# A question about the equivalence principle.

by Nick20
Tags: theory of relativity, time dilation
P: 3,783
 Quote by Austin0 So have you calculated the differential due to velocity actually/ or is it just a logical guess? ;-0
At the time it was a estimate, but I have now done the calculations just to check and can confirm that clock speed differential between front and back increases over time if we use the velocity time dilation approach. For example if we have a constant proper acceleration of 1 at the back of the rocket and 0.1 at the front and use units of c=1, then initially the time dilation ratio (back/front) is 1 at time t=0, 0.71 at t=1 and 0.456 at t=2. This is not what would be observed on board the rocket(s) because the time (t) here is the simultaneous coordinate time in the launch frame and the notion of what is simultaneous is different on board the rocket(s). This is the crux of the matter that I shall come back to.
 Quote by Austin0 i agree with your assumption that it appears that the relative dilation due to coordinate velocity would increase. That was my intuition also but after thought i became less sure.
After doing the calculations I am now sure that is what would be observed as coordinate velocity increases relative to any given MCIRF, as measured relative to the simultaneous coordinate time in that given MCIRF. As above, it is the notion of simultaneity that is important and this changes with reference frames and you have not been careful to specify one.
 Quote by Austin0 That was partly my interest in the analysis I described. But if we are right and the dilation differential would increase up the velocity curve this would seem to present a major problem.
We are right but it does not present a problem ;)
 Quote by Austin0 As I said, the two clocks in question could be unconnected. Independently accelerating clocks that would maintain the proper contracted separation simply as a consequence of the acceleration differential.
Agree.
 Quote by Austin0 It seems clear that the fundamental SR principles must pertain.I.e. The gamma relation between velocity and proper time rates and deltas. And the clock hypothesis.
Agree, these fundamental SR principles do hold.
 Quote by Austin0 So if the results predicted by the Rindler coordinates do not agree with the results predicted by the fundamental principles of SR there would be a real question. What possible physics would account for this ? i.e. what would prevent the velocity dilation from occurring?
Nothing. Velocity time dilation still occurs, but the effect cancels out. Consider two signals emitted simultaneously from the front and back of a rocket at coordinate time t=0 as measured in a given MCIRF. To give this MCIRF a label I will call it the launch frame (LF), but there is nothing special about this MCIRF and I could do an identical analysis in any other MCIRF. When the signal from the front arrives at the back let us say the velocity in the LF is vb and when the signal from the back arrives at the front, the velocity in the LF is vf. It turns out that with Born rigid acceleration, vb = vf. This means that sqrt(1-(vb)^2) = sqrt(1-(vf)^2) so at the time of the reception of the signals the velocities and velocity based time dilation is identical at back and front and so there is no differential time dialtion between back and front due to velocity time dilation. All the apparent differential time dilation observed on board the rocket is due to classical Doppler shift according to the observer in the inertial LF.
 Quote by Austin0 Which prediction should be considered valid?
They are both valid. One is the interpretation according to the simultaneity of an inertial reference frame and the other is the interpretation due to a different notion of simultaneity in an accelerating reference frame.
 Quote by Austin0 Would you agree there would be a question?
Not really. We are used to observer dependent quantities such as coordinate velocity, time, distance, time dilation, length contraction, simultaneity etc. being different in different inertial reference frame (by definition) in regular unaccelerated SR.
 Quote by Austin0 You missed this one. regarding Steve's derivative approach: "But what I don't understand is why the simultaneity relative to the MCIRF's would be relevant to the accelerating system?"
Well now we are back to the crux of the problem. I have been putting this off because while I feel I understand it intuitively in my head, putting it into words is not so easy :) The first difficulty is how to define a notion of simultaneity for the observers on board the rocket, when there own reference clocks appear to be running at different rates in different locations on board the rocket. If we ask the observers at the front and back of the rocket to send signals simultaneously in there own reference frame, how are they going to arrange that? One way to do this would be to agree a convention that defines "simultaneous" events at the front and back of the rocket as being events that are both simultaneous in a shared MCIRF. If they do this they will observe that the signals are received at the back and front "simultaneously" because the reception events will be simultaneous in a shared MCIRF, (but this shared MCIRF will not be the same shared MCIRF that the signals were emitted simultaneously in). For a practical example, lets say we an inertial rocket ir1 moving at 0.6c and another ir2 moving at 0.8c both relative to the LF. After launch we tell the observers at the front and back to send a signal at the moment they are at rest with ir1, then for a suitable length accelerating rocket, they will both receive signals when they are momentarily at rest with ir2. You might argue that I have cheated here, because I have simply defined, rather than derived a notion of simultaneous for the accelerating rocket observers.

Another approach is to speed up the rear clock (as mentioned in an earlier post) so that it appears to be running at the same rate as the front clock. Now we have an unequivocal method of defining simultaneous in the accelerating rocket reference frame and can synchronise clocks using the usual Einstein clock synchronisation convention. Now when we send signals from the front and back using the on board rocket reference clocks, the signals arrive simultaneously at the back and front respectively according to the rocket clocks and the elapsed time between sending and receiving is equal according to both the front and back accelerating rocket observers. At no point here have we had to refer to the MCIRF's but if we compare results, simultaneous emission and simultaneous reception of signals as measured by the accelerating rocket observers, agrees with simultaneous as measured in the MCIRF's. Now the rocket observers know they had to speed up the rear clock to make it run at the same rate as the front clock so they know the real proper time rate of the rear clock must be slower than the proper time rate of the front clock. If they consider themselves to be stationary in their own reference frame they would ascribe this differential clock rate to a gravitational field and this coincides with the proper acceleration they can feel and and measure. The observers in a given inertial reference frame outside the rocket would ascribe the differential clock rates observed by the rocket observers to classical Doppler shift.

 Quote by harrylin Good one! Yes, there is a very slight difference due to length contraction - and it's always the same for a given distance between clocks and given acceleration. However, if one allows the Earth sensor to free-fall inside or next to the rocket, it can detect no gradient at all, not even in principle. And I now wonder if one would nevertheless interpret the (incredibly small) difference in accelerometer readings as due to a real field, with what distance to the gravitation source that would correspond, and with what mass - could one really be fooled?
The differences between the Born rigid accelerating rocket and a real gravitational field, due to tidal effects, are significant enough, that we would be only be fooled if limited to measurements in a very restricted local region.
P: 1,396
 Quote by Austin0 But what I don't understand is why the simultaneity relative to the MCIRF's would be relevant to the accelerating system?
I'm not sure I understand your question. You're asking why the inertial coordinate system is relevant to the accelerated observer? In some ways, it's just a matter of convention for what the accelerated observer considers two "simultaneous" events. One convention is to use the same notion of simultaneity as a comoving inertial observer.

 I would think that given the proper acceleration values for front and back for the initial length , that simply calculating velocities from coordinate acceleration would give relative gamma between the front and back directly.
No, that's not correct. If you wait a short time δt after launch, the rear clock will be traveling at speed vrear = grear δt, and the front clock will be traveling at speed vfront = gfront δt. So just based on that, you would expect a relative rate of the clocks to be
√(1- (grear δt)2/c2))/√(1- (gfront δt)2/c2))
but the actual ratio of rates is
gfront /grear
those aren't close at all. The first goes to 1 as δt → 0, but the second doesn't.
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P: 4,764
 Quote by stevendaryl No, that's not correct. If you wait a short time δt after launch, the rear clock will be traveling at speed vrear = grear δt, and the front clock will be traveling at speed vfront = gfront δt. So just based on that, you would expect a relative rate of the clocks to be √(1- (grear δt)2/c2))/√(1- (gfront δt)2/c2)) but the actual ratio of rates is gfront /grear those aren't close at all. The first goes to 1 as δt → 0, but the second doesn't.
I haven't been following this thread closely, and I can see a lot of formulas have been written down that I have not looked at in detail, but this doesn't look right; the ratio of clock rates *should* be governed by the first formula, based on the two velocities. The ratio of clock rates is certainly *not* equal to the ratio of accelerations. And the ratio *should* go to 1 as delta t -> 0, because at launch, which is what delta t -> 0 means, the two clocks are at rest relative to each other, so their clock rates are the same.

Some web pages that seem to me to be relevant are Greg Egan's page on the Rindler horizon:

http://gregegan.customer.netspace.ne...erHorizon.html

And the Usenet Physics FAQ pages on SR and acceleration (which links to the page on the relativistic rocket equation, another good resource), and on the clock hypothesis:

http://math.ucr.edu/home/baez/physic...eleration.html

http://math.ucr.edu/home/baez/physic.../SR/clock.html
P: 1,396
 Quote by PeterDonis I haven't been following this thread closely, and I can see a lot of formulas have been written down that I have not looked at in detail, but this doesn't look right; the ratio of clock rates *should* be governed by the first formula, based on the two velocities. The ratio of clock rates is certainly *not* equal to the ratio of accelerations. And the ratio *should* go to 1 as delta t -> 0, because at launch, which is what delta t -> 0 means, the two clocks are at rest relative to each other, so their clock rates are the same.
But that doesn't give the correct answer, which is that, according to measurements performed by observers aboard the rocket, the front clock always runs faster than the rear clock, the ratio of their rates is always the same: 1 + gL/c2.
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 Quote by stevendaryl But that doesn't give the correct answer, which is that, according to measurements performed by observers aboard the rocket, the front clock always runs faster than the rear clock, the ratio of their rates is always the same: 1 + gL/c2.
It's possible that I'm misunderstanding the scenario you are talking about. When you say that a short time after launch, each clock will be traveling at a velocity g delta t, where g is the acceleration of the clock, that tells me that exactly at time t = 0, i.e., before the time delta t elapses, each clock is at rest in the same "launch frame". If the two clocks are at rest relative to each other at some instant, then their clock rates, at that instant, are the same. See the web page I linked to on the clock postulate. My understanding is that that is the scenario you are talking about.

The formula 1 + gL/c describes something a bit different than what I would call the "ratio of instantaneous clock rates", but I'd rather not get into that until I'm sure I correctly understand the scenario you're talking about.
P: 1,396
 Quote by PeterDonis It's possible that I'm misunderstanding the scenario you are talking about. When you say that a short time after launch, each clock will be traveling at a velocity g delta t, where g is the acceleration of the clock, that tells me that exactly at time t = 0, i.e., before the time delta t elapses, each clock is at rest in the same "launch frame". If the two clocks are at rest relative to each other at some instant, then their clock rates, at that instant, are the same.
To talk about clock "rates", you have to refer to two different points in time. To talk about the relative rates of two clocks, you need 4 events, and you need a notion of simultaneity.

e1 = the rear clock shows time 0.
Let F = the instantaneous rest frame of the rear clock at this event.

e2 = some event at the front clock that is simultaneous with e1 in frame F . Let T2 = the time showing on the front clock at event e2.

e3 = the rear clock shows time δT
Let F' be the instantaneous rest frame of the rear clock at this event.

e4 = some event at the front clock that is simultaneous with e3 in frame F'. Let T4 = the time showing on the front clock at event e4.

So from the point of view of an observer in the rear of the rocket, the rear clock advanced by δT, and the front clock advanced by: T4 - T2

Then the ratio of clock rates, as measured by the rear clock is:

Rfront/Rrear
= (T4 - T2)/δT

As δT → 0, this ratio does not go to 1, but goes to 1 + gL/c2, where L is the length of the rocket, and g is the acceleration of the rear.
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 Quote by stevendaryl To talk about clock "rates", you have to refer to two different points in time. To talk about the relative rates of two clocks, you need 4 events, and you need a notion of simultaneity.
I agree with the second sentence given the first. I take the first as your definition of how you are using the term "clock rate"; it is not the only possible definition, but I have no problem using yours for this discussion.

 Quote by stevendaryl e1 = the rear clock shows time 0. Let F = the instantaneous rest frame of the rear clock at this event. e2 = some event at the front clock that is simultaneous with e1 in frame F . Let T2 = the time showing on the front clock at event e2.
You've left out a key piece of information here: is the front clock at rest, in frame F, at event e2? (By hypothesis the rear clock is at rest in frame F at event e1.)

 Quote by stevendaryl e3 = the rear clock shows time δT Let F' be the instantaneous rest frame of the rear clock at this event. e4 = some event at the front clock that is simultaneous with e3 in frame F'. Let T4 = the time showing on the front clock at event e4.
Same question here. I ask these, again, to make sure I understand the scenario you are talking about. I think the answer to both is "yes", but I would really like confirmation before going into details, since there's no point in my talking about a different scenario than you are talking about.
P: 1,396
 Quote by PeterDonis I agree with the second sentence given the first. I take the first as your definition of how you are using the term "clock rate"; it is not the only possible definition, but I have no problem using yours for this discussion. You've left out a key piece of information here: is the front clock at rest, in frame F, at event e2? (By hypothesis the rear clock is at rest in frame F at event e1.)
That's not really relevant to the question of what is the ratio of clock rates, as measured by an observer in the rear of the rocket. But in the way I've set things up, the answer is "yes".

 Same question here. I ask these, again, to make sure I understand the scenario you are talking about. I think the answer to both is "yes", but I would really like confirmation before going into details, since there's no point in my talking about a different scenario than you are talking about.
Whether the front clock is at rest at events e2 and e4 doesn't seem relevant to me. But if you work out the details for the constant proper acceleration, it turns out to be true, that in frame F, the front clock is at rest at event e2, and in frame F', the front clock is at rest at event e4.
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 Quote by stevendaryl That's not really relevant to the question of what is the ratio of clock rates, as measured by an observer in the rear of the rocket. But in the way I've set things up, the answer is "yes". Whether the front clock is at rest at events e2 and e4 doesn't seem relevant to me. But if you work out the details for the constant proper acceleration, it turns out to be true, that in frame F, the front clock is at rest at event e2, and in frame F', the front clock is at rest at event e4.
Thanks for the confirmation. The reason it's relevant (other than giving me confirmation that I was correctly understanding your scenario) is the clock postulate, which I referred to before. Consider what you have here: you have two pairs of events, at each of which two objects are mutually at rest. By the clock postulate, then, their "instantaneous clock rates" should be the same, because their velocities, relative to any inertial frame, are the same, therefore their "time dilation factors", relative to any inertial frame, are the same. Yet between their respective events, one object (the front clock) experiences more elapsed time than the other (the rear clock).

In fact it's even weirder than that. Consider the entire worldlines of the two clocks; I'll write their equations in the "launch frame" as follows:

Rear clock: x^2 - t^2 = R^2

Front clock: x^2 - t^2 = (R + L)^2

Now draw any straight line through the origin, in the "launch frame", t = vx, where v >= 0. This is either a horizontal line (for v = 0) or a line sloping up and to the right at less than 45 degrees (for v > 0). Call the event where the line intersects the rear clock's worldline Rv, and the event where the line intersects the front clock's worldline Fv. Then all of the following are true:

(1) For any v, the line t = vx is a "line of simultaneity" in the instantaneous rest frame of the rear clock at Rv *and* of the front clock at Fv. Therefore, Rv and Fv are simultaneous as seen by both the front and the rear clocks.

(2) For any v, relative to the launch frame, the rear clock at Rv and the front clock at Fv are both moving at velocity v. Therefore, the front and rear clocks both see each other as at mutual rest at these events, and they both have the same "time dilation" factor at these events (because of the clock postulate).

(3) For any v, the proper time experienced by the front clock between F0 and Fv is greater, by the ratio (R+L)/R, than the proper time experienced by the rear clock between R0 and Rv.

So we have two clocks, which remain a constant distance apart at mutual rest, and have the same "time dilation factor" at any pair of corresponding events, and yet experience different proper times between corresponding events. I realize all this is true; I just point it out to explain why one has to be careful using the term "clock rate" without more explanation. Many people will think "time dilation factor", as in 1/sqrt(1 - v^2), when they see "clock rate" (after all, that's what I initially thought when I saw it), but that's not what you mean by the term. In particular, I think this may be one source of confusion in some of Austin0's posts.
P: 1,396
 Quote by PeterDonis Thanks for the confirmation. The reason it's relevant (other than giving me confirmation that I was correctly understanding your scenario) is the clock postulate, which I referred to before. Consider what you have here: you have two pairs of events, at each of which two objects are mutually at rest. By the clock postulate, then, their "instantaneous clock rates" should be the same, because their velocities, relative to any inertial frame, are the same, therefore their "time dilation factors", relative to any inertial frame, are the same. Yet between their respective events, one object (the front clock) experiences more elapsed time than the other (the rear clock).
For problems involving acceleration, time dilation is not the only consideration. Let's consider a variant of the old twin paradox. We have two twins that are the same age, but live on different planets, light-years apart. One twin on his 20th birthday hops into a rocket and accelerates rapidly to nearly the speed of light and gets to his other twin in just one year, according to the clock in his rocket. But the other twin has aged 20 years during the trip. How did that happen? From the point of view of the traveling twin, the trip only lasted a year. How could the other twin age 20 years?

Well, what you have to take into account is that simultaneity is relative. Let e1 be the traveling twin celebrated his 20th birthday. Let e2 be the event at which the distant twin celebrated his 20th birthday. Let F be the "launch" frame of the traveling twin, and let F' be the "traveling" frame. Events e1 and e2 are simultaneous in frame F, but not in frame F'; in frame F', e2 took place long, long before e1. So in jumping from frame F to frame F', the traveling twin changed his notion of what time "now" is on the distant planet, and so he changed his notion of how old the distant twin is.

So for an accelerated observer, the time on a distant clock changes both because that clock advances, and also because the observer's notion of what is "now" for the distant clock changes.
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 Quote by stevendaryl So for an accelerated observer, the time on a distant clock changes both because that clock advances, and also because the observer's notion of what is "now" for the distant clock changes.
Yes, the "simultaneity lines" t = vx that I described illustrate this: they change slope as v increases, meaning that the direction between the rear clock's "now" and the front clock's "now" changes. That means a given segment of the rear clock's worldline includes a set of simultaneity lines that sweep over a larger segment of the front clock's worldline. It's the hyperbolic equivalent of concentric circles, where the same angle sweeps out a longer arc on the circle with a larger radius.
P: 3,783
 Quote by PeterDonis I haven't been following this thread closely, and I can see a lot of formulas have been written down that I have not looked at in detail, ...
Just to summarise, the various formulas written down by Dalespam, stevendaryl and myself for the relative red shift of clock rates on board the accelerating rocket, as measured by observers on board the accelerating rocket are all equivalent, i.e.:

$$\frac{\Delta\tau_2}{\Delta\tau_1} = (1+g_1L/c^2) = \frac{R_2}{R_1} = \frac{g_1}{g_2}$$

To the accelerated observers on board the rocket, the clocks remain at constant distance from each other, so from their point of view, none of the observed red shift is due to classical Doppler shift caused by relative velocities.
P: 1,162
 Quote by yuiop This is not what would be observed on board the rocket(s) because the time (t) here is the simultaneous coordinate time in the launch frame and the notion of what is simultaneous is different on board the rocket(s). This is the crux of the matter that I shall come back to. After doing the calculations I am now sure that is what would be observed as coordinate velocity increases relative to any given MCIRF, as measured relative to the simultaneous coordinate time in that given MCIRF. As above, it is the notion of simultaneity that is important and this changes with reference frames and you have not been careful to specify one.
Only one. The launch frame.

 Quote by yuiop Nothing. Velocity time dilation still occurs, but the effect cancels out. Consider two signals emitted simultaneously from the front and back of a rocket at coordinate time t=0 as measured in a given MCIRF. To give this MCIRF a label I will call it the launch frame (LF), but there is nothing special about this MCIRF and I could do an identical analysis in any other MCIRF. When the signal from the front arrives at the back let us say the velocity in the LF is vb and when the signal from the back arrives at the front, the velocity in the LF is vf. It turns out that with Born rigid acceleration, vb = vf.
i am not sure if this actually tracks. If the signals are simultaneous in the emission MCIRF it appears unlikely the reception ,in a frame which would be many frames and spatial distance removed, would be simultaneous. I will have to work on this.

 Quote by yuiop This means that sqrt(1-(vb)^2) = sqrt(1-(vf)^2) so at the time of the reception of the signals the velocities and velocity based time dilation is identical at back and front and so there is no differential time dialtion between back and front due to velocity time dilation. All the apparent differential time dilation observed on board the rocket is due to classical Doppler shift according to the observer in the inertial LF.
even if your assumption is correct and the reception is simultaneous wrt the MCIRF how does this imply no velocity dilation?? by definition there is no motion relative to the MCIRF's
so any calculation based on the MCIRF couldn't reveal relative velocity between front and back.

When you say here the observer in the LF do you mean the initial launch frame or the current mCIRF ,,,,earlier you were refering to MCIrfs as LF
In any case Doppler shift is as you say ,apparent dilation, so not really relevant

 Quote by yuiop Well now we are back to the crux of the problem. I have been putting this off because while I feel I understand it intuitively in my head, putting it into words is not so easy :) The first fdifficulty is how to define a notion of simultaneity for the observers on board the rocket, when there own reference clocks appear to be running at different rates in different locations on board the rocket. If we ask the observers at the front and back of the rocket to send signals simultaneously in there own reference frame, how are they going to arrange that? One way to do this would be to agree a convention that defines "simultaneous" events at the front and back of the rocket as being events that are both simultaneous in a shared MCIRF. If they do this they will observe that the signals are received at the back and front "simultaneously" because the reception events will be simultaneous in a shared MCIRF, (but this shared MCIRF will not be the same shared MCIRF that the signals were emitted simultaneously in). For a practical example, lets say we an inertial rocket ir1 moving at 0.6c and another ir2 moving at 0.8c both relative to the LF. After launch we tell the observers at the front and back to send a signal at the moment they are at rest with ir1, then for a suitable length accelerating rocket, they will both receive signals when they are momentarily at rest with ir2. You might argue that I have cheated here, because I have simply defined, rather than derived a notion of simultaneous for the accelerating rocket observers
I think you misunderstood my question. I am aware of the problems implementing simultaneity in this circumstance , i mentioned a few earlier.
the question was why take an approach which had these problems and the inherent ambiguity of the result due to these problems.??

 Quote by yuiop Another approach is to speed up the rear clock (as mentioned in an earlier post) so that it appears to be running at the same rate as the front clock. Now we have an unequivocal method of defining simultaneous in the accelerating rocket reference frame and can synchronise clocks using the usual Einstein clock synchronisation convention. Now when we send signals from the front and back using the on board rocket reference clocks, the signals arrive simultaneously at the back and front respectively according to the rocket clocks and the elapsed time between sending and receiving is equal according to both the front and back accelerating rocket observers.
well I have to disagree here. Simply scaling the clocks does not make it an inertial frame.
It is still an accelerating system.
Even disregarding the acceleration/velocity differential, a synchronization which works for one velocity cannot work for other different velocities. Yeah?? How could it?

 Quote by yuiop If they consider themselves to be stationary in their own reference frame they would ascribe this differential clock rate to a gravitational field and this coincides with the proper acceleration they can feel and and measure. The observers in a given inertial reference frame outside the rocket would ascribe the differential clock rates observed by the rocket observers to classical Doppler shift.
 Quote by yuiop The snag is that the by calculating the relative time dilation by using the relative velocities at the front and back of the rocket one comes to the conclusion that the relative rates of clocks on board the rocket as measured by observers on board the rocket increases over time which is simply not true. The relative rates of clocks on board the rocket as measured by observers on board the rocket is constant over time.
I think you have misunderstood both my approach and point.

regarding measurement of relative clock rates at the front and back from the launch frame. LF accelerating system AF
Simultaneity is not an issue. measurement of clock rates of course requires some interval of time a single event doesn't work.

so in LF at (bxo,t0) (fx0 ,t0) we get

observations bT'0 and fT'0 of AF ..

at some later point at (bx1, t1), (fx1,t1)

we get observations bT'1 and fT'1 of AF

(bx) t1 - (bx)t0 = (bx)dt

(fx)t1 -(fx)t0 = (fx)dt

bT1-bT0 =dbT'

fT1 - fT0= dfT

then (fx)dt/$\gamma$ =dfT' and (bx)dt/$\gamma$=dbT' or not

Would you agree that in this circumstance the simultaneity of the LF clocks at either measurement is not important because there is no direct comparison between the observations between them?
The comparison is between an observation of the back clock with a later observation of the back clock.Etc.

Of course it would be necessary to calculate the proper times for these events using the Rindler coordinates to make these comparisons.

Also the measurement points could be widely separated say 0.7c and 0.8c

More and more i suspect that the velocity dilation would be insignificant and might possibly agree with the Rindler predictions. I.e. would not increase with greater velocities. The acceleration magnitudes you used were totally unrealistic. The back of the rocket quickly passing the front and leaving it in the dust ;-) so I am still unsure.
thanks
P: 3,783
 Quote by Austin0 i am not sure if this actually tracks. If the signals are simultaneous in the emission MCIRF it appears unlikely the reception ,in a frame which would be many frames and spatial distance removed, would be simultaneous. I will have to work on this.
It does track. See the attached chart showing two rockets with Born rigid acceleration. Each time signals (black diagonal lines) are emitted simultaneously in one MCIRF, they are received simultaneously in a subsequent MCIRF. In the chart green curves represent equal proper time for the rocket clocks, red lines represent lines of equal velocity relative to the LF (or subsequent MCIRFs) and the blue lines represent the worldlines of the rockets in the LF. Note the constant nature of the redshift. Signals sent every 3 ticks from the rear rocket are received every 4 ticks by the leading rocket.
 Quote by Austin0 even if your assumption is correct and the reception is simultaneous wrt the MCIRF how does this imply no velocity dilation?? by definition there is no motion relative to the MCIRF's so any calculation based on the MCIRF couldn't reveal relative velocity between front and back.
You have the reasoning reversed. The MCIRFs reveal that there is no relative velocity in the reference frames of the accelerating rockets. You are right that simultaneously at given instant (other than t=0) in the LF, the velocities of the front and rear rockets are different, but this simultaneity does not apply to the accelerating reference frame. We can for example show that if there is no length contraction and the rockets are accelerating identically in the LF, that differential time dilation is still occurring in the rocket reference frame.
 Quote by Austin0 When you say here the observer in the LF do you mean the initial launch frame or the current mCIRF
Here I meant LF, as in pick one MCIRF and stick with it, rather than keep switching to subsequent MCIRFs.
 Quote by Austin0 ,,,,earlier you were refering to MCIrfs as LF
I just meant you could pick any arbitrary MCIRF as a LF. There is nothing particularly special about the LF. The rockets are still accelerating even in the LF.
 Quote by Austin0 In any case Doppler shift is as you say ,apparent dilation, so not really relevant
This is the bizarre aspect. The inertial observer attributes the redshift to classical Doppler shift due to velocity differential between emission and reception and yet the time dilation observed by the accelerating observers is real (as in physical) and an observer in the rear rocket really ages slower than an observer in the front rocket.
 Quote by Austin0 the question was why take an approach which had these problems and the inherent ambiguity of the result due to these problems.??
.. because nature does not give us much choice and there is no natural way to have synchronised clocks and static coordinate system in an accelerating reference frame or in a gravitational field.
 Quote by Austin0 well I have to disagree here. Simply scaling the clocks does not make it an inertial frame. It is still an accelerating system.
Yes, it is still an accelerating system and I never claimed to make it an inertial reference frame. I only claimed it gave us a way to have a sensible coordinate system and a way to synchronise clocks that gives a static reference system with coordinate axes that are not changing over time. Can you suggest another way to synchronise clocks that are running at different rates?
 Quote by Austin0 Even disregarding the acceleration/velocity differential, a synchronization which works for one velocity cannot work for other different velocities. Yeah?? How could it?
It does work for other velocities. Have another look at the posted chart. The rear rocket sends signals every 3 ticks and the front rocket receives those signals every 4 ticks, consistently even as the relative rocket velocities constantly change over time. If we speed up the rear clock by a factor of 4/3 then the front observer will see the rear clock ticking at the same rate as his own clock for all time.
 Quote by Austin0 Also the measurement points could be widely separated say 0.7c and 0.8c More and more i suspect that the velocity dilation would be insignificant and might possibly agree with the Rindler predictions. I.e. would not increase with greater velocities. The acceleration magnitudes you used were totally unrealistic. The back of the rocket quickly passing the front and leaving it in the dust ;-) so I am still unsure. thanks
Again, if you look at the attached chart you see that the front rocket sends a signal when the rocket velocities in the LF are approximately 0.69c and the rear rocket receives that signal when the rocket velocities are approximately 0.81c in the LF and there are no problems with the rear rocket overtaking the front rocket. (Another curious aspect is that a light signal sent from left of the origin can never catch up with the accelerating rockets, even though they never attain light speed. That should bake your noodle! ).

P.S. I think Peter has a pretty good handle on it all in post #81.
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 Quote by yuiop [..] This is the bizarre aspect. The inertial observer attributes the redshift to classical Doppler shift due to velocity differential between emission and reception and yet [..] an observer in the rear rocket really ages slower than an observer in the front rocket.
That looks bizarre because it's wrong. Indeed the redshift of accelerating rockets is almost purely (semi)classical Doppler shift, insofar as there is no or negligible difference in acceleration. I see no reason for the claim that in reality it's not so, as I also stressed in post #65:
 Quote by harrylin - The equivalence principle has that the observable effect will be the same. Einstein calculated (predicted) what the observable effect will be due to gravitation, basing himself on the observable Doppler effect due to acceleration. Thus the clock at the bottom of the ship will only *appear* to slow down by the gravitational time dilation factor *if* you assume that the ship is not accelerating but at rest in a gravitational field. If instead you assume, as you do, that there is negligible gravitational field, then the clock at the rear will really *not* slow down (at least, by far not by that amount) compared to the one at the front.
It depends on your choice of inertial reference system if you deem that the clock in the rear ages slower or not; according to the launch frame's POV they age equally.
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 Quote by harrylin That looks bizarre because it's wrong. Indeed the redshift of accelerating rockets is almost purely (semi)classical Doppler shift, insofar as there is no or negligible difference in acceleration. I see no reason for the claim that in reality it's not so, as I also stressed in post #65:
Are you talking about the case with no length contraction or the case with length contraction as used in in Rindler coordinates or Born rigid acceleration? In the latter case there is most definitely significant difference in proper acceleration as measured on board the rockets and a significant difference in coordinate acceleration as measured simultaneously in the launch frame
 Quote by harrylin It depends on your choice of inertial reference system if you deem that the clock in the rear ages slower or not; according to the launch frame's POV they age equally.
They only age equally from the launch frame's POV if there is no length contraction. Even then, the rear rocket observers will see blue shift of clocks at the front of the rocket and any experiment they carry out will convince them that the rear clocks are tangibly running slower than the front clocks. In this thread we are mainly discussing the equivalence between measurements in an artificially accelerated system and a gravitational system, so we are more interested in what the accelerated observers measure. Also, the case where there is length contraction such that the accelerated observers consider themselves to be at constant distance from each other, is more relevant to a typical gravitational field such as that of the Earth. When we stand on top of a tower and look towards the base, we generally consider the height of the tower to remain constant.

P.S. Some reservations about the none length contracting case have occurred to me. I will try to analyse that in detail later.
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 Quote by yuiop Are you talking about the case with no length contraction or the case with length contraction as used in in Rindler coordinates or Born rigid acceleration? [..]
I thought that you were talking about the case with no length contraction as seen from the launch pad frame. Else a little correction is needed, as I also hinted at in my post #65 which I also cited again. I thought (and still think) that you were not talking about that small effect of length contraction when you made your claim about "really slower aging", as that redshift is very small compared to "the redshift" that you discussed. If I misunderstood you, please clarify.
 [...] any experiment they carry out will convince them that the rear clocks are tangibly running slower than the front clocks.[..]
Only if, as I pointed out, he is fooling himself into thinking that he his not accelerating. However, that would not be reasonable for someone in a rocket with firing rocket engines - as you also seemed to realise in your answer in post #73. For some reason that escapes me, you replaced "fooled"(=not real) by "real" (=true) between that post and post #86. Someone's instrument reading is not necessarily identical to "what really happens", nor does a smart rocket pilot accept everything at face value.
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 Quote by harrylin I thought that you were talking about the case with no length contraction as seen from the launch pad frame. Else a little correction is needed, as I also hinted at in my post #65 which I also cited again. I thought (and still think) that you were not talking about that small effect of length contraction when you made your claim about "really slower aging", as that redshift is very small compared to "the redshift" that you discussed. If I misunderstood you, please clarify.
I'm not sure exactly what you are claiming. The differential aging of someone in the front and rear of a rocket is real. We can make it operational as follows:
1. Take a pair of clocks to the rear of the rocket.
2. Set them to the same time, t=0.
3. Move one of the clocks to the front of the rocket.
4. Wait a year.
5. Move it back to the rear.
6. Compare the two clocks.

The prediction is that if we allow length contraction (Born rigid acceleration) then the moving clock will be ahead of the clock that was always in the rear by a factor of 1+gL/c2. So it's not simply some kind of illusion.

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