Physics 150 Fall 2003
Solution to HW 6
1. Question C,
2. Question A, Ch. 9 of March: You should write a short description with the basic idea that sending light from the middle guarantees that it arrives at the two end at the same time according to anyone on the train. This defines “same time”.
Place a light in the center of a train car and place the two timers at each end of the train. Flash the light and instruct each timer to set their clocks when they see the light. Since the light travels the same distance and travels at the same speed towards each timer, it will take the same amount of time for the light to reach each timer and thus their clocks will be synchronized in their frame, regardless of whether the train moves or not. However, an observer on the ground sees the light travel the same speed as the people on the train, despite the fact that the train moves (classically, we would think that the light moves at the speed of light + the speed of the train relative to a ground observer, but in relativity this is incorrect). This is why relativity matters; the ground observer now sees that the light going towards the front of the train has to travel a farther distance to reach the timer at the front of the car, since he is moving away from the light. The timer at the rear moves toward the light and therefore gets the sync signal first. Thus her watch is ahead of the timer at the front.
3. Question B, Ch. 9 of March: You should write a short description with the key point being that NOTHING can travel faster than light. The logic is essentially the same as the next question.
4. Question E, Ch. 9 of March: This is closely related to the previous question. This is really a logic question. If one speed is maximum, the other cannot be greater.
5. The important point is that one can measure the width with meter sticks on either the train or the ground. It is NOT crucial that the measurements be made at the same time. Observers at either side of the train (on the train or on the ground) can check the width of the train and the report later the measured position of the each side of the train. All observers must agree. This is DIFFERENT from the length where one had to define a procedure to measure the length by measuring the positions of the front and back “at the same time”.
6. Exercise 1, Ch. 10 of March: The act that you think her clocks are slow means that you find more time has elapsed on your clock, i.e., you think your clock is faster. Thus your clock reads gamma x 10 min = 1.25 x 10 min = 12.5 min. (This is given in the solutions)
Exercise 2, Ch. 10 of March: For v = 0.6c, the factor of gamma is the same as above so that the length appears to 10m/gamma = 8m.
For v = 0.8c, gamma = 1.67 and the length appears to 10m/gamma = 6 m.