Mastering Physics: Moving Light Clock Ticks Slower

In this problem, we consider a light clock--a clock that ticks every time light makes a round trip between two mirrors separated by a distance . The key point is that we can deduce the rate at which this clock ticks, when it moves relative to us as well as when it is at rest, directly from the postulates of special relativity. The postulates of special relativity are as follows:

1. The laws of physics are the same in any coordinate system that moves at constant velocity (i.e., any inertial reference frame).
2. The speed of light is when measured with respect to any coordinate system moving at a constant velocity.

As shown in the figure , the light clock is based on the propagation time of light between mirrors spaced a distance apart (in the rest frame of the clock). As the light bounces back and forth between the mirrors, a small bit of light is allowed to escape through the partially silvered lower mirror. These transmitted pulses of light hit a detector, which therefore emits evenly spaced (in time) "ticks" each round-trip time of the pulse. (New pulses are injected in phase with the detected pulses to keep the clock going.)

In this problem, we show that if the clock is moving with speed relative to a reference frame S, then the ticks from the clock appear slowed when viewed from S. The key is to calculate explicitly the path length of the light pulses as seen in the frame S. From that you can find the length of time that an observer in S will measure between "ticks" of the relatively moving clock.
We consider the case where the clock is moving perpendicular to its long axis as shown. Considering a "thought experiment" ("Gedankenexperiment" in German) like this allows us to calculate the tick rate of the clock when viewed by the observer in S. Einstein used Gedanken experiments like this to clarify his thinking as he strove to remove inconsistencies in the basic formulations of physics.

Part A
What is the period between successive ticks of the clock in its rest frame? Express your answer in terms of variables given in the introduction. ANSWER:    =

We now imagine that the clock is moving with velocity relative to the observer in S. We know that the pulse of light will travel at speed in S. Therefore, to calculate the tick rate of the clock, we have to figure out the path of the light in S. The first step is to consider whether the length appears to change due to relative motion perpendicular to its long axis (i.e., motion perpendicular to ). The situation is shown in the figure, where the mirror is shown displaced a distance to where the light pulse hits it, and the source-mirror-detector system is shown where the source emits the pulse and also where it detects the pulse after its first round trip.

Part B

Imagine that pens were mounted beside the mirrors that are apart in the moving frame. How does the separation of the lines that they would draw on a piece of paper lying on the ground in S compare to , the separation measured in the rest frame of the mirrors? ANSWER: It is larger than . It is equal to . It is less than . Imagine that the mirror separation in S appears some factor larger than (i.e., the separation in S is ). Then, by the relativity postulate, the lines drawn on the paper would appear the same factor larger back in the frame of the clock (since the paper appears to move relative to the clock). The clock would then have to have length , which is ridiculous unless (i.e., the lines are separated by ).

Part C

We now want to consider the length of the path the light traverses in the clock from the point of view of an observer in reference frame S. One expression for the length of this path is simply: --that is, the length of the path is simply the distance that light travels in one half-tick. Use geometry to find another expression for the length of a one-way trip (e.g., from the source to the top mirror) according to this observer. Express your answer in terms of the time , , and . ANSWER:    =

Part D

What is the time between the ticks of the light clock as viewed from reference frame S? Express the time of these ticks in terms of (the ticks in the frame of the clock), the relative speed , and the speed of light . ANSWER:    = This result shows that the tick period of a moving clock ( ) appears longer than that of the clock in its rest frame ( ) by the factor . This phenomenon is called time dilation. This factor occurs frequently in expressions from special relativity, in particular from the Lorentz transformations. This applet shows the situation described in the problem. On the left is the frame of reference in which the light clock is stationary; on the right is the frame of reference in which the light clock is moving. You can adjust the relative speed of the frames by changing the value of . When thinking about time dilation, it is important to recognize that the time interval in the moving frame appears longer ("dilates") when viewed from the stationary frame only becuase the measurement compares a clock at a fixed location in the moving frame with an array of clocks in the stationary frame. If the measurement were made at a fixed location in the stationary frame, the time in the frame of the moving observer (measured by an array of clocks) would appear to go faster than a clock stationary in frame S. Of course, the observer in S would explain this by saying that the clocks in the moving frame were not properly synchronized. The main point is that comparison of a clock that's fixed in one frame must be made against an array of clocks synchronized by an observer in the other frame (and this array may not appear synchronized to an observer in the frame of the single clock).