# What happens when a light is shone across a spaceship in its direction of motion? [closed]

Imagine a space ship in S frame moving with speed v (in the x direction) relative to S'. An astronaut on board shines a light from the the rear of the spacecraft to the front (along the x-direction). The astronaut measures the time taken for the light to reach the other end to be T. The proper length of the spacecraft is L.

The astronaut would then determine the speed of light to be:

$$c = L/T\tag{1}$$

From the S' frame, a stationary observer would measure the length of the spacecraft to be:

$$l' = L/\gamma$$

Then the total distance travelled by the light ray would be:

$$d = vt' + l' = vT\gamma + L/\gamma$$

This is because the light ray would have to traverse the contracted length of the spacecraft ($$l'$$), but also cover the additional distance that the craft has moved from the starting point ($$vt'$$) as seen from the $$S'$$ frame.

and $$d$$ would also be:

$$d = ct' = cT\gamma$$

which leads to $$cT\gamma = vT\gamma + L/\gamma$$

from which follows $$c = (1/\gamma^2)(L/T) + vT\gamma\tag{2}$$

Now, if we compare equations $$(1)$$ and $$(2)$$ we would find that the two observers would disagree on the speed of light.

So, my question is where has this argument gone wrong?

My first thoughts would be that it is incorrect to just simply add on the extra distance that the spacecraft travels to the total distance measured from $$S'$$, but I can't seem to understand why that would be the case.

The mistake is in your formula for $$d$$, which involves a $$t'$$. I don't think you've been careful about what $$t'$$ represents.

So let's start over:

1. In the frame of the spacecraft, the light leaves location $$x=0$$ at time $$t=0$$ and arrives at location $$x=L$$ at time $$t=L$$.

2. Lorentz transform this to the earth-frame (which is moving at velocity $$-v$$ with respect to the ship). You'll find that the light leaves location $$x'=0$$ at time $$t'=0$$ and ultimately arrives at location $$x'=\beta L$$ at time $$t'=\beta L$$ where $$\beta=\sqrt{(1+v)/(1-v)}$$.

3. Along the way (calculating in the earth frame), the light passes the original location of the front of the ship, which is to say $$x'=L'$$. This must happen at time $$t'=L'$$.

4. So: in the earth frame, we can break the light's journey up into two parts. PART ONE: The light travels to where the front of the ship used to be --- distance $$L'$$. PART TWO: The light travels from there to its ultimate destination. According to paragraph 2), the entire journey has length $$\beta L$$. So PART TWO has length $$\beta L - L'$$.

5. You've got the length of PART TWO as $$vt'$$, but you haven't told us what $$t'$$ is. Apparently you mean for it to represent the time taken (in the earth frame) for PART TWO of the journey. If so, $$t'=\beta L-L'$$, but you seem to have assumed otherwise.

Edited to add: I haven't worked through the details of your calculations, but I'm pretty sure I know what led you astray:

1. The length of a rod (in your frame) is the distance between two events that happen at the same time (in your frame). Those events are, for example, "left end of rod at noon" and "right end of rod at noon".

2. The distance between the completions of PARTS ONE and TWO of the journey is the distance between two events that happen at different times (in your frame). Those events are the arrival of the light where, according to you, the front of the ship used to be and the arrival of the light at the actual front of the ship.

3. You have a formula for computing the length of a rod in your frame, given the length in some other frame. You are (I think, though I haven't fully worked through it) trying to apply that formula to the two events in point 2) above. But those events are not the sort of events to which the formula applies.

4. One way to avoid such mistakes is to be very careful to memorize all the conditions in which you can use your formula. A better way is to forget the formula entirely and work things out from first principles and Lorentz transformations, as I've done above.