FREE TAPATALK APP

Vorpal · 9th April 2012, 09:39 AM

Originally Posted by DeiRenDopa

And why is this? I mean, there's nothing particularly unusual in what Vorpal writes, is there? Nothing that you won't find in a suitable, contemporary physics textbook, is there?

As far as I'm aware, the notable deviation from many GTR textbooks is that on the subject of what the "gravitational field" actually is, the stance most of them seem to take is roughly "we don't talk about that." Sometimes very literally: MTW effectively spend an entire section explaining their reasons why they leave the term "gravitational field" undefined and hence why they're not going to use it, ever. (That's certainly a very consistent view: they still have all the physically relevant quantities in their mathematics anyway; they just refuse to give any of them that label.)

Under that (non-)interpretation, Synge's comment doesn't mean anything at all, so we need to define what "gravitational field" means. It so happens that Einstein did define it to be the connection coefficients/Christoffel symbols Γ^α_μν and the gravitational potential to be the metric components. There are good reasons why this is sensible, and also a reason to reject it--Γ's are not tensorial as a whole. But it remains the case that there isn't a better candidate, and its coordinate-dependent nature actually embraces the equivalence principle: gravity would be explicitly an inertial ('fictitious') force.

And then Farsight seems to be arguing for an almost-schizophrenic position that Einstein's "potential = metric, Γ's = gravitational field" identification is very obviously correct (I don't know why else he would be harping about how I'm a woo-woo for using the term "almost eerie" to describe just how well it works) and yet completely denying its implication: that curvature vanishing at an event does not mean the field vanishes there.

P.S. I really should have made the last bit more explicit in my previous post.

Farsight · 9th April 2012, 11:07 AM

Originally Posted by Vorpal

Nope, not in that frame, I wouldn't. "It's not the fall that kills you--it's the sudden stop at the end." My terminal ends is because the ground puts a tremendous force on me.

When you're in a box falling to earth your inertial reference frame is continually changing but you can't tell. When that change stops occuring all of a sudden, then you can.

Originally Posted by Vorpal

I don't think you understand what local means.

Of course I do.

Originally Posted by Vorpal

Of course--what is or isn't rest is frame-dependent. But the major difference is that in Newtonian physics, making a frame that comoves with a gravitationally freefalling particle gives an accelerated frame--it doesn't follow a geodesic. In GTR, it does: the acceleration four-vector along the trajectory identically vanishes. That's the point of EP.

And again, it only works for a region of infinitesimal size.

Originally Posted by Vorpal

Synge's argument is about the relationship between the gravitational field and the gravitational tidal forces. You're simply claiming some relevance without even a word as to why it is so.

I've said it already - you cannot transform away the Riemann curvature. All you can do is take a small region of a curve, a region so small that it looks flat. But we know it isn't flat.

Originally Posted by Vorpal

Since you yourself specified the case where the charge is not moving in our local inertial frame, the electric field is what you get to measure. Remember Lorentz force and it'll be obvious.

The Lorentz force isn't relevant. If you set a charged particle down and find it doesn't move, you say there's no electromagnetic field present. It's that simple.

Originally Posted by Vorpal

For the purposes of this discussion, it doesn't matter whether there are or aren't any point charges. Read what I wrote again: I'm saying that even if that particular statement of yours was 100% absolute truth, nothing about my point changes. Because what you said literally adds nothing more than a side comment of "... and the field comes from the potential." Yes, everyone knows that. I don't mind if you consider it the mostest fundamentalest thing there ever was. What I said was:
(*) the gravitational field and gravitational tidal forces are different things
and the electromagnetic field was brought in to illustrate why Synge's argument is mistaken: because we can define a tensor that describes electromagnetic tidal forces, in the sense of deviation of equally-charged test particles, and directly confirm that its vanishing does not imply that the electromagnetic field vanishes.

Forget your equally-charged test particles. Use a single test particle. When you set it down and find it doesn't move, then you say there's no electromagnetic field present.

Originally Posted by Vorpal

Or, more bluntly: Synge's implication is very clearly false for other fields.

No it isn't, you've misunderstood the scenario.

Originally Posted by Vorpal

Why should we believe him about the gravitational field?

Because he's right. You can't make the Riemann curvature vanish. All you do when you take an infinitesimal region is sweep it under the carpet. Take that too literally and you end up with a flat hill and a misunderstanding of the gravitational field.

Originally Posted by Vorpal

Evidently, you don't agree with it, because if you accepted it as valid, you'd see that Synge is wrong immediately. Hence my question about your reasons against it.

No I don't see that Synge is wrong immediately. Instead what I see immediately is that you got the scenario wrong.

Originally Posted by Vorpal

It's just poking fun at the fact that for someone who incessantly insistent that he was "with Einstein", you spend extraordinary amounts of time arguing against him.

No I don't. I spend far more time saying Einstein said x. And I've said already that he wasn't perfect.

Originally Posted by Vorpal

And I'll put this last:

Originally Posted by Farsight

I've said it already. When gravitational tidal forces vanish you've got a gravitational field that doesn't diminish with distance. That's a nonsense implication.

What's wrong with that?

You drop an object from a great distance, and it accelerates until it's going faster than light. That can't happen.

Originally Posted by Vorpal

If you believe the analogy to be valid, then the gravitational field is the connection, and it's trivial that all you need for a homogeneous but nonvanishing gravitational field is to take flat spacetime in a uniformly accelerated frame. So I ask yet again: do you have any reasons to consider the analogy wrong?

Yes. A gravitational field is curved spacetime. It isn't flat spacetime, and accelerating through space does not turn flat spacetime into curved spacetime. The principle of equivalence is a principle that shows you why the two are similar, but it doesn't tell you that they're identical.

Farsight · 9th April 2012, 11:19 AM

Originally Posted by Vorpal

...And then Farsight seems to be arguing for an almost-schizophrenic position that Einstein's "potential = metric, Γ's = gravitational field" identification is very obviously correct (I don't know why else he would be harping about how I'm a woo-woo for using the term "almost eerie" to describe just how well it works) and yet completely denying its implication: that curvature vanishing at an event does not mean the field vanishes there.

It isn't schizophrenic, it's really important. When spacetime is flat, no gravitational field is present. A beam of light doesn't curve. Accelerating past that light beam only changes how you see it, not the light beam itself. And I was scathing of your "eerie" because you don't understand why that comparison can be made.

DeiRenDopa · 9th April 2012, 12:23 PM

Originally Posted by Farsight

... A beam of light doesn't curve. Accelerating past that light beam only changes how you see it, not the light beam itself. ...

Funny how Farsight's 'absolute time and space' fundamental worldview keeps appearing, in his posts, isn't it?

dafydd · 9th April 2012, 12:51 PM

Originally Posted by DeiRenDopa

Funny how Farsight's 'absolute time and space' fundamental worldview keeps appearing, in his posts, isn't it?

If beams of light do not bend then how does gravitational lensing occur? I'm looking forward to Farsight's explanation.

Brian-M · 9th April 2012, 09:54 PM

Originally Posted by Farsight

That's wrong. If the clock at the ceiling stayed synchronised with the clock at the floor, things wouldn't fall down.

You're affirming the consequent.

The normal point of view is that ~~time~~ light passes at different speeds because the gravitational attraction is different. Your point of view reverses this, claiming that gravitational attraction is different because ~~time~~ light is passing at different speeds.

You can't know that if ~~time~~ light wasn't passing at different speeds at different heights then "things wouldn't fall down" until you first demonstrate that your point of view is correct.

Originally Posted by Farsight

I've said it already. When gravitational tidal forces vanish you've got a gravitational field that doesn't diminish with distance. That's a nonsense implication.

Is it nonsense?

Increase the distance of your floating lab/city from the planet, and then increase the mass of the planet to compensate so that you're still experiencing the same amount of gravity. Suddenly, the tidal forces in your lab are smaller than they were before.

Repeat this process indefinitely. As distance approaches infinity, tidal force (difference in gravity) approaches at zero.

Sure, you can't actually get infinite distance. But at some finite distance you're going to reach a point where no practical test inside your lab can detect the presence of any gravitational variation with distance.

Now here's something to think about. If things fall down because light is moving slower at the lower elevation as you claim, shouldn't differences in the speed of light over the same elevation (as in my example above) cause things to fall with different force? But if gravity follows the inverse square law (as I assumed in my example above), then this can't possibly be true.

W.D.Clinger · 9th April 2012, 10:17 PM

Originally Posted by Farsight

Originally Posted by Vorpal

...And then Farsight seems to be arguing for an almost-schizophrenic position that Einstein's "potential = metric, Γ's = gravitational field" identification is very obviously correct (I don't know why else he would be harping about how I'm a woo-woo for using the term "almost eerie" to describe just how well it works) and yet completely denying its implication: that curvature vanishing at an event does not mean the field vanishes there.

It isn't schizophrenic, it's really important. When spacetime is flat, no gravitational field is present. A beam of light doesn't curve. Accelerating past that light beam only changes how you see it, not the light beam itself. And I was scathing of your "eerie" because you don't understand why that comparison can be made.

It isn't schizophrenic so much as ignorant. Farsight doesn't realize that the coordinate-dependent Christoffel symbols (which, according to Einstein, describe the gravitational field) can be nonzero even in flat spacetime.

Had Farsight made an effort to understand the metric form posted by sol invictus, he might not have made that mistake. On the other hand, Farsight would have had to learn calculus first...

Farsight · 10th April 2012, 02:12 AM

Originally Posted by dafydd

If beams of light do not bend then how does gravitational lensing occur? I'm looking forward to Farsight's explanation.

Huh? I said this:

When spacetime is flat, no gravitational field is present. A beam of light doesn't curve. Accelerating past that light beam only changes how you see it, not the light beam itself.

Gravitational lensing occurs when beams of light bend. They bend because they move through a region of space where a gravitational field is present. When this happens we call it curved spacetime, when it doesn't and the beams of light go straight, we call it flat spacetime.

Farsight · 10th April 2012, 03:04 AM

Originally Posted by Brian-M

You're affirming the consequent.

No, you are.

Originally Posted by Brian-M

The normal point of view is that ~~time~~ light passes at different speeds because the gravitational attraction is different. Your point of view reverses this, claiming that gravitational attraction is different because ~~time~~ light is passing at different speeds.

My point of view is Einstein's point of view. He said a curvature of rays of light can only take place when "die Ausbreitungs-geschwindigkeit des Lichtes mit dem Orte variiert", which translates into "the speed of light varies with the locality". When it doesn't, there is no gravitational attraction.

Originally Posted by Brian-M

You can't know that if ~~time~~ light wasn't passing at different speeds at different heights then "things wouldn't fall down" until you first demonstrate that your point of view is correct.

Einstein was the author of general relativity, his point of view was correct. And you know it's correct from the parallel-mirror gif. And you know about the wave nature of matter don't you? You know we can make an electron out of light via pair production? And that the electron has spin angular momentum and a magnetic dipole moment? And that it can be diffracted? So think of the electron as a standing wave of light. A circle of light, as it were. Put it in a gravitational field and think it through. Divide the circle into four flat quadrants to make it even simpler:

..←
↓....↑
..→

Starting from the left and going anticlockwise, at a given instant we have light travelling down like this ↓. There’s a gradient in c from top to bottom, but all it does is make the light look blueshifted. A little while later the light is moving like this → and the lower portion of the wave-front is subject to a slightly lower c than the upper portion. So it bends down a little. Later it’s going this way ↑ and looks redshifted, and later still it’s going this way ← and bends down again. These bends translate into a different position for the electron. It falls down:

○
↓

The reducing c bleeds rotational motion out into linear motion. But only half the cycle got bent, so only half the reduced c goes into kinetic energy aka relativistic mass. That’s why light is deflected twice as much as matter, and gravity is not the sort of force that increases the relativistic mass. Simple.

Originally Posted by Brian-M

Originally Posted by Farsight

I've said it already. When gravitational tidal forces vanish you've got a gravitational field that doesn't diminish with distance. That's a nonsense implication.

Is it nonsense?

Yes. You've got a reducing speed of light that keeps on reducing, so much so that it and ends up being negative. There's no such thing as a negative speed. A negative speed is nonsense.

Originally Posted by Brian-M

Increase the distance of your floating lab/city from the planet, and then increase the mass of the planet to compensate so that you're still experiencing the same amount of gravity. Suddenly, the tidal forces in your lab are smaller than they were before.

No problem with that.

Originally Posted by Brian-M

Repeat this process indefinitely. As distance approaches infinity, tidal force (difference in gravity) approaches at zero. Sure, you can't actually get infinite distance. But at some finite distance you're going to reach a point where no practical test inside your lab can detect the presence of any gravitational variation with distance.

But things in your lab still fall down, and the parallel-mirror light-clock at the ceiling still runs faster than the one at the floor.

Originally Posted by Brian-M

Now here's something to think about. If things fall down because light is moving slower at the lower elevation as you claim, shouldn't differences in the speed of light over the same elevation (as in my example above) cause things to fall with different force?

Yes. If there's a big difference between the speed of light at the ceiling as compared to the speed of light at the floor, the force of gravity is large. If there's a lesser difference, the force of gravity is less. When there's no difference, things don't fall down.

Originally Posted by Brian-M

But if gravity follows the inverse square law (as I assumed in my example above), then this can't possibly be true.

It is true. When you measure the difference between the speed of light at the ceiling as opposed to the floor in a very very tall building, you find that on floor 100,000 the difference is less than it is in the basement. If you could take an equatorial slice through the planet and measure the speed of light at various locations, when you plot them out what you end up with is the upturned hat. When you look at one small region of this there's still a gradient in gravitational potential and a gradient in c. There's no discernible curvature of the gradient, you can't detect any tidal force, but things still fall down, and light still curves when it moves through space, so we say spacetime is curved. Way out a zillion miles to the left or right where there's no discernible gradient, your measurements can't detect any curvature of light, or things falling down, and then we say spacetime is flat.

Vorpal · 10th April 2012, 03:14 AM

Farsight, you're simply repeating irrelevancies.
1) Absolutely no one is claiming that you can transform away Riemann curvature at events where it's nonzero. Repeating it incessantly is completely pointless, because no one is questioning this fact.
2) The real issue is whether the vanishing of Riemann curvature at some event implies that the non-presence of the gravitational field there. This is something you, and Synge, simply assert as true, and provide no reasoning for that claim.
You've simply made no connection between (1), which everybody knows, and (2), which is what we're trying to have a discussion about. And failing.

Originally Posted by Farsight

The Lorentz force isn't relevant. If you set a charged particle down and find it doesn't move, you say there's no electromagnetic field present. It's that simple.
...When you set it down and find it doesn't move, then you say there's no electromagnetic field present.

You're mistaken. If you have a charged test particle that starts at rest and it doesn't move, it in no way implies that there's no electromagnetic field present. It only implies that there is no electric field present.

Originally Posted by Farsight

You drop an object from a great distance, and it accelerates until it's going faster than light. That can't happen.

Nonsense. This is a basic calculus exercise for STR. Suppose a particle starts from rest and undergoes constant force:
$F = \frac{dp}{dt} = m\frac{d(\gamma v)}{dt} = mc\left(\cosh\alpha\right)\frac{d\alpha}{dt}$
Where I've applied the substitution v/c = tanh α, so γv/c = sinh α and γ = cosh α. Just integrate: Ft = sinh α, and therefore:
$v = c\tanh\left(\sinh^{-1}\left(\frac{Ft}{mc}\right)\right) = \frac{Ft/m}{\sqrt{1 + (Ft/mc)^2}}$
Note that the tanh form makes it obvious at the speed is bounded by c.

Originally Posted by Farsight

Yes. A gravitational field is curved spacetime. It isn't flat spacetime, and accelerating through space does not turn flat spacetime into curved spacetime.

I asked for reasons, not mere repetition of baseless assertion.

dafydd · 10th April 2012, 04:23 AM

Originally Posted by Farsight

Huh? I said this:

When spacetime is flat, no gravitational field is present. A beam of light doesn't curve. Accelerating past that light beam only changes how you see it, not the light beam itself.

Gravitational lensing occurs when beams of light bend. They bend because they move through a region of space where a gravitational field is present. When this happens we call it curved spacetime, when it doesn't and the beams of light go straight, we call it flat spacetime.

Thank you. You didn't mention flat spacetime in your previous post.

Brian-M · 10th April 2012, 04:39 AM

Originally Posted by Farsight

My point of view is Einstein's point of view. He said a curvature of rays of light can only take place when "die Ausbreitungs-geschwindigkeit des Lichtes mit dem Orte variiert", which translates into "the speed of light varies with the locality". When it doesn't, there is no gravitational attraction.

Can you provide a link to the source of this quote? I'd like to see if the actual text supports your assertion that it's the change in the speed of light that causes objects to exhibit gravitation attraction.

Originally Posted by Farsight

Einstein was the author of general relativity, his point of view was correct.

That's the assumption we're operating from. Whether your point of view is an accurate representation of his point of view is a different subject altogether.

Originally Posted by Farsight

And you know it's correct from the parallel-mirror gif.

I don't know anything is correct from the parallel-mirror GIF. It's just an animation, and animations aren't necessarily an accurate representation of reality. Anything can happen in an animation.

But since it was created as a graphical representation of our pre-existing understanding that clocks lose synchronization at different elevations, we can assume that clocks (whether atomic clocks, or parallel-mirror light clocks) do lose synchronization at different elevations with or without the GIF.

But even given that, we can't say that Einstein's theory is correct from this alone. (But we do have other reasons separate from this to assume Einstein's theory is correct.)

Originally Posted by Farsight

And you know about the wave nature of matter don't you? You know we can make an electron out of light via pair production? And that the electron has spin angular momentum and a magnetic dipole moment? And that it can be diffracted?

I don't see what this has to do with relativity.

Originally Posted by Farsight

So think of the electron as a standing wave of light.

WTF?

Originally Posted by Farsight

A circle of light, as it were. Put it in a gravitational field and think it through. Divide the circle into four flat quadrants to make it even simpler:

..←
↓....↑
..→

Starting from the left and going anticlockwise, at a given instant we have light travelling down like this ↓. There’s a gradient in c from top to bottom, but all it does is make the light look blueshifted. A little while later the light is moving like this → and the lower portion of the wave-front is subject to a slightly lower c than the upper portion. So it bends down a little. Later it’s going this way ↑ and looks redshifted, and later still it’s going this way ← and bends down again. These bends translate into a different position for the electron. It falls down:

○
↓

The reducing c bleeds rotational motion out into linear motion. But only half the cycle got bent, so only half the reduced c goes into kinetic energy aka relativistic mass. That’s why light is deflected twice as much as matter, and gravity is not the sort of force that increases the relativistic mass. Simple.

Originally Posted by Farsight

If there's a big difference between the speed of light at the ceiling as compared to the speed of light at the floor, the force of gravity is large. If there's a lesser difference, the force of gravity is less. When there's no difference, things don't fall down.

I'm going to take some time to think of an appropriate response to this.

Farsight · 10th April 2012, 06:57 AM

Originally Posted by Brian-M

Can you provide a link to the source of this quote? I'd like to see if the actual text supports your assertion that it's the change in the speed of light that causes objects to exhibit gravitation attraction.

Sure. See Über die spezielle und allgemeine Relativitätstheorie and look at section 22. It's on page 51, about three-quarters of the way down.

Originally Posted by Brian-M

That's the assumption we're operating from. Whether your point of view is an accurate representation of his point of view is a different subject altogether.

He said what he said.

Originally Posted by Brian-M

I don't know anything is correct from the parallel-mirror GIF. It's just an animation, and animations aren't necessarily an accurate representation of reality. Anything can happen in an animation. But since it was created as a graphical representation of our pre-existing understanding that clocks lose synchronization at different elevations, we can assume that clocks (whether atomic clocks, or parallel-mirror light clocks) do lose synchronization at different elevations with or without the GIF.

OK. The gif just brings it home I guess.

Originally Posted by Brian-M

But even given that, we can't say that Einstein's theory is correct from this alone. (But we do have other reasons separate from this to assume Einstein's theory is correct.)

Yes, it's a well-tested theory.

Originally Posted by Brian-M

I don't see what this has to do with relativity.

It isn't relativity, but it is physics, and it is relevant.

Originally Posted by Brian-M

WTF?

Yes, think of the electron as a standing wave of light. You can make it from light along with a positron in pair production. You can diffract an electron. Electrons interfere with one another. You can annihilate an electron with a positron and what you get is light. Read up on atomic orbitals and see the bit that says "The electrons do not orbit the nucleus in the sense of a planet orbiting the sun, but instead exist as standing waves".

Originally Posted by Brian-M

I'm going to take some time to think of an appropriate response to this.

OK noted.

Farsight · 10th April 2012, 07:55 AM

Originally Posted by Vorpal

Farsight, you're simply repeating irrelevancies.
1) Absolutely no one is claiming that you can transform away Riemann curvature at events where it's nonzero. Repeating it incessantly is completely pointless, because no one is questioning this fact.
2) The real issue is whether the vanishing of Riemann curvature at some event implies that the non-presence of the gravitational field there. This is something you, and Synge, simply assert as true, and provide no reasoning for that claim.
You've simply made no connection between (1), which everybody knows, and (2), which is what we're trying to have a discussion about. And failing.

How many more times do I have to tell you about flat hills? Find some plot of gravitational potential like the upturned hat, and just look at it. Even this picture will do. If you have no Reimann curvature then when you start from a zillion miles to the right you've got a totally flat line. Start from the left at the surface of the sun and without Riemann curvature you've got a constant slope that goes on forever. The former is where there is no gradient in gravitational potential, so light doesn't curve and nothing falls down. The latter is where the coordinate speed of light goes on increasing forever, and the force of gravity doesn't reduce with distance. Neither is a description of a gravitational field. You need that Riemann curvature. When you take an infinitesimal region you're ignoring the change in the gradient but you aren't ignoring the gradient! How much simpler can I make it?

Originally Posted by Vorpal

You're mistaken. If you have a charged test particle that starts at rest and it doesn't move, it in no way implies that there's no electromagnetic field present. It only implies that there is no electric field present.

There is no such thing as an electric field! It's the electromagnetic field! For God's sake Vorpal, go and read Minkowski’s Space and Time. Go and look at this bit two pages from the back:

"Then in the description of the field produced by the electron we see that the separation of the field into electric and magnetic force is a relative one with regard to the underlying time axis; the most perspicious way of describing the two forces together is on a certain analogy with the wrench in mechanics, though the analogy is not complete".

Did you catch that? It's the field. It exerts force in two ways, resulting in linear and/or rotational motion. If you're a charged particle and I set you down somewhere in the electron's electromagnetic field, you move linearly towards it or away from it, and the electron similarly moves towards or away from you. If I throw you through it, you go round and round as well. You only experience the rotational force when you have relative motion.

Originally Posted by Vorpal

Nonsense. This is a basic calculus exercise for STR. Suppose a particle starts from rest and undergoes constant force:
$F = \frac{dp}{dt} = m\frac{d(\gamma v)}{dt} = mc\left(\cosh\alpha\right)\frac{d\alpha}{dt}$
Where I've applied the substitution v/c = tanh α, so γv/c = sinh α and γ = cosh α. Just integrate: Ft = sinh α, and therefore:
$v = c\tanh\left(\sinh^{-1}\left(\frac{Ft}{mc}\right)\right) = \frac{Ft/m}{\sqrt{1 + (Ft/mc)^2}}$
Note that the tanh form makes it obvious at the speed is bounded by c.

Gravity isn't a force in the usual sense. Forget your constant force, it's a constant acceleration, and escape velocity is the flip side of the speed of an infalling body dropped from "a great distance". When you drop your body into a black hole, it supposedly ends up going at the speed of light. You cannot contrive a gravitational field where it continues to accelerate at 9.8 m/s/s ad infinitum and gets to the speed of light when it's only halfway there. That's why the infinite wall falls down. That's why I said the problem was that you drop an object from a great distance, and it accelerates until it's going faster than light. That can't happen. Because the reducing c bleeds rotational motion out into linear motion. That's why it's obvious that the speed is bounded by c.

Originally Posted by Vorpal

Originally Posted by Farsight

Yes. A gravitational field is curved spacetime. It isn't flat spacetime, and accelerating through space does not turn flat spacetime into curved spacetime.

I asked for reasons, not mere repetition of baseless assertion.

It's not a base assertion. If we're all in space where light doesn't curve, Vorpal putting the pedal to the metal doesn't make it curve one iota. Learn to tell the difference between what you see and what's there.

Farsight · 10th April 2012, 07:58 AM

Originally Posted by dafydd

Thank you. You didn't mention flat spacetime in your previous post.

My pleasure.

Hellbound · 10th April 2012, 08:00 AM

Originally Posted by Farsight

Yes, think of the electron as a standing wave of light. You can make it from light along with a positron in pair production. You can diffract an electron. Electrons interfere with one another. You can annihilate an electron with a positron and what you get is light. Read up on atomic orbitals and see the bit that says "The electrons do not orbit the nucleus in the sense of a planet orbiting the sun, but instead exist as standing waves".

Just because I like to throw in wrenches...

I do hope you realize that you can annihilate a proton with an anti-proton, and also get light.

And you can annihilate a neutron with an anti-neutron, and get light.

So are you suggesting that all particles are standing light waves?

edd · 10th April 2012, 08:07 AM

Originally Posted by Farsight

There is no such thing as an electric field! It's the electromagnetic field! For God's sake Vorpal, go and read Minkowski’s Space and Time. Go and look at this bit two pages from the back:

I'm really not sure you got Vorpal's point. You take a charged particle and place it at rest and it doesn't move. You'd seem to be saying that that demonstrates the absence of any magnetic field, since you're saying that it shows there's no electromagnetic field.

DeiRenDopa · 10th April 2012, 08:28 AM

Originally Posted by Hellbound

Just because I like to throw in wrenches...

I do hope you realize that you can annihilate a proton with an anti-proton, and also get light.

And you can annihilate a neutron with an anti-neutron, and get light.

So are you suggesting that all particles are standing light waves?

By ~~Jove~~ Democritus!

That's it! The TRUE SECRET TO THE UNIVERSE!!!!!

Atoms are made up of neutrons, protons, and electrons. Each of which, in turn, per Farsight, is a standing wave of light.

A crystal, of an element such as iron, is then also just a bunch of standing waves of light. Molecules are also just mixtures of standing waves of light.

A virus, which is made up of molecules, is, then a complex mixture of standing waves of light.

So is a tree, you, me, an asteroid, a planet, a star, ...

And if you put enough standing waves of light together, and scrunch them down, you get ... wait for it ... a black hole!

What a waste, 25 pages of largely irrelevant posts, when we could have just considered black holes as yet another manifestation of light ...

sol invictus · 10th April 2012, 08:33 AM

Originally Posted by Farsight

You need that Riemann curvature. When you take an infinitesimal region you're ignoring the change in the gradient but you aren't ignoring the gradient! How much simpler can I make it?

You can make it as simple as you like, but it will remain wrong. There simply isn't a direct analog of "gravitational field" in GR, but the closest is the connection, not the curvature.

Of course that's meaningless to you, since you have no clue what those words even mean, let alone what the difference is.

Quote:

There is no such thing as an electric field! It's the electromagnetic field! For God's sake Vorpal, go and read Minkowski’s Space and Time.

You asserted "If you set a charged particle down and find it doesn't move, you say there's no electromagnetic field present. It's that simple.
...When you set it down and find it doesn't move, then you say there's no electromagnetic field present."

That's false, as Vorpal just tried to explain to you. If the electromagnetic field strength tensor in the rest frame of the particle has zero time-space components, but non-zero space-space components (that's a long-winded way of saying "there's a magnetic field, but zero electric field") then the particle won't move when you set it down.

So you're wrong. Admit it, learn from it, and move on. Can you do that?

ben m · 10th April 2012, 08:48 AM

Originally Posted by Hellbound

So are you suggesting that all particles are standing light waves?

Hellbound, Farsight will talk to you as though he knows that some sort of topological field theory is true, and as though you're behind-the-times for NOT knowing that all particles are differently-knotted configurations of some underlying ur-field.

Farsight · 10th April 2012, 09:13 AM

Originally Posted by Hellbound

Just because I like to throw in wrenches...

I do hope you realize that you can annihilate a proton with an anti-proton, and also get light.

And you can annihilate a neutron with an anti-neutron, and get light.

So are you suggesting that all particles are standing light waves?

Yes, that's where you end up. Shocking isn't it? But you can diffract neutrons too. And buckyballs.

We've had this before, but I'll say it again. See Is the electron a photon with a toroidal topology? by Williamson and van der Mark, Annales de la Fondation Louis de Broglie, Volume 22, no.2, 133 (1997). Here's a picture from it. Inflate the torus until it's more like a sphere, and note that 4π denotes a sphere. Also see The nature of the electron by Qiu-Hong Hu, which is similar. Then there's Harmonic quintessence and the derivation of the charge and mass of the electron and the proton and quark masses by Andrew Worsley, Physics Essays Jun 2011, Vol. 24, No. 2 pp. 240-253. The latter applies spherical harmonics, usually applied to electron orbitals, to the particles themselves, wherein the electron Compton wavelength is given as λ = 4π / n c^1½ metres, where n is a dimensionality conversion factor with a value of 1. The light is going round at c one way and at half c the other, a bit like a moebius strip, and c dictates the structure. He gives the proton/electron mass ratio r = c^½ / 3π. Both expressions are subject to small binding-energy adjustments, but here's the raw numbers:

4π = 12.566370
c = 299792458
c^½ = 17314.5158177
4π / c^1½ = 12.566370 / (299792458 * 17314.5158177)
λ = 2.420910 x 10ˉ¹² m
Actual = 2.426310 x 10ˉ¹² m

c^½ = 17314.5158177
3π = 9.424778
c^½ / 3π = 17314.5158177 / 9.424778
r = 1837.12717877
Actual = 1836.15267245

ben m · 10th April 2012, 10:07 AM

Originally Posted by Farsight

We've had this before, but I'll say it again. See Is the electron a photon with a toroidal topology? by Williamson and van der Mark, Annales de la Fondation Louis de Broglie, Volume 22, no.2, 133 (1997). Here's a picture from it. Inflate the torus until it's more like a sphere, and note that 4π denotes a sphere. Also see The nature of the electron by Qiu-Hong Hu, which is similar. Then there's Harmonic quintessence and the derivation of the charge and mass of the electron and the proton and quark masses by Andrew Worsley, Physics Essays Jun 2011, Vol. 24, No. 2 pp. 240-253.

Speculative nonsense, pseudo-published in specialized crackpot journals. (Physics Essays is well known, but Annals de la Fondation Louis de Broglie is basically the same thing.)

Farsight, are you utterly unable to tell the difference between beyond-the-mainstream, utterly-unchecked speculation, and actual particle physics? Is there something preventing you from using qualifiers, like "A few isolated authors have speculated ...", rather than pretending that this is all known, or true, or established?

Anyway, your crackpot claims about toroidal-particle-substructure are off-topic in this thread about black holes. I recommend you start a new thread if you want to discuss that.

Vorpal · 10th April 2012, 04:35 PM

Originally Posted by Farsight

There is no such thing as an electric field! It's the electromagnetic field! For God's sake Vorpal, go and read Minkowski’s Space and Time. Go and look at this bit two pages from the back:
...
Did you catch that? It's the field.

And I say yet again: Take a look at the Lorentz force and it will be obvious:
$\frac{dp^\mu}{d\tau} = qF^\mu{}_\nu u^\nu$
In the frame of the particle, it is at rest and only the time component u⁰ of the four-velocity is nonzero. The electromagnetic tensor is antisymmetric, so it has six independent components, and in particular F₀₀ = 0. As a result, the condition of remaining stationary (i.e., Lorentz force vanishing) means the time-space components F_0a = -F_a0 vanish. What is that? The electric field. What are the three components that are missing from this conditions? The magnetic field.

This is EM 101: F = q(E + v×B). If the particle is does not start out moving, then v = 0 and you get no information whatsoever about the magnetic field. This is neatly packaged using an antisymmetric rank-2 tensor, but it doesn't change the fact that you will have both electric and magnetic components to the electromagnetic field. Spacetime doesn't make basic EM go away.

Originally Posted by Farsight

Gravity isn't a force in the usual sense. Forget your constant force, it's a constant acceleration, and escape velocity is the flip side of the speed of an infalling body dropped from "a great distance".

I've talked about the simpler STR case because I know you can't understand geodesic motion and willfully ignore the Rindler coordinate chart. So to elaborate on my previous post:
1) In flat spacetime, a uniformly accelerated observer has a hyperbolic worldline, with speed bounded by the speed of light.
2) A uniform gravitational field must have no tidal forces, and so have flat spacetime. Hence, the uniform gravitational field is flat spacetime in the frame of a uniformly accelerated observer.
3) Hence, as any inertial particle passes the observer, the measured speed will be less than the speed of light, no matter the initial conditions.

Originally Posted by Farsight

When you drop your body into a black hole, it supposedly ends up going at the speed of light.

In what frame? In Schwarzschild coordinates, it does not. The freefalling particle gets asymptotically close to lightspeed in the sense that a sequence of stationary observers will measure it going to lightspeed as those observers get closer to the horizon.

Incidentally, that's just what happens for the uniform gravitational field: there is an acceleration horizon. Because accelerated observers have hyperbolic worldlines and hyperbolas have asymptotes.

Originally Posted by Farsight

It's not a base assertion. If we're all in space where light doesn't curve, Vorpal putting the pedal to the metal doesn't make it curve one iota. Learn to tell the difference between what you see and what's there.

Learn to actually check your assertions mathematically. You're completely wrong throughout, despite the fact that the calculations are pretty trivial.

Originally Posted by Farsight

How many more times do I have to tell you about flat hills? Find some plot of gravitational potential like the upturned hat, and just look at it. ...

Are you ever going to actually calculate a geodesic to prove your assertions? Because that's how gravitational freefall works: geodesic motion.

ben m · 10th April 2012, 05:59 PM

Originally Posted by Vorpal

1) In flat spacetime, a uniformly accelerated observer has a hyperbolic worldline, with speed bounded by the speed of light.
2) A uniform gravitational field must have no tidal forces, and so have flat spacetime. Hence, the uniform gravitational field is flat spacetime in the frame of a uniformly accelerated observer.

Yep. This reasoning works both ways. Suppose that you found yourself in a sealed box, in which you (and your portable gravimeter) always find yourselves accelerating in the -x direction. Since it's a very large box, you can explore it and look for gradients---you find none. The acceleration vector is of the same magnitude, and pointing in the same direction, no matter where you are.

So, you want to interpret that by guessing what's outside the box? Here are two guesses, both of which are consistent with the in-box spacetime:

a) Flat, empty spacetime, but there's a rocket attached to your box. Perfectly good explanation of the observations.

b) You're sitting still on the surface of a flat slab, infinite in x-y as far as you can tell, of constant areal density. That this matches the observations.

c) You're sitting still a distance r far above the same nearly-infinite slab, which must be large enough (R > r) to avoid inducing gradients. This matches the in-car observations too.

The a-c mapping is where the two intuitions match. A large slab, of radius R, has M = R^2. Therefore for some value of r the slab exceeds M/R = 1 and is a black hole with a horizon. Therefore, the spacetime geometry inside an accelerating car (case a) is exactly the same as the spacetime geometry inside a car hovering over a black hole, with a horizon (case c).

Brian-M · 10th April 2012, 06:45 PM

Farsight, I'm trying to figure out the practicalities of your idea for gravity, and it just isn't working out.

Since we're replacing the traditional concept of gravity as an attractive force with a new concept of a ~~time~~ light retardant force, I'll call it Farsight Force to distinguish it from standard gravitational force.

I'm assuming that the attraction towards the source of this Farsight Force is directly proportional to the gradient of the Farsight Force (if this isn't the case, please provide details), and this is the source of the problem.

If you take a force diminishing by distance according to the inverse square law, the gradient of that force is diminishing proportional the inverse cube of distance (and damn you for making me do math in order to figure this out

).

So the Farsight Force would have to diminish by less than the inverse square law in order for us to perceive gravity as diminishing according to the inverse square law. How is it possible for a force radiating evenly in all directions in three dimensional space to diminish at a rate less than the inverse square law would predict?

Farsight · 11th April 2012, 12:31 AM

Originally Posted by edd

I'm really not sure you got Vorpal's point. You take a charged particle and place it at rest and it doesn't move. You'd seem to be saying that that demonstrates the absence of any magnetic field, since you're saying that it shows there's no electromagnetic field.

Yes, that's right. If you set down a charged particle in a region of space and it doesn't move, there's no electromagnetic field present. If you set down an uncharged particle in a region of space and it doesn't move, there's no gravitational field present. If you set down a charged particle and an uncharged particle and only the charged particle moves, an electromagnetic field is present. If you set down a charged particle and an uncharged particle and they both move linearly, a gravitational field is present. If you set down a charged particle and it moves rotationally, you might say that a magnetic field is present, but you should always remember that the magnetic field is merely how you see an electromagnetic field when you have motion relative to it. So it's like throwing a charged particle through an electromagnetic field.

Farsight · 11th April 2012, 12:35 AM

Originally Posted by DeiRenDopa

By ~~Jove~~ Democritus!

That's it! The TRUE SECRET TO THE UNIVERSE!!!!!

Atoms are made up of neutrons, protons, and electrons. Each of which, in turn, per Farsight, is a standing wave of light.

A crystal, of an element such as iron, is then also just a bunch of standing waves of light. Molecules are also just mixtures of standing waves of light.

You got it. Hence this physicsworld article:

"The first real-time movie of large molecules creating an interference pattern after passing through two slits has been made by an international team of physicists".

Farsight · 11th April 2012, 12:50 AM

Originally Posted by sol invictus

You can make it as simple as you like, but it will remain wrong. There simply isn't a direct analog of "gravitational field" in GR, but the closest is the connection, not the curvature. Of course that's meaningless to you, since you have no clue what those words even mean, let alone what the difference is.

Huff puff, sol really doesn't like it when I explain the gravitational field with reference to Einstein. He behaves like a witchdoctor faced with a pharmacist, all outraged, saying "you don't understand it". Tough. I do.

Originally Posted by sol invictus

That's false, as Vorpal just tried to explain to you. If the electromagnetic field strength tensor in the rest frame of the particle has zero time-space components, but non-zero space-space components (that's a long-winded way of saying "there's a magnetic field, but zero electric field") then the particle won't move when you set it down.
So you're wrong. Admit it, learn from it, and move on. Can you do that?

I'm not wrong. See Electron magnetic dipole moment. The magnetic field exerts a torque on the electron. It "indeed behaves like a tiny bar magnet". Or a compass needle if you prefer. It moves. And you're wrong again.

Farsight · 11th April 2012, 01:04 AM

Originally Posted by ben m

Speculative nonsense, pseudo-published in specialized crackpot journals. (Physics Essays is well known, but Annals de la Fondation Louis de Broglie is basically the same thing.)

Farsight, are you utterly unable to tell the difference between beyond-the-mainstream, utterly-unchecked speculation, and actual particle physics? Is there something preventing you from using qualifiers, like "A few isolated authors have speculated ...", rather than pretending that this is all known, or true, or established?

Anyway, your crackpot claims about toroidal-particle-substructure are off-topic in this thread about black holes. I recommend you start a new thread if you want to discuss that.

I answered a direct question related to gravity and why things fall down. These people have to settle for back-street journals because of attitudes like yours. You dismiss the patent evidence and you have no counter-argument. Your contribution to this thread is negligible, all you offer is outraged abuse. These aren't crackpot claims, we really can make an electron from light in pair production, it really does have a magnetic dipole moment, we really can diffract it, and atomic orbitals really do feature standing waves. What do you think the electron is made of? Cheese? No, they aren't crackpot claims, but they have been censored by people like you.

Farsight · 11th April 2012, 02:05 AM

Originally Posted by Vorpal

And I say yet again: Take a look at the Lorentz force and it will be obvious:
$\frac{dp^\mu}{d\tau} = qF^\mu{}_\nu u^\nu$
In the frame of the particle, it is at rest and only the time component u⁰ of the four-velocity is nonzero. The electromagnetic tensor is antisymmetric, so it has six independent components, and in particular F₀₀ = 0. As a result, the condition of remaining stationary (i.e., Lorentz force vanishing) means the time-space components F_0a = -F_a0 vanish. What is that? The electric field. What are the three components that are missing from this conditions? The magnetic field.

This is EM 101: F = q(E + v×B). If the particle is does not start out moving, then v = 0 and you get no information whatsoever about the magnetic field.

Electrons behave like little compass needles. You do get information about the magnetic field.

Originally Posted by Vorpal

This is neatly packaged using an antisymmetric rank-2 tensor, but it doesn't change the fact that you will have both electric and magnetic components to the electromagnetic field. Spacetime doesn't make basic EM go away.

No motion, no field. The potential is uniform. That doesn't go away. Like I said, the potential is more fundamental than the field.

Originally Posted by Vorpal

I've talked about the simpler STR case because I know you can't understand geodesic motion and willfully ignore the Rindler coordinate chart.

I understand geodesic motion, and I don't "wilfully ignore" the Rindler coordinate chart. I just find it irrelevant, a portrayal of what you see when you accelerate. I still don't seem to be able to get this across to you: a light beam doesn't bend just because you stepped on the gas. You just see it as bent. In similar vein rain doesn't slant just because you start running.

Originally Posted by Vorpal

So to elaborate on my previous post:
1) In flat spacetime, a uniformly accelerated observer has a hyperbolic worldline, with speed bounded by the speed of light.

Yes, but a gravitational field is curved spacetime, so flat spacetime just isn't relevant.

Originally Posted by Vorpal

2) A uniform gravitational field must have no tidal forces, and so have flat spacetime. Hence, the uniform gravitational field is flat spacetime in the frame of a uniformly accelerated observer.

And there is no such thing as a uniform gravitational field! A gravitational field is curved spacetime, so what you're proposing is a contradiction in terms. No wonder you end up with impossible nonsense scenarios like the sky falling in around a black hole and neverneverland descriptions of the infalling observer. Come on Vorpal, work it out. If you accelerate at 9.8m/s/s and keep on accelerating at a uniform 9.8 m/s/s you end up exceeding the speed of light, and you've contradicted special relativity.

Originally Posted by Vorpal

3) Hence, as any inertial particle passes the observer, the measured speed will be less than the speed of light, no matter the initial conditions.

An accelerating observer always measures that speed as less than the speed of light. And he always measures the speed of light to be the same. Now go and read The Other Meaning of Special Relativity and understand why. When you're made of waves along with everything else, you always measure wave speed to be the same. And nothing made from waves can exceed that speed.

Originally Posted by Vorpal

In what frame? In Schwarzschild coordinates, it does not. The freefalling particle gets asymptotically close to lightspeed in the sense that a sequence of stationary observers will measure it going to lightspeed as those observers get closer to the horizon.

In my frame. If you keep on accelerating at a uniform 9.8m/s/s whilst traversing the starry sky, I end up seeing you going from one star to another faster than light can go.

Originally Posted by Vorpal

Incidentally, that's just what happens for the uniform gravitational field: there is an acceleration horizon. Because accelerated observers have hyperbolic worldlines and hyperbolas have asymptotes.

Aaaagh! There is no uniform gravitational field. You end up accelerating a falling body less than you accelerate a body that you placed at some elevation. If it's 9.8 m/s/s, that's what it is for all bodies.

Originally Posted by Vorpal

Learn to actually check your assertions mathematically. You're completely wrong throughout, despite the fact that the calculations are pretty trivial.

I'm not wrong. You've used mathematics to confuse a constant force and a constant change of speed. When you're already moving at 299,792,457 m/s, a second later you're moving at 299,792,466.8 m/s. It isn't going to happen.

Originally Posted by Vorpal

Are you ever going to actually calculate a geodesic to prove your assertions? Because that's how gravitational freefall works: geodesic motion.

No. Because that isn't how it works. I told you how it works. Get used to it. It isn't in your textbooks yet, but it will be.

Farsight · 11th April 2012, 02:21 AM

Originally Posted by Brian-M

Farsight, I'm trying to figure out the practicalities of your idea for gravity, and it just isn't working out.

Since we're replacing the traditional concept of gravity as an attractive force with a new concept of a ~~time~~ light retardant force, I'll call it Farsight Force to distinguish it from standard gravitational force.

Don't think of it as a force, the Newtonian F=ma and E=F x d doesn't apply because conservation of energy applies instead. Just think of it as inhomogeneous space that causes light to bend as per gravitational lensing. Take your cue for that from Einstein's Leyden address:

"This space-time variability of the reciprocal relations of the standards of space and time, or, perhaps, the recognition of the fact that “empty space” in its physical relation is neither homogeneous nor isotropic, compelling us to describe its state by ten functions (the gravitation potentials gμν), has, I think, finally disposed of the view that space is physically empty".

Originally Posted by Brian-M

I'm assuming that the attraction towards the source of this Farsight Force is directly proportional to the gradient of the Farsight Force (if this isn't the case, please provide details), and this is the source of the problem.

No, the "force" of attraction is directly proportional to the gradient in the potential.

Originally Posted by Brian-M

If you take a force diminishing by distance according to the inverse square law, the gradient of that force is diminishing proportional the inverse cube of distance (and damn you for making me do math in order to figure this out

).

No no, it's the gradient in the potential. You measure the speed of light at different floors in your very very tall building, and then draw a graph to plot the potential. It's a curve that starts out with a steep gradient which gradually gets shallower with altitude. Where the gradient in the potential is steeper, the force is greater.

Originally Posted by Brian-M

So the Farsight Force would have to diminish by less than the inverse square law in order for us to perceive gravity as diminishing according to the inverse square law. How is it possible for a force radiating evenly in all directions in three dimensional space to diminish at a rate less than the inverse square law would predict?

You need to start again with this I'm afraid Brian.

Dancing David · 11th April 2012, 04:36 AM

Originally Posted by Vorpal

I don't think you understand what local means.

I sense a pattern here.

Private idiomatic definitions to reinforce tautological fallacies on the part of FS.

Dancing David · 11th April 2012, 04:41 AM

Originally Posted by Farsight

Yes, that's right. If you set down a charged particle in a region of space and it doesn't move, there's no electromagnetic field present. If you set down an uncharged particle in a region of space and it doesn't move, there's no gravitational field present. If you set down a charged particle and an uncharged particle and only the charged particle moves, an electromagnetic field is present. If you set down a charged particle and an uncharged particle and they both move linearly, a gravitational field is present. If you set down a charged particle and it moves rotationally, you might say that a magnetic field is present, but you should always remember that the magnetic field is merely how you see an electromagnetic field when you have motion relative to it. So it's like throwing a charged particle through an electromagnetic field.

Oh my.

W.D.Clinger · 11th April 2012, 05:05 AM

Once again, Farsight has stated an important difference between FGR (Farsight general relativity) and the general theory of relativity put forth by Einstein:

Originally Posted by Farsight

When spacetime is flat, no gravitational field is present.

Originally Posted by Farsight

When spacetime is flat, no gravitational field is present.

Originally Posted by Farsight

Originally Posted by sol invictus

You can make it as simple as you like, but it will remain wrong. There simply isn't a direct analog of "gravitational field" in GR, but the closest is the connection, not the curvature. Of course that's meaningless to you, since you have no clue what those words even mean, let alone what the difference is.

Huff puff, sol really doesn't like it when I explain the gravitational field with reference to Einstein. He behaves like a witchdoctor faced with a pharmacist, all outraged, saying "you don't understand it". Tough. I do.

Farsight can huff and puff all he likes, but he's arguing with Einstein here. From the English translation of "The Foundation of the General Theory of Relativity":

Originally Posted by Albert Einstein

C. THEORY OF THE GRAVITATIONAL FIELD

§ 13. Equations of Motion of a Material Point in the Gravitational Field. Expression for the Field-components of Gravitation

A freely movable body not subjected to external forces moves, according to the special theory of relativity, in a straight line and uniformly. This is also the case, according to the general theory of relativity, for a part of four-dimensional space in which the system of co-ordinates K₀, may be, and is, so chosen that they have the special constant values given in (4).

Einstein's formula (4) is the 4x4 matrix of coefficients for the Minkowski metric in the familiar coordinate system for flat spacetime. Einstein continues by talking about the gravitational field that appears when you use a different coordinate system to describe exactly the same locally flat spacetime:

Originally Posted by Albert Einstein

If we consider precisely this movement from any chosen system of co-ordinates K₁, the body, observed from K₁, moves, according to the considerations in § 2, in a gravitational field....

If the Γ^τ_μν vanish, then the point moves uniformly in a straight line. These quantities therefore condition the deviation of the motion from uniformity. They are the components of the gravitational field.

According to Einstein, therefore, the 64 coordinate-dependent Christoffel symbols "are the components of the gravitational field". (Because the Christoffel symbols are symmetric in their lower indices, not all of those components are independent.)

Let's do an example. As this example shows, Einstein's "components of the gravitational field" can be nonzero even in flat spacetime, contrary to Farsight's repeated baseless assertions.

Let's use the metric form posted by sol invictus. In that coordinate system, the four nonzero components of the covariant metric tensor are

$ \[ \begin{align*} g_{00} &= - (x^1 - r_0) \\ g_{11} &= \frac{1}{x^1 - r_0} \\ g_{22} &= 1 \\ g_{33} &= 1 \end{align*} \] $

(I'm using MTW notational conventions, and I'm writing the r coordinate as x¹ to prepare for the index and Einstein summation notations used below.)

To calculate the 64 components of the gravitational field, we'll need the components of the contravariant form as well:

$ \[ \begin{align*} g^{00} &= - \frac{1}{(x^1 - r_0)} \\ g^{11} &= (x^1 - r_0) \\ g^{22} &= 1 \\ g^{33} &= 1 \end{align*} \] $

Einstein's equation (45) states his formula for the components of the gravitational field. Expanding Einstein's notation into modern notation, Einstein's equation becomes

$ \[ \begin{align*} \Gamma^\lambda_{\mu\nu} &= \frac{1}{2} g^{\lambda\alpha} \left( \partial_{\nu} g_{\alpha\mu} + \partial_{\mu} g_{\alpha\nu} - \partial_{\alpha} g_{\mu\nu} \right) \end{align*} \] $

If I've done that calculation correctly, only 4 of the 64 Christoffel symbols are nonzero:

$ \[ \begin{align*} \Gamma^0_{01} &= \Gamma^0_{10} = \frac{1}{2 (x^1 - r_0)} \\ \Gamma^1_{00} &= \frac{1}{2} (x^1 - r_0) \\ \Gamma^1_{11} &= - \frac{1}{2 (x^1 - r_0)} \end{align*} \] $

With those Christoffel symbols in hand, it's straightforwardly tedious to calculate the 256 components of the Riemann curvature tensor via Einstein's equation (43). In modern notation, that equation becomes

$ \[ \begin{align*} R^{\alpha}_{\beta\gamma\delta} &= \partial_{\gamma} \Gamma^{\alpha}_{\beta\delta} - \partial_{\delta} \Gamma^{\alpha}_{\beta\gamma} + \Gamma^{\alpha}_{\kappa\gamma} \Gamma^{\kappa}_{\beta\delta} - \Gamma^{\alpha}_{\kappa\delta} \Gamma^{\kappa}_{\beta\gamma} \end{align*} \] $

That calculation is greatly simplified by the following observations. As can be seen from the Christoffel symbols calculated above, the partial derivatives of the first two terms are zero unless the subscript on the derivative operator is 1. Similarly, every component of the Riemann tensor that has a 2 or a 3 as an index will be zero; that observation reduces the number of components from 256 to 16, even before we take advantage of symmetries.

I calculated the components of the Riemann tensor by hand, and found all of its 256 components are zero.

That means sol invictus has indeed given us an example of flat spacetime (because the Riemann curvature tensor is zero) that has a nonzero gravitational field (because four of the Christoffel symbols, which Einstein regards as the components of the gravitational field, are nonzero).

In his response to sol invictus, quoted above, Farsight said he understands all this stuff. That's great. I invite Farsight to check my calculations.

The bottom line is that Einstein disagreed with Farsight's repeated bare assertion that flat spacetime implies a zero gravitational field.

Brian-M · 11th April 2012, 06:16 AM

Originally Posted by Farsight

Don't think of it as a force, the Newtonian F=ma and E=F x d doesn't apply because conservation of energy applies instead. Just think of it as inhomogeneous space that causes light to bend as per gravitational lensing. Take your cue for that from Einstein's Leyden address:

Force, potential, curvature of space, whatever. It's irrelevant to the point. If it's effect is evenly distributed and omnidirectional in three dimensional space, and it diminishes at less than the inverse square law, then this means that the total effect of this force is increasing over distance (even if the localized effect is decreasing with distance because the effect is spread over a larger area).

And the total effect of the force increasing with distance is what doesn't make any sense.

Originally Posted by Farsight

No, the "force" of attraction is directly proportional to the gradient in the potential.

Wait, you disagree with me then you basically repeat what I just said? By Farsight Force I'm referring to this potential you mention.

Originally Posted by Farsight

No no, it's the gradient in the potential. You measure the speed of light at different floors in your very very tall building, and then draw a graph to plot the potential. It's a curve that starts out with a steep gradient which gradually gets shallower with altitude. Where the gradient in the potential is steeper, the force is greater.

But that's what I'm talking about. I figured it out by making up an example of objects experiencing different gravitational forces experienced at points different distances from a large mass, and then estimated the local gradient of this "potential" over a distance of a meter by taking the gravitational force half a meter closer to the mass from the point and subtracting the gravitational force thats half a meter further away.

I found that where gravitational force diminishes with the inverse square law, the gravitational gradient diminishes according to the the inverse cubed law. (This is the math I referred to in my previous post.) From this I must conclude that any force (or potential, or whatever) which diminishes according to the inverse square law must have a gradient which diminishes according to the inverse cubed law.

So if the experienced (or directly measured) force of gravity is directly proportional to the gradient of this potential, and this potential diminishes with distance according to the inverse square law, then we should be measuring gravity as diminishing according to the inverse cubed law... but we don't!

For us to experience or measure gravity as diminishing with the inverse square law, then the potential (whose gradient produces what we experience as gravity) must be diminishing at less than the inverse square law.

And that doesn't make any sense.

Originally Posted by Farsight

You need to start again with this I'm afraid Brian.

Where is my understanding of your physics flawed? The way I'm reading this you're disagreeing with me, but your corrections are the exact same thing I'm talking about, only worded slightly differently.

Farsight · 11th April 2012, 06:57 AM

Originally Posted by Brian-M

Force, potential, curvature of space, whatever. It's irrelevant to the point. If it's effect is evenly distributed and omnidirectional in three dimensional space, and it diminishes at less than the inverse square law, then this means that the total effect of this force is increasing over distance (even if the localized effect is decreasing with distance because the effect is spread over a larger area).

No. The effect diminishes with distance in line with the inverse square law, and the total effect tends to a limit.

Originally Posted by Brian-M

And the total effect of the force increasing with distance is what doesn't make any sense.

That's what I've been trying to say to Vorpal about his uniform gravitational field.

Originally Posted by Brian-M

Wait, you disagree with me then you basically repeat what I just said? By Farsight Force I'm referring to this potential you mention.

Don't. The force of gravity at some location is a measure of the gradient in the potential at that location. Look at the upturned hat again. The gradient reduces with distance.

Originally Posted by Brian-M

But that's what I'm talking about. I figured it out by making up an example of objects experiencing different gravitational forces experienced at points different distances from a large mass, and then estimated the local gradient of this "potential" over a distance of a meter by subtracting the gravitational force half a meter further away from the gravitational force half a meter closer.

That's wrong. The force of gravity at some location tells you the local gradient at that location. The difference between the force of gravity at one location and another tells you how the gradient changes. That's the tidal force.

Originally Posted by Brian-M

I found that where gravitational force diminishes with the inverse square law, the gravitational gradient diminishes according to the the inverse cubed law.

No. The force of gravity diminishes with the inverse law, and the gradient in gravitational potential diminishes with the inverse square law.

Originally Posted by Brian-M

(This is the math I referred to in my previous post.) From this I must conclude that any force (or potential, or whatever) which diminishes according to the inverse square law must have a gradient which diminishes according to the inverse cubed law.

The force of gravity is only there because there's a gradient in the gravitational potential. Where there's no gradient, light goes straight and things don't fall down.

Originally Posted by Brian-M

So if the experienced (or directly measured) force of gravity is directly proportional to the gradient of this potential, and this potential diminishes with distance according to the inverse square law, then we should be measuring gravity as diminishing according to the inverse cubed law... but we don't!

The force of gravity is there because of the gradient of the potential. Both diminish according to the inverse square law.

Originally Posted by Brian-M

For us to experience or measure gravity as diminishing with the inverse square law, then the potential (whose gradient produces what we experience as gravity) must be diminishing at less than the inverse square law.

Look at the upturned hat to see how the potential varies. Invert it if you wish.

Originally Posted by Brian-M

Where is my understanding of your physics flawed? The way I'm reading this you're disagreeing with me, but your corrections are the exact same thing I'm talking about, only worded slightly differently.

You're getting the gradient in the potential and the potential mixed up, and thinking that things are more complicated than they are. Look, here's the upturned hat. It's a plot of gravitational potential. Where the gradient is steep, the force of gravity at that location is high. It diminishes with distance from the centre. The potential increases with distance from the centre, but doesn't keep on increasing ad infinitum, it tends to a limit like a curve tending up to a horizontal line.

CCASA image by AllenMc see wikipedia

sol invictus · 11th April 2012, 07:06 AM

Originally Posted by Farsight

Huff puff, sol really doesn't like it when I explain the gravitational field with reference to Einstein.

Except that you're lying about Einstein. Einstein directly and explicitly contradicts you, as Clinger's quote shows.

Quote:

He behaves like a witchdoctor faced with a pharmacist, all outraged, saying "you don't understand it". Tough. I do.

Why is it that your posts are always full of such bizarre fantasies? No one else needs to resort to such delusions - why you, Farsight?

Quote:

I'm not wrong. See Electron magnetic dipole moment. The magnetic field exerts a torque on the electron. It "indeed behaves like a tiny bar magnet". Or a compass needle if you prefer. It moves. And you're wrong again.

No, Farsight - you're wrong. Your statement was a general unqualified assertion about charged particles, and it was false. As for electrons (or anything else with a mdm) - in a uniform magnetic field they do not move. Their quantum spin might or might not precess (depending on the initial quantum state and the orientation of the field), but they don't change position. Even in a non-uniform field they may or may not move, again depending on the state and the orientation of the field.

Farsight · 11th April 2012, 08:54 AM

I'm not lying about Einstein, I'm quoting him. And I'm the one shattering the bizarre fantasies here, like the waterfall and neverneverland and the uniform gravitational field. The change in spin orientation is a movement, just as it is in the hyperfine transition where a photon is emitted. Go look up ferromagnetism and the Einstein-de Haas effect:

"The effect corresponds to the mechanical rotation that is induced in a ferromagnetic material (of cylindrical shape and originally at rest), suspended with the aid of a thin string inside a coil, on driving an impulse of electric current through the coil.[1] To this mechanical rotation of the ferromagnetic material (say, iron) is associated a mechanical angular momentum, which, by the law of conservation of angular momentum, must be compensated by an equally large and oppositely directed angular momentum inside the ferromagnetic material. Given the fact that an external magnetic field, here generated by driving electric current through the coil, leads to magnetisation of electron spins in the material (or to reversal of electron spins in an already magnetised ferromagnet — provided that the direction of the applied electric current is appropriately chosen), the Einstein–de Haas effect demonstrates that spin angular momentum is indeed of the same nature as the angular momentum of rotating bodies as conceived in classical mechanics. This is remarkable, since electron spin, being quantized, cannot be described within the framework of classical mechanics".

It isn't quantum magic. It's just movement. And as it happens electron spin can be described within the framework of classical mechanics, as a biaxial rotation where one rate is half the other. However when people do this they are met with opprobium and abuse and censorship by those who peddle surpasseth-all-human-understanding mysticism.

Farsight · 11th April 2012, 09:24 AM

Originally Posted by W.D.Clinger

...Let's use the metric form posted by sol invictus. In that coordinate system, the four nonzero components of the covariant metric tensor are

$ \[ \begin{align*} g_{00} &= - (x^1 - r_0) \\ g_{11} &= \frac{1}{x^1 - r_0} \\ g_{22} &= 1 \\ g_{33} &= 1 \end{align*} \] $

You lost me, and everybody else. What does sol's metric represent again? All I've managed to squeeze out of him is "flat spacetime", and more recently "constant proper acceleration". You can't make a gravitational field by zooming through space, Clinger. Things don't fall down as you pass by. Look, go back to what Einstein said and think it through then contribute sensibly to the discussion instead of hiding behind mathematics. Remember this:

"This space-time variability of the reciprocal relations of the standards of space and time, or, perhaps, the recognition of the fact that “empty space” in its physical relation is neither homogeneous nor isotropic, compelling us to describe its state by ten functions (the gravitation potentials gμν), has, I think, finally disposed of the view that space is physically empty".

Do you get it yet? Space isn't nothing. And you don't change space by moving through it.

Farsight · 11th April 2012, 09:28 AM

Originally Posted by Dancing David

I sense a pattern here.

And I sense a troll. Now start making a sincere contribution to this discussion, or sling your hook.

9th April 2012, 09:39 AM	#1001
Vorpal Extrapolate! Join Date: Jan 2005 Posts: 1,013	Originally Posted by DeiRenDopa And why is this? I mean, there's nothing particularly unusual in what Vorpal writes, is there? Nothing that you won't find in a suitable, contemporary physics textbook, is there? As far as I'm aware, the notable deviation from many GTR textbooks is that on the subject of what the "gravitational field" actually is, the stance most of them seem to take is roughly "we don't talk about that." Sometimes very literally: MTW effectively spend an entire section explaining their reasons why they leave the term "gravitational field" undefined and hence why they're not going to use it, ever. (That's certainly a very consistent view: they still have all the physically relevant quantities in their mathematics anyway; they just refuse to give any of them that label.) Under that (non-)interpretation, Synge's comment doesn't mean anything at all, so we need to define what "gravitational field" means. It so happens that Einstein did define it to be the connection coefficients/Christoffel symbols Γ^α_μν and the gravitational potential to be the metric components. There are good reasons why this is sensible, and also a reason to reject it--Γ's are not tensorial as a whole. But it remains the case that there isn't a better candidate, and its coordinate-dependent nature actually embraces the equivalence principle: gravity would be explicitly an inertial ('fictitious') force. And then Farsight seems to be arguing for an almost-schizophrenic position that Einstein's "potential = metric, Γ's = gravitational field" identification is very obviously correct (I don't know why else he would be harping about how I'm a woo-woo for using the term "almost eerie" to describe just how well it works) and yet completely denying its implication: that curvature vanishing at an event does not mean the field vanishes there. P.S. I really should have made the last bit more explicit in my previous post.
Vorpal Extrapolate! Join Date: Jan 2005 Posts: 1,013	__________________ For every philosopher, there exists an equal and opposite philosopher. They're both wrong.

9th April 2012, 12:51 PM	#1005
dafydd Penultimate Amazing Join Date: Feb 2008 Location: Belgium (Flatland) Posts: 31,680	Originally Posted by DeiRenDopa Funny how Farsight's 'absolute time and space' fundamental worldview keeps appearing, in his posts, isn't it? If beams of light do not bend then how does gravitational lensing occur? I'm looking forward to Farsight's explanation.
	__________________ Yesterday upon the stairs I met a man who wasn't there He wasn't there again today I wish that he would go away.

9th April 2012, 09:54 PM	#1006
Brian-M Daydreamer Join Date: Jul 2008 Location: Downunder Posts: 4,387	Originally Posted by Farsight That's wrong. If the clock at the ceiling stayed synchronised with the clock at the floor, things wouldn't fall down. You're affirming the consequent. The normal point of view is that ~~time~~ light passes at different speeds because the gravitational attraction is different. Your point of view reverses this, claiming that gravitational attraction is different because ~~time~~ light is passing at different speeds. You can't know that if ~~time~~ light wasn't passing at different speeds at different heights then "things wouldn't fall down" until you first demonstrate that your point of view is correct. Originally Posted by Farsight I've said it already. When gravitational tidal forces vanish you've got a gravitational field that doesn't diminish with distance. That's a nonsense implication. Is it nonsense? Increase the distance of your floating lab/city from the planet, and then increase the mass of the planet to compensate so that you're still experiencing the same amount of gravity. Suddenly, the tidal forces in your lab are smaller than they were before. Repeat this process indefinitely. As distance approaches infinity, tidal force (difference in gravity) approaches at zero. Sure, you can't actually get infinite distance. But at some finite distance you're going to reach a point where no practical test inside your lab can detect the presence of any gravitational variation with distance. Now here's something to think about. If things fall down because light is moving slower at the lower elevation as you claim, shouldn't differences in the speed of light over the same elevation (as in my example above) cause things to fall with different force? But if gravity follows the inverse square law (as I assumed in my example above), then this can't possibly be true.
	__________________ "That is just what you feel, that isn't reality." - hamelekim

10th April 2012, 03:14 AM	#1010
Vorpal Extrapolate! Join Date: Jan 2005 Posts: 1,013	Farsight, you're simply repeating irrelevancies. 1) Absolutely no one is claiming that you can transform away Riemann curvature at events where it's nonzero. Repeating it incessantly is completely pointless, because no one is questioning this fact. 2) The real issue is whether the vanishing of Riemann curvature at some event implies that the non-presence of the gravitational field there. This is something you, and Synge, simply assert as true, and provide no reasoning for that claim. You've simply made no connection between (1), which everybody knows, and (2), which is what we're trying to have a discussion about. And failing. Originally Posted by Farsight The Lorentz force isn't relevant. If you set a charged particle down and find it doesn't move, you say there's no electromagnetic field present. It's that simple. ...When you set it down and find it doesn't move, then you say there's no electromagnetic field present. You're mistaken. If you have a charged test particle that starts at rest and it doesn't move, it in no way implies that there's no electromagnetic field present. It only implies that there is no electric field present. Originally Posted by Farsight You drop an object from a great distance, and it accelerates until it's going faster than light. That can't happen. Nonsense. This is a basic calculus exercise for STR. Suppose a particle starts from rest and undergoes constant force: $F = \frac{dp}{dt} = m\frac{d(\gamma v)}{dt} = mc\left(\cosh\alpha\right)\frac{d\alpha}{dt}$ Where I've applied the substitution v/c = tanh α, so γv/c = sinh α and γ = cosh α. Just integrate: Ft = sinh α, and therefore: $v = c\tanh\left(\sinh^{-1}\left(\frac{Ft}{mc}\right)\right) = \frac{Ft/m}{\sqrt{1 + (Ft/mc)^2}}$ Note that the tanh form makes it obvious at the speed is bounded by c. Originally Posted by Farsight Yes. A gravitational field is curved spacetime. It isn't flat spacetime, and accelerating through space does not turn flat spacetime into curved spacetime. I asked for reasons, not mere repetition of baseless assertion.
Vorpal Extrapolate! Join Date: Jan 2005 Posts: 1,013	__________________ For every philosopher, there exists an equal and opposite philosopher. They're both wrong. Last edited by Vorpal; 10th April 2012 at 03:15 AM.

10th April 2012, 04:23 AM	#1011
dafydd Penultimate Amazing Join Date: Feb 2008 Location: Belgium (Flatland) Posts: 31,680	Originally Posted by Farsight Huh? I said this: When spacetime is flat, no gravitational field is present. A beam of light doesn't curve. Accelerating past that light beam only changes how you see it, not the light beam itself. Gravitational lensing occurs when beams of light bend. They bend because they move through a region of space where a gravitational field is present. When this happens we call it curved spacetime, when it doesn't and the beams of light go straight, we call it flat spacetime. Thank you. You didn't mention flat spacetime in your previous post.
	__________________ Yesterday upon the stairs I met a man who wasn't there He wasn't there again today I wish that he would go away.

9th April 2012, 11:07 AM	#1002
Farsight Master Poster Join Date: Mar 2008 Location: Poole, UK Posts: 2,008	Originally Posted by Vorpal Nope, not in that frame, I wouldn't. "It's not the fall that kills you--it's the sudden stop at the end." My terminal ends is because the ground puts a tremendous force on me. When you're in a box falling to earth your inertial reference frame is continually changing but you can't tell. When that change stops occuring all of a sudden, then you can. Originally Posted by Vorpal I don't think you understand what local means. Of course I do. Originally Posted by Vorpal Of course--what is or isn't rest is frame-dependent. But the major difference is that in Newtonian physics, making a frame that comoves with a gravitationally freefalling particle gives an accelerated frame--it doesn't follow a geodesic. In GTR, it does: the acceleration four-vector along the trajectory identically vanishes. That's the point of EP. And again, it only works for a region of infinitesimal size. Originally Posted by Vorpal Synge's argument is about the relationship between the gravitational field and the gravitational tidal forces. You're simply claiming some relevance without even a word as to why it is so. I've said it already - you cannot transform away the Riemann curvature. All you can do is take a small region of a curve, a region so small that it looks flat. But we know it isn't flat. Originally Posted by Vorpal Since you yourself specified the case where the charge is not moving in our local inertial frame, the electric field is what you get to measure. Remember Lorentz force and it'll be obvious. The Lorentz force isn't relevant. If you set a charged particle down and find it doesn't move, you say there's no electromagnetic field present. It's that simple. Originally Posted by Vorpal For the purposes of this discussion, it doesn't matter whether there are or aren't any point charges. Read what I wrote again: I'm saying that even if that particular statement of yours was 100% absolute truth, nothing about my point changes. Because what you said literally adds nothing more than a side comment of "... and the field comes from the potential." Yes, everyone knows that. I don't mind if you consider it the mostest fundamentalest thing there ever was. What I said was: () the gravitational field and gravitational tidal forces are different things and the electromagnetic field was brought in to illustrate* why Synge's argument is mistaken: because we can define a tensor that describes electromagnetic tidal forces, in the sense of deviation of equally-charged test particles, and directly confirm that its vanishing does not imply that the electromagnetic field vanishes. Forget your equally-charged test particles. Use a single test particle. When you set it down and find it doesn't move, then you say there's no electromagnetic field present. Originally Posted by Vorpal Or, more bluntly: Synge's implication is very clearly false for other fields. No it isn't, you've misunderstood the scenario. Originally Posted by Vorpal Why should we believe him about the gravitational field? Because he's right. You can't make the Riemann curvature vanish. All you do when you take an infinitesimal region is sweep it under the carpet. Take that too literally and you end up with a flat hill and a misunderstanding of the gravitational field. Originally Posted by Vorpal Evidently, you don't agree with it, because if you accepted it as valid, you'd see that Synge is wrong immediately. Hence my question about your reasons against it. No I don't see that Synge is wrong immediately. Instead what I see immediately is that you got the scenario wrong. Originally Posted by Vorpal It's just poking fun at the fact that for someone who incessantly insistent that he was "with Einstein", you spend extraordinary amounts of time arguing against him. No I don't. I spend far more time saying Einstein said x. And I've said already that he wasn't perfect. Originally Posted by Vorpal And I'll put this last: Originally Posted by Farsight I've said it already. When gravitational tidal forces vanish you've got a gravitational field that doesn't diminish with distance. That's a nonsense implication. What's wrong with that? You drop an object from a great distance, and it accelerates until it's going faster than light. That can't happen. Originally Posted by Vorpal If you believe the analogy to be valid, then the gravitational field is the connection, and it's trivial that all you need for a homogeneous but nonvanishing gravitational field is to take flat spacetime in a uniformly accelerated frame. So I ask yet again: do you have any reasons to consider the analogy wrong? Yes. A gravitational field is curved spacetime. It isn't flat spacetime, and accelerating through space does not turn flat spacetime into curved spacetime. The principle of equivalence is a principle that shows you why the two are similar, but it doesn't tell you that they're identical.

9th April 2012, 11:19 AM	#1003
Farsight Master Poster Join Date: Mar 2008 Location: Poole, UK Posts: 2,008	Originally Posted by Vorpal ...And then Farsight seems to be arguing for an almost-schizophrenic position that Einstein's "potential = metric, Γ's = gravitational field" identification is very obviously correct (I don't know why else he would be harping about how I'm a woo-woo for using the term "almost eerie" to describe just how well it works) and yet completely denying its implication: that curvature vanishing at an event does not mean the field vanishes there. It isn't schizophrenic, it's really important. When spacetime is flat, no gravitational field is present. A beam of light doesn't curve. Accelerating past that light beam only changes how you see it, not the light beam itself. And I was scathing of your "eerie" because you don't understand why that comparison can be made.

9th April 2012, 12:23 PM	#1004
DeiRenDopa Master Poster Join Date: Feb 2008 Posts: 2,381	Originally Posted by Farsight ... A beam of light doesn't curve. Accelerating past that light beam only changes how you see it, not the light beam itself. ... Funny how Farsight's 'absolute time and space' fundamental worldview keeps appearing, in his posts, isn't it?

9th April 2012, 10:17 PM	#1007
W.D.Clinger Master Poster Join Date: Oct 2009 Posts: 2,463	Originally Posted by Farsight Originally Posted by Vorpal ...And then Farsight seems to be arguing for an almost-schizophrenic position that Einstein's "potential = metric, Γ's = gravitational field" identification is very obviously correct (I don't know why else he would be harping about how I'm a woo-woo for using the term "almost eerie" to describe just how well it works) and yet completely denying its implication: that curvature vanishing at an event does not mean the field vanishes there. It isn't schizophrenic, it's really important. When spacetime is flat, no gravitational field is present. A beam of light doesn't curve. Accelerating past that light beam only changes how you see it, not the light beam itself. And I was scathing of your "eerie" because you don't understand why that comparison can be made. It isn't schizophrenic so much as ignorant. Farsight doesn't realize that the coordinate-dependent Christoffel symbols (which, according to Einstein, describe the gravitational field) can be nonzero even in flat spacetime. Had Farsight made an effort to understand the metric form posted by sol invictus, he might not have made that mistake. On the other hand, Farsight would have had to learn calculus first...

10th April 2012, 02:12 AM	#1008
Farsight Master Poster Join Date: Mar 2008 Location: Poole, UK Posts: 2,008	Originally Posted by dafydd If beams of light do not bend then how does gravitational lensing occur? I'm looking forward to Farsight's explanation. Huh? I said this: When spacetime is flat, no gravitational field is present. A beam of light doesn't curve. Accelerating past that light beam only changes how you see it, not the light beam itself. Gravitational lensing occurs when beams of light bend. They bend because they move through a region of space where a gravitational field is present. When this happens we call it curved spacetime, when it doesn't and the beams of light go straight, we call it flat spacetime.

10th April 2012, 03:04 AM	#1009
Farsight Master Poster Join Date: Mar 2008 Location: Poole, UK Posts: 2,008	Originally Posted by Brian-M You're affirming the consequent. No, you are. Originally Posted by Brian-M The normal point of view is that ~~time~~ light passes at different speeds because the gravitational attraction is different. Your point of view reverses this, claiming that gravitational attraction is different because ~~time~~ light is passing at different speeds. My point of view is Einstein's point of view. He said a curvature of rays of light can only take place when "die Ausbreitungs-geschwindigkeit des Lichtes mit dem Orte variiert", which translates into "the speed of light varies with the locality". When it doesn't, there is no gravitational attraction. Originally Posted by Brian-M You can't know that if ~~time~~ light wasn't passing at different speeds at different heights then "things wouldn't fall down" until you first demonstrate that your point of view is correct. Einstein was the author of general relativity, his point of view was correct. And you know it's correct from the parallel-mirror gif. And you know about the wave nature of matter don't you? You know we can make an electron out of light via pair production? And that the electron has spin angular momentum and a magnetic dipole moment? And that it can be diffracted? So think of the electron as a standing wave of light. A circle of light, as it were. Put it in a gravitational field and think it through. Divide the circle into four flat quadrants to make it even simpler: ..← ↓....↑ ..→ Starting from the left and going anticlockwise, at a given instant we have light travelling down like this ↓. There’s a gradient in c from top to bottom, but all it does is make the light look blueshifted. A little while later the light is moving like this → and the lower portion of the wave-front is subject to a slightly lower c than the upper portion. So it bends down a little. Later it’s going this way ↑ and looks redshifted, and later still it’s going this way ← and bends down again. These bends translate into a different position for the electron. It falls down: ○ ↓ The reducing c bleeds rotational motion out into linear motion. But only half the cycle got bent, so only half the reduced c goes into kinetic energy aka relativistic mass. That’s why light is deflected twice as much as matter, and gravity is not the sort of force that increases the relativistic mass. Simple. Originally Posted by Brian-M Originally Posted by Farsight I've said it already. When gravitational tidal forces vanish you've got a gravitational field that doesn't diminish with distance. That's a nonsense implication. Is it nonsense? Yes. You've got a reducing speed of light that keeps on reducing, so much so that it and ends up being negative. There's no such thing as a negative speed. A negative speed is nonsense. Originally Posted by Brian-M Increase the distance of your floating lab/city from the planet, and then increase the mass of the planet to compensate so that you're still experiencing the same amount of gravity. Suddenly, the tidal forces in your lab are smaller than they were before. No problem with that. Originally Posted by Brian-M Repeat this process indefinitely. As distance approaches infinity, tidal force (difference in gravity) approaches at zero. Sure, you can't actually get infinite distance. But at some finite distance you're going to reach a point where no practical test inside your lab can detect the presence of any gravitational variation with distance. But things in your lab still fall down, and the parallel-mirror light-clock at the ceiling still runs faster than the one at the floor. Originally Posted by Brian-M Now here's something to think about. If things fall down because light is moving slower at the lower elevation as you claim, shouldn't differences in the speed of light over the same elevation (as in my example above) cause things to fall with different force? Yes. If there's a big difference between the speed of light at the ceiling as compared to the speed of light at the floor, the force of gravity is large. If there's a lesser difference, the force of gravity is less. When there's no difference, things don't fall down. Originally Posted by Brian-M But if gravity follows the inverse square law (as I assumed in my example above), then this can't possibly be true. It is true. When you measure the difference between the speed of light at the ceiling as opposed to the floor in a very very tall building, you find that on floor 100,000 the difference is less than it is in the basement. If you could take an equatorial slice through the planet and measure the speed of light at various locations, when you plot them out what you end up with is the upturned hat. When you look at one small region of this there's still a gradient in gravitational potential and a gradient in c. There's no discernible curvature of the gradient, you can't detect any tidal force, but things still fall down, and light still curves when it moves through space, so we say spacetime is curved. Way out a zillion miles to the left or right where there's no discernible gradient, your measurements can't detect any curvature of light, or things falling down, and then we say spacetime is flat.

10th April 2012, 04:39 AM	#1012
Brian-M Daydreamer Join Date: Jul 2008 Location: Downunder Posts: 4,387	Originally Posted by Farsight My point of view is Einstein's point of view. He said a curvature of rays of light can only take place when "die Ausbreitungs-geschwindigkeit des Lichtes mit dem Orte variiert", which translates into "the speed of light varies with the locality". When it doesn't, there is no gravitational attraction. Can you provide a link to the source of this quote? I'd like to see if the actual text supports your assertion that it's the change in the speed of light that causes objects to exhibit gravitation attraction. Originally Posted by Farsight Einstein was the author of general relativity, his point of view was correct. That's the assumption we're operating from. Whether your point of view is an accurate representation of his point of view is a different subject altogether. Originally Posted by Farsight And you know it's correct from the parallel-mirror gif. I don't know anything is correct from the parallel-mirror GIF. It's just an animation, and animations aren't necessarily an accurate representation of reality. Anything can happen in an animation. But since it was created as a graphical representation of our pre-existing understanding that clocks lose synchronization at different elevations, we can assume that clocks (whether atomic clocks, or parallel-mirror light clocks) do lose synchronization at different elevations with or without the GIF. But even given that, we can't say that Einstein's theory is correct from this alone. (But we do have other reasons separate from this to assume Einstein's theory is correct.) Originally Posted by Farsight And you know about the wave nature of matter don't you? You know we can make an electron out of light via pair production? And that the electron has spin angular momentum and a magnetic dipole moment? And that it can be diffracted? I don't see what this has to do with relativity. Originally Posted by Farsight So think of the electron as a standing wave of light. WTF? Originally Posted by Farsight A circle of light, as it were. Put it in a gravitational field and think it through. Divide the circle into four flat quadrants to make it even simpler: ..← ↓....↑ ..→ Starting from the left and going anticlockwise, at a given instant we have light travelling down like this ↓. There’s a gradient in c from top to bottom, but all it does is make the light look blueshifted. A little while later the light is moving like this → and the lower portion of the wave-front is subject to a slightly lower c than the upper portion. So it bends down a little. Later it’s going this way ↑ and looks redshifted, and later still it’s going this way ← and bends down again. These bends translate into a different position for the electron. It falls down: ○ ↓ The reducing c bleeds rotational motion out into linear motion. But only half the cycle got bent, so only half the reduced c goes into kinetic energy aka relativistic mass. That’s why light is deflected twice as much as matter, and gravity is not the sort of force that increases the relativistic mass. Simple. Originally Posted by Farsight If there's a big difference between the speed of light at the ceiling as compared to the speed of light at the floor, the force of gravity is large. If there's a lesser difference, the force of gravity is less. When there's no difference, things don't fall down. I'm going to take some time to think of an appropriate response to this.
	__________________ "That is just what you feel, that isn't reality." - hamelekim

10th April 2012, 06:57 AM	#1013
Farsight Master Poster Join Date: Mar 2008 Location: Poole, UK Posts: 2,008	Originally Posted by Brian-M Can you provide a link to the source of this quote? I'd like to see if the actual text supports your assertion that it's the change in the speed of light that causes objects to exhibit gravitation attraction. Sure. See Über die spezielle und allgemeine Relativitätstheorie and look at section 22. It's on page 51, about three-quarters of the way down. Originally Posted by Brian-M That's the assumption we're operating from. Whether your point of view is an accurate representation of his point of view is a different subject altogether. He said what he said. Originally Posted by Brian-M I don't know anything is correct from the parallel-mirror GIF. It's just an animation, and animations aren't necessarily an accurate representation of reality. Anything can happen in an animation. But since it was created as a graphical representation of our pre-existing understanding that clocks lose synchronization at different elevations, we can assume that clocks (whether atomic clocks, or parallel-mirror light clocks) do lose synchronization at different elevations with or without the GIF. OK. The gif just brings it home I guess. Originally Posted by Brian-M But even given that, we can't say that Einstein's theory is correct from this alone. (But we do have other reasons separate from this to assume Einstein's theory is correct.) Yes, it's a well-tested theory. Originally Posted by Brian-M I don't see what this has to do with relativity. It isn't relativity, but it is physics, and it is relevant. Originally Posted by Brian-M WTF? Yes, think of the electron as a standing wave of light. You can make it from light along with a positron in pair production. You can diffract an electron. Electrons interfere with one another. You can annihilate an electron with a positron and what you get is light. Read up on atomic orbitals and see the bit that says "The electrons do not orbit the nucleus in the sense of a planet orbiting the sun, but instead exist as standing waves". Originally Posted by Brian-M I'm going to take some time to think of an appropriate response to this. OK noted.

10th April 2012, 07:55 AM	#1014
Farsight Master Poster Join Date: Mar 2008 Location: Poole, UK Posts: 2,008	Originally Posted by Vorpal Farsight, you're simply repeating irrelevancies. 1) Absolutely no one is claiming that you can transform away Riemann curvature at events where it's nonzero. Repeating it incessantly is completely pointless, because no one is questioning this fact. 2) The real issue is whether the vanishing of Riemann curvature at some event implies that the non-presence of the gravitational field there. This is something you, and Synge, simply assert as true, and provide no reasoning for that claim. You've simply made no connection between (1), which everybody knows, and (2), which is what we're trying to have a discussion about. And failing. How many more times do I have to tell you about flat hills? Find some plot of gravitational potential like the upturned hat, and just look at it. Even this picture will do. If you have no Reimann curvature then when you start from a zillion miles to the right you've got a totally flat line. Start from the left at the surface of the sun and without Riemann curvature you've got a constant slope that goes on forever. The former is where there is no gradient in gravitational potential, so light doesn't curve and nothing falls down. The latter is where the coordinate speed of light goes on increasing forever, and the force of gravity doesn't reduce with distance. Neither is a description of a gravitational field. You need that Riemann curvature. When you take an infinitesimal region you're ignoring the change in the gradient but you aren't ignoring the gradient! How much simpler can I make it? Originally Posted by Vorpal You're mistaken. If you have a charged test particle that starts at rest and it doesn't move, it in no way implies that there's no electromagnetic field present. It only implies that there is no electric field present. There is no such thing as an electric field! It's the electromagnetic field! For God's sake Vorpal, go and read Minkowski’s Space and Time. Go and look at this bit two pages from the back: "Then in the description of the field produced by the electron we see that the separation of the field into electric and magnetic force is a relative one with regard to the underlying time axis; the most perspicious way of describing the two forces together is on a certain analogy with the wrench in mechanics, though the analogy is not complete". Did you catch that? It's the field. It exerts force in two ways, resulting in linear and/or rotational motion. If you're a charged particle and I set you down somewhere in the electron's electromagnetic field, you move linearly towards it or away from it, and the electron similarly moves towards or away from you. If I throw you through it, you go round and round as well. You only experience the rotational force when you have relative motion. Originally Posted by Vorpal Nonsense. This is a basic calculus exercise for STR. Suppose a particle starts from rest and undergoes constant force: $F = \frac{dp}{dt} = m\frac{d(\gamma v)}{dt} = mc\left(\cosh\alpha\right)\frac{d\alpha}{dt}$ Where I've applied the substitution v/c = tanh α, so γv/c = sinh α and γ = cosh α. Just integrate: Ft = sinh α, and therefore: $v = c\tanh\left(\sinh^{-1}\left(\frac{Ft}{mc}\right)\right) = \frac{Ft/m}{\sqrt{1 + (Ft/mc)^2}}$ Note that the tanh form makes it obvious at the speed is bounded by c. Gravity isn't a force in the usual sense. Forget your constant force, it's a constant acceleration, and escape velocity is the flip side of the speed of an infalling body dropped from "a great distance". When you drop your body into a black hole, it supposedly ends up going at the speed of light. You cannot contrive a gravitational field where it continues to accelerate at 9.8 m/s/s ad infinitum and gets to the speed of light when it's only halfway there. That's why the infinite wall falls down. That's why I said the problem was that you drop an object from a great distance, and it accelerates until it's going faster than light. That can't happen. Because the reducing c bleeds rotational motion out into linear motion. That's why it's obvious that the speed is bounded by c. Originally Posted by Vorpal Originally Posted by Farsight Yes. A gravitational field is curved spacetime. It isn't flat spacetime, and accelerating through space does not turn flat spacetime into curved spacetime. I asked for reasons, not mere repetition of baseless assertion. It's not a base assertion. If we're all in space where light doesn't curve, Vorpal putting the pedal to the metal doesn't make it curve one iota. Learn to tell the difference between what you see and what's there.
	Last edited by Farsight; 10th April 2012 at 07:58 AM.

10th April 2012, 07:58 AM	#1015
Farsight Master Poster Join Date: Mar 2008 Location: Poole, UK Posts: 2,008	Originally Posted by dafydd Thank you. You didn't mention flat spacetime in your previous post. My pleasure.

10th April 2012, 08:00 AM	#1016
Hellbound Abiogenic Spongiform Join Date: Sep 2002 Location: In a handbasket Posts: 9,030	Originally Posted by Farsight Yes, think of the electron as a standing wave of light. You can make it from light along with a positron in pair production. You can diffract an electron. Electrons interfere with one another. You can annihilate an electron with a positron and what you get is light. Read up on atomic orbitals and see the bit that says "The electrons do not orbit the nucleus in the sense of a planet orbiting the sun, but instead exist as standing waves". Just because I like to throw in wrenches... I do hope you realize that you can annihilate a proton with an anti-proton, and also get light. And you can annihilate a neutron with an anti-neutron, and get light. So are you suggesting that all particles are standing light waves?

10th April 2012, 08:07 AM	#1017
edd Graduate Poster Join Date: Nov 2007 Posts: 1,616	Originally Posted by Farsight There is no such thing as an electric field! It's the electromagnetic field! For God's sake Vorpal, go and read Minkowski’s Space and Time. Go and look at this bit two pages from the back: I'm really not sure you got Vorpal's point. You take a charged particle and place it at rest and it doesn't move. You'd seem to be saying that that demonstrates the absence of any magnetic field, since you're saying that it shows there's no electromagnetic field.
edd Graduate Poster Join Date: Nov 2007 Posts: 1,616	__________________ When I look up at the night sky and think about the billions of stars out there, I think to myself: I'm amazing. - Peter Serafinowicz

10th April 2012, 08:28 AM	#1018
DeiRenDopa Master Poster Join Date: Feb 2008 Posts: 2,381	Originally Posted by Hellbound Just because I like to throw in wrenches... I do hope you realize that you can annihilate a proton with an anti-proton, and also get light. And you can annihilate a neutron with an anti-neutron, and get light. So are you suggesting that all particles are standing light waves? By ~~Jove~~ Democritus! That's it! The TRUE SECRET TO THE UNIVERSE!!!!! Atoms are made up of neutrons, protons, and electrons. Each of which, in turn, per Farsight, is a standing wave of light. A crystal, of an element such as iron, is then also just a bunch of standing waves of light. Molecules are also just mixtures of standing waves of light. A virus, which is made up of molecules, is, then a complex mixture of standing waves of light. So is a tree, you, me, an asteroid, a planet, a star, ... And if you put enough standing waves of light together, and scrunch them down, you get ... wait for it ... a black hole! What a waste, 25 pages of largely irrelevant posts, when we could have just considered black holes as yet another manifestation of light ...

10th April 2012, 08:33 AM	#1019
sol invictus Philosopher Join Date: Oct 2007 Location: Nova Roma Posts: 8,434	Originally Posted by Farsight You need that Riemann curvature. When you take an infinitesimal region you're ignoring the change in the gradient but you aren't ignoring the gradient! How much simpler can I make it? You can make it as simple as you like, but it will remain wrong. There simply isn't a direct analog of "gravitational field" in GR, but the closest is the connection, not the curvature. Of course that's meaningless to you, since you have no clue what those words even mean, let alone what the difference is. Quote: There is no such thing as an electric field! It's the electromagnetic field! For God's sake Vorpal, go and read Minkowski’s Space and Time. You asserted "If you set a charged particle down and find it doesn't move, you say there's no electromagnetic field present. It's that simple. ...When you set it down and find it doesn't move, then you say there's no electromagnetic field present." That's false, as Vorpal just tried to explain to you. If the electromagnetic field strength tensor in the rest frame of the particle has zero time-space components, but non-zero space-space components (that's a long-winded way of saying "there's a magnetic field, but zero electric field") then the particle won't move when you set it down. So you're wrong. Admit it, learn from it, and move on. Can you do that?

10th April 2012, 08:48 AM	#1020
ben m Illuminator TLA Winner Join Date: Jul 2006 Posts: 4,712	Originally Posted by Hellbound So are you suggesting that all particles are standing light waves? Hellbound, Farsight will talk to you as though he knows that some sort of topological field theory is true, and as though you're behind-the-times for NOT knowing that all particles are differently-knotted configurations of some underlying ur-field.

10th April 2012, 09:13 AM	#1021
Farsight Master Poster Join Date: Mar 2008 Location: Poole, UK Posts: 2,008	Originally Posted by Hellbound Just because I like to throw in wrenches... I do hope you realize that you can annihilate a proton with an anti-proton, and also get light. And you can annihilate a neutron with an anti-neutron, and get light. So are you suggesting that all particles are standing light waves? Yes, that's where you end up. Shocking isn't it? But you can diffract neutrons too. And buckyballs. We've had this before, but I'll say it again. See Is the electron a photon with a toroidal topology? by Williamson and van der Mark, Annales de la Fondation Louis de Broglie, Volume 22, no.2, 133 (1997). Here's a picture from it. Inflate the torus until it's more like a sphere, and note that 4π denotes a sphere. Also see The nature of the electron by Qiu-Hong Hu, which is similar. Then there's Harmonic quintessence and the derivation of the charge and mass of the electron and the proton and quark masses by Andrew Worsley, Physics Essays Jun 2011, Vol. 24, No. 2 pp. 240-253. The latter applies spherical harmonics, usually applied to electron orbitals, to the particles themselves, wherein the electron Compton wavelength is given as λ = 4π / n c^1½ metres, where n is a dimensionality conversion factor with a value of 1. The light is going round at c one way and at half c the other, a bit like a moebius strip, and c dictates the structure. He gives the proton/electron mass ratio r = c^½ / 3π. Both expressions are subject to small binding-energy adjustments, but here's the raw numbers: 4π = 12.566370 c = 299792458 c^½ = 17314.5158177 4π / c^1½ = 12.566370 / (299792458 * 17314.5158177) λ = 2.420910 x 10ˉ¹² m Actual = 2.426310 x 10ˉ¹² m c^½ = 17314.5158177 3π = 9.424778 c^½ / 3π = 17314.5158177 / 9.424778 r = 1837.12717877 Actual = 1836.15267245
	Last edited by Farsight; 10th April 2012 at 09:49 AM. Reason: Wrong link to paper

10th April 2012, 10:07 AM	#1022
ben m Illuminator TLA Winner Join Date: Jul 2006 Posts: 4,712	Originally Posted by Farsight We've had this before, but I'll say it again. See Is the electron a photon with a toroidal topology? by Williamson and van der Mark, Annales de la Fondation Louis de Broglie, Volume 22, no.2, 133 (1997). Here's a picture from it. Inflate the torus until it's more like a sphere, and note that 4π denotes a sphere. Also see The nature of the electron by Qiu-Hong Hu, which is similar. Then there's Harmonic quintessence and the derivation of the charge and mass of the electron and the proton and quark masses by Andrew Worsley, Physics Essays Jun 2011, Vol. 24, No. 2 pp. 240-253. Speculative nonsense, pseudo-published in specialized crackpot journals. (Physics Essays is well known, but Annals de la Fondation Louis de Broglie is basically the same thing.) Farsight, are you utterly unable to tell the difference between beyond-the-mainstream, utterly-unchecked speculation, and actual particle physics? Is there something preventing you from using qualifiers, like "A few isolated authors have speculated ...", rather than pretending that this is all known, or true, or established? Anyway, your crackpot claims about toroidal-particle-substructure are off-topic in this thread about black holes. I recommend you start a new thread if you want to discuss that.

10th April 2012, 04:35 PM	#1023
Vorpal Extrapolate! Join Date: Jan 2005 Posts: 1,013	Originally Posted by Farsight There is no such thing as an electric field! It's the electromagnetic field! For God's sake Vorpal, go and read Minkowski’s Space and Time. Go and look at this bit two pages from the back: ... Did you catch that? It's the field. And I say yet again: Take a look at the Lorentz force and it will be obvious: $\frac{dp^\mu}{d\tau} = qF^\mu{}_\nu u^\nu$ In the frame of the particle, it is at rest and only the time component u⁰ of the four-velocity is nonzero. The electromagnetic tensor is antisymmetric, so it has six independent components, and in particular F₀₀ = 0. As a result, the condition of remaining stationary (i.e., Lorentz force vanishing) means the time-space components F_0a = -F_a0 vanish. What is that? The electric field. What are the three components that are missing from this conditions? The magnetic field. This is EM 101: F = q(E + v×B). If the particle is does not start out moving, then v = 0 and you get no information whatsoever about the magnetic field. This is neatly packaged using an antisymmetric rank-2 tensor, but it doesn't change the fact that you will have both electric and magnetic components to the electromagnetic field. Spacetime doesn't make basic EM go away. Originally Posted by Farsight Gravity isn't a force in the usual sense. Forget your constant force, it's a constant acceleration, and escape velocity is the flip side of the speed of an infalling body dropped from "a great distance". I've talked about the simpler STR case because I know you can't understand geodesic motion and willfully ignore the Rindler coordinate chart. So to elaborate on my previous post: 1) In flat spacetime, a uniformly accelerated observer has a hyperbolic worldline, with speed bounded by the speed of light. 2) A uniform gravitational field must have no tidal forces, and so have flat spacetime. Hence, the uniform gravitational field is flat spacetime in the frame of a uniformly accelerated observer. 3) Hence, as any inertial particle passes the observer, the measured speed will be less than the speed of light, no matter the initial conditions. Originally Posted by Farsight When you drop your body into a black hole, it supposedly ends up going at the speed of light. In what frame? In Schwarzschild coordinates, it does not. The freefalling particle gets asymptotically close to lightspeed in the sense that a sequence of stationary observers will measure it going to lightspeed as those observers get closer to the horizon. Incidentally, that's just what happens for the uniform gravitational field: there is an acceleration horizon. Because accelerated observers have hyperbolic worldlines and hyperbolas have asymptotes. Originally Posted by Farsight It's not a base assertion. If we're all in space where light doesn't curve, Vorpal putting the pedal to the metal doesn't make it curve one iota. Learn to tell the difference between what you see and what's there. Learn to actually check your assertions mathematically. You're completely wrong throughout, despite the fact that the calculations are pretty trivial. Originally Posted by Farsight How many more times do I have to tell you about flat hills? Find some plot of gravitational potential like the upturned hat, and just look at it. ... Are you ever going to actually calculate a geodesic to prove your assertions? Because that's how gravitational freefall works: geodesic motion.
Vorpal Extrapolate! Join Date: Jan 2005 Posts: 1,013	__________________ For every philosopher, there exists an equal and opposite philosopher. They're both wrong.

10th April 2012, 05:59 PM	#1024
ben m Illuminator TLA Winner Join Date: Jul 2006 Posts: 4,712	Originally Posted by Vorpal 1) In flat spacetime, a uniformly accelerated observer has a hyperbolic worldline, with speed bounded by the speed of light. 2) A uniform gravitational field must have no tidal forces, and so have flat spacetime. Hence, the uniform gravitational field is flat spacetime in the frame of a uniformly accelerated observer. Yep. This reasoning works both ways. Suppose that you found yourself in a sealed box, in which you (and your portable gravimeter) always find yourselves accelerating in the -x direction. Since it's a very large box, you can explore it and look for gradients---you find none. The acceleration vector is of the same magnitude, and pointing in the same direction, no matter where you are. So, you want to interpret that by guessing what's outside the box? Here are two guesses, both of which are consistent with the in-box spacetime: a) Flat, empty spacetime, but there's a rocket attached to your box. Perfectly good explanation of the observations. b) You're sitting still on the surface of a flat slab, infinite in x-y as far as you can tell, of constant areal density. That this matches the observations. c) You're sitting still a distance r far above the same nearly-infinite slab, which must be large enough (R > r) to avoid inducing gradients. This matches the in-car observations too. The a-c mapping is where the two intuitions match. A large slab, of radius R, has M = R^2. Therefore for some value of r the slab exceeds M/R = 1 and is a black hole with a horizon. Therefore, the spacetime geometry inside an accelerating car (case a) is exactly the same as the spacetime geometry inside a car hovering over a black hole, with a horizon (case c).

10th April 2012, 06:45 PM	#1025
Brian-M Daydreamer Join Date: Jul 2008 Location: Downunder Posts: 4,387	Farsight, I'm trying to figure out the practicalities of your idea for gravity, and it just isn't working out. Since we're replacing the traditional concept of gravity as an attractive force with a new concept of a ~~time~~ light retardant force, I'll call it Farsight Force to distinguish it from standard gravitational force. I'm assuming that the attraction towards the source of this Farsight Force is directly proportional to the gradient of the Farsight Force (if this isn't the case, please provide details), and this is the source of the problem. If you take a force diminishing by distance according to the inverse square law, the gradient of that force is diminishing proportional the inverse cube of distance (and damn you for making me do math in order to figure this out ). So the Farsight Force would have to diminish by less than the inverse square law in order for us to perceive gravity as diminishing according to the inverse square law. How is it possible for a force radiating evenly in all directions in three dimensional space to diminish at a rate less than the inverse square law would predict?
	__________________ "That is just what you feel, that isn't reality." - hamelekim Last edited by Brian-M; 10th April 2012 at 07:19 PM.

11th April 2012, 12:31 AM	#1026
Farsight Master Poster Join Date: Mar 2008 Location: Poole, UK Posts: 2,008	Originally Posted by edd I'm really not sure you got Vorpal's point. You take a charged particle and place it at rest and it doesn't move. You'd seem to be saying that that demonstrates the absence of any magnetic field, since you're saying that it shows there's no electromagnetic field. Yes, that's right. If you set down a charged particle in a region of space and it doesn't move, there's no electromagnetic field present. If you set down an uncharged particle in a region of space and it doesn't move, there's no gravitational field present. If you set down a charged particle and an uncharged particle and only the charged particle moves, an electromagnetic field is present. If you set down a charged particle and an uncharged particle and they both move linearly, a gravitational field is present. If you set down a charged particle and it moves rotationally, you might say that a magnetic field is present, but you should always remember that the magnetic field is merely how you see an electromagnetic field when you have motion relative to it. So it's like throwing a charged particle through an electromagnetic field.

11th April 2012, 12:35 AM	#1027
Farsight Master Poster Join Date: Mar 2008 Location: Poole, UK Posts: 2,008	Originally Posted by DeiRenDopa By ~~Jove~~ Democritus! That's it! The TRUE SECRET TO THE UNIVERSE!!!!! Atoms are made up of neutrons, protons, and electrons. Each of which, in turn, per Farsight, is a standing wave of light. A crystal, of an element such as iron, is then also just a bunch of standing waves of light. Molecules are also just mixtures of standing waves of light. You got it. Hence this physicsworld article: "The first real-time movie of large molecules creating an interference pattern after passing through two slits has been made by an international team of physicists".

11th April 2012, 12:50 AM	#1028
Farsight Master Poster Join Date: Mar 2008 Location: Poole, UK Posts: 2,008	Originally Posted by sol invictus You can make it as simple as you like, but it will remain wrong. There simply isn't a direct analog of "gravitational field" in GR, but the closest is the connection, not the curvature. Of course that's meaningless to you, since you have no clue what those words even mean, let alone what the difference is. Huff puff, sol really doesn't like it when I explain the gravitational field with reference to Einstein. He behaves like a witchdoctor faced with a pharmacist, all outraged, saying "you don't understand it". Tough. I do. Originally Posted by sol invictus That's false, as Vorpal just tried to explain to you. If the electromagnetic field strength tensor in the rest frame of the particle has zero time-space components, but non-zero space-space components (that's a long-winded way of saying "there's a magnetic field, but zero electric field") then the particle won't move when you set it down. So you're wrong. Admit it, learn from it, and move on. Can you do that? I'm not wrong. See Electron magnetic dipole moment. The magnetic field exerts a torque on the electron. It "indeed behaves like a tiny bar magnet". Or a compass needle if you prefer. It moves. And you're wrong again.

11th April 2012, 01:04 AM	#1029
Farsight Master Poster Join Date: Mar 2008 Location: Poole, UK Posts: 2,008	Originally Posted by ben m Speculative nonsense, pseudo-published in specialized crackpot journals. (Physics Essays is well known, but Annals de la Fondation Louis de Broglie is basically the same thing.) Farsight, are you utterly unable to tell the difference between beyond-the-mainstream, utterly-unchecked speculation, and actual particle physics? Is there something preventing you from using qualifiers, like "A few isolated authors have speculated ...", rather than pretending that this is all known, or true, or established? Anyway, your crackpot claims about toroidal-particle-substructure are off-topic in this thread about black holes. I recommend you start a new thread if you want to discuss that. I answered a direct question related to gravity and why things fall down. These people have to settle for back-street journals because of attitudes like yours. You dismiss the patent evidence and you have no counter-argument. Your contribution to this thread is negligible, all you offer is outraged abuse. These aren't crackpot claims, we really can make an electron from light in pair production, it really does have a magnetic dipole moment, we really can diffract it, and atomic orbitals really do feature standing waves. What do you think the electron is made of? Cheese? No, they aren't crackpot claims, but they have been censored by people like you.

11th April 2012, 02:05 AM	#1030
Farsight Master Poster Join Date: Mar 2008 Location: Poole, UK Posts: 2,008	Originally Posted by Vorpal And I say yet again: Take a look at the Lorentz force and it will be obvious: $\frac{dp^\mu}{d\tau} = qF^\mu{}_\nu u^\nu$ In the frame of the particle, it is at rest and only the time component u⁰ of the four-velocity is nonzero. The electromagnetic tensor is antisymmetric, so it has six independent components, and in particular F₀₀ = 0. As a result, the condition of remaining stationary (i.e., Lorentz force vanishing) means the time-space components F_0a = -F_a0 vanish. What is that? The electric field. What are the three components that are missing from this conditions? The magnetic field. This is EM 101: F = q(E + v×B). If the particle is does not start out moving, then v = 0 and you get no information whatsoever about the magnetic field. Electrons behave like little compass needles. You do get information about the magnetic field. Originally Posted by Vorpal This is neatly packaged using an antisymmetric rank-2 tensor, but it doesn't change the fact that you will have both electric and magnetic components to the electromagnetic field. Spacetime doesn't make basic EM go away. No motion, no field. The potential is uniform. That doesn't go away. Like I said, the potential is more fundamental than the field. Originally Posted by Vorpal I've talked about the simpler STR case because I know you can't understand geodesic motion and willfully ignore the Rindler coordinate chart. I understand geodesic motion, and I don't "wilfully ignore" the Rindler coordinate chart. I just find it irrelevant, a portrayal of what you see when you accelerate. I still don't seem to be able to get this across to you: a light beam doesn't bend just because you stepped on the gas. You just see it as bent. In similar vein rain doesn't slant just because you start running. Originally Posted by Vorpal So to elaborate on my previous post: 1) In flat spacetime, a uniformly accelerated observer has a hyperbolic worldline, with speed bounded by the speed of light. Yes, but a gravitational field is curved spacetime, so flat spacetime just isn't relevant. Originally Posted by Vorpal 2) A uniform gravitational field must have no tidal forces, and so have flat spacetime. Hence, the uniform gravitational field is flat spacetime in the frame of a uniformly accelerated observer. And there is no such thing as a uniform gravitational field! A gravitational field is curved spacetime, so what you're proposing is a contradiction in terms. No wonder you end up with impossible nonsense scenarios like the sky falling in around a black hole and neverneverland descriptions of the infalling observer. Come on Vorpal, work it out. If you accelerate at 9.8m/s/s and keep on accelerating at a uniform 9.8 m/s/s you end up exceeding the speed of light, and you've contradicted special relativity. Originally Posted by Vorpal 3) Hence, as any inertial particle passes the observer, the measured speed will be less than the speed of light, no matter the initial conditions. An accelerating observer always measures that speed as less than the speed of light. And he always measures the speed of light to be the same. Now go and read The Other Meaning of Special Relativity and understand why. When you're made of waves along with everything else, you always measure wave speed to be the same. And nothing made from waves can exceed that speed. Originally Posted by Vorpal In what frame? In Schwarzschild coordinates, it does not. The freefalling particle gets asymptotically close to lightspeed in the sense that a sequence of stationary observers will measure it going to lightspeed as those observers get closer to the horizon. In my frame. If you keep on accelerating at a uniform 9.8m/s/s whilst traversing the starry sky, I end up seeing you going from one star to another faster than light can go. Originally Posted by Vorpal Incidentally, that's just what happens for the uniform gravitational field: there is an acceleration horizon. Because accelerated observers have hyperbolic worldlines and hyperbolas have asymptotes. Aaaagh! There is no uniform gravitational field. You end up accelerating a falling body less than you accelerate a body that you placed at some elevation. If it's 9.8 m/s/s, that's what it is for all bodies. Originally Posted by Vorpal Learn to actually check your assertions mathematically. You're completely wrong throughout, despite the fact that the calculations are pretty trivial. I'm not wrong. You've used mathematics to confuse a constant force and a constant change of speed. When you're already moving at 299,792,457 m/s, a second later you're moving at 299,792,466.8 m/s. It isn't going to happen. Originally Posted by Vorpal Are you ever going to actually calculate a geodesic to prove your assertions? Because that's how gravitational freefall works: geodesic motion. No. Because that isn't how it works. I told you how it works. Get used to it. It isn't in your textbooks yet, but it will be.

11th April 2012, 02:21 AM	#1031
Farsight Master Poster Join Date: Mar 2008 Location: Poole, UK Posts: 2,008	Originally Posted by Brian-M Farsight, I'm trying to figure out the practicalities of your idea for gravity, and it just isn't working out. Since we're replacing the traditional concept of gravity as an attractive force with a new concept of a ~~time~~ light retardant force, I'll call it Farsight Force to distinguish it from standard gravitational force. Don't think of it as a force, the Newtonian F=ma and E=F x d doesn't apply because conservation of energy applies instead. Just think of it as inhomogeneous space that causes light to bend as per gravitational lensing. Take your cue for that from Einstein's Leyden address: "This space-time variability of the reciprocal relations of the standards of space and time, or, perhaps, the recognition of the fact that “empty space” in its physical relation is neither homogeneous nor isotropic, compelling us to describe its state by ten functions (the gravitation potentials gμν), has, I think, finally disposed of the view that space is physically empty". Originally Posted by Brian-M I'm assuming that the attraction towards the source of this Farsight Force is directly proportional to the gradient of the Farsight Force (if this isn't the case, please provide details), and this is the source of the problem. No, the "force" of attraction is directly proportional to the gradient in the potential. Originally Posted by Brian-M If you take a force diminishing by distance according to the inverse square law, the gradient of that force is diminishing proportional the inverse cube of distance (and damn you for making me do math in order to figure this out ). No no, it's the gradient in the potential. You measure the speed of light at different floors in your very very tall building, and then draw a graph to plot the potential. It's a curve that starts out with a steep gradient which gradually gets shallower with altitude. Where the gradient in the potential is steeper, the force is greater. Originally Posted by Brian-M So the Farsight Force would have to diminish by less than the inverse square law in order for us to perceive gravity as diminishing according to the inverse square law. How is it possible for a force radiating evenly in all directions in three dimensional space to diminish at a rate less than the inverse square law would predict? You need to start again with this I'm afraid Brian.

11th April 2012, 04:36 AM	#1032
Dancing David Penultimate Amazing Join Date: Mar 2003 Location: Central Illinois Posts: 34,919	Originally Posted by Vorpal I don't think you understand what local means. I sense a pattern here. Private idiomatic definitions to reinforce tautological fallacies on the part of FS.
	__________________ Hell, dynamiting fish in a barrel is more challenging. - Ladewig I suspect you are a sandwich, metaphorically speaking. -Donn And a shot rang out. Now Space is doing time... -Ben Burch You built the toilet - don't complain when people crap in it. _Kid Eager

11th April 2012, 04:41 AM	#1033
Dancing David Penultimate Amazing Join Date: Mar 2003 Location: Central Illinois Posts: 34,919	Originally Posted by Farsight Yes, that's right. If you set down a charged particle in a region of space and it doesn't move, there's no electromagnetic field present. If you set down an uncharged particle in a region of space and it doesn't move, there's no gravitational field present. If you set down a charged particle and an uncharged particle and only the charged particle moves, an electromagnetic field is present. If you set down a charged particle and an uncharged particle and they both move linearly, a gravitational field is present. If you set down a charged particle and it moves rotationally, you might say that a magnetic field is present, but you should always remember that the magnetic field is merely how you see an electromagnetic field when you have motion relative to it. So it's like throwing a charged particle through an electromagnetic field. Oh my.
	__________________ Hell, dynamiting fish in a barrel is more challenging. - Ladewig I suspect you are a sandwich, metaphorically speaking. -Donn And a shot rang out. Now Space is doing time... -Ben Burch You built the toilet - don't complain when people crap in it. _Kid Eager

11th April 2012, 05:05 AM	#1034
W.D.Clinger Master Poster Join Date: Oct 2009 Posts: 2,463	Farsight versus Einstein, part 1 Once again, Farsight has stated an important difference between FGR (Farsight general relativity) and the general theory of relativity put forth by Einstein: Originally Posted by Farsight When spacetime is flat, no gravitational field is present. Originally Posted by Farsight When spacetime is flat, no gravitational field is present. Originally Posted by Farsight Originally Posted by sol invictus You can make it as simple as you like, but it will remain wrong. There simply isn't a direct analog of "gravitational field" in GR, but the closest is the connection, not the curvature. Of course that's meaningless to you, since you have no clue what those words even mean, let alone what the difference is. Huff puff, sol really doesn't like it when I explain the gravitational field with reference to Einstein. He behaves like a witchdoctor faced with a pharmacist, all outraged, saying "you don't understand it". Tough. I do. Farsight can huff and puff all he likes, but he's arguing with Einstein here. From the English translation of "The Foundation of the General Theory of Relativity": Originally Posted by Albert Einstein C. THEORY OF THE GRAVITATIONAL FIELD § 13. Equations of Motion of a Material Point in the Gravitational Field. Expression for the Field-components of Gravitation A freely movable body not subjected to external forces moves, according to the special theory of relativity, in a straight line and uniformly. This is also the case, according to the general theory of relativity, for a part of four-dimensional space in which the system of co-ordinates K₀, may be, and is, so chosen that they have the special constant values given in (4). Einstein's formula (4) is the 4x4 matrix of coefficients for the Minkowski metric in the familiar coordinate system for flat spacetime. Einstein continues by talking about the gravitational field that appears when you use a different coordinate system to describe exactly the same locally flat spacetime: Originally Posted by Albert Einstein If we consider precisely this movement from any chosen system of co-ordinates K₁, the body, observed from K₁, moves, according to the considerations in § 2, in a gravitational field.... If the Γ^τ_μν vanish, then the point moves uniformly in a straight line. These quantities therefore condition the deviation of the motion from uniformity. They are the components of the gravitational field. According to Einstein, therefore, the 64 coordinate-dependent Christoffel symbols "are the components of the gravitational field". (Because the Christoffel symbols are symmetric in their lower indices, not all of those components are independent.) Let's do an example. As this example shows, Einstein's "components of the gravitational field" can be nonzero even in flat spacetime, contrary to Farsight's repeated baseless assertions. Let's use the metric form posted by sol invictus. In that coordinate system, the four nonzero components of the covariant metric tensor are $<br /> \[<br /> \begin{align}<br /> g_{00} &= - (x^1 - r_0) \\<br /> g_{11} &= \frac{1}{x^1 - r_0} \\<br /> g_{22} &= 1 \\<br /> g_{33} &= 1<br /> \end{align}<br /> \]<br />$ (I'm using MTW notational conventions, and I'm writing the r coordinate as x¹ to prepare for the index and Einstein summation notations used below.) To calculate the 64 components of the gravitational field, we'll need the components of the contravariant form as well: $<br /> \[<br /> \begin{align}<br /> g^{00} &= - \frac{1}{(x^1 - r_0)} \\<br /> g^{11} &= (x^1 - r_0) \\<br /> g^{22} &= 1 \\<br /> g^{33} &= 1<br /> \end{align}<br /> \]<br />$ Einstein's equation (45) states his formula for the components of the gravitational field. Expanding Einstein's notation into modern notation, Einstein's equation becomes $<br /> \[<br /> \begin{align}<br /> \Gamma^\lambda_{\mu\nu}<br /> &= \frac{1}{2} g^{\lambda\alpha} \left(<br /> \partial_{\nu} g_{\alpha\mu}<br /> + \partial_{\mu} g_{\alpha\nu}<br /> - \partial_{\alpha} g_{\mu\nu} \right)<br /> \end{align}<br /> \]<br />$ If I've done that calculation correctly, only 4 of the 64 Christoffel symbols are nonzero: $<br /> \[<br /> \begin{align}<br /> \Gamma^0_{01} &= \Gamma^0_{10} = \frac{1}{2 (x^1 - r_0)} \\<br /> \Gamma^1_{00} &= \frac{1}{2} (x^1 - r_0) \\<br /> \Gamma^1_{11} &= - \frac{1}{2 (x^1 - r_0)}<br /> \end{align}<br /> \]<br />$ With those Christoffel symbols in hand, it's straightforwardly tedious to calculate the 256 components of the Riemann curvature tensor via Einstein's equation (43). In modern notation, that equation becomes $<br /> \[<br /> \begin{align}<br /> R^{\alpha}_{\beta\gamma\delta} &=<br /> \partial_{\gamma} \Gamma^{\alpha}_{\beta\delta}<br /> - \partial_{\delta} \Gamma^{\alpha}_{\beta\gamma}<br /> + \Gamma^{\alpha}_{\kappa\gamma}<br /> \Gamma^{\kappa}_{\beta\delta}<br /> - \Gamma^{\alpha}_{\kappa\delta}<br /> \Gamma^{\kappa}_{\beta\gamma}<br /> \end{align}<br /> \]<br />$ That calculation is greatly simplified by the following observations. As can be seen from the Christoffel symbols calculated above, the partial derivatives of the first two terms are zero unless the subscript on the derivative operator is 1. Similarly, every component of the Riemann tensor that has a 2 or a 3 as an index will be zero; that observation reduces the number of components from 256 to 16, even before we take advantage of symmetries. I calculated the components of the Riemann tensor by hand, and found all of its 256 components are zero. That means sol invictus has indeed given us an example of flat spacetime (because the Riemann curvature tensor is zero) that has a nonzero gravitational field (because four of the Christoffel symbols, which Einstein regards as the components of the gravitational field, are nonzero). In his response to sol invictus, quoted above, Farsight said he understands all this stuff. That's great. I invite Farsight to check my calculations. The bottom line is that Einstein disagreed with Farsight's repeated bare assertion that flat spacetime implies a zero gravitational field.
	Last edited by W.D.Clinger; 11th April 2012 at 05:13 AM. Reason: base => baseless

11th April 2012, 06:16 AM	#1035
Brian-M Daydreamer Join Date: Jul 2008 Location: Downunder Posts: 4,387	Originally Posted by Farsight Don't think of it as a force, the Newtonian F=ma and E=F x d doesn't apply because conservation of energy applies instead. Just think of it as inhomogeneous space that causes light to bend as per gravitational lensing. Take your cue for that from Einstein's Leyden address: Force, potential, curvature of space, whatever. It's irrelevant to the point. If it's effect is evenly distributed and omnidirectional in three dimensional space, and it diminishes at less than the inverse square law, then this means that the total effect of this force is increasing over distance (even if the localized effect is decreasing with distance because the effect is spread over a larger area). And the total effect of the force increasing with distance is what doesn't make any sense. Originally Posted by Farsight No, the "force" of attraction is directly proportional to the gradient in the potential. Wait, you disagree with me then you basically repeat what I just said? By Farsight Force I'm referring to this potential you mention. Originally Posted by Farsight No no, it's the gradient in the potential. You measure the speed of light at different floors in your very very tall building, and then draw a graph to plot the potential. It's a curve that starts out with a steep gradient which gradually gets shallower with altitude. Where the gradient in the potential is steeper, the force is greater. But that's what I'm talking about. I figured it out by making up an example of objects experiencing different gravitational forces experienced at points different distances from a large mass, and then estimated the local gradient of this "potential" over a distance of a meter by taking the gravitational force half a meter closer to the mass from the point and subtracting the gravitational force thats half a meter further away. I found that where gravitational force diminishes with the inverse square law, the gravitational gradient diminishes according to the the inverse cubed law. (This is the math I referred to in my previous post.) From this I must conclude that any force (or potential, or whatever) which diminishes according to the inverse square law must have a gradient which diminishes according to the inverse cubed law. So if the experienced (or directly measured) force of gravity is directly proportional to the gradient of this potential, and this potential diminishes with distance according to the inverse square law, then we should be measuring gravity as diminishing according to the inverse cubed law... but we don't! For us to experience or measure gravity as diminishing with the inverse square law, then the potential (whose gradient produces what we experience as gravity) must be diminishing at less than the inverse square law. And that doesn't make any sense. Originally Posted by Farsight You need to start again with this I'm afraid Brian. Where is my understanding of your physics flawed? The way I'm reading this you're disagreeing with me, but your corrections are the exact same thing I'm talking about, only worded slightly differently.
	__________________ "That is just what you feel, that isn't reality." - hamelekim Last edited by Brian-M; 11th April 2012 at 06:20 AM.