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Let's say a light clock consists of two parallel mirrors, some photons bouncing between the mirrors, and a spring that pulls the mirrors together with the same force that the photons push them apart.

Now we start accelerating the mirrors and the spring, in the direction parallel to the mirrors. Whenever a photon falls off from the end of the light clock, we put a similar photon at the front part of the light clock.

Now my question is: does the distance between the mirrors change?

(Actually I'm interested whether there are any changes in the forces in the spring)

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You're asking if there are any "changes in the forces in the spring". There are. If, in addition, you're interested in just what those changes are, it's easy to calculate from the answer I've given below. – Carl Brannen Feb 25 '12 at 3:16

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The way I understand your description, the motion of the light and the spring tension are perpendicular to the direction of acceleration. Consider the problem in the (instantaneous) rest frame of the mirrors and spring.

Let's use coordinates where x is the direction of the photon's motion (in the unmoving rest frame), and y is in the direction of acceleration (and motion) of the mirrors. The relevant equations are spring force $F = kx$, photon momentum in the mirror (x) direction = $P_x = \hbar k_x$, photon rate $1/T$ where $T$ is the time between photons, and the photon force = change in momentum per unit time = $2\hbar k_x / T$. The requirement for no change to the mirror is that $kx = 2\hbar k_x/T$.

Clearly the spring is just a spring in the moving rest frame and so there's no change to the spring force $kx$.

As for the photons, their momentum perpendicular to the direction of velocity (or acceleration) is unchanged, so there's no change to the momentum any single photon imparts to the mirror $2\hbar k_x$. Uh, note that since the photon reverses its direction, it imparts twice its momentum to the mirror. Another way of saying the same thing is to note that stationary and moving observers agree on the number of photon wavelengths $\lambda_x$ between the two mirrors. They also agree on the distance between the two mirrors. Therefore they agree on the wave number $k_x = 2\pi/\lambda_x$ for the photon's momentum in the x direction.

So what's left is the rate at which the photons impact the mirror $1/T$. This rate does not depend on the acceleration (relativity problems rarely do). Instead it depends on the velocity of the mirror.

As the mirror velocity increases, the distance traveled by the photons in the moving frame increases. Thus the time between photons increases and the photon force decreases.

Of course in the unmoving rest frame the photon rate is unchanged, it is only in the moving frame that the photon rate changes. This effect is called time dilation. Consequently, the non moving observer, on noting the force on the mirror, must conclude that the spring constant $k$ has changed due to the motion of the spring. For a discussion of this interesting effect, see: http://www.mathpages.com/home/kmath068/kmath068.htm

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All quantities that have time in the units should be expected to change with reference frame anyway because of the time dilation. I'm more curious exactly how the momentum changes. A photon can have unlimited momentum so it's not a problem to just add a y-axis component to the momentum until the cows come home. They are then blue-shifted by accelerating sideways. Is this the correct view? – AlanSE Feb 25 '12 at 4:24
@Carl Thank you. This could also be derived by considering a perpendicular pulling directions rope pulling contest of identical twins in space ship moving 0.8 c. Twin pulling perpendicular to the velocity pulls with 50% force a 200% distance compared to the other twin. – kartsa Feb 26 '12 at 2:17
Zass; that seems correct to me. The photons are blue shifted, but they're no longer traveling in the direction that makes all their momentum switch at the mirror. Only the perpendicular momentum changes direction at the mirror and this momentum is not changed by the acceleration. By the way, I think that the same result would have been obtained if we'd analyzed the problem with the mirrors aligned perpendicular to the direction of motion (so the photon would be alternately blue and red shifted). – Carl Brannen Feb 27 '12 at 0:07
@kartsa; That's a nice way of looking at it. Seems like the kind of problem to expect to see on a qualifying exam or the physics GRE. – Carl Brannen Feb 27 '12 at 0:08

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