# Time kept by accelerated extended (non-ideal) "light-clock"

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#### TubbaBlubba

1. A light-clock (a photon travelling between two mirrors) has proper length l and moves longitudinally through an inertial frame with proper acceleration a (ignore any variation of a along the rod). By looking at the time it takes the photon to make one to-and-fro bounce in the instantaneous rest-frame, show that the frequency and proper frequency are related, in lowest approximation, by F = fγ-1(1 + al/2c2). (so the deviation from idealness is proportional to a and l).

2. Homework Equations
x^2 - c^2 t^2 = c^4 / a^2 = X^2

3. The Attempt at a Solution
So the proper frequency is proportional to 1 / l, meaning that under uniform motion the factor would simply be 1/γ. I'm really quite boggled at it - I've tried a lot of approaches, but I can't really figure out where the al/c^2 factor comes from (I just know it as a measure of the "magnitude" of relativistic acceleration), or how to get both that parenthesis (a taylor expansion of... what?) AND a gamma in there. I get the basic idea, it's like Bell's Spaceships - if you begin accelerating uniformly, the front end will accelerate more quickly than the back end. However, if you "ignore the variation of a" then the factor of stretching is simply γ, as for an ideal clock, irrespective of the acceleration. The key, therefore, must be the fact that it takes some finite end for the photon to make the "round trip", and that we have to account for both the back end moving toward it and the front end moving away from it, over some small time interval, but I can't figure out a good way to work that out. The taylor expansion within the parentheses looks like it could be cosh(√(al) / c), but again, not sure where that would come from. The physical interpretation of the dimensionless quantity al/c^2 is actually not totally clear to me.

I realize my inadequacy with regard to hyperbolic geometry plays a part in this, and that's kind of why I've been obsessing over this problem for more than a day with little progress. The text is Wolfgang Rindler's 2006 "Realtivity: Special and General", for what it's worth.

I'm thankful for some hints on how to approach this.

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#### TSny

Homework Helper
Gold Member
To work out the frequency of the light clock in the instantaneous rest frame, you do not need to use relativity. It's just a kinematics problem where you need to take into account the effect of the acceleration of the clock relative to the instantaneous rest frame in working out the time for the photon to go out and back. You are only asked to determine the result to first order in the dimensionless quantity $\frac{aL}{c^2}$. This quantity will show up naturally in your analysis.

• TubbaBlubba
T

#### TubbaBlubba

To work out the frequency of the light clock in the instantaneous rest frame, you do not need to use relativity. It's just a kinematics problem where you need to take into account the effect of the acceleration of the clock relative to the instantaneous rest frame in working out the time for the photon to go out and back. You are only asked to determine the result to first order in the dimensionless quantity $\frac{aL}{c^2}$. This quantity will show up naturally in your analysis.
I've worked similar problems kinematically without acceleration, but somehow this just doesn't click for me. I'll have to look again tomorrow, I probably confused myself by digging too deep into hyperbolic geometry, rapidity, etc.

I guess it's something like, while the photon travels length l that takes l/c seconds, so in that time the clock has time to accelerate, its velocity increases by al/c so it has time to move about 1/2 × al^2/c^2 meters... that's a start.

#### TSny

Homework Helper
Gold Member
You could try to get an equation for the position of the photon as a function of time on the outward journey and also an equation for the position of the far mirror as a function of time.

• TubbaBlubba
T

#### TubbaBlubba

You could try to get an equation for the position of the photon as a function of time on the outward journey and also an equation for the position of the far mirror as a function of time.
OK, after some sleep, let's give this another go. If the mirrors travel inertially the photon has length 2l to travel, giving frequency c/2l = f. In the instantaneously inertial frame, both the photon and the "back" mirror start at 0, the "front" mirror starts at l. To a first-order approximation, the photon has to travel l(1 + al/2c^2) to reach the front mirror. Now, we assume the mirror is small enough to accelerate uniformly, so the distance between the mirrors is still l. Now the back mirror already moves with speed al/c and it accelerates to about 2al/c, so it travels about 3al^2/2c^2, shortening the path by that much. So the photon has to make the round trip l(1 + al/2c^2 + 1 - 3al/2c^2) = l(2 - al/c^2) or according to our comoving frame f' = c/(l(2 - al/2c^2)) = f(1 - al/2c^2)^-1. We can Taylor expand that (1/(1 - x) ≈ 1 + x about x = 0) to f(1 + al/2c^2). Finally, since the frame is inertially comoving, we get a γ^-1 from length contraction of the measuring sticks used to measure the length of the clock.

So finally: F = f'γ^-1 = f(1 + al/2c^2)γ^-1 which was to be demonstrated.

#### TSny

Homework Helper
Gold Member
If the mirrors travel inertially the photon has length 2l to travel, giving frequency c/2l = f. In the instantaneously inertial frame, both the photon and the "back" mirror start at 0, the "front" mirror starts at l. To a first-order approximation, the photon has to travel l(1 + al/2c^2) to reach the front mirror. Now, we assume the mirror is small enough to accelerate uniformly, so the distance between the mirrors is still l. Now the back mirror already moves with speed al/c and it accelerates to about 2al/c, so it travels about 3al^2/2c^2, shortening the path by that much. So the photon has to make the round trip l(1 + al/2c^2 + 1 - 3al/2c^2) = l(2 - al/c^2) or according to our comoving frame f' = c/(l(2 - al/2c^2)) = f(1 - al/2c^2)^-1. We can Taylor expand that (1/(1 - x) ≈ 1 + x about x = 0) to f(1 + al/2c^2).
This is all very nicely done. I agree with your result for the frequency in the instantaneous rest frame. Edit: I think there is a typo concerning a factor of 2 that I colored red, but your final answer looks good.
Finally, since the frame is inertially comoving, we get a γ^-1 from length contraction of the measuring sticks used to measure the length of the clock.

So finally: F = f'γ^-1 = f(1 + al/2c^2)γ^-1 which was to be demonstrated.
Here I'm having some difficulty. I don't follow the length contraction argument for switching to the moving frame. I don't see why you can assume that you can get the frequency in the moving frame by just throwing in a length contraction factor.

To me, it is more natural to think in terms of time dilation. In the frame in which the clock has an instantaneous speed v, the time between ticks of the clock should be dilated relative to the time between ticks in the instantaneous rest frame. At first, I thought the gamma factor was just coming from the time dilation formula Δt = γΔto where Δt is the period of the clock in the frame in which the clock is moving at speed v and Δto is the time in the instantaneous rest frame of the clock.

However, I ran into a snag. The time dilation formula Δt = γΔto requires that Δto be a proper time interval. But the time interval Δto measured in the instantaneous rest frame is not a proper time interval. The event of the photon starting out from the first mirror and the event of the returning of the photon to the first mirror do not take place at the same point in the instantaneous rest frame. This is because the first mirror accelerates to a new position by the time the photon returns.

If I use the Lorentz transformation equations to obtain the time interval between the two events in the non-instantaneous rest frame, I don't get Rindler's result. I find that the frequency of the clock in the frame in which the clock is instantaneously moving at velocity $v$ is given by $F = (f/\gamma)\left( 1 + \frac{aL}{2c^2} (1-2v/c) \right)$. Here, $v$ is positive if the velocity of the clock is in the same direction as the acceleration of the clock.

So, I do not get Rindler's result. I could be overlooking something.

[EDIT 02/27/2018: I believe both the result I got above and Rindler's result are in agreement. In my result, the gamma factor and $v$ correspond to the speed of the rocket relative to the observer at the instant the light leaves the first mirror. I get Rindler's result if I express the answer in terms of the gamma factor corresponding to the speed of the rocket relative to the observer at the instant the light reflects off of the second mirror. Choosing this instant of time seems more natural as it is a "midpoint" instant for the round trip of the light and it's nice to see that Rindler's expression is simpler in form than what I got. The whole calculation is easier if you choose the instantaneous rest frame to be the inertial frame in which the rocket is at rest at the instant the light reflects off of the second mirror. The two legs of the round trip of the light take the same time in this frame. Also the two events corresponding to the emission and return of the light at the first mirror take place at the same location in this frame of reference.]

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• TubbaBlubba
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#### TubbaBlubba

This is all very nicely done. I agree with your result for the frequency in the instantaneous rest frame. Edit: I think there is a typo concerning a factor of 2 that I colored red, but your final answer looks good.
Here I'm having some difficulty. I don't follow the length contraction argument for switching to the moving frame. I don't see why you can assume that you can get the frequency in the moving frame by just throwing in a length contraction factor.
I had to think for a while about that. Yes, of course, the rest frame will not consider the clock to be LONGER than the comoving frame does!

I think the problem is simply elementary enough that the difference between measured time and proper time and such can be ignored. But thanks for pointing that out!

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