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This is an analysis of how a classic Einstein light clock behaves when accelerated orthogonal to its main axis. The results might be considered slightly controversial, so I am putting the details of the calculation here so that they can be checked.

Consider a light clock that has its long axis parallel to the y axis that is being accelerated in the x direction as per this diagram:

https://www.physicsforums.com/attachment.php?attachmentid=62152&stc=1&d=1380051964

Let the emitter end of the light clock start at x=y=t=0 and accelerate with constant proper acceleration g.

At times t1, t2, t3 the light clock is at x1, x2, x3 respectively. To find the round trip time of the photon (or the time of one tick of light clock) we need to find ##(\Delta t_2 + \Delta t_1) = (t_3-t_1) ##

Given t1 we can find t2 starting from the obvious Pythagorus relationship:

##L^2 + (\Delta x_1)^2 = c^2( \Delta t_1)^2##

##L^2 + (x_2-x_1)^2 = c^2 (t_2-t_1)^2##

Using the equations of relativistic accelerating motion given here the spatial coordinates can be expressed in terms of the time coordinates via the relationship (using units such that c=1), ##x =\sqrt{1/g^2 + t^2}## giving :

##L^2 + \left(\sqrt{1/g^2+t_2^2}-\sqrt{1/g^2+t_1^1}\right)^2 = (t_2-t_1)^2##

This can rearranged to give t2 in quadratic form:

##t_2^2 - t_2*t_1(2+L^2g^2) + (t_1^2- L^4g^2/4 - L^2) = 0##

and solved in the usual way to give:

##t_2 = t_1(1+L^2g^2/2) \pm \sqrt{t_1^2((1+L^2g^2/2)^2 -1) + L^2(1+L^2g^2/4)}##

There are two possible solutions. When the sign of the second term is positive t2>t1 and when it is negative t2<t1 so the positive term is chosen. (Note that by swapping t2 for t1 and vice versa we can find t1 using the solution with the negative second term, if we are given t2.)

The above expression is generic and we can find t3 by substituting t3 for t2 and t2 for t1 in the above expression:

##t_3 = t_2(1+L^2g^2/2) + \sqrt{t_2^2((1+L^2g^2/2)^2 -1) + L^2(1+L^2g^2/4)}##

The above calculations show that given t1 we can find t2 and t3 and therefore the round trip time (t3-t1). We can find t1 given any initial conditions such as initial position x1 or initial velocity v1 at the time the photon is emitted at the start of a tick, using the relativistic equations of accelerating motion, given in the link above.

The round trip time (t3-t1) is the coordinate time in the initial inertial frame. The time interval of the light clock (its tick rate) according to a comoving (accelerating) observer is given by:

##(t_3' - t_1') = (1/g)\left(\sinh^{-1}(g*t_3) - \sinh^{-1}(g*t_1)\right)##

Assuming I have done the calculations correctly, the proper frequency of the light clock does not keep time with a co-accelerating ideal clock. The equations are time dependent and therefore also dependent on the instantaneous velocity or position of the clock. This is a slightly curious result, and demonstrates that while the ruler length of a measurement orthogonal to the direction of motion is constant and not subject to length contraction, the radar length is constantly changing, (except for the special case when g/c=1). It turns out that when g/c<1 the frequency of the light clock increases with increasing velocity and conversely the frequency slows down with increasing velocity of the light clock when g/c>1.

Before anyone gets too excited and thinks the accelerating light clock indicates absolute motion, bear in mind that it only records its velocity relative to the time it started accelerating. If the light clock stops accelerating and starts again it restarts from v=0. Effectively it does nothing more than multiplying the output of a traditional accelerometer by the elapsed time to estimate the instantaneous velocity.

Anyway, I won't dwell too long on the implications of the results found here, just in case I have made a glaring mistake. Hopefully someone here will do the honours and check the calculations.

P.S. From some old discussions on PF, we know that a light clock that has constant proper acceleration parallel to its main axis (undergoing Born rigid motion), does maintain constant proper frequency. It follows that a Michelson-Morley type arrangement of two light clocks at right angles to each other will show an interference fringe shift under constant proper acceleration. There is no reason to think that a MMX type arrangement would show a fringe shift when stationary in a gravitational field due to the acceleration of gravity.

This is an interesting exception to the equivalence principle which is not due to tidal effects. How does a light clock distinguish between acceleration in flat space, which causes an actual change in location and velocity, and gravitational acceleration when there is no actual change in location and velocity relative to the gravitational field?

Consider a light clock that has its long axis parallel to the y axis that is being accelerated in the x direction as per this diagram:

https://www.physicsforums.com/attachment.php?attachmentid=62152&stc=1&d=1380051964

Let the emitter end of the light clock start at x=y=t=0 and accelerate with constant proper acceleration g.

At times t1, t2, t3 the light clock is at x1, x2, x3 respectively. To find the round trip time of the photon (or the time of one tick of light clock) we need to find ##(\Delta t_2 + \Delta t_1) = (t_3-t_1) ##

Given t1 we can find t2 starting from the obvious Pythagorus relationship:

##L^2 + (\Delta x_1)^2 = c^2( \Delta t_1)^2##

##L^2 + (x_2-x_1)^2 = c^2 (t_2-t_1)^2##

Using the equations of relativistic accelerating motion given here the spatial coordinates can be expressed in terms of the time coordinates via the relationship (using units such that c=1), ##x =\sqrt{1/g^2 + t^2}## giving :

##L^2 + \left(\sqrt{1/g^2+t_2^2}-\sqrt{1/g^2+t_1^1}\right)^2 = (t_2-t_1)^2##

This can rearranged to give t2 in quadratic form:

##t_2^2 - t_2*t_1(2+L^2g^2) + (t_1^2- L^4g^2/4 - L^2) = 0##

and solved in the usual way to give:

##t_2 = t_1(1+L^2g^2/2) \pm \sqrt{t_1^2((1+L^2g^2/2)^2 -1) + L^2(1+L^2g^2/4)}##

There are two possible solutions. When the sign of the second term is positive t2>t1 and when it is negative t2<t1 so the positive term is chosen. (Note that by swapping t2 for t1 and vice versa we can find t1 using the solution with the negative second term, if we are given t2.)

The above expression is generic and we can find t3 by substituting t3 for t2 and t2 for t1 in the above expression:

##t_3 = t_2(1+L^2g^2/2) + \sqrt{t_2^2((1+L^2g^2/2)^2 -1) + L^2(1+L^2g^2/4)}##

The above calculations show that given t1 we can find t2 and t3 and therefore the round trip time (t3-t1). We can find t1 given any initial conditions such as initial position x1 or initial velocity v1 at the time the photon is emitted at the start of a tick, using the relativistic equations of accelerating motion, given in the link above.

The round trip time (t3-t1) is the coordinate time in the initial inertial frame. The time interval of the light clock (its tick rate) according to a comoving (accelerating) observer is given by:

##(t_3' - t_1') = (1/g)\left(\sinh^{-1}(g*t_3) - \sinh^{-1}(g*t_1)\right)##

Assuming I have done the calculations correctly, the proper frequency of the light clock does not keep time with a co-accelerating ideal clock. The equations are time dependent and therefore also dependent on the instantaneous velocity or position of the clock. This is a slightly curious result, and demonstrates that while the ruler length of a measurement orthogonal to the direction of motion is constant and not subject to length contraction, the radar length is constantly changing, (except for the special case when g/c=1). It turns out that when g/c<1 the frequency of the light clock increases with increasing velocity and conversely the frequency slows down with increasing velocity of the light clock when g/c>1.

Before anyone gets too excited and thinks the accelerating light clock indicates absolute motion, bear in mind that it only records its velocity relative to the time it started accelerating. If the light clock stops accelerating and starts again it restarts from v=0. Effectively it does nothing more than multiplying the output of a traditional accelerometer by the elapsed time to estimate the instantaneous velocity.

Anyway, I won't dwell too long on the implications of the results found here, just in case I have made a glaring mistake. Hopefully someone here will do the honours and check the calculations.

P.S. From some old discussions on PF, we know that a light clock that has constant proper acceleration parallel to its main axis (undergoing Born rigid motion), does maintain constant proper frequency. It follows that a Michelson-Morley type arrangement of two light clocks at right angles to each other will show an interference fringe shift under constant proper acceleration. There is no reason to think that a MMX type arrangement would show a fringe shift when stationary in a gravitational field due to the acceleration of gravity.

This is an interesting exception to the equivalence principle which is not due to tidal effects. How does a light clock distinguish between acceleration in flat space, which causes an actual change in location and velocity, and gravitational acceleration when there is no actual change in location and velocity relative to the gravitational field?

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