Change in time due to accelerated light-clock

  • #1

Homework Statement


A light-clock (a photon travelling between two mirrors) has proper length l and moves longitudinally through an inertial frame with proper acceleration ##\alpha## (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 ##\nu = \nu_0 \gamma^{-1}(1 + \frac{\alpha l}{2c^2})##. (so the deviation from idealness is proportional to ##\alpha## and ##l##).[/B]


Homework Equations



Basic kinematics equations: i.e. ##x(t) = x_0 + vt + \frac{a t^2}{2}## and so on...

The Attempt at a Solution


[/B]
Note: This is a problem that someone already posted: https://www.physicsforums.com/threa...erated-extended-non-ideal-light-clock.893252/, however, I don't quite understand their solution.

I began by trying to figure out the time it takes the photon to go from the left mirror (@ ##x = 0##) and the right mirror (@ ##x = l##) in the instantaneous rest frame. In this frame, you use the basic kinematic equation giving us ##cT = l + \frac{\alpha T^2}{c^2}##. Solving for ##T## gives us ##\frac{c \pm c\sqrt{1 - \frac{2\alpha l}{c^2}}}{\alpha}##. We can do the same for the opposite direction, just solving the equation with ##-ct## instead and we get ##\frac{-c \pm c\sqrt{1 - \frac{2\alpha l}{c^2}}}{\alpha}##. Adding these two together we get a total period of ##\pm \frac{2c}{\alpha} \sqrt{1-\frac{2 \alpha l}{c^2}}##. We can invert this to find the frequency and expand ##\frac{1}{\sqrt{1-x}}## as ##1 + \frac{x}{2}## giving us an approximate frequency of ##\frac{\alpha}{2c}(1+ \frac{\alpha l}{c^2})##. My issue now is that I'm not quite sure what to do up to after here. I have a few ideas: namely getting the velocity in the rest frame and then setting up a similar kinematical equation and solving for the period. However, I don't think this problem is supposed to be that complicated.... Any tips would be greatly appreciated!
 

Answers and Replies

  • #2
TSny
Homework Helper
Gold Member
12,953
3,305
Looks like you calculated the light travel time from the left mirror to the right mirror in the frame of reference in which the light clock is at rest at the instant the light signal leaves the left mirror. Let ##A## denote this particular frame of reference and let the left-to-right travel time of the light in this frame be denoted by ##\Delta t^A_1##. You got ##\Delta t^A_1 = \frac{c \pm c\sqrt{1 - \frac{2\alpha l}{c^2}}}{\alpha}##. This looks right. But which sign are you going to choose in the ##\pm##? You can approximate ##\Delta t^A_1## to first order in the quantity ##z = \frac{\alpha l}{c^2}##.

At the instant the light signal reaches the right mirror, the clock is moving relative to reference frame ##A##. You will need to take this into account when finding the return light-travel time, ##\Delta t^A_2##, (from the right mirror to the left mirror) in the same reference frame ##A##.
 
  • #3
Looks like you calculated the light travel time from the left mirror to the right mirror in the frame of reference in which the light clock is at rest at the instant the light signal leaves the left mirror.
Is this not the right approach? I figured that since we are in the instantaneous rest frame, the light clock will be at rest when the light signal leaves the left mirror and thus are justified in using this kinematic equation.
But which sign are you going to choose in the ##\pm##?
So I don't choose a sign until I add it to the time it takes to come back, ##t_{2}^{A}##, which I now see that I got the wrong value for. What's tripping me up is the use of the 'instantaneous rest frame.' I thought that the idea behind the instantaneous rest frame is that you have inertial frames at each instant that travel with the accelerating object and so with respect to this object, the instantaneous rest frame has a velocity of ##0##. However, if I'm not understanding it correctly, what we're actually doing is choosing an initial instantaneous reference frame which we denote as ##A## and then calculating the time it takes for the photon to make it to the right mirror in this frame and this frame only. When the photon is on it's return trip we are still calculating things in this frame and thus must take into account the it's velocity at this time. Is this the correct understanding?
 
  • #4
TSny
Homework Helper
Gold Member
12,953
3,305
Is this not the right approach? I figured that since we are in the instantaneous rest frame, the light clock will be at rest when the light signal leaves the left mirror and thus are justified in using this kinematic equation.
Yes, this is fine. I just wanted to be clear on your choice of reference frame for calculating the left-to-right light-travel time.

What's tripping me up is the use of the 'instantaneous rest frame.' I thought that the idea behind the instantaneous rest frame is that you have inertial frames at each instant that travel with the accelerating object and so with respect to this object, the instantaneous rest frame has a velocity of ##0##. However, if I'm not understanding it correctly, what we're actually doing is choosing an initial instantaneous reference frame which we denote as ##A## and then calculating the time it takes for the photon to make it to the right mirror in this frame and this frame only. When the photon is on it's return trip we are still calculating things in this frame and thus must take into account the it's velocity at this time. Is this the correct understanding?
Ultimately, you want the time it takes the light to make one round trip between the mirrors as measured in the given inertial frame of reference through which the light clock is moving longitudinally (with relativistic speed, perhaps). Let ##B## denote this reference frame. For me, it's less confusing to work with only one "instantaneously co-moving" inertial frame of reference, such as frame ##A## which is co-moving with the clock at the particular instant when the light leaves the left mirror. (This frame is co-moving only at the instant when the light leaves the left mirror.) Suppose ##E1## is the event when the light leaves the left mirror and ##E3## is the event when the light arrives back at the left mirror. Then if you can find the space-time coordinates of these two events in frame ##A##, you can use these coordinates to find the time interval between the two events in frame ##B##.
 
  • #5
Ultimately, you want the time it takes the light to make one round trip between the mirrors as measured in the given inertial frame of reference through which the light clock is moving longitudinally (with relativistic speed, perhaps). Let ##B## denote this reference frame.
Isn't the light clock accelerating though? How can it be an inertial frame? I guess that's what I'm not understanding - between which two frames are we measuring the frequencies? I think what you're suggesting is to analyze the problem in the frame ##A## then transform to ##B##, but I don't see how ##B## is an inertial frame unless it is always moving with the light clock and in this case ##B## would be changing at every instant.
 
  • #6
TSny
Homework Helper
Gold Member
12,953
3,305
Isn't the light clock accelerating though? How can it be an inertial frame?
The light clock is accelerating. So, a frame of reference that is always at rest with respect to the clock is not an inertial frame.

I think the purpose of this problem is to investigate how the rate of a light-clock is affected by being accelerated longitudinally. It is often assumed in special relativity that you can have clocks for which acceleration doesn't affect the "inner workings" of the clock. These ideal clocks are often used when considering accelerating reference frames in special relativity. For example, such an ideal clock carried along by an accelerated observer will measure proper time along the world line of the accelerated observer. Rindler is asking you to show that a longitudinally accelerating light-clock is not ideal. That is, the frequency of the clock is affected by acceleration. The "proper frequency" of the clock is the frequency that it would have if it were not accelerating and if you were to measure the frequency in the rest frame of the (non-accelerating) clock. When the clock is accelerating and also zooming past you at high speed, the clock will have a frequency that differs from the proper frequency. Part of the difference is due to the velocity of the clock (the gamma factor in the result) and part of the difference is due to the effect of acceleration. The presence of the acceleration in the answer means that the light-clock is not acting like one of those "ideal" clocks (which is assumed to not be affected by acceleration).

I guess that's what I'm not understanding - between which two frames are we measuring the frequencies?
You want to deduce the frequency of the clock in an inertial frame of reference in which the clock is zooming past you and accelerating. This is frame ##B## that I mentioned in my previous post. You want to express this frequency in terms of the "proper frequency" mentioned above. To help do this, Rindler is suggesting using "the instantaneous rest frame". This wording is confusing to me. If he means a non-inertial frame that is always co-moving with the clock, then there would be no need for the word "instantaneous". If he means an instantaneously co-moving inertial frame, then he should not have said "the instantaneous rest frame" since there is a different instantaneously co-moving inertial frame at each instant of time.

I think what you're suggesting is to analyze the problem in the frame ##A## then transform to ##B##,
Yes. But if you see a different way that you think would be better, then by all means try it.

but I don't see how ##B## is an inertial frame unless it is always moving with the light clock and in this case ##B## would be changing at every instant.
Frame ##B## is a single inertial frame as mentioned above. It is not moving with the clock. We are interested in the frequency of the clock in frame ##B## at the time the clock has a speed relative to ##B## corresponding to the gamma factor that occurs in the answer.
 
  • #7
TSny
Homework Helper
Gold Member
12,953
3,305
Things will work out more easily if you choose the instantaneous rest frame to be the frame that is at rest with respect to the rocket at the instant the light signal reflects off of the mirror on the right. Thus, in this frame the rocket is instantaneously at rest at an instant of time halfway between the beginning and end of the round trip of the light. Your work that you showed in your first post looks like it might apply to this frame.

You should then get Rindler's answer where the gamma factor ##\gamma## corresponds to the velocity of the rocket relative to frame B at the instant the light reflects off of the mirror on the right.

When I worked out the problem using frame A (i.e., the frame instantaneously at rest with respect to the rocket when the light leaves the mirror on the left) I got the answer given in the thread that you linked to in the first post. That is, I got ##\nu = \large \frac{\nu_0}{\gamma '} ## ##\left( 1 + \frac{aL}{2c^2} (1-2v'/c) \right)##. Here the gamma factor ##\gamma '## and ##v '## correspond to the velocity of the rocket with respect to frame B at the instant the light leaves the left mirror. I thought I was getting an answer that disagrees with Rindler's answer. But it turns out that it is equivalent to Rindler's answer (to first order in ##\frac{aL}{2c^2}##) if you take into account the differences in the gamma factors ##\gamma## and ##\gamma '##.
 

Related Threads on Change in time due to accelerated light-clock

Replies
6
Views
854
  • Last Post
Replies
1
Views
4K
  • Last Post
Replies
10
Views
815
  • Last Post
Replies
7
Views
2K
Replies
1
Views
677
Replies
0
Views
3K
Replies
2
Views
2K
  • Last Post
Replies
13
Views
723
Replies
4
Views
8K
Replies
10
Views
5K
Top