Does traveling at relativistic speeds change distances?

[Bemused]

@naita

Try this. You’re interested in travelling 1 ly from Earth, as measured in Earth’s reference frame, but timed in your travelling reference frame. Your time t at 0.707 c will be 0.707 t[sub]0[/sub] (Earth time).

Your mixed reference frame speed will then be distance in Earth’s frame / time in yours: (0.707 c * t[sub]0[/sub])/(0.707 t[sub]0[/sub]) = c

So you are saying you solved it by inspection?

shrug OK.

[Chronos]

And, since this is a 4th order equation, there are 4 roots. Only one of them will turn out to be the right one, and it should be obvious. This equation might have a doubled root. Not sure.

Yes, and those four roots are 1, 1, -1, and -1. The equation factors to (v^2-1)*(v^2-1) = 0. This is not a difficult problem.

[Bemused]

@Chronos

Yes, and those four roots are 1, 1, -1, and -1. The equation factors to (v^2-1)*(v^2-1) = 0. This is not a difficult problem.

Yeah, that’s right. c = 1 light-year/year. I should have seen that.

[naita]

@Bemused

So you are saying you solved it by inspection?

shrug OK.

No. I’m not saying that, I’m simply demonstrating that my solution was correct. The solution to which you responded “No, that’s not right.”

What I actually did was solve the equation a * c * t / (t * sqrt(1 - ^(a*c)^2/c^2)) = c