# Thought experiment (SR)

1. Aug 10, 2008

### Chatt

I was thinking of this thought experiment the other day and it didnt quite seem to make sense to me. I hope some of you can explain it. I have drawn a quick example of the setup:

and sorry for my english

(c = 1 m/s)

Imagine a 1 meter long spaceship with a velocity of 0,5 m/s (c/2). The spacesip has two light sensors attached to it, one in the front and one in the back. Each of these sensors has clocks attached to them (A and B on the drawing) and when the sensors register a photon the clocks will stop. The clocks are synchronized and they tick once each second.

Now imagine a photon coming towards the spaceship ofcourse with a velocity of c (1 m/s). It hits the first sensor and clock (B) stops. The photon continues through the spaceship and hits the other sensor and clock (A) stops. Now since the speed of light must always be the same for all observers the two clocks must now differ by one second.

But here is my question,.. according to SR the clocks on the spaceship ticks slower (from our point of view) because they are moving, and if the clocks has to differ by one second, time must go 'faster' (and definately not slower) for the moving spaceship..

thx in advance for an answer,, and sorry for not expressing me perfectly.. pls ask if you have any questions yourself

2. Aug 10, 2008

### JesseM

Remember, the ship's length in its own rest frame is different from its length in our frame because of Lorentz contraction. If the ship's speed is 0.5c in our frame, then its length in our frame is shrunk by $$\sqrt{1 - 0.5^2}$$ = 0.866 relative to its rest length, so if the length is 1 meter in its own frame, it must only be 0.866 meters in our frame.
In the ship's own frame they will differ by one second. But are you familiar with the relativity of simultaneity? If the clocks were synchronized and a distance d apart in the ship's rest frame, then in our frame if the ship is moving at speed b along the axis between the clocks, according to our frame's definition of simultaneity the clocks were out-of-sync by vd/c^2, with the trailing clock's reading being ahead of the front clock's reading by this amount. So, in this case, before the clocks stopped we would observer the clock at the back of the ship (A) to be ahead of the clock at the front (B) by (0.5 meters/second)*(1 meter)/(1 meter/second) = 0.5 seconds.

So, suppose the light hits the front (B) at time t=0 in our frame, and at position x=0.866 meters. Since the ship is 0.866 meters long in our frame, this means the back clock (A) is at position x = 0 meters at time t=0 in our frame. If clock B shows a time of T=10 seconds when the light hits it, that means clock A shows a time of T=10.5 seconds at the same moment in our frame (t=0) because of the relativity of simultaneity.

Now if the light's position at t=0 is x=0.866 meters, then the light's position as a function of time will be x(t) = 0.866 - 1*t, since the light is moving in the -x direction at 1 meter/second. Meanwhile, if the back clock's position at t=0 is x=0, then the back clock's position as a function of time will be x(t) = 0.5*t, since the back clock is moving in the +x direction at 0.5 meters/second. So, to figure out when the light hits the back clock in our frame, we set these equal and solve for t:

0.866 - 1*t = 0.5*t
1.5*t = 0.866
t = 0.577.

So in our frame there are only 0.577 seconds between the light hitting B and the light hitting A, since B and A were less than 1 meter apart in our frame and A was rushing forward to meet the light. According to the time dilation formula, clock A will only have ticked forward by 0.866 * 0.577 seconds in this time (since it's ticking slower by a factor of 0.866), which works out to 0.5 seconds. And as I mentioned, because of the relativity of simultaneity, in our frame A already read 10.5 seconds at the "same moment" (according to our frame's definition of simultaneity) that B read 10 seconds and was stopped by the light hitting it, so by the time the light reaches A and stops it, it will have advanced forward to 10.5 + 0.5 = 11 seconds. So you see, if you take all these factors into account--the fact that the rod is shrunk in our frame, the fact that A is moving towards the light in our frame, the fact that the clocks A and B are out-of-sync by 0.5 seconds in our frame prior to either one being stopped, and the fact that the clocks are running slow in our frame--then you end up with exactly the same prediction as you made in the rod's rest frame, namely that B reads 10 seconds when it stops while A reads 11 seconds.

3. Aug 10, 2008

### Janus

Staff Emeritus
The answer is in the Relativity of Simultaneity.

While from the ships point of view the Clocks are synchronized, they are not from our point of view. From our point of view, clock A actually reads a time later than clock B does. So even though clock A ticks off less than one sec from the time clock B stops, when it stops, it will read a one second difference from clock B.

4. Aug 10, 2008

### Chatt

Ah,,, I forgot to take into account that when the spaceship accelerated from 0 to 0,5c the clocks got out of synch..

Well, many thx JesseM for the quick yet very detailed answer ^^ you too Janus!