- #36
JesseM
Science Advisor
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You're being kind of vague, "apply the transformation" to what, exactly? You can only apply the Lorentz transformation to a pair of x,t coordinates (or x',t' coordinates) representing a single event, to find the coordinates of that same event in another frame. So what event are you interested in, and what coordinates does it have in B, that would give x=12 in A when you apply the transformation? Note that when you go back from B to A, since in B's frame A is moving in the -x' direction rather than the +x' direction, the Lorentz transformation will look a little different:Austin0 said:[
So are you saying that given the coincidence of( t=0 at x=0 )and (t'=0 at x'=0) and assuming the same, at rest ,metric, you could NOT apply the trasnsformation and determine that x'=15 would at that T be positioned at x=12 in A ?
x = gamma*(x' + vt')
t' = gamma*(t' + vx'/c^2)
The only two events I was interested in where the event of the tachyon signal being sent and the event of it being received, and neither had position coordinate x=12 in A.
Whose clock? In B's frame, all of B's own clocks--which are used to define the t' coordinate of events that take place next to any given clock of B's--are synchronized. Are you asking what B's clock at x'=15 would see if it looked at A's clock right next to it at time t'=0? If so you're not expressing yourself very clearly, and I also don't really see the relevance of that question, but it can certainly be calculated using the Lorentz transformation.Austin0 said:Are you saying that you could NOT calculate the degree of desynchronization at time t'=0 at x'=0 for a clock located at x'=15 in B ?
By the way, I also realized I messed up the math somewhat before, if the velocity is 0.8c then gamma should be 1.666..., not 1.25 (I was thinking of the fact that with a velocity of 0.6c you get sqrt(1 - 0.6^2) = 0.8, so gamma = 1/0.8 = 1.25 in this case). So if the event of the signal being received has coordinates x=20, t=10 in A, the coordinates of this event in B are:
x' = 1.666...*(20 - 0.8*10) = 20
t' = 1.666...*(10 - 0.8*20) = -10
So if you restate your question in terms of these corrected coordinates, are you asking what B's clock at x'=20 will see at t'=0 if it looks at the clock of A right next to it? If so, applying the Lorentz transformation to this case gives:
x = 1.666...*(20 + 0.8*0) = 33.333...
t = 1.666...*(0 + 0.8*20) = 26.666...
So, if B's clock at x'=20 looks at the clock of A that's right next to it at t'=0, it'll see the clock attached to the x=33.333... mark on A's ruler, and that clock will be reading t=26.666... seconds.
I certainly don't think an individual FTL signal is objectively going forwards or backwards in time; different frames disagree about whether the event of it being received happens before or after the event of it being sent, because of different definitions of simultaneity, and there can be no objective truth in relativity about frame-dependent questions. However, if you assume the first postulate of relativity holds (that the laws of physics work identically in all the different frames given by the Lorentz transform), then it must be possible to send a pair of FTL signals between two observers A and B moving away from each other such that the first one goes forwards in time in A's frame but back in time in B's frame, then B's reply goes forward in time in B's frame but backwards in time in A's frame, such that both frames agree that A gets B's reply before A sent the original signal. This is a widely recognized consequence of FTL signalling + the first postulate of relativity, it's a thought-experiment sometimes known as the "tachyonic anti-telephone", see my post #30 to Max where I linked to a physics paper that mentions it.Austin0 said:Ay, there's the rub. Proving that FTL signalling would go back in time in ANY frame.
It appears to me that you and others who have proposed this hypothesis have set out with two a priori assumptions:
(1) FTL signals go back in time.
No philosophical assumptions are required, it's just a straightforward consequence of the fact that some observers will measure an FTL signal to be received at an earlier time than it's sent in their coordinate system, combined with the assumption of the first postulate of relativity which says that any measurements obtained in one coordinate system during a given experiment can be repeated in a different coordinate system with an analogous experiment. The first postulate says that if you perform any experiment, then whatever observations you get about the results of the experiment in one frame (in terms of that frame's coordinates), it must always be possible to perform an analogous experiment which gives the same results in some other frame (in terms of that frame's own coordinates). See my comments to RandallB earlier:Austin0 said:(2) That block time is an accepted physical reality.
Do you agree that this is an accurate description of the implications of the first postulate?Just to be clear, when I say "same experiment", I mean that whatever the coordinates of events that take place in the first experiment as seen in the first frame, one can run a second experiment such that identical events happen at the same coordinates in the second frame. For instance, suppose in frame A you have a clock which starts at position x=5 light-seconds, t=10 seconds with the clock reading T=30 seconds, and the clock is moving at 0.8c in the +x direction of frame A so it'll be at position x=13 light-seconds at time t=20 seconds, and because of time dilation it only reads T=36 seconds at that point. Do you agree that the first postulate automatically implies we can have a different clock moving at 0.8c in the +x' direction of frame B, which starts out reading T=30 seconds at x'=5 light-seconds and t'=10 seconds (in the x', t' coordinates of frame B), and then it must be true that the clock will read T=36 seconds at coordinates x'=13 light-seconds and time t'=20 seconds? If so, do you agree that any experiment in frame A can be replicated in this way in frame B, and vice versa?
I have never said anything about the "absolute chronology" of any two events. Again, there is no objective fact of the matter about whether the event of a given FTL signal being received happens at an earlier or later time than the event of that signal being sent, different frames disagree. However, if A can send a signal to B which travels forward in time at 2c as measured in A's frame, then the first postulate says it must be possible for B to send a reply to A which which travels forward in time at 2c as measured in B's frame. The consequence of this is that A's original message was traveling backwards in time in B's frame, and B's reply was traveling backwards in time in A's frame, and if the relative velocity between A and B is large enough then both frames agree that B's reply reaches A before A sent the original signal. This conclusion is an unavoidable result of the assumptions that FTL signaling is possible plus the first postulate of relativity; you can have FTL signaling without such causality violations only if you drop the first postulate of relativity and say that one of these observers is unable to send signals forward in time at 2c as measured in their own coordinates.Austin0 said:Forget for the moment the question of FTL signalling. Assume that A sends up a flare at the moment of transmission and B likewise sends up a flare on reception.
How is this situation any different from two lightning bolts in the classic demonstration of the relativity of simultaneity? How is the absolute chronology of these two events [flares] any more determinable than the absolute chronology of the two lightning bolts,, which appear simultaneous to a track observer while the train observer perceives the one at the front occurring before the back and some other observer would see the back occurring before the front?
Maybe I should come up with a full numerical example to illustrate. To make the numbers a little easier, I'll assume that two observers A and B can both send signals forward in time at 5c in their own frames, and they are moving apart from one another at 0.6c. So, say A sends a signal at x=0,t=0 in the A frame and the signal is received by B at x=100,t=20 in the A frame. This means that in the B frame, the event of the signal being sent was x'=0,t'=0, while the event of it being received was:
x' = 1.25*(100 - 0.6*20) = 110
t' = 1.25*(20 - 0.6*100) = -50
In the B frame, we know A is moving in the -x' direction at 0.6c and was at position x'=0 at t'=0, so in this frame A's position as a function of time is x'(t') = -0.6*t'. If B sends an FTL signal in the -x direction at 5c from position x'=110 at t'=-50, then 50 seconds later at t'=0 the signal will have moved 5*50 = 250 in the -x direction, so it'll be at x'=110 - 250 = -140. So, the signal's position as a function of time is x'(t') = -5*t' - 140. So to figure out the coordinates in B that the reply catches up with A, just set -5*t' - 140 = -0.6*t', which gives -4.4*t' = 140, or t' = -31.8181... Since A's position as a function of time is -0.6*t' in this frame, the position of A in this frame when the reply reaches him must be x' = -0.6*-31.8181... = 19.0909...
So, in the B frame, A receives B's reply at (x'=19.0909..., t'=-31.8181...) which is earlier than the time A sent the original message at (x'=0, t'=0). You can use the Lorentz transform to find the time that A receives B's reply in the A frame:
x = 1.25*(19.0909.. + 0.6*-31.8181...) = 0
t = 1.25(-31.8181... + 0.6*19.0909...) = -25.4545...
So, you can see it's also true in the A frame that A received B's reply before A sent the original message. The only assumption being made here is that both of them can send a signal at 5c in their own frame.