# Spaceship Chase Scene

1. Sep 16, 2011

### frerelupin

I have been pondering but cannot quite resolve the following question:

Imagine two spaceships, separated by 4 light-seconds initially. Spaceship A is traveling at 0.8c and Spaceship B (the one with the lead) is traveling at 0.4c.

Clearly an external observer will see this "chase" last for 10 seconds.
Spaceship A determines that the chase lasts ~6 seconds.
Spaceship B determines that the chase lasts ~9.17 seconds.

Calculating their velocities relative to one another is trivial and will be 0.588c and -0.588c.

Here is my conundrum: I want to be able to show that all frames are relative thus doing Einstein a solid. I feel I should be able to do this just using d=vt if I can answer the following question:

How far does Spaceship B appear to travel from A's perspective, and vice versa.

I can't quite get this to work out, but I feel that Spaceship A should see Spaceship B coming at them for 6 seconds at -0.588c, which means they covered 3.53 light-seconds, and that Spaceship B should see Spaceship A coming at them for 9.17 seconds at 0.588c meaning they covered 5.39 light-seconds.

There should be a manner in which I can determine the distance each appeared to travel using length contraction; that is to say, without using the times and velocities as derived above.

Am I off somewhere? Help!

2. Sep 16, 2011

### HallsofIvy

Staff Emeritus
It may be trivial but that is not what I get for their relative velocities. I get
$$\frac{.8c- .4c}{(1+ .8(.4)}= (.4c)/(1.32)= 0.303c$$
and -.303c

3. Sep 16, 2011

### frerelupin

Shouldn't the function in the denominator be subtraction? They are both traveling in the positive direction, so I would think there would be a subtraction in the numerator as well as the denominator.

4. Sep 16, 2011

### PAllen

I agree with +/- .588 c relative speed.

However, the problem is you cannot just use length contraction. Like it or not, you must use relativity of simultaneity as well. The initial distance of 4 light seconds is using the 'external' observer's simultaneity. The .8c rocket not only disagrees with the 4 light seconds, but it also disagrees with the external observer as when the .4 rocket was contracted 4 light seconds light seconds away.

The easiest way to work such a problem is to ignore the breakdown into velocity addition, contraction, simultaneity, and just Lorentz transform the whole scenario from 'external' to .8c rockect, then from external to .4 rocket.

5. Sep 18, 2011

### frerelupin

PAllen: Hmmmm... if I understand what you're saying (and I'm not certain I do!) you mean to say just do the velocity addition anyway? I'm trying to find a way around that part. As I said in my original post, I want to be able to show that v=d/t works in the rest frame, the 0.8c frame, and the 0.4c frame as long as we use the properly framed d and t values.

If you are saying I could do this, could you elaborate on which transforms you are talking about? Do you mean to say use:

x'=\frac{x-ut}{\sqrt{1-\frac{u^2}{c^2}}}

and

t'=\frac{t-\left(\frac{u}{c^2}\right)x}{\sqrt{1-\frac{u^2}{c^2}}}

6. Sep 18, 2011

### frerelupin

Woot! PAllen, you the man! Here's what I did:

v'=\frac{x'}{t'}=\frac{\frac{x-ut}{\sqrt{1-\frac{u^2}{c^2}}}}{\frac{t-\left(\frac{u}{c^2}\right)x}{\sqrt{1-\frac{u^2}{c^2}}}}=\frac{x-ut}{t-\frac{ux}{c^2}=0.588c

Done! Thank you so much! This has been bothering me for a while now.

Cheers!