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A question about the equivalence principle.

 Quote by stevendaryl The formula I gave was for the length of the rocket as viewed in the launch frame, under the assumption that the original length L is very small compared with the characteristic length c2/g, where g is the acceleration of the rear of the rocket. In this case, the length of the rocket in the comoving frame will be L, and you can transform back to get the length in the original frame as L/gamma. Of course, the various parts of the rocket are not at rest relative to each other if the rocket is accelerating, so the "comoving frame" is slightly ambiguous, but if the rocket is small enough compared with c2/g, that doesn't make much difference.
They are not at rest wrt each other in a given inertial reference frame, when the MCIRF of accelerating rocket has significant relative velocity to that given inertial reference frame, but momentarily in any MCIRF they are at rest wrt each other and in the accelerating reference frame of the rocket they are permanently at rest wrt each other. On board the rocket, any part of the rocket has a constant radar (and ruler) distance from any other part of the rocket so in that sense they are at rest wrt each other. The ambiguity only arises if you fail to specify which reference frame the measurements are made from.

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 Quote by stevendaryl I'm not sure that we are disagreeing. I said that long after launching, the velocity approaches c, and so the desynchronization approaches the constant L/c.
I am not sure what you are getting at here. Let's say the clocks are initially synchronised in the launch frame. After launch lets say that when 1 second elapses on the rear clock, 4 seconds elapses on the front clock in the MCIRF. It then follows that after 2 seconds elapses on the rear clock, 8 seconds will have elapsed on the front clock in the new MCIRF in which the rocket is momentarily at rest. The ratio between the time elapsed on the front clock versus the rear clock is always constant in any given MCIRF. If we stay with the measurements made in the initial inertial launch frame, then the desynchronisation is always increasing.

 Quote by stevendaryl I'm not sure whether we disagree, or not. In the comoving frame, the front clock always runs at a rate (R+L)/R faster than the rear clock, where L is the length of the rocket, and R is the characteristic length c2/g (where g is the acceleration of the rear).
This approximation is good if R is large and L is small, but becomes very inaccurate if R is small and L is large.

 Quote by yuiop They are not at rest wrt each other in a given inertial reference frame, when the MCIRF of accelerating rocket has significant relative velocity to that given inertial reference frame, but momentarily in any MCIRF they are at rest wrt each other
You are exactly right! Somehow, either I never did the calculation, or forgot it, but I just convinced myself of that this morning. The description of the rocket's position and velocity as a function of time in the launch frame I always knew was this:
1. xrear = √((ct)2 + R2)
2. vrear = c2 t/√((ct)2 + R2)
3. xfront = √((ct)2 + (R+L)2)
4. vfront = c2 t/√((ct)2 + (R+L)2)

(So at time t=0, both the front and the rear have speed 0.) But what's amazing is that equations 1-4 aren't just true in the launch frame, it's true in every inertial frame in which the rocket is momentarily at rest. That's pretty amazing; the rocket launches at time t=0 in frame F. At some time later, the rocket is at rest in some frame F' moving at speed v relative to F'. That time is t'=0, according to frame F', and equations 1-4 still hold, with x replaced by x' and t replaced by t'.

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Quote by stevendaryl
.... In the comoving frame, the front clock always runs at a rate (R+L)/R faster than the rear clock, where L is the length of the rocket, and R is the characteristic length c2/g (where g is the acceleration of the rear).
 Quote by yuiop This approximation is good if R is large and L is small, but becomes very inaccurate if R is small and L is large.
I may have to retract this last claim and you may be right here! The analysis I did with graphical software may have had limitations due to the accuracy of the software.

Here is a small "derivation" of the ratio of the clock rates being proportional to the ratio (R+L)/R.

The velocity of a clock on the rocket after proper time T is given as:

$$v = c - tanh(aT/c)$$

This can be rearranged to:

$$T = atanh(c-v)*c/a$$

Since the acceleration is equal to c^2/r this becomes:

$$T = atanh(c-v)*r/c$$

The ratio of the clock rates of two clocks at r1 and r2 respectively is then given by:

$$\frac{T_2}{T_1} = \frac{ atanh(c-v_2)*r_2/c}{atanh(c-v_1)*r_1/c}$$

In any given MCIRF, v is equal for all clocks at rest in that MCIRF, from the point of view of any other inertial reference frame, so we can assume v1 = v2 and can conclude:

$$\frac{T_2}{T_1} = \frac{r_2}{r_1} = \frac{r_1+L}{r_1}$$

This is how more time elapses (from launch) on the front clock relative to the rear clock as seen in any MCIRF and is also a measure of the constant redshift seen by an observer at r2 of a signal at r1.

 Quote by yuiop I am not sure what you are getting at here. Let's say the clocks are initially synchronised in the launch frame. After launch lets say that when 1 second elapses on the rear clock, 4 seconds elapses on the front clock in the MCIRF. It then follows that after 2 seconds elapses on the rear clock, 8 seconds will have elapsed on the front clock in the new MCIRF in which the rocket is momentarily at rest. The ratio between the time elapsed on the front clock versus the rear clock is always constant in any given MCIRF. If we stay with the measurements made in the initial inertial launch frame, then the desynchronisation is always increasing. This approximation is good if R is large and L is small, but becomes very inaccurate if R is small and L is large.
I misunderstood how you were using the word "desynchronization". I was meaning "relativity of simultaneity", the fact that clocks that are synchronized in one frame are not synchronized in another frame. You are using to mean the discrepancy between the two clocks (regardless of the cause).

Let me try to explain what I meant by the two causes for the discrepancy between the two clocks.

Let e1 be some event taking place at the rear clock. Let T1 be the time on that clock at that event.

Let e2 be some event taking place at the front clock that is simultaneous with e1, according to the launch frame. Let T2 be the time on that clock at that event.

Let e3 be some event taking place at the front clock that is simultaneous with e1, according to the instantaneous inertial reference frame of the rocket. Let T3 be the time on that clock at that event.

Let δT1 = T2 - T1
Let δT2 = T3 - T2

δT1 is completely due to length contraction; the front travels less than the rear, and so experiences less time dilation.

δT2 is an additional effect due to relativity of simultaneity.

In the instantaneous rest frame of the rocket, the discrepancy between the times on the two clocks is the sum of the two δs. What I've been saying is that soon after launch, δT2 is much bigger than δT1. Long after launch, δT1 eventually becomes much bigger than δT2. (The latter approaches a maximum value of L/c). In the accelerated reference frame of the rocket itself, the only relevant number is the sum of the two, which grows steadily at a constant rate.

Mentor
 Quote by Austin0 Could you perhaps be a little more exspansive in explaining what may be to you, obvious?
OK, so what I was talking about was using the metric to get an expression for time dilation as follows:

Time dilation in Minkowski spacetime:
$$c^2 d\tau^2=c^2 dt^2-dx^2-dy^2-dz^2$$$$\frac{d\tau^2}{dt^2}=1-\frac{1}{c^2}\left(\frac{dx^2}{dt^2}+\frac{dy^2}{dt^2}+\frac{dz^2}{dt^2 }\right)$$$$\frac{d\tau}{dt}=\sqrt{1-\frac{v^2}{c^2}}=1/\gamma$$

Time dilation in Schwarzschild spacetime:
$$c^2 d\tau^2 = \left(1-\frac{R}{r}\right) c^2 dt^2 - \left(1-\frac{R}{r}\right)^{-1} dr^2 - r^2 \left( d\theta^2 + sin^2\theta d\phi^2 \right)$$$$\frac{d\tau^2}{dt^2} = \left(1-\frac{R}{r}\right) - \frac{1}{c^2}\left(1-\frac{R}{r}\right)^{-1} \frac{dr^2}{dt^2} - \frac{r^2}{c^2} \left( \frac{d\theta^2}{dt^2} + sin^2\theta \frac{d\phi^2}{dt^2} \right)$$for an object at rest in the Schwarzschild coordinates all of the spatial derivatives are 0 leaving$$\frac{d\tau}{dt} = \sqrt{\left(1-\frac{R}{r}\right) }$$

Time dilation in Rindler coordinates:
$$c^2 d\tau^2 = \frac{g^2 x^2}{c^2}dt^2 - dx^2 - dy^2 - dz^2$$$$\frac{d\tau^2}{dt^2} = \frac{g^2 x^2}{c^4} -\frac{1}{c^2}\left( \frac{dx^2}{dt^2} + \frac{dy^2}{dt^2} + \frac{dz^2}{dt^2} \right)$$$$\frac{d\tau}{dt} = \sqrt{\frac{g^2 x^2}{c^4} -\frac{v^2}{c^2}}$$for an object at rest in the Rindler coordinates v=0 leaving$$\frac{d\tau}{dt} = \frac{g x}{c^2}$$

 Quote by yuiop I may have to retract this last claim and you may be right here! The analysis I did with graphical software may have had limitations due to the accuracy of the software. Here is a small "derivation" of the ratio of the clock rates being proportional to the ratio (R+L)/R. The velocity of a clock on the rocket after proper time T is given as: $$v = c - tanh(aT/c)$$
Are you sure about that? The formulas that I use are:
v = c tanh(aT/c)
not
v = c - tanh(aT/c)

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 Quote by yuiop ... so we can assume v1 = v2 and can conclude: $$\frac{T_2}{T_1} = \frac{r_2}{r_1} = \frac{r_1+L}{r_1}$$ This is how more time elapses (from launch) on the front clock relative to the rear clock as seen in any MCIRF and is also a measure of the constant redshift seen by an observer at r2 of a signal at r1.
I think it is interesting to expand on this relation T2/T1 = R2/R1 (which appears to be an exact result) by comparing it to the gravitational time dilation in the Schwarzschild metric.

First the time dilation ratio in the Born rigid acceleration case, as a function of constant proper acceleration is given by:

$$\frac{T_2}{T_1} = \frac{r_2}{r_1} = \frac{a_1}{a_2}$$

In the Schwarzschild metric the time dilation ratio of two stationary clocks in the field is given by:

$$\frac{T_2}{T_1} = \frac{\sqrt{1-2GM/(r_2c^2)}}{\sqrt{1-2GM/(r_1c^2)}}$$

and when we express this in terms of proper acceleration it becomes:

$$\frac{T_2}{T_1} = \sqrt{\frac{(c^2-2a_2r_2)}{(c^2-2a_1 r_1)}}$$

This appears to be very different to the Born rigid acceleration case with possibly a poor equivalence principle correlation, but maybe I am missing a gamma cubed factor in there somewhere between proper acceleration and coordinate acceleration. Any ideas?

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 Quote by stevendaryl Are you sure about that? The formulas that I use are: v = c tanh(aT/c) not v = c - tanh(aT/c)
Oops, I slipped up in quoting the given formula. Here is the derivation again using your correct expression. (The end result is still the same).

===============================

Here is a small "derivation" of the ratio of the clock rates being proportional to the ratio (R+L)/R.

The velocity of a clock on the rocket after proper time T is given as:

$$v = c * tanh(aT/c)$$

This can be rearranged to:

$$T = atanh(v/c)*c/a$$

Since the acceleration is equal to c^2/r this becomes:

$$T = atanh(v/c)*r/c$$

The ratio of the clock rates of two clocks at r1 and r2 respectively is then given by:

$$\frac{T_2}{T_1} = \frac{ atanh(v_2/c)*r_2/c}{atanh(v_1/c)*r_1/c}$$

In any given MCIRF, v is equal for all clocks at rest in that MCIRF, from the point of view of any other inertial reference frame, so we can assume v1 = v2 and can conclude:

$$\frac{T_2}{T_1} = \frac{r_2}{r_1} = \frac{r_1+L}{r_1}$$

This is how much more time elapses (from launch) on the front clock relative to the rear clock as seen in any MCIRF and is also a measure of the constant redshift seen by an observer at r2 of a signal at r1

===============================

Hopefully I got it right this time :)

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 Quote by DaleSpam Time dilation in Rindler coordinates: ... for an object at rest in the Rindler coordinates v=0 leaving$$\frac{d\tau}{dt} = \frac{g x}{c^2}$$
From the Wikipedia article on Rindler coordinates we get
 If all Rindler observers set their clocks to zero at T=0, then when defining a Rindler coordinate system we have a choice of which Rindler observer's proper time will be equal to the coordinate time t in Rindler coordinates, and this observer's proper acceleration defines the value of g above (for other Rindler observers at different distances from the Rindler horizon, the coordinate time will equal some constant multiple of their own proper time).[1] It is a common convention to define the Rindler coordinate system so that the Rindler observer whose proper time matches coordinate time is the one who has proper acceleration g=1, so that g can be eliminated from the equations
From this we can note that g is the same for all observers in Rindler coordinates, so we can rewrite the ratio based on Dalespam's equation:

$$\frac{d\tau_2}{dt_2} * \frac{dt_1}{d\tau_1} = \frac{g x_2}{g x_1}$$

as:

$$\frac{d\tau_2}{d\tau_1} = \frac{x_2}{x_1}$$

which agrees with equation quoted by stevendaryl and myself.

 Quote by stevendaryl No. There actually is not very much difference between those two, if the length of the rocket is sufficiently small. I was comparing the comoving inertial frame to the launch frame. I'm not sure that we are disagreeing. I said that long after launching, the velocity approaches c, and so the desynchronization approaches the constant L/c.
I agree , we are not disagreeing. I see now you were saying that the rate of desynchronization increase was diminishing , not that it didn't continue to increase. Yeah?

 Quote by stevendaryl I'm not sure whether we disagree, or not. In the comoving frame, the front clock always runs at a rate (R+L)/R faster than the rear clock, where L is the length of the rocket, and R is the characteristic length c2/g (where g is the acceleration of the rear). In the launch frame, the rates for both clocks start out the same, but the ratio gradually approaches the same value.
Agreed, that was what I was saying also.
 I haven't yet grokked the significance of this fact, but I recently discovered (I'm sure it's already well-known, but I didn't know it) an interesting characterization of constant proper acceleration. First, we pick a launch frame F, and set up an infinite collection of inertial frames, all related by the Lorentz transformation. For every possible value v, there is a corresponding frame Fv with coordinates xv, tv defined by: xv = (x - vt)/√(1-(v/c)2) tv = (t - vx/c2)/√(1-(v/c)2) where x and t are the coordinates for frame F. Now, suppose we have a rocket and we want (for whatever reason) to accelerate in such a way that: We are momentarily at rest in frame F at time t=0. For every possible value of v, we come to rest in frame Fv at time tv = 0. So we're traveling in such a way that the "local" time is always t=0. It's sort of like traveling west so that it takes you 1 hour to pass through a time zone, so you're always entering a time zone at 12:00. I can't think of any good reason to want to travel this way, but the interesting fact is that if you travel this way, you will be undergoing constant proper acceleration. This gives a more direct route to the equations for proper acceleration (the usual way requires knowing how acceleration transforms, and also knowing how to integrate hyperbolic trigonometric functions). If we want to come to rest in frame Fv at time tv = 0, then using the Lorentz transformations, we find: tv = (t - vx/c2)/√(1-(v/c)2) In order for tv to be zero, we need t - vx/c2 = 0 Now we can write this (multiplying by -1/2): d/dt (-1/2 t2 + 1/2 x2/c2) = 0 which has the solution x2 - (ct)2 = R2 where R = some constant of integration. This is hyperbolic motion.

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 Quote by stevendaryl ... I can't think of any good reason to want to travel this way, but the interesting fact is that if you travel this way, you will be undergoing constant proper acceleration.
and if all parts of an extended rocket follow this rule then you will be undergoing Born rigid acceleration.

It took me a while to realise that the equation you derived in post 39 ...
 Quote by stevendaryl 1-(v2/c)2 = (1-(v/c)2)(1+2(v/c)2gL/c2) (ignoring higher powers of L) √(1-(v2/c)2) = √(1-(v/c)2)(1+(v/c)2gL/c2) Letting T2 be the time on the front clock, and T be the time on the rear clock, we find: T2 = ∫√(1-(v/c)2)(1+(v/c)2gL/c2) dt In the limit v→c (long after launch), this becomes T2 = T(1 + gL/c2)
... which I assumed to be an approximation is actually the same as the exact result T2/T = r2/r, as follows:

T2/T = 1 + gL/c2

Using the notation r2 and r for x2 and x respectively when t=0 in any given MCIRF and noting that r2 = r+L and that g is defined as c^2/r, then:

T2/T = 1 + L/r

T2/T = (r + L)/r

T2/T = r2/r

This is surprising to me because you discarded the higher powers of L presumably introducing a small error and ended up with an exact result. I guess the approximation self cancelled out somewhere.

 Quote by 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 space ship that is accelerating. One at the bottom and one at the top of the space ship. Now think that the space ship 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?
Hi Nick, welcome to physicsforums.

Before the discussion here went off into very technical (and very interesting) details, you received two "yes" answers which possibly missed the very point that you were missing. Thus here's my slightly different answer.

No, not really:

- 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: the effect is then at the start (zero speed) explained as just due to Doppler effect. Next, when picking up speed, there is in addition the effect of time dilation of *both* clocks due to speed, in combination with length contraction which makes that the clocks do not exactly slow down the same; this is where things get tricky. But that is all purely special relativity, without any gravitational time dilation.

- Different from fake gravitational fields as caused by acceleration, real gravitational fields get weaker at greater distance from the source, and that gradient can be used for Earth sensors in a "free-falling" satellite that indicate where the earth is.
 Blog Entries: 6 Following on from #64, another perhaps interesting observation is that if we speed up the clock at the rear of a Born rigid accelerating rocket so that the radar length of the rocket as measured at the rear, agrees with the radar length of the rocket as measured at the front, then the clocks at the front and rear will remain permanently synchronised with each other and read the same as each other in any MCIRF.

 Quote by stevendaryl Let me try to explain what I meant by the two causes for the discrepancy between the two clocks. Let e1 be some event taking place at the rear clock. Let T1 be the time on that clock at that event. Let e2 be some event taking place at the front clock that is simultaneous with e1, according to the launch frame. Let T2 be the time on that clock at that event. Let e3 be some event taking place at the front clock that is simultaneous with e1, according to the instantaneous inertial reference frame of the rocket. Let T3 be the time on that clock at that event. Let δT1 = T2 - T1 Let δT2 = T3 - T2 δT1 is completely due to length contraction; the front travels less than the rear, and so experiences less time dilation. δT2 is an additional effect due to relativity of simultaneity. In the instantaneous rest frame of the rocket, the discrepancy between the times on the two clocks is the sum of the two δs. What I've been saying is that soon after launch, δT2 is much bigger than δT1. Long after launch, δT1 eventually becomes much bigger than δT2. (The latter approaches a maximum value of L/c). In the accelerated reference frame of the rocket itself, the only relevant number is the sum of the two, which grows steadily at a constant rate.
I agree with everything here.
The increase from simultaneity starts out at a maximum rate wrt coordinate time in the LF and diminishes at a factor of 1/ $\gamma$3
The discrepancy due to relative velocity increases over time by, I would guess, by a factor of $\gamma$3 ,as coordinate acceleration diminished, coordinate time for increased velocity would increase so the cumulative increase of proper time on the front clock due to the differential would comparably increase.
But what I don't understand is why the simultaneity relative to the MCIRF's would be relevant to the accelerating system?

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.

For vf,vb
$\gamma$b= 1/√ 1-v b2/c2

and $\gamma$ f = 1/√ 1-v f[SUP]2[/SU/c2 for any coordinate t in the LF ,,,so $\gamma$f/$\gamma$b should be it , no?

with no necessity of considering length contraction directly or including it in calculations. Likewise for simultaneity as you are only interested in the proper rates and delta t's of the two clocks which can be regarded as independent entities in this circumstance.
SO is this idea too out of the park??
Whats the hidden snag??? Of course I have no idea of how to do the math to make this work ;-)

 Quote by harrylin Hi Nick, welcome to physicsforums. Before the discussion here went off into very technical (and very interesting) details, you received two "yes" answers which possibly missed the very point that you were missing. Thus here's my slightly different answer. No, not really: - 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: the effect is then at the start (zero speed) explained as just due to Doppler effect. Next, when picking up speed, there is in addition the effect of time dilation of *both* clocks due to speed, in combination with length contraction which makes that the clocks do not exactly slow down the same; this is where things get tricky. But that is all purely special relativity, without any gravitational time dilation. - Different from fake gravitational fields as caused by acceleration, real gravitational fields get weaker at greater distance from the source, and that gradient can be used for Earth sensors in a "free-falling" satellite that indicate where the earth is.
Thank you. I really find physics quite interresting, and I have been doing some thinking of my own. I don't have much of an education though. I'll be 20 years old in october, and I've had 2 years of basic physics. Now this particular task puzzled me and really got me to wonder what was behind it all, because it made no sense to me that the "bottom" clock would experience time dilation as if it was in a gravitational field. So my answer to that task was that there will be no time dilation (concidering that the space ship is not yet travelling in a relativistic speed.) But appearently that was not correct. I do not believe that they thought too proper about it when they wrote that one.

Anyway I want to thank you all. I have gotten many interresting answers to my question. Reaching over 65 answers is a good first impression to a new forum

Nick

 Tags theory of relativity, time dilation