PDA

View Full Version : Faster Than Light data transmission cable devices

aggle_rithm
13th June 2003, 10:32 AM
Wouldn't such data arrive before it was sent? At two times the speed of light, how much time before it is sent would it be received? Would that allow us to receive messages about future events?

No, really, I want to know.

CurtC
13th June 2003, 11:01 AM
With a 10 m cable, you'd normally expect around a 45 ns delay. Twice as fast as the speed of light would be about 17 ns. It's less than half, because a standard cable propagates slower than the speed of light.

The difference between 45 and 17 ns is very easy to measure, provided you have the right equipment, which is quite commonly available.

Skeptical Greg
13th June 2003, 11:22 AM
Originally posted by aggle_rithm
Wouldn't such data arrive before it was sent? At two times the speed of light, how much time before it is sent would it be received? Would that allow us to receive messages about future events?

No, really, I want to know.

If something is transmitted at the speed of light, it would travel one light second ( ~186,000 miles ) in .. prepare yourself ..... one second..

At twice the speed of light, it would cover the same distance in 1/2 second or twice that distance in one second.. It would arrive sooner at any distance less than that, approaching zero as the transmitter and receiver are moved closer together.

It ( the data ) wouldn't go anywhere, until it was sent and the transit time would never be less than zero..

aggle_rithm
13th June 2003, 12:09 PM
Originally posted by Diogenes

If something is transmitted at the speed of light, it would travel one light second ( ~186,000 miles ) in .. prepare yourself ..... one second..

At twice the speed of light, it would cover the same distance in 1/2 second or twice that distance in one second.. It would arrive sooner at any distance less than that, approaching zero as the transmitter and receiver are moved closer together.

It ( the data ) wouldn't go anywhere, until it was sent and the transit time would never be less than zero..

According to relativity, time speeds up as an object approaches the speed of light. At the speed of light, time moves infinitely fast from the object's point of view. Faster than the speed of light (theoretically impossible), time moves backwards -- beyond "infinitely fast".

It began to occur to me that, once time begins moving backwards, as the object once again approaches the speed of light from the OTHER direction (from our viewpoint, 2 X c). Would time reverse once again, (because it is now moving backwards in time infinitely fast) or would it keep moving backwards at an ever increasing rate?

Since it is impossible for anything to move faster than light or for time to go by faster than infinitely fast, I guess this is an absurd discussion. Theoretically, though, if we could remove that constraint, what would really happen?

WooBot
13th June 2003, 12:21 PM
Originally posted by aggle_rithm

According to relativity, time speeds up as an object approaches the speed of light. At the speed of light, time moves infinitely fast from the object's point of view. Faster than the speed of light (theoretically impossible), time moves backwards -- beyond "infinitely fast".

Not unless you mean that, viewed from the object, the rest of the galaxy is moving infinitely fast. Is that what you mean? From everyone else's point of view, time slows down on the object.

Skeptical Greg
13th June 2003, 12:29 PM
Originally posted by aggle_rithm

......................
According to relativity, time speeds up as an object approaches the speed of light..............

I think you have this backwards...........;)

WooBot
13th June 2003, 12:46 PM
Originally posted by Diogenes

If something is transmitted at the speed of light, it would travel one light second ( ~186,000 miles ) in .. prepare yourself ..... one second..

At twice the speed of light, it would cover the same distance in 1/2 second or twice that distance in one second.. It would arrive sooner at any distance less than that, approaching zero as the transmitter and receiver are moved closer together.

It ( the data ) wouldn't go anywhere, until it was sent and the transit time would never be less than zero..

I think he's talking about the idea that things traveling faster than light would move backwards in time. This is as impossible as the speed itself, because as you point out, it would violate causality.

I just know some real physicist is going to nail me here, but my understanding is that even if you twist the equations out of shape and use them to "see" what would happen at these speeds, no one takes the possibility seriously because of problems like these.

Another problem is that even if there are particles that can travel faster than light, they will always travel at that speed - they can no more slow down through lightspeed than we can speed up through it. We cannot interact with them in any way.

The barrier at lightspeed is pretty easy to understand. As you approach lightspeed, your mass increases asymptotically, faster and faster and gets closer and closer to infinity. At lightspeed, your mass would be infinite. Therefore it would take an infinite amount of energy to accelerate you any more. This is as true for an electron as it is for a spaceship; it's true for anything with mass. Photons are massless but they don't help us, they always travel at exactly lightspeed, no matter who's looking at them.

69dodge
13th June 2003, 03:46 PM
If in one reference frame a signal travels faster than c, there's some other reference frame in which it arrives before it was sent.

In relativity, the order in which events occur may be different in different reference frames. This is the case if the two events are far enough away from each other in space and close enough to each other in time. On the other hand, if two events are close enough in space and far enough in time, then all reference frames agree about which happened first.

How far is far enough? How close is close enough? Here's the rule: If a light beam starting at the time and place of the first event can reach the location of the second event before it happens, then everyone agrees the second event really did happen after the first. If the distance between the events is so great, and the time between them so short, that even a light beam can't make it from one event to the other, then the order of the events is frame-dependent.

Kess
14th June 2003, 08:48 AM
Assuming this is the cable described at http://www.ultra-faster-than-light.com it doesn't work anyway.

The experiment helpfully described in detail by the manufacturer to "prove" that the cable sends signals faster-than-light gives it all away (see http://www.ultra-faster-than-light.com/hwodoesitwork.htm). The experiment will appear convincing ... but it's an illusion and I've proved to myself that the same result could be achieved with any cable.

The danger, of course, is that the trick could easily fool the unwary, those unfamiliar with transmission line theory, patent examiners, etc. Perhaps it could even puzzle Randi...

It's hard to explain what's really happening in a few words. I've started to write a (technical) explanation of how the trick works - should I send this to Randi?

BillyJoe
15th June 2003, 01:06 AM
Originally posted by WooBot
Not unless you mean that, viewed from the object, [for] the rest of the galaxy [time] is moving infinitely fast. Is that what you mean? From everyone else's point of view, time slows down on the object.:)

regards,
BillyJoe.

PS:
Did you understand 69dodge's post?
He has the correct answer to algorithm's question, although a bit awkwardly put. :cool:

BillyJoe
15th June 2003, 03:36 AM
Back again.

Actually I lost sight of the question in making the last post...

I've just read it again and it occurred to me that algorithm's question was based on a misunderstanding.....

He was probably thinking about the situation where an object travelling towards you at faster than light speed seems to arrive before it departs. This is because the light from the arriving object reaches you before the light from the departing object.

But it only seems to arrive before it departs.

In the case of the message sent along the cable at faster than light speed, you can't see the message, so you can't see it arriving before it departs. You can't see it depart at all. It just arrives sooner the faster it travels. If it was transmitted as a visible signal travelling though space, it would seem to arrive before it departs.

But, again, it would only seem to arrive before it departs.

BJ.

SteveGrenard
15th June 2003, 04:21 AM
For a useful Q&A on this subject from Scientific American visit:

http://www.physics.hku.hk/~tboyce/sf/topics/lightfreeze/0701haubox1.html

aggle_rithm
16th June 2003, 05:20 AM
Originally posted by BillyJoe
Back again.

Actually I lost sight of the question in making the last post...

I've just read it again and it occurred to me that algorithm's question was based on a misunderstanding.....

He was probably thinking about the situation where an object travelling towards you at faster than light speed seems to arrive before it departs. This is because the light from the arriving object reaches you before the light from the departing object.

But it only seems to arrive before it departs.

In the case of the message sent along the cable at faster than light speed, you can't see the message, so you can't see it arriving before it departs. You can't see it depart at all. It just arrives sooner the faster it travels. If it was transmitted as a visible signal travelling though space, it would seem to arrive before it departs.

But, again, it would only seem to arrive before it departs.

BJ.

THIS was the answer I was looking for. Thanks.

It did seem a little counterintuitive that the message would arrive before it was sent, because it would only move backwards in time from it's viewpoint, not ours.

aggle_rithm
16th June 2003, 05:23 AM
Originally posted by WooBot

I just know some real physicist is going to nail me here, but my understanding is that even if you twist the equations out of shape and use them to "see" what would happen at these speeds, no one takes the possibility seriously because of problems like these.

There are other cases where physicists twist equations out of shape just to see what happens -- that's where imaginary numbers come from (those, for instance, where the product of two negative numbers is negative). The use of imaginary numbers actually solves some problems that are unsolvable in "real" numbers.

I don't know why they wouldn't do the same thing with faster-than-light equations. Just to see what happens.

xouper
16th June 2003, 09:04 AM
aggle_rithm: -- that's where imaginary numbers come from (those, for instance, where the product of two negative numbers is negative).Pardon my asking for clarification here, are you saying the square root of negative one is a negative number?

aggle_rithm
16th June 2003, 10:49 AM
Originally posted by xouper
Pardon my asking for clarification here, are you saying the square root of negative one is a negative number?

Not in true mathematics, no. Perhaps, in the mathematics of imaginary numbers.

I don't really know that much about imaginary numbers, except that it is possible in this system for the product of two negative numbers to be negative. Any mathematicians want to field this one?

xouper
16th June 2003, 11:20 AM
aggle_rithm: -- that's where imaginary numbers come from (those, for instance, where the product of two negative numbers is negative).

xouper: Pardon my asking for clarification here, are you saying the square root of negative one is a negative number?

aggle_rithm: Not in true mathematics, no. Perhaps, in the mathematics of imaginary numbers. I don't really know that much about imaginary numbers, except that it is possible in this system for the product of two negative numbers to be negative. This is false. The square root of negative one is neither negative nor positive. It is not possible with imaginary numbers that the product of two negatives can be negative.

WooBot
16th June 2003, 11:27 AM
Originally posted by aggle_rithm

Not in true mathematics, no. Perhaps, in the mathematics of imaginary numbers.

I don't really know that much about imaginary numbers, except that it is possible in this system for the product of two negative numbers to be negative. Any mathematicians want to field this one?

"Imaginary" is a terrible name for these numbers, leading you to believe they don't have real-world applications. Our engineer friends here could tell you that "imaginary" numbers are very real indeed, when it comes to solving equations they use in their work every day.

Mathematics expanded to include the imaginaries, in order to answer questions such as "what is the square root of -1". At first glance you might say "why bother", but the expansion has a host of real-world results, just as the previous "expansion" into negative numbers ("How can a number be less than zero", some asked) suddenly allowed you to calculate how much debt you owed.

BobM
16th June 2003, 12:17 PM
This is false. The square root of negative one is neither negative nor positive. It is not possible with imaginary numbers that the product of two negatives can be negative.

The square root of -1 is i or -i. That is pretty much the definition of the imaginary number i. Without "i" you can't take the square root of a negative number, with it, you can.

xouper
16th June 2003, 01:26 PM
BobM: The square root of -1 is i or -i.Which says nothing about whether i itself is negative or positive. For example, which of the following statements are true (if any)?

i > 0

i < 0

-i > 0

-i < 0
This demonstrates why complex numbers (with a non-zero imaginary part) do not have the ordering property of real numbers.

BillyJoe
17th June 2003, 04:05 AM
Originally posted by aggle_rithm
.....that's where imaginary numbers come from (those, for instance, where the product of two negative numbers is negative). I think you meant.....

The product of the square root of two negative numbers is a negative number

Specifically

sqrt(-1) X sqrt(-1) = -1

BillyJoe
17th June 2003, 04:14 AM
Originally posted by xouper
...which of the following statements are true (if any)?

i > 0

i < 0

-i > 0

-i < 0At first glance, either 1 and 4 or 2 and 3 must be correct. But the following suggest none of them are correct.....
Originally posted by xouper
The square root of negative one is neither negative nor positive But who knows what effect negating a number which is neither positive nor negative has. :confused:

Suspected Idiot
17th June 2003, 07:36 AM
If I remember correctly from my complex analysis days, none of those equations are correct, or even vaild. the "&gt;" and "&lt;" operators are only defined for real numbers.
Saying "i &gt; 0" makes as much sense as saying "banana &gt; France".

Suspected Idiot
17th June 2003, 07:39 AM
On topic though, think of the low pings you could get playing online games if the whole internet was made up of these.

Reb
17th June 2003, 07:52 AM
Originally posted by Suspected Idiot
If I remember correctly from my complex analysis days, none of those equations are correct, or even vaild. the "&gt;" and "&lt;" operators are only defined for real numbers.
Saying "i &gt; 0" makes as much sense as saying "banana &gt; France".

Minor tangent:
Actually, that's not a nonsense comparison in the world of SQL queries. Think of sorting a list of words: some words are "greater" than others.

Reb

WooBot
17th June 2003, 08:12 AM
Originally posted by Reb

Minor tangent:
Actually, that's not a nonsense comparison in the world of SQL queries. Think of sorting a list of words: some words are "greater" than others.

Reb

Oh my God. A drive by SQL attack. Someone call a cop.

Goryus
20th June 2003, 08:32 PM
Doesn't the fitzgerald ratio become imaginary - not negative - at faster-than-light speeds? Doesn't that mean that an object traveling at above light speeds would move through imaginary time? :confused:

BillyJoe
21st June 2003, 05:11 AM
That's the thing about imaginary numbers, they are not actually imaginary. They are real, except that they are not real numbers either. It's so confusing, I can't imagine anyone really understands it.

BillyJoe :cool:

xouper
21st June 2003, 12:25 PM
Suspected Idiot: If I remember correctly from my complex analysis days, none of those equations are correct, or even vaild. the "&gt;" and "&lt;" operators are only defined for real numbers.That was exactly my point when I said that the square root of negative one is neither positive nor negative.

BillyJoe
21st June 2003, 10:02 PM
xouper, you understand imaginary numbers then?

xouper
21st June 2003, 10:49 PM
BillyJoe: xouper, you understand imaginary numbers then?Maybe a little, in an abstract way. I'm not sure what kind of understanding you are asking about. I don't claim to be an expert on them, of course.

BillyJoe
22nd June 2003, 01:39 AM
I can't get a handle on them at all so I can't even begin to ask a sensible question on them - imaginary numbers and imaginary time

tonygraham
1st July 2003, 05:16 AM
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaarrrrgggghhhh!

Solitaire
1st July 2003, 03:11 PM
Originally posted by xouper
Which says nothing about whether i itself is negative or positive. For example, which of the following statements are true (if any)?

i > 0

i < 0

-i > 0

-i < 0
This demonstrates why complex numbers (with a non-zero imaginary part) do not have the ordering property of real numbers. http://www.mathnstuff.com/math/spoken/here/3essay/eimg8.gif

Okay.
(1) True (2) False (3) False (4) True.
An imaginary number just occupies a different axis than real numbers.

P.S. I balance my checkbook using complex numbers. :)
More imaginary number stuff. (http://home1.gte.net/simres/k1-imagi.htm#Real)

Beleth
1st July 2003, 03:28 PM
If you think of 0 as being equal to the complex number 0 + 0i, then it's demonstrable that Synchronicity's answers are correct.

But then you get into questions like "Is 3 - 3i > 0?" which really can't be answered.

Maybe this diagram will help explain:

http://www.timecube.com/graphic2.gif
(sorry, couldn't resist)

xouper
1st July 2003, 07:29 PM
i > 0
i < 0
-i > 0
-i < 0Synchronicity: (1) True (2) False (3) False (4) True.Interesting that you would propose those answers, whereas aggle_rithm proposed that (2) is true. Actually, all four are false.

Let's look at case #1, i > 0, which you proposed is true.

If we assume i > 0, then multiplying both sides by i does not change the direction of the inequality and we get <nobr>-1 > 0.</nobr> This contradiction means our initial assumption, <nobr>i > 0,</nobr> must be false.

Similarly for case #4, -i < 0, which you also proposed is true. Multiplying both sides by negative one changes the direction of the inequality, which brings us back to case #1.

In the diagram you posted, the coefficients of i are represented along the vertical axis, but this says nothing about whether i itself is greater or less than zero.

Beleth: If you think of 0 as being equal to the complex number 0 + 0i, then it's demonstrable that Synchronicity's answers are correct.I'd be interested to see that demonstration, if you would.

BTW, the Time Cube reference is always good for a laugh when we need it. :)

gnome
6th July 2003, 07:39 AM
The best way I ever came up with to explain why "imaginary" numbers are as real and significant as any other is this:

(by "Imaginary" I mean the mathematical jargon, by "real" I mean common usage, as "Real" also has a jargon meaning)

i is a perfectly valid number, the deal is that you can't count or measure quantities with it, which is what we're used to using numbers for. It's just a number used for something besides what we're used to.

BillyJoe
7th July 2003, 03:40 AM
I imagine that would be really helpful for some, gnome. :cool:

fooboy
10th July 2003, 10:07 AM
My specialty at the University was Electromagnetics, I got a M.S.E.E, which for me focused on a lot of computational electromagnetics, and high speed digital signal transmission.

You can't go FTL in a cable, in fact, most cables actually SLOW DOWN the velocity of propagation of a wave. We call it the "Phase Velocity".

I read with alarm that this guy invokes the name of Lorentz, and Maxwell, and am amused by his thought experiments. I saw no good, AFAIAC, refutation so I registered to this forum and am posting here now.

Obviously, the three "mind" experiments were flawed. In EM theory we often speak of retarded time. Let me modify any of the magnets, and make them switchible. They are identical to any nomal permenant magnet, but the field can be shut down by an operator.

The field will collapse starting from the center, and propagating outward at c, the free space velocity, or as we all say, the speed of light. Any change at the magnetic sources are not felt at the coil side until a short time later, which we can measure and determine. Rotating, moveing the coil or anything else doesn't change this, the proof is not complicated but would require an understanding of vector theory.

The coaxial and single wire observation is a first year lab student's experiment in cable and EM. A cable with a correct resistor at the end, and a source at the other end is called a matched load. This cable will propagate any signal down it, and dissapate (burn) the power in the load.

When you remove the load, the power has to go somewhere. The power find the end of the cable but when the load is removed it has to go somewhere. It will go back up the cable to the source. This causes a phase shift. What he reports is actually not a change at the loaded end of the cable but the power returning to the source and changing the measurement there. If this guy has any RF/EM backgrouod he should know that! Unless he is lying. :)

Hopefully I am not sharing my thoughts out of turn. It is just so maddening to see someone use soemthing as poorly understood but easily calculated if you have the tools, to play a con.

fooboy
10th July 2003, 10:40 AM
DANG! Doublepost!

CurtC
10th July 2003, 11:31 AM
fooboy wrote:
We call it the "Phase Velocity".I agree with everything else you said, but you're applying the term "phase velocity" to the wrong thing. Phase velocity is something else entirely. A cable's propagation velocity, in my experience, would just be called is propagation velocity. Usually, though, we're talking about a specific length of a cable, so "propagation delay" or just "prop delay" is used. Cables can be "phase matched," which means that their length is selected for, or trimmed, so that two cables will have the same delay (and thus the same phase shift), but "phase velocity" is something else.

fooboy
11th July 2003, 03:37 AM
I agree with everything else you said, but you're applying the term "phase velocity" to the wrong thing. Phase velocity is something else entirely. A cable's propagation velocity, in my experience, would just be called is propagation velocity.

To give out some exposition to those who don't do any RF work, there are three things in propagation that are important, Phase Velocity, Group Velocity, and Propagation Velocity. Phase velocity is the speed of one wave, group is the speed of a set of waves or a pulse in a group, and the propagation velocity is the speed of a signal in a cable. Most of the time they are indeed the same thing.

Many cable manifacturers use phase and prop velocity interchangably. In optical and some RF waveguides, group and phase are not always equal.

I keep saying phase vel, which in the case of a single wire in space, and a coaxial cable are the same thing. Its often about .66c in many commercial cables I use.

Still, the guy using the scope is either lying, or has made a triggering error.

The best way to meadure the speed in a cable is to use a Time Domain Reflectometer or TDR.

(I wish I spent more time learning the english language instead of engineering language and spelling)