unforum

[Dust]

[baf]

Certainly.

[Dust]

Thanks very much.

[trystero]

Baf, it seems to me that your solution proves that A is moving away from the disk, in violation of the problem statement.

[baf]

Blast! It looks like you're right.

But I also don't see anything wrong with my math. I suspect that the problem results from my simplifying assumption, and that if you throw that out, the resulting (more complicated) equations would give us a range of velocities for both A and B for which the numbers come out right and both ships are actually moving in the right direction. But I don't feel like doing that right now. Maybe tomorrow, when I'm fresher of mind.

[Ashin]

[baf]

I don't mean a range of solutions. I just mean a range of velocities that each ship could have. The solution is the relative speed of the two ships -- that is, the difference in their individual speeds. I'll be surprised if the difference varies as the individual speeds do.

[LoneWolf]

[cubivore]

I'm late to the party.

AngusA's solution is convincing. I need to either derive or find a reference that talks about that tan (theta) = sqrt((c-v)/(c+v)) tan (theta). I'm sure it's true.

I was seeing Adamek as going faster because I was blaming the increase in apparent size of the disk on a length contraction in their z-axis. A's z was 5 times shorter than B's, so gamma would be 5 between them. That neglected the relativistic effect on the triangle's hypotenuse (edge of the disk to the observer), which is probably why I was having such trouble.

I also got hung up on the ratio of the perceived distances being tan(theta)/tan(5*theta). That is overcome when one takes into consideration that the disc is "small."

It's been a while since my special relativity classes at university. I seem to recall us avoiding problems like this, too. Ah well.

Always learning!