# Homework Help: Time dilation rocket ship speed problem

1. Aug 19, 2007

### ehrenfest

A rocket ship leaves earth at a speed of 3/5 c. When a clock on the rocket says 1 hour has elapsed, the rocket sends a light signal back to earth.

When was the signal sent according to clocks on earth?

So we have T_earth and T_rocket and they are proportional. I am confused about which side to put the Lorentz factor on. Is it:

T_earth = gamma T_rocket

or

gamma T_earth = T_rocket

?

Apparently only earth is an inertial reference frame (see below), so does gamma go with the inertial reference frame?

If earth is the inertial reference frame and the rocket is not inertial it must have been accelerating, in which case the problem is inherently wrong because it implied the velocity of the rocket is ALWAYS 3/5c. Am I right?

2. Aug 19, 2007

### Staff: Mentor

The rule of thumb is that moving clocks are measured to run slow according to observers in an inertial frame seeing it move. In this case, the rocket clock is the single moving clock for which you are given an elapsed time, so earth observers will deduce that the actual time according to them was longer.

Both frames are inertial. (Ignore the initial acceleration of the rocket.)

Just ignore the inconvenient fact that the rocket had to have started at speed zero. Assume that it takes off at 3/5c and always has that speed.

A cleaner way of describing the problem would have been to have the rocket pass by earth at the exact moment that the rocket clock reads 1:00 pm (say) and then have it send the signal when the rocket clock reads 2:00 pm. This way you can have the rocket always moving at constant speed.

3. Aug 19, 2007

### ehrenfest

OK. So, the main issue I am having is that I think you MUST assume that the rocket accelerated from earth for the problem to make sense.

If you assume that the rocket just passed earth at t = 0 with its velocity 3/5c, then there is absolutely no reason why that rule of thumb should be applied because an observer on the rocket is at rest in his own intertial reference frame and the earth is moving away from him with velocity 3/5c.

Thus the problem is symmetric and you have both

T_earth = gamma T_rocket

and

gamma T_earth = T_rocket.

There is no way to decide which to use unless you assume the rocket accelerates from earth and the picture is asymmetric, right?

4. Aug 19, 2007

### learningphysics

You don't have to assume acceleration... Remember that although seeing the clock on the earth at gammaT and the clock on the rocket at T are simultaneous in the earth's frame of reference, they are not simulatenous in the rocket's frame of reference. clock on earth at gammaT is a separate space-time event from the clock on the rocket at T. The order of events can be reversed in a different frame of reference. So from the rocket's point of view, the earth's clock is moving slower that its own clock... it may seem paradoxical at first, but it isn't.

Just analyze the problem from the earth's frame of reference. The question asks when the rocket sent the signal, from the earth's frame of reference. So the clock on the rocket is the moving clock so it's going slower than the earth's clock, from the earth's frame of reference. So just use the time dilation formula accordingly.

5. Aug 19, 2007

### Staff: Mentor

It is certainly true that each frame views the other as moving at 3/5c. It's further true that each frame will measure the other's clocks as running slow.

Ah, but you only have data for one actual clock: the rocket clock. You are told what it reads, so you can use the "moving clocks run slow" rule to figure out what another frame will measure for the time interval. You have no data about earth clocks.

T_rocket is the time measured by a single moving clock; T_earth represents a deduction (involving multiple clocks and synchronization).

No, incorrect.

To make this clearer, consider this. Imagine a space station at rest with respect to the earth some distance away in space. Let's further assume that the earth and space station clocks are synchronized (according to them), so they are part of the same inertial frame. Now have the rocket pass by earth exactly as the rocket clock and earth clock both read 1:00pm, then have it pass by the space station exactly as the rocket clock reads 2:00pm. What can you deduce?

According to earth-space station measurements, which involve comparing multiple clock readings, the time of flight will be longer that 1 hour by the factor gamma. (Moving clocks run slow.) In fact, you can deduce that the space station clock must read 1 hour*gamma at the exact moment that the rocket passes by the space station.

What does the rocket observer think of all this? He agrees that the space station clock reads 1 hour*gamma as he passes it. But according to the rocket observer, the earth and space station clocks aren't synchronized. He also knows that the clock on earth (a moving clock according to him), must run slow per his measurements. If he measures his flight to take 1 hour, he can deduce that the earth clock must read only 1 hour/gamma (according to him) at the moment he passes the space station.

To have everything make sense, you just can't apply time dilation--you also need length contraction and the relativity of simultaneity. All relativistic effects work together to form a coherent picture. Furthermore, all the effects (considered separately) are completely symmetric.

6. Aug 19, 2007

### ehrenfest

I understand now.

If the rocket was dragging a clock 3/5*c* 60^2 meters behind it, and it passed earth at 1:00 (according to the rocket clock and the earth clock), then it could take the time of the earth clock when it passes the space station, multiply by gamma, and get the time the clock dragging behind it. Cool.

So, according to the earth clock, the signal was sent at t_earth = 4500s and the signal will arrive at earth at t_earth = 4500 + 2160=7660s and t_rocket = 6300s, right?

To get that last number, you have to add the 60^2 with the time change for the rocket in motion during an earth time interval of 2160, right? Or is there a simpler way to do it?

7. Aug 19, 2007

### learningphysics

The 2160 is wrong. It should be 2700. The distance the rocket travels is:
(3/5)c*4500 m. Divide that by c to get the time it takes the signal to get to earth, so that's 2700s. So it arrives at t_earth = 4500+2700 = 7200s.

Also, it is problematic to talk about the time on the rocket's clock, when the signal arrives at earth... it depends on which frame of reference you are in. The time on the earth clock when the signal hits the earth is always the same in any frame of reference... That's because the signal hitting the earth and the time on the earth clock at that moment are a single spacetime event... they both happen at the same position at the same time.

For example, in the earth's frame of reference, when the signal hits earth, the clock on the rocket is at: 0.8*7200 = 5760s. So in the earth's frame, when the signal hits the earth, the earth clock is at 7200s and the rocket clock is at 5760s

However, in the rocket's frame, when the signal hits the earth, the clock on the rocket is at 9000s. Analyzing the problem from the rocket's frame of reference... the distance between the earth and the rocket when the signal is sent is (3/5)c*3600. The relative velocity of light and the earth is c-(3/5)c = 2/5c... Distance/speed gives 5400s. 3600+5400s = 9000s. And the earth's clock is at 0.8*9000s = 7200s... so earth's clock is at 7200s (as expected) and the rocket's clock is at 9000s.

So the rocket's clock is at different times depending on which frame you're talking about.

Last edited: Aug 19, 2007