Solving Spaceship Chase Scene Relative Velocity Problem

[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!

 

[HallsofIvy]

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

 

[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.

 

[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.

 

[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:

\begin{equation}
x'=\frac{x-ut}{\sqrt{1-\frac{u^2}{c^2}}}
\end{equation}
and
\begin{equation}
t'=\frac{t-\left(\frac{u}{c^2}\right)x}{\sqrt{1-\frac{u^2}{c^2}}}
\end{equation}

 

[frerelupin]

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

\begin{equation}
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
\end{equation}

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

Cheers!