A question about the equivalence principle.

[Nick20]

I had a physics test at school recently. One of the questions was based on the equivalence principle, going something like this: Two clocks in a spaceship that is accelerating. One at the bottom and one at the top of the space ship. Now think that the spaceship is so far away from any object in space, that it is not affected by any gravitational force.

It is my understanding that according to the equivalence principle, one can not be able to do any test that suggests that you are no longer in a gravitational field, but in an accelerated system. And according to the theory of relativity, the clock furthest down in the gravitational field will go slower than any clock higher up. Yet it makes no sense to me that the same rules would apply in an accelerated system. The correct answer supposedly is that the clock at the bottom of the ship will slow down. What am I missing? Would the clock at the bottom of the ship really slow down? Making the rules of time dilation apply in an accelerated system, the same way it does in an gravitational field?

 

[A.T.]

Would the clock at the bottom of the ship really slow down?

Yes. Time flows non-uniformly in non-inertial frames, like the rest frame of the ship.

 

[Dale]

The correct answer supposedly is that the clock at the bottom of the ship will slow down. What am I missing? Would the clock at the bottom of the ship really slow down? Making the rules of time dilation apply in an accelerated system, the same way it does in an gravitational field?

Yes. In fact, the fact that the clock "lower" in the accelerating system slows down can be derived purely using SR without any reference to the equivalence principle or the Einstein field equations.

The elegant way to do it is to determine the transform between the inertial and accelerating systems and then evaluate the spacetime interval for a stationary object.

The brute force way to do it is to calculate the Doppler shift between the upper and lower clocks.

 

[yuiop]

Consider the point of view of an inertial (non accelerating) observer who watches the rocket accelerating. If the length of the rocket appears constant to the accelerating observers on board the rocket, then the rocket appears to be length contracting according to the inertial observer. This means the back end of the rocket is accelerating faster than the front end and the back end of the rocket is moving faster than the front end at any given time according to that inertial observer, so the back end experiences more time dilation relative to the front end.

 

[stevendaryl]

It is my understanding that according to the equivalence principle, one can not be able to do any test that suggests that you are no longer in a gravitational field, but in an accelerated system. And according to the theory of relativity, the clock furthest down in the gravitational field will go slower than any clock higher up. Yet it makes no sense to me that the same rules would apply in an accelerated system. The correct answer supposedly is that the clock at the bottom of the ship will slow down. What am I missing? Would the clock at the bottom of the ship really slow down? Making the rules of time dilation apply in an accelerated system, the same way it does in an gravitational field?

The fact that the clock in the front of a rocket will tick faster than a clock in the rear of a rocket follows from SR alone. Here's a heuristic argument: From the point of view of the launch frame of the rocket, as the rocket accelerates, it becomes length-contracted. That means that the distance between the front of the rocket and the rear of the rocket decreases. That means that the rear of the rocket is traveling slightly faster than the front of the rocket. That means that time dilation affects the rear of the rocket more than the front of the rocket. That means that the rear clock runs slower than the front clock.

 

[stevendaryl]

Yes. In fact, the fact that the clock "lower" in the accelerating system slows down can be derived purely using SR without any reference to the equivalence principle or the Einstein field equations.

The elegant way to do it is to determine the transform between the inertial and accelerating systems and then evaluate the spacetime interval for a stationary object.

The brute force way to do it is to calculate the Doppler shift between the upper and lower clocks.

The Doppler shift derivation is very nice, but there is something a little puzzling about it, which is this:

We can compute the discrepancy between the clock rates at the front and the rear of the rocket using SR time dilation and length contraction. But the Doppler shift derivation works whether or not the rocket undergoes length-contraction: Imagine, instead of a single rocket with two clocks, we have two identical rockets, one above the other. They launch identically, and the distance between the rockets remains constant (as measured in
the launch frame). Then even though the distance between the rockets remains constant, the Doppler shift explanation still holds: if the rear rocket sends signals at rate one per second up toward the front, the front will receive them at a rate slower than one per second.

I'm just leaving this as a puzzle---I'm not saying the Doppler shift explanation is wrong, but it seems weird that it works whether or not there is length contraction, while the other explanation relies only on length contraction (and differential time dilation).

 

[A.T.]

We can compute the discrepancy between the clock rates at the front and the rear of the rocket using SR time dilation and length contraction. But the Doppler shift derivation works whether or not the rocket undergoes length-contraction: Imagine, instead of a single rocket with two clocks, we have two identical rockets, one above the other. They launch identically, and the distance between the rockets remains constant (as measured in
the launch frame). Then even though the distance between the rockets remains constant, the Doppler shift explanation still holds: if the rear rocket sends signals at rate one per second up toward the front, the front will receive them at a rate slower than one per second.

But the frequency shift will be different than for a single rocket with constant proper length, right?

 

[stevendaryl]

But the frequency shift will be different than for a single rocket with constant proper length, right?

Not immediately; the shifts will start out the same as in the case of a single rocket.

 

[jartsa]

Consider the point of view of an inertial (non accelerating) observer who watches the rocket accelerating. If the length of the rocket appears constant to the accelerating observers on board the rocket, then the rocket appears to be length contracting according to the inertial observer. This means the back end of the rocket is accelerating faster than the front end and the back end of the rocket is moving faster than the front end at any given time according to that inertial observer, so the back end experiences more time dilation relative to the front end.

I have always thought that when somebody carries the clock from the back end of the rocket to the front end of the rocket, that is when the clock experiences velocity time dilation.

But now I see that if I use the non-contracted lenght, then the time dilation is too large.
If I use the contracted length, then the time dilation is too small.

 

[yuiop]

But the Doppler shift derivation works whether or not the rocket undergoes length-contraction: Imagine, instead of a single rocket with two clocks, we have two identical rockets, one above the other. They launch identically, and the distance between the rockets remains constant (as measured in the launch frame). Then even though the distance between the rockets remains constant, the Doppler shift explanation still holds: if the rear rocket sends signals at rate one per second up toward the front, the front will receive them at a rate slower than one per second.

I'm just leaving this as a puzzle---I'm not saying the Doppler shift explanation is wrong, but it seems weird that it works whether or not there is length contraction ...

This works because the velocity of the rocket at the rear at the time the signal is emitted, is lower than the velocity of the rocket at the front at the time the signal is received. This is actually a better explanation to the OP than the length contraction explanation, that largely ignores the relativity of simultaneity.

 

[m4r35n357]

This works because the velocity of the rocket at the rear at the time the signal is emitted, is lower than the velocity of the rocket at the front at the time the signal is received. This is actually a better explanation to the OP than the length contraction explanation, that largely ignores the relativity of simultaneity.

Particularly since there should be no length contraction in the rocket frame, or have I missed the point badly here?

 

[yuiop]

Particularly since there should be no length contraction in the rocket frame, or have I missed the point badly here?

I think you are along the right tracks. If there is no change of length in the rocket frame then the red shift observed in the rocket frame can only be explained by time dilation, because in that reference frame the ends of the rocket are stationary with respect to each other so there is no classical Doppler shift.

However, in the particular example we gave where the rocket is not length contracting in the inertial launch reference frame, the rocket will actually be length expanding in the rocket frame, but that is probably an unnecessary distraction and there will still be time dilation red shift over and above the classical Doppler shift.

P.S. I think you are also right to have concerns about the length contraction explanation (when no length contraction is measured in the rocket frame undergoing Born rigid acceleration) so that makes the explanation a bit weak. It does explain how the inertial observer in the launch frame concludes that the clocks at the back and front are running at different rates, and if we allow for the change in simultaneity when transforming to the rocket reference frame the conclusion still holds, but the explanation becomes more convoluted then while the Doppler explanation seems more direct. However, we still have to refer to the inertial reference frame to observe the different velocities of the back and front ends of the rocket, even for the Doppler explanation.

 

[Dale]

The Doppler shift derivation is very nice, but there is something a little puzzling about it, which is this:

We can compute the discrepancy between the clock rates at the front and the rear of the rocket using SR time dilation and length contraction. But the Doppler shift derivation works whether or not the rocket undergoes length-contraction: Imagine, instead of a single rocket with two clocks, we have two identical rockets, one above the other. They launch identically, and the distance between the rockets remains constant (as measured in
the launch frame). Then even though the distance between the rockets remains constant, the Doppler shift explanation still holds: if the rear rocket sends signals at rate one per second up toward the front, the front will receive them at a rate slower than one per second.

I'm just leaving this as a puzzle---I'm not saying the Doppler shift explanation is wrong, but it seems weird that it works whether or not there is length contraction, while the other explanation relies only on length contraction (and differential time dilation).

I have actually never seen the length contraction argument worked out quantitatively, and I suspect that it will give the wrong value for exactly the reason that you mention. There is time dilation between two clocks that are undergoing identical accelerations in some inertial frame so that the time dilation is the same in that frame.

 

[yuiop]

I have actually never seen the length contraction argument worked out quantitatively, and I suspect that it will give the wrong value for exactly the reason that you mention. There is time dilation between two clocks that are undergoing identical accelerations in some inertial frame so that the time dilation is the same in that frame.

The OP was asking how to demonstrate that the clock at the back of the ship runs slower than the clock at the front of the ship. Stevendaryl in his explanation that considers the case that has no length contraction in the inertial launch frame stated "Then even though the distance between the rockets remains constant, the Doppler shift explanation still holds: if the rear rocket sends signals at rate one per second up toward the front, the front will receive them at a rate slower than one per second." He seems to implying that we can conclude that the rear clock runs slower than the front clock from this observation. However, I think the reverse is true. If the front rocket sends signals at the rate of once per second, the rear will receive them at a rate slower than once per second. I may be wrong, as this is just my qualitative hunch. If that is true, then the red shift between the observers at the front at rear is reciprocal and it cannot be stated that the rear clock runs slower than the front clock with any certainty. The observed mutual redshift would be accounted for by the fact that they both see the other as going away from them (i.e. the observed redshift is mainly due to classical Newtonian Doppler shift.) If that is the case, then it would be necessary to invoke length contraction to explain the difference in clock rates at the front and rear.

 

[Austin0]

The OP was asking how to demonstrate that the clock at the back of the ship runs slower than the clock at the front of the ship. Stevendaryl in his explanation that considers the case that has no length contraction in the inertial launch frame stated "Then even though the distance between the rockets remains constant, the Doppler shift explanation still holds: if the rear rocket sends signals at rate one per second up toward the front, the front will receive them at a rate slower than one per second." He seems to implying that we can conclude that the rear clock runs slower than the front clock from this observation. However, I think the reverse is true. If the front rocket sends signals at the rate of once per second, the rear will receive them at a rate slower than once per second. I may be wrong, as this is just my qualitative hunch. If that is true, then the red shift between the observers at the front at rear is reciprocal and it cannot be stated that the rear clock runs slower than the front clock with any certainty. The observed mutual redshift would be accounted for by the fact that they both see the other as going away from them (i.e. the observed redshift is mainly due to classical Newtonian Doppler shift.) If that is the case, then it would be necessary to invoke length contraction to explain the difference in clock rates at the front and rear.

I think the signals received at the rear would be blue shifted. The acceleration between time of emission and reception would mean greater velocity at the reception point. In addition there would be greater time dilation by some non zero factor due to that increased velocity which would add blue shift. If you consider the ships separated by a great distance it is obvious there would be significant increase in the rear ships velocity relative to the ealrier velocity of emission.
It would seem to also apply in the other direction with the exception that in that case the difference in time dilation would cancel out part of the red shift due to the increased velocity
It also seems like there would be more red shift due to velocity measured in the front ship than blue shift due purely to velocity measured in the rear ship.
With the dilation factor working oppositely I guess it would take calculations for specific conditions to determine what the relationship would be.
Perhaps

 

[yuiop]

I think the signals received at the rear would be blue shifted. The acceleration between time of emission and reception would mean greater velocity at the reception point. In addition there would be greater time dilation by some non zero factor due to that increased velocity which would add blue shift. If you consider the ships separated by a great distance it is obvious there would be significant increase in the rear ships velocity relative to the ealrier velocity of emission.

I have had a closer look and it looks like I was a bit hasty and you (and stevendaryl) are right. The rear observer does observe blue shift from signals coming from the leading rocket, even if there is no length contraction. This can be seen from examining the attached graph "f2b" which plots the paths (blue) of two rockets with equal and constant proper acceleration as seen in an inertial reference frame. The horizontal green lines are lines of equal proper time on the accelerating rockets at one unit time intervals. On this graph the diagonal black lines are signals sent from the front rocket to the rear rocket at regular unit time intervals. One interesting aspect is the the blue shift seen by the rear rocket reduces asymptotically towards unity as time progresses and the proper separation of the rockets increases.

It would seem to also apply in the other direction with the exception that in that case the difference in time dilation would cancel out part of the red shift due to the increased velocity
It also seems like there would be more red shift due to velocity measured in the front ship than blue shift due purely to velocity measured in the rear ship.
With the dilation factor working oppositely I guess it would take calculations for specific conditions to determine what the relationship would be.
Perhaps

This is correct too. The leading rocket does observe red shift of signals coming from the rear rocket with or without length contraction. The attached chart "b2f" charts the no length contraction case where the rear rocket sends signals at regular unit time intervals to the front rocket. Again there is an interesting aspect, in that the red shift increases dramatically over time, which is the exact opposite of how the blue shift seen by the rear rocket evolves.

The final attached chart is the length contraction case which is associated with the Rindler Metric and Born Rigid acceleration. In this case the distance between rockets and the blue or red shift from neighbouring rockets, remains constant over time, in an accelerating rocket reference frame. The green curves are the "lines of equal proper time" for the rocket and plotted at 1/2 unit time intervals. For those that might be interested, these curves are plotted using these parametric equations:

$$x = r * \cosh (T/r)$$
$$t = r * sinh(T/r)$$
where r is the parametric variable and T is a constant (The proper time). r is also the nominal radius and is inversely proportional to the constant proper acceleration of a given rocket.

 

[Austin0]

Not immediately; the shifts will start out the same as in the case of a single rocket.

Hi WHy do you thnk there would be a difference between a single rocket and two rockets if the distance remained constant in the reference frame in both cases ?

 

[Austin0]

I have had a closer look and it looks like I was a bit hasty and you (and stevendaryl) are right. The rear observer does observe blue shift from signals coming from the leading rocket, even if there is no length contraction. This can be seen from examining the attached graph "f2b" which plots the paths (blue) of two rockets with equal and constant proper acceleration as seen in an inertial reference frame. The horizontal green lines are lines of equal proper time on the accelerating rockets at one unit time intervals. On this graph the diagonal black lines are signals sent from the front rocket to the rear rocket at regular unit time intervals. One interesting aspect is the the blue shift seen by the rear rocket reduces asymptotically towards unity as time progresses and the proper separation of the rockets increases.
This is correct too. The leading rocket does observe red shift of signals coming from the rear rocket with or without length contraction. The attached chart "b2f" charts the no length contraction case where the rear rocket sends signals at regular unit time intervals to the front rocket. Again there is an interesting aspect, in that the red shift increases dramatically over time, which is the exact opposite of how the blue shift seen by the rear rocket evolves.

The final attached chart is the length contraction case which is associated with the Rindler Metric and Born Rigid acceleration. In this case the distance between rockets and the blue or red shift from neighbouring rockets, remains constant over time, in an accelerating rocket reference frame. The green curves are the "lines of equal proper time" for the rocket and plotted at 1/2 unit time intervals. For those that might be interested, these curves are plotted using these parametric equations:

$$x = r * \cosh (T/r)$$
$$t = r * sinh(T/r)$$
where r is the parametric variable and T is a constant (The proper time). r is also the nominal radius and is inversely proportional to the constant proper acceleration of a given rocket.

Hi Thanks for doing the diagrams for us. Interesting.
Having thought a little more it appears that in any case the overall observed Doppler shift would not be a meaningful indication of the relative dilation, as a large part of the effect is purely classical doppler and even if you went through the business of separating out this component, you would be left with a pure dilation evaluation that was comparing the clock rates at two different intervals on the respective world lines and still have the problem of interpreting that into relative rates at the "same" time in the accelerating frame.
Regarding the Rindler clocks and the constant relative Doppler: I find this curious and would like to know more but maybe not in this thread. ;-)

 

[jartsa]

Let's say we have a non-length-contracting accelerating rocket.

If a clock is thown from the rear of the rocket to the front of the rocket, it will be time dilated during the flight an amount that makes the reading of the clock to be the same as if the clock would have been in a gravity well.

If this same clock is thrown back to the rear from the front, it will be time dilated less than everything else in the rocket during the flight, and this effect is larger than the effect of the first throw, and the reading of the clock will be the same as if the clock had visited upper regions of a gravity well.

 

[Austin0]

Let's say we have a non-length-contracting accelerating rocket.

If a clock is thown from the rear of the rocket to the front of the rocket, it will be time dilated during the flight an amount that makes the reading of the clock to be the same as if the clock would have been in a gravity well.

If this same clock is thrown back to the rear from the front, it will be time dilated less than everything else in the rocket during the flight, and this effect is larger than the effect of the first throw, and the reading of the clock will be the same as if the clock had visited upper regions of a gravity well.

Hi Take a look at yuiop's diagrams a couple of posts back. Compare the relative path lengths between forward and backward travel and see if you still think the front to back travel would have greater effect on elapsed time than the travel from back to front.

 

[jartsa]

Hi Take a look at yuiop's diagrams a couple of posts back. Compare the relative path lengths between forward and backward travel and see if you still think the front to back travel would have greater effect on elapsed time than the travel from back to front.

We did have an accelerating and a non-length-contracting rocket, so in this rocket distances stay the same, while everything is slowing down, because of time dilation ... so the travel bacwards takes a little bit more time, or a lot more time if there is a large rocket velocity difference.

Interestingly, if the clock is thrown at relativistic speed, this doesn't seem to work.

Well I'll check the diagrams now.

 

[stevendaryl]

Hi WHy do you thnk there would be a difference between a single rocket and two rockets if the distance remained constant in the reference frame in both cases ?

In the two rocket case, the distance between the two rockets remains constant (in the launch frame), while in the one rocket case, the distance from the bottom of the rocket to the top decreases with time (in the launch frame). But in the latter case, the rate of decrease starts off zero, so it makes no difference to the Doppler shift initially.

 

[Austin0]

In the two rocket case, the distance between the two rockets remains constant (in the launch frame), while in the one rocket case, the distance from the bottom of the rocket to the top decreases with time (in the launch frame). But in the latter case, the rate of decrease starts off zero, so it makes no difference to the Doppler shift initially.

Gotcha. I thought we were talking about the no contraction case for both instances as is sometimes done in the Bell or Born scenarios.

 

[Austin0]

Yes. In fact, the fact that the clock "lower" in the accelerating system slows down can be derived purely using SR without any reference to the equivalence principle or the Einstein field equations.

The elegant way to do it is to determine the transform between the inertial and accelerating systems and then evaluate the spacetime interval for a stationary object.

The brute force way to do it is to calculate the Doppler shift between the upper and lower clocks.

Hi I encountered this question in an old thread and I later tried a ballpark calculation like this:
c=300,000km/s
A rocket R with a rest length of 1 km in the launch frame . With a constant coordinate acceleration of 1000g= 10km/s² in LaunchF

Accelerating between 0.6 c and 0.7 c the displacement arising from contraction is 0.09km
in 3,000s Average v=10^-10 c
with an Additive avg. v =1.7316 e^-10 within the ship.

The gamma comes up NaN for the relative velocity in the ship frame (HypPhy calculater). back then I got a gamma of 1 +( 2.9484 x 10 ^-21 ) but I can no longer remember how I arrived at it.

It is 1.315903389919538 for the average gamma for the front of the ship and LF

It is 1.3159033900676484 for the average gamma for the back of the ship and LF

With a difference of 1.48110401x 10^-10 between the two in the LF

I tried to get a handle on the Rindler math to work out a comparison (actually you tried to help me with that) but never actually got a figure.

Looking at the Pound-Rebka results for 22.5 m and 1 g ,,giving a factor on the orderof 5

x 10^-15 it is hard to tell if 1,000m and 1,000g with a resulting difference

factor of 1.48110401 x 10^-10is equivalent or not? Or even close.

Any hints?

Is this aything along the lines of what you were talking about above??

Where did the original Rindler calculations come from, what was the basis?

Do you know of any treatment done through Doppler that has been worked out for review??

The brute force approach as you called it. It certainly seems messy in principle and hard to interpret but I would like to know the method of approach..


Thanks

 

[stevendaryl]

Okay, I did some calculations and I think I have a good grasp now on the clock discrepancies in the accelerated rocket. There are two effects going on:

1. Because of relativity of simultaneity, if two clocks are synchronized in the "launch" frame, then the forward clock will be farther ahead, as measured in the instantaneous rest frame.
2. The front clock is traveling slightly slower than the rear clock (as measured in any inertial frame), and so experiences slightly less time dilation.

Here's what's interesting: Early on, effect number 1 is the most important, and effect number 2 is negligible. Much later, when the rocket is traveling very close to the speed of light, effect number 2 is most important, and effect number 1 is negligible.

 

[Austin0]

Okay, I did some calculations and I think I have a good grasp now on the clock discrepancies in the accelerated rocket. There are two effects going on:

1. Because of relativity of simultaneity, if two clocks are synchronized in the "launch" frame, then the forward clock will be farther ahead, as measured in the instantaneous rest frame.
2. The front clock is traveling slightly slower than the rear clock (as measured in any inertial frame), and so experiences slightly less time dilation.

Here's what's interesting: Early on, effect number 1 is the most important, and effect number 2 is negligible. Much later, when the rocket is traveling very close to the speed of light, effect number 2 is most important, and effect number 1 is negligible.

If by 1) you mean relative to a MCIRF I think your right but that doesn't really address the question of the mechanism of why the clocks would have different rates, which relative velocity does. It also brings up the complication that the system in a sense doesn't have a single momentarily comoving frame
Just looking at synchronization alone let's assume the there is no differential dilation.
The ship starts out with a certain synch. If there is no resynchronization it will maintain that relationship throughout the course , right?
So initially there will be small discrepancy between it and succeeding MCIRF's But as velocity increases that discrepancy will also increase because the relationship of the ship to the MCIRF is the same as the relationship of that MCIRF to the initial frame. So adding in the time dilation acts to increase that discrepancy , the front clock keeps running even further ahead of the rear and the front of the MCIRF
So to figure out dilation you would have to eliminate the ever increasing effect of desynchronization if you are going to use a MCIRF as a reference. Or so it seems to me.

 

[stevendaryl]

If by 1) you mean relative to a MCIRF I think your right but that doesn't really address the question of the mechanism of why the clocks would have different rates, which relative velocity does.

Well, consider the events
e_1: the rear clock shows time 12:00
e_2: the front clock shows time 12:00

If there is no length contraction, then e_1 and e_2 are simultaneous in the launch frame. But in the frame in which the rocket is momentarily at rest, e_2 happens before e_1. So in that frame, the front clock runs ahead of the rear clock.

If there is length contraction, then it means that even in the launch frame, event e_2 happens before e_1.

 

[stevendaryl]

The ship starts out with a certain synch. If there is no resynchronization it will maintain that relationship throughout the course , right?
So initially there will be small discrepancy between it and succeeding MCIRF's But as velocity increases that discrepancy will also increase because the relationship of the ship to the MCIRF is the same as the relationship of that MCIRF to the initial frame. So adding in the time dilation acts to increase that discrepancy , the front clock keeps running even further ahead of the rear and the front of the MCIRF
So to figure out dilation you would have to eliminate the ever increasing effect of desynchronization if you are going to use a MCIRF as a reference. Or so it seems to me.

I'm not exactly sure what you are saying. But if I did the calculation correctly, the exact result for the time τ on a clock aboard a Born-rigid rocket of constant proper acceleration is this:

τ = R/c arcsinh(ct/R)

where t is the time in the launch frame, and where R is a distance computed by:

R = c²/a + x

where x = the height of the clock above the rear of the rocket, and a = the acceleration felt by the rear of the rocket.

 

[Austin0]

Well, consider the events
e_1: the rear clock shows time 12:00
e_2: the front clock shows time 12:00

If there is no length contraction, then e_1 and e_2 are simultaneous in the launch frame. But in the frame in which the rocket is momentarily at rest, e_2 happens before e_1. So in that frame, the front clock runs ahead of the rear clock.

If there is length contraction, then it means that even in the launch frame, event e_2 happens before e_1.

I thought we were no longer talking about the case of non contraction.
And I agreed with your thought regarding the desynchronization relative to the momentarily comoving frame. I disagreed with your assumption that that effect would dimish over time.
IMO it would increase throughout the complete course of acceleration. Why do you think it would diminish?

 

[Austin0]

I'm not exactly sure what you are saying. But if I did the calculation correctly, the exact result for the time τ on a clock aboard a Born-rigid rocket of constant proper acceleration is this:

τ = R/c arcsinh(ct/R)

where t is the time in the launch frame, and where R is a distance computed by:

R = c²/a + x

where x = the height of the clock above the rear of the rocket, and a = the acceleration felt by the rear of the rocket.

I wish I could handle the math. I have essentially completely forgotten basic trig and never knew the hyperbolic functions.
So did you derive this from your calculations involving the momentarily comoving inertial frames? Aren't these the Rindler coordinate functions?

 

[jartsa]

Hi Take a look at yuiop's diagrams a couple of posts back. Compare the relative path lengths between forward and backward travel and see if you still think the front to back travel would have greater effect on elapsed time than the travel from back to front.

I have no idea what the diagrams mean.

I think we can say this:

In all cases when the clock did make extra ticks, because of the visit it made upstairs, in these cases the bacward travel had greater effect on elapsed time.

 

[jartsa]

Okay, I did some calculations and I think I have a good grasp now on the clock discrepancies in the accelerated rocket. There are two effects going on:

1. Because of relativity of simultaneity, if two clocks are synchronized in the "launch" frame, then the forward clock will be farther ahead, as measured in the instantaneous rest frame.
2. The front clock is traveling slightly slower than the rear clock (as measured in any inertial frame), and so experiences slightly less time dilation.

Here's what's interesting: Early on, effect number 1 is the most important, and effect number 2 is negligible. Much later, when the rocket is traveling very close to the speed of light, effect number 2 is most important, and effect number 1 is negligible.

OK. But effect 1 is not a time dilation effect. Clocks aren't in anyway effected by effect 1.

In our accelerating rocket case clocks are effected only by velocity time dilation.


I have tried to suggest a very simpe thing: After the rocket has accelerated and stopped accelerating, somebody carries a clock from the rear of the rocket to the front of the rocket, where he compares the clock to a clock that was at the front of the rocket all the time, and he observes that the clocks have ticked at different rates, which he may explain by gravitational time dilation, while an outside observer must explain it by velocity time dilation, which was caused by two things: 1: length contraction 2: carrying

 

[stevendaryl]

I thought we were no longer talking about the case of non contraction.

Well, I was trying to see the contributions of length contraction and relativity of simultaneity to the discrepancy between the front and rear clocks in the comoving frame. Obviously, if there is no length contraction, then it is all due to relativity of simultaneity.

And I agreed with your thought regarding the desynchronization relative to the momentarily comoving frame. I disagreed with your assumption that that effect would dimish over time.
IMO it would increase throughout the complete course of acceleration. Why do you think it would diminish?

Okay, let's pick a time t after the two clocks have been accelerating. Let the event e₁ be an event taking place at the rear clock at time t, and let e₂ be an event taking place at the front clock at time t. Let the coordinates of e₁ be (x₁,t) and let the coordinates of e₂ be (x₂,t), as measured in the launch frame. Let the corresponding coordinates in the momentary inertial rest frame be (x₁',t₁') and (x₂',t₂'). Letting δt' be the difference between t₁' and t₂', and letting δx be the difference between x₁ and x₂, the Lorentz transforms tell us that:

δt' = γ (δt - v/c² δx)

We chose the two events so that they are simultaneous in the launch frame, so δt = 0. So we have:

δt' = -γ v/c² δx

But δx is the distance between the front and the rear, as measured in the launch frame. So by length contraction, that is L/γ, where L is the length of the rocket in its comoving frame. So we have:

δt' = -γ v/c² L/γ
= - v/c² L

As time goes on, v→c, so this expression approaches

δt' = -L/c

So the desynchronization effect doesn't keep growing, it approaches a fixed constant (which happens to be the length of time required for light to travel from the rear to the front, in the comoving frame; hmm, not sure what the significance of that is).

In contrast, the discrepancy due to length contraction keeps getting bigger and bigger.

 

[Dale]

Do you know of any treatment done through Doppler that has been worked out for review??

The brute force approach as you called it. It certainly seems messy in principle and hard to interpret but I would like to know the method of approach.

I posted it a few years back in this post, but the LaTeX has changed and so now it is all math processing errors:
https://www.physicsforums.com/showthread.php?t=236880&page=4

I will try to re-work it and re-post it here.

 

[stevendaryl]

I wish I could handle the math. I have essentially completely forgotten basic trig and never knew the hyperbolic functions.
So did you derive this from your calculations involving the momentarily comoving inertial frames? Aren't these the Rindler coordinate functions?

Yeah, it's the Rindler coordinates. The Rindler coordinates X and T relate to the coordinates x and t of the launch frame via:

X = √(x² - (ct)²)

T = c/a arctanh(ct/x)
= c/a arcsinh(ct/X)

where a = the acceleration felt by the rear of the rocket, and where x is not measured from the rear of the rocket, but is instead measured so that the rear of rocket is at x=c²/a; that just makes the formulas come out nicer).

T is not the same as the time shown on clocks aboard the accelerated rocket; that time, τ, is related to T through

τ = T aX/c²
= X/c arcsinh(ct/X)