Learning Goal: To understand length contraction and time dilation.
In classical physics, and in your everyday experience, lengths and times seem to be the same no matter who measures them. In fact, the notion that lengths or time intervals might be different depending on who measures them can seem profoundly disturbing (or just plain silly), because measuring space and time are two of the most fundamental ideas in physics. However, Einstein's special theory of relativity shows that space and time are not as fundamental or as absolute as you are accustomed to believing.
To discuss measurements more precisely, you must consider both the thing being measured and the person (technically the frame of reference) doing the measuring. An inertial frame of reference is a frame of reference (i.e., a consistent way of measuring distances and times) that is not accelerating. An inertial frame of reference moves at a constant speed, relative to other inertial frames of reference.
When making a measurement of some object's length, you would like to make the measurement while whatever you are measuring is in the same inertial frame as you (i.e., at rest relative to you). This sort of measurement gives the proper length lo of an object. Similarly, it is preferred that you measure the time interval between two events when the events occur in the inertial reference frame that you are in. This sort of measurement yields the proper time t0 When making measurements of events in an inertial reference frame different from your own, what you measure will be related to the proper time and length, as well as the relative speed between the two frames. The measured length and time are given by the following two equations:
t = to/√1-u2/c2 and l = lo √1-u2/c2
where c is the speed of light, t is the measured time, and l is the measured length in the direction of motion.
Part A
Suppose that you measure the length of a spaceship, at rest relative to you, to be 400m . How long will you measure it to be if it flies past you at a speed of μ = .75c?
Express the length l in meters to three significant figures.
Part B
The spaceship from Part A has a large clock attached to its side. This clock ran at the same rate as your watch when you were in the same reference frame. How much time t will pass on your watch as 80s passes on the clock?
Two spaceships, unimaginatively named A and B, are flying toward each other with relative speed .
Part C
If the captain of ship A fires a missile, counts 10.0s on his watch, and then fires a second missile, how much time t will the captain of ship B measure to have passed between the firing of the two missiles?
Part D
The captain of ship B knows that ship A uses 2-m-long missiles. She measures the length of the first missile, once it has finished accelerating, and finds it to be only 0.872m long. What is the speed u of the missile, relative to ship B?
Express your answer in meters per second to three significant figures. Use c= 3*108m/s
It should be noted that the same equations apply to events in your everyday life. The reason that you don't notice them is that things in your everyday life move so much slower than the speed of light. Now let's look at the differences in measurements between two frames moving relative to one another at a speed of 30m/s (108kph or about 67mph ). Your calculator may not be able to store enough digits to work these problems accurately, so you may need to use the approximations from the binomial expansion:
Part E
What would be the difference between the time t measured by an observer moving at 30m/s and the proper time T0 for a proper time interval of 1 hour (3600 s)? The answer is small but nonzero. You will need to find an expression for the time difference using the approximation given in this problem before you substitute in the numbers; otherwise your calculator will just give zero.