000 years later... now thats something...
"Hey B whats hapening 300000 years into the future,?"
"Not much Richie still hasn't re-married Brook and Homer Simpson is still an idiot"
"ahh so nothing to look forward too..pity, but thanks for the drop":D

UncaYimmy

28th October 2008, 07:24 PM

That is wrong. Observer B is '300000 years in the furture of A' but still has the same HSP as when they departed A.
Observer A never includes any other HSP. Observers just have 1 HSP.

Observer A can include observer B in observer A's spacetime diagram. Any light from observer B will take 300,000 years (according to your figures) to reach observer A.

But (wait for it)....

If all photons in existence exist simultaneously then it follows that absolutely all observations of photons must also.

The Man

28th October 2008, 07:26 PM

If observer A did a light cone for his hsp he will according to SRT have to include in his list of simultaneous observer/photon events another HSP which is not simultaneous with his. but he would be forced to include it because he can observe B in his hsp/phton sphere.
which means that B must be in A's future by 300000 years and yet A can see him.
http://bizzymate.com/uploads/images/physics/hsp02.gif
B is receiving light information that is 300000 years in the furture of A yet A can see B in his zero duration moment.

Ah, as I figured you are confusing "time separation" as different reference frames. Your HSPB, in that diagram, is just the HSPA 300,000 year in the future, it is just the same reference frame at a different time.

If you remember Fredrik mentioned (but did not precisely specify) a reference frame moving relative to A’s reference frame might share the same origin as A sometime in the future but would have a different slope then A’s reference frame.

ETA: This is not your example of A and B moving away at 0.5c.

ozziemate

28th October 2008, 07:27 PM

But (wait for it)....

If all photons in existence exist simultaneously then it follows that absolutely all observations of photons must also.

but only....if Richie marries Brook and Homer wins the nobel
:D

Fredrik

28th October 2008, 07:30 PM

fredrik,
With this gedanken I am not tryng to prove simulatenity in fact quite the opposite. I am asking SRT to prove non-simultaneity in a way that clarifies the degree of that non-simultaneiity.

OK, that's a good idea. There is however no way you can do that without understanding the definition of simultaneity that I posted in #436 and #446, so I suggest that you go back to those posts and make an effort to understand them.

Also note that all my other comments in my previous posts are still valid, even though I didn't understand what your purpose with the post was at the time. You should pay extra attention to the stuff about how it doesn't make sense to talk about the distance between two different hypersurfaces that aren't parallel.

If Iam not mistaken using an old excel spread sheet:
if according to observer 'A' 1,000,000 years passes, observer 'B' will experience those 1,000,000 years as 700,000 years {from A's perspective}

I get 866025 years, but OK, the numbers aren't really important here.

What's important is that you understand what this number represents. A million years into the future, observer A would call a completely different surface his "HSP". It would be parallel to the original, but it would be defined by t=1000000 instead of t=0 (so it's another horizontal plane in the diagram, above the old one). The number we're trying to find is the time coordinate that B would assign to the event where B's world line intersects A's new hypersurface of simultaneity.

which means the time separation or our HSPs zero duration moment will be approximately 300,000 years into the future or past of either frame depending on observer used.

That sentence makes no sense.

But because Observer A can see Observer B in his HSP [ say B flies past A with only inches of distance iseparation]

Nope. They were at the same place in space at t=0, but at t=1M, they are 500000 light-years apart (in A's frame).

which places Observer B as a simultaneious observer/photon in his hSP.

This makes no sense. If you mean that B's world line intersects one of his hypersurfaces of simultaneity, then you should say so.

observer B is still part of A's HSP yet according to B his zero moment is 300000 years away into the future or past of observer A's zero duration HSP...

This makes no sense. An observer can't be a part of a hypersurface, and the rest...I just don't know what you're trying to say.

ozziemate

28th October 2008, 07:32 PM

say after 700000 years observer A hangs out a light wire mesh that is incredibly flexible and B flies into this mesh at his .05c v and delivers information to A only 10 meters away. [ by hull to mesh contact ] [ no light involved]
info reads:
"Ritchie does marry Brook again but Homer misses out on the nobel by a one hair"

Reality Check

28th October 2008, 07:37 PM

Fredrik

28th October 2008, 07:38 PM

If observer A did a light cone for his hsp he will according to SRT have to include in his list of simultaneous observer/photon events another HSP which is not simultaneous with his. but he would be forced to include it because he can observe B in his hsp/phton sphere.
which means that B must be in A's future by 300000 years and yet A can see him.
http://bizzymate.com/uploads/images/physics/hsp02.gif
B is receiving light information that is 300000 years in the furture of A yet A can see B in his zero duration moment.
As The Man said in #707, the upper horizontal line in that picture represents the set of events that have t=300000. That's just what A would call his HSP 300000 years into the future. What we're interested in is what a hypersurface of constant t' looks like (not another hypersurface of constant t).

ozziemate

28th October 2008, 07:43 PM

ozziemate

28th October 2008, 07:52 PM

As The Man said in #707, the upper horizontal line in that picture represents the set of events that have t=300000. That's just what A would call his HSP 300000 years into the future. What we're interested in is what a hypersurface of constant t' looks like (not another hypersurface of constant t).
The problem is Observer A can alwasy see Observer B in his HSP even though Observer B is travelin into the future.
The upper hsp [B]is concurrent with the lower [A] [ this is what non-simultaneity means in the terms of a zero duration HSP]

http://bizzymate.com/uploads/images/physics/hsp02.gif
so B's HSP is co-existing in A's hsp in A's zero duration moment.

The Man

28th October 2008, 07:58 PM

Nope. They were at the same place in space at t=0, but at t=1M, they are 500000 light-years apart (in A's frame).

I think this is the primary problem, Fredrik, Ozziemate’s focus on temporal displacement without considering spatial displacement, particularly in his A moving away from B example. I tried to explain it, but he may take it as more significant from you. This consideration of the same reference frame in the future being some how a different reference frame seems to be the consistent element of his assertions to refute SRT.

Fredrik

28th October 2008, 07:58 PM

Does Observer B ever leave observer A's HSP during the total flight time of 100000 years using A's clock?

B's world line intersects the HSP at exactly one point. That's the only event where he can be said to be "in" the HSP. However, every point on B's world line is an intersection point between the world line and some hypersurface of simultaneity of A (each one associated with a different value of t).

Does Observer A ever leave B's HSP during the total flight time of 100000years using B's clock?

Same answer (with A and B swapped).

Can Observer 'A' see observer 'B' in a fly by and pass information to A via material means [ a pseudo Schrödinger's handshake if you like]

Only at points where their world lines intersect. In the scenario you have described, there's only one such point.

When Observer B flies by observer A is 'B' recieving light information that is 300000 years into A's future and if so how is it that 'A' cannot see that light info directly even though B is more or less right next to him?

You lost me. I don't know what scenario you're describing now.

Reality Check

28th October 2008, 08:01 PM

The problem is Observer A can alwasy see Observer B in his HSP even though Observer B is travelin into the future.
The upper hsp [b]is concurrent with the lower [A] [ this is what non-simultaneity means in the terms of a zero duration HSP]

http://bizzymate.com/uploads/images/physics/hsp02.gif
so B's HSP is co-existing in A's hsp in A's zero duration moment.
It is not a problem. Both A and B are travelling into the future.
The upper HSP does not exist because the diagam is what A can observe and there is only 1 HSP for A.
In that diagram B is not an observer. B is a possible event for A. If B never emits light then A never sees B. If B emits light then A will see that 300,000 years later.

Fredrik

28th October 2008, 08:07 PM

The problem is Observer A can alwasy see Observer B in his HSP even though Observer B is travelin into the future.
The upper hsp [B]is concurrent with the lower [A] [ this is what non-simultaneity means in the terms of a zero duration HSP]

http://bizzymate.com/uploads/images/physics/hsp02.gif
so B's HSP is co-existing in A's hsp in A's zero duration moment.
None of that makes any sense. I really can't understand what you're saying in any of those sentences.

Previously when you have referred to "non-simultaneity", you meant relativity of simultaneity, i.e. that observers disagree about which events are simultaneous. I can tell you what relativity of simultaneity really means: It's that B's hypersurfaces of simultaneity are not parallel to A's hypersurfaces of simultaneity.

If they were parallel, A and B would agree about which events are simultaneous. Please try to understand why.

ozziemate

28th October 2008, 08:08 PM

It is not a problem. Both A and B are travelling into the future.
The upper HSP does not exist because the diagam is what A can observe and there is only 1 HSP for A.
In that diagram B is not an observer. B is a possible event for A. If B never emits light then A never sees B. If B emits light then A will see that 300,000 years later.
could you respond to post #713
the questions and thus the answers clarify the situation for me and for you.

ozziemate

28th October 2008, 08:11 PM

None of that makes any sense. I really can't understand what you're saying in any of those sentences.
I shall post a clearer set up shortly. Hopefully it will make better sense then.

Reality Check

28th October 2008, 08:17 PM

questions:
Does Observer B ever leave observer A's HSP during the total flight time of 100000 years using A's clock?
Does Observer A ever leave B's HSP during the total flight time of 100000years using B's clock?
If the anwser is no to both questions:

Can Observer 'A' see observer 'B' in a fly by and pass information to A via material means [ a pseudo Schrödinger's handshake if you like]
When Observer B flies by observer A is 'B' recieving light information that is 300000 years into A's future and if so how is it that 'A' cannot see that light info directly even though B is more or less right next to him?

What do you mean by 'leave observer A's HSP'?
B was never in A's HSP. That would make observer B observer A.
Observer A and Observer B start with their own HSPs. That is the definition od HSP - the hypersurface of an observer's present time. The fact that A and B have syncronized clocks and so they measure the same time does not make their HSP's the same HSP.

ozziemate

28th October 2008, 08:21 PM

In this set up the moving orange dot is observer B and the rest of the assembly is observer A
Observer A and Observer B were initially at same V and after a period of acceleration we have what the animation below is showing.

Observer B is traveling at .O5 c and miraculously manages to travel the circuit with out slowing or stopping for 1,000,000 years according to observer A's clocks.

http://bizzymate.com/uploads/images/physics/aniobab.gif
Questions:
Can we accept this animation as a vehicle to discuss this issue?
Is there any other information needed ?

The Man

28th October 2008, 08:27 PM

Let’s take a different consideration: HSPA then 300,000 years in the future HSPB traveling at 0.5c relative to A collides with HSPA. At that point in time and space they share the same origin but as Fredrik mentioned HSPB will subtend an angle of 25.6 degrees with HSPA.

sol invictus

28th October 2008, 08:27 PM

In this set up the moving orange dot is observer B and the rest of the assembly is observer A
Observer A and Observer B were initially at same V and after a period of acceleration we have what the animation below is showing.

You're aware that observer B is accelerating continually (or at least every time she goes around those corners), right?

That takes this out of the domain of vanilla special relativity, but it's still perfectly OK to ask about.

What would you like to know?

ozziemate

28th October 2008, 08:28 PM

if we can agree I'll start another thread so that others can participate with out 19 pages to wade through

ozziemate

28th October 2008, 08:34 PM

You're aware that observer B is accelerating continually (or at least every time she goes around those corners), right?

That takes this out of the domain of vanilla special relativity, but it's still perfectly OK to ask about.

What would you like to know?
yes we have to assume that for the sake of this gedanken that the observer B [ orange dot] manages to go round the corners with out any loss to the overall set ups objectives...deaccelerations and accellerations to compensate for the loss of cornering but end up with the same overall picture as a a "constant" journey at 0.5c woudl show.

I shall start another thread if others agree that it will suffice.

Titled something like "SRT gedanken - non-simultaeity"

The Man

28th October 2008, 08:35 PM

Is there any other information needed ?

Yes you need to specify A spatial origin or d=0 for A, but we can take it as the center of the diagram. Again you seem not to consider spatial aspects in your assertions yet maintain the consistency of the seed of light which is just a proportion of unit space per unit time that remains consistent for all observers.

sol invictus

28th October 2008, 08:39 PM

yes we have to assume that for the sake of this gedanken that the observer B [ orange dot] manages to go round the corners with out any loss to the overall set ups objectives...deaccelerations and accellerations to compensate for the loss of cornering but end up with the same overall picture as a a "constant" journey at 0.5c woudl show.

That's impossible. You can't move in a trajectory like that without acceleration. Acceleration is by definition a change in velocity, and the velocity of B changes every time she goes around a corner. (Remember that velocity is a vector - it has a magnitude, called speed, and a direction. If either changes that's acceleration.)

Again, that's not a problem - it's a perfectly valid gedanken experiment - but it's NOT equivalent to B moving at constant velocity.

ozziemate

28th October 2008, 08:41 PM

ozziemate

28th October 2008, 08:43 PM

back in 6

Reality Check

28th October 2008, 08:48 PM

if we can agree I'll start another thread so that others can participate with out 19 pages to wade through
I would say no - unless you have given up on your Logical proof of SRT invalidity. I think it is best to keep all of the postings together until you actually develop that logical proof.

The Man

28th October 2008, 08:56 PM

ok , I'll add a point of origon of gedanken start and sychronisation point and clean it up a little, and then I'll post it again to get opinion before starting another thread...in other words I hope that the OP for the thread will be correct and not confusing. [If possible]

You can do that but it will not change Sol’s assertions, or mine which is not specifically related to your pervious diagram. Instead of going around for another 729 posts trying to explain the elements of relative motion and SRT to you, I am trying to get at the crux of your assertions, which seems to me to be your attempt to separate space and time. Time being some absolute reference that you relate to c, yet ignore the spatial aspect of that proportion, as you seem to focus only on temporal displacement, particularly of one given HSP yet do not consider spatial displacements of some other HSPs moving relative to your given HSP.

Reality Check

28th October 2008, 10:31 PM

Hi ozziemate, All these posts about simultaneity reminded me of an arXiv paper:
The challenge of changing deeply-held student beliefs about the relativity of simultaneity (http://arxiv.org/ftp/physics/papers/0207/0207081.pdf)
This is a Physics Education paper but it has several exercises that might interest you.

UncaYimmy

28th October 2008, 10:35 PM

That's impossible. You can't move in a trajectory like that without acceleration. Acceleration is by definition a change in velocity, and the velocity of B changes every time she goes around a corner. (Remember that velocity is a vector - it has a magnitude, called speed, and a direction. If either changes that's acceleration.)

Again, that's not a problem - it's a perfectly valid gedanken experiment - but it's NOT equivalent to B moving at constant velocity.

Couldn't he just eliminate the corners entirely and have the dot move from left to right once and call that 0.5c for a million years or whatever?

Fredrik

28th October 2008, 11:23 PM

I think starting a new thread sounds like a terrible idea. Do you think the problem with this thread is that you're not getting comments from the right people?

I have some advice for you that I think you should consider before you get started on your next attempt to describe your ideas:

You should always draw a spacetime diagram. You don't have to post it here. Just draw it on a piece of paper, or even in your mind, so that you know that your ideas make sense. When you do this, you have to keep in mind what sort of geometrical objects represent the various concepts you're talking about:

Events are represented by points.

A spacetime diagram represents one inertial observer's point of view. Don't forget that. The numbers that you can read off the t axis are precisely the times that this observer's clock is showing at those events on the t axis. The numbers you can read off the x axis are precisely the lengths that can be measured by a long ruler oriented in the direction in space that's represented by the x axis. The same goes for the y axis, and would go for the z axis if we could draw all three.

Observers are represented by curves. (Actually the motion of any small massive object is represented by a curve, but human observers are in that category). The slope of these curves is always >1. What this means is that at any point on the curve, the tangent of the curve at that point makes an angle with the x-y plane that's >45 degrees.

Inertial observers are represented by straight lines. The time axis is of special importance, since it represents the motion of the observer whose point of view the diagram represents.

What this observer thinks of as "space at time t" is represented by a horizontal plane that intersects the t axis at (t,0,0,0). Actually, that's not just what he thinks of as space for no good reason. It's what he's forced to think of as space at time t, if he adopts the definition of simultaneity that I posted in #436 and #446.

If you feel that you already understand these things, my advice to you is that you try much harder than before to only express ideas that are consistent with these very basic geometrical representations. (Note that no assumptions from special relativity except that coordinates can be assigned to events were used in this construction). I'm asking you to do this because you keep talking about things that make no sense at all.

Fredrik

28th October 2008, 11:33 PM

Couldn't he just eliminate the corners entirely and have the dot move from left to right once and call that 0.5c for a million years or whatever?
I think he wants B to meet A at least twice for some reason. He could let A be a dot on the left wall and have the moving dot (i.e. B) start at the same location, move to the right, bounce off the right wall and end up back at the start. This would of course make sure that B's world line isn't inertial (i.e. straight), so there's no natural way to associate one inertial frame with it.

It might be a bad idea to even mention these things here, because once you let B go off in one direction and bounce back to his original position, you end up in twin "paradox" territory very fast, and I really don't want to discuss the twin paradox with him.

The Man

29th October 2008, 12:09 AM

Let’s consider two reference frames, with synchronized clocks, RFA and RFB with no relative motion between them and only separated by, say, 300,000,000 meters in the x axis. Thus they have the same HSP or they will regard the same events as simultaneous or non- simultaneous, but they do not have the same spatial origins. Which means that although they might agree that two events are simultaneous they will disagree about what time those simultaneous events occurred, if those events have occurred along the x axis. Given the RFA spatial origin is 300,000,000 meters closer to those events then the RFB spatial origin. When RFA observes those events at say 13:00:00, RFB will observer them at 13:00:01. So, even though they agree that the events are simultaneous and their clocks remain synchronized, there is still a “time separation” of 1 second between the RFA observation and the RFB observation of those events, even though they still remain at the same HSP. It is just that RFB will observe those simultaneous events 1 second into the future (on a space time diagram) from when RFA observed those events (due to the 45 degree angle of the edge of the light cone) . This is the consideration you need to work into your thinking, Ozzimate, and it seems to be what you are lacking, temporal displacement (or “time separation”) based solely on spatial origin differences, which is a result of the speed of light c (or where the edge of the light cone intersects the spatial and temporal origins of each of those reference frames), which you fully accept. Your focus on temporal origins (and “time separation”) without considering spatial origins is your confusion, and your professed logic, to refute SRT you must consider both as not only SRT but also your acceptance of the speed of light, is dependent on both.

ozziemate

29th October 2008, 01:00 AM

deleted

The Man

29th October 2008, 02:29 AM

OK...I have had a think on it and realised the need to get rid of accelleration issues and stick to something really simple.

The following animation uses a platform with two light gates, one at either end as displayed:

http://bizzymate.com/uploads/images/physics/aniobab01.gif

The light gates are a laser beam that is broken when obsrever B passes through the gates. In breaking the laser beam A's clocks A or A1 are either started or stopped, In having contact with the "perpedicular to vector" laser beam Observer B's Clocks also start and stop.

We then have clocks synchronisation at t=0 at clock A' and when the gate at A1 is interupted clock A1 should see a recorded elapsed time of 2,000,000 years as the ship has been travelling at rel.v = .05c for 2,000,000 years

Summary:

For observer A, when B passes through gate A1, t=2,000,000 years

<>
Do we need any further information to start the gedanken?

I was going to mention the distance between the emitter and the receivers of the light gates as that would introduce a delay between when the beam was broken and when the receiver stops receiving photons (due to the speed of light). However the real problem would be your lack of a spatial reference (or origin) for the A frame or in other words even if you place the clock at the center of the path (within A) it will still take time for the signal (at the speed of light) from the light gate receivers to reach the clock to start and stop it. Not much of a consideration in our normal experience but quite relevant given the distances and velocity you present. You can not consider A’s spatial origin to be distributive along its entire length, you need some spatial origin (where its clock is) for A as is evident with B, otherwise it is very easy to misinterpret. You might consider placing the clock, emitters and receivers at the mid point along of the length, on the edge of A and have the A clock start and stop upon the return of the reflected laser beam off of B. this would minimize the signal travel time for the activation and deactivation of the A clock. However, that precise spatial location to the path of B would be quite relevant to the experiment. There are several such experiments already worked out, you might consider searching for, looking at them and presenting one of them for discussion, as these considerations would have already been, well, considered.

ozziemate

29th October 2008, 02:39 AM

I was going to mention the distance between the emitter and the receivers of the light gates as that would introduce a delay between when the beam was broken and when the receiver stops receiving photons (due to the speed of light). However the real problem would be your lack of a spatial reference (or origin) for the A frame or in other words even if you place the clock at the center of the path (within A) it will still take time for the signal (at the speed of light) from the light gate receivers to reach the clock to start and stop it. Not much of a consideration in our normal experience but quite relevant given the distances and velocity you present. You can not consider A’s spatial origin to be distributive along its entire length, you need some spatial origin (where its clock is) for A as is evident with B, otherwise it is very easy to misinterpret. You might consider placing the clock, emitters and receivers at the mid point along of the length, on the edge of A and have the A clock start and stop upon the return of the reflected laser beam off of B. this would minimize the signal travel time for the activation and deactivation of the A clock. However, that precise spatial location to the path of B would be quite relevant to the experiment. There are several such experiments already worked out, you might consider searching for, looking at them and presenting one of them for discussion, as these considerations would have already been, well, considered.
regarding the light gates:
the length of the laser beam would be say set tp be no more than 10 meters in that observer B would fit quite easilly between transmitter and reciever of that beam so the delays in signalling would be normally a lot less as it is an interupter beam so the distance from the hull of obserevre Bs ship and the clock gate receiver could be millimetres, thus inconsequestial.
So for the sake of this gedanken info delays on the gates interupion would be inconsequential.

ozziemate

29th October 2008, 02:51 AM

ahhhh! nope this form is even more confusing than the other....

back to the drawing board....damn
have deleted the post to rework the image

Reality Check

29th October 2008, 03:00 AM

deleted since ozziemate is confused...as usual:rolleyes:

The Man

29th October 2008, 03:19 AM

regarding the light gates:
the length of the laser beam would be say set tp be no more than 10 meters in that observer B would fit quite easilly between transmitter and reciever of that beam so the delays in signalling would be normally a lot less as it is an interupter beam so the distance from the hull of obserevre Bs ship and the clock gate receiver could be millimetres, thus inconsequestial.
So for the sake of this gedanken info delays on the gates interupion would be inconsequential.

It is not “delays on the gates interruption” but more the delay in the time that the on and off single is received by A’s clock (wherever it is). With the A clock at the mid point it will start 500,000 years after B has crossed the first light gate and will stop 500,000 years after B has crossed the second light gate (providing the signal is traveling at the speed of light). Although the total elapsed time for clock A will be 2,000,000 years, the starting and stopping of A’s and B’s clocks will not be simultaneous or there will be a 500,000 year “time separation” between them from A’s perspective. Are you abandoning your assertion of absolute time and that all light events are simultaneous in a given (A’s) HSP?

ETA: Given the width of ten meters you provide, someone at 1/8 the distance from the start on A could see B passing 250,000 years before A’s clock started.

ozziemate

29th October 2008, 03:28 AM

It is not “delays on the gates interruption” but more the delay in the time that the on and off single is received by A’s clock (wherever it is). With the A clock at the mid point it will start 500,000 years after B has crossed the first light gate and will stop 500,000 years after B has crossed the second light gate (providing the signal is traveling at the speed of light). Although the total elapsed time for clock A will be 2,000,000 years, the starting and stopping of A’s and B’s clocks will not be simultaneous or there will be a 500,000 year “time separation” between them from A’s perspective. Are you abandoning your assertion of absolute time and that all light events are simultaneous in a given (A’s) HSP?

yes you are correct about the clock thingo...just trying to recall all the gedankens I ran about 6 years ago on another forum... none of which touched at the level I wish this one to.
and no I haven't abandoned the issue of non-simultaneity but am tryng to devise a gedanken that will demonstrate the issue clearly.

basically as already described a few posts back where we end up with that cone diagram with the hsp's 300000 years apart.
http://bizzymate.com/uploads/images/physics/hsp02.gif