- #26

- 33

- 2

I'm pretty sure this isn't right. For example, it's possible that [itex]A = (0, 0)[/itex], and then [itex]v = 0/0[/itex], undefined.Assume event [tex]A(x_A,t_A)[/tex] . Then, event [tex]B(t_B=t_A,x_B)[/tex]. So, A and B are simultaneous in frame F.

Assume that there is a frame F' moving at speed v wrt F. In frame F', A has the coordinates [tex]A'(t'_A,x'_A)[/tex] and it is simultaneous with [tex]C(t'_A,x'_C)[/tex]

where:

[tex]t'_A=\gamma(v)(t_A-\frac{vx_A}{c^2})[/tex]

If you want for B and C to be simultaneous, this simply means [tex]t_A=t'_A[/tex]

i.e.

[tex]t_A=\gamma(v)(t_A-\frac{vx_A}{c^2})[/tex]

The above is a simple equation in v with the solution:

[tex]v=\frac{2x_At_A}{t_A^2+(x_A/c)^2}<c[/tex]

So, you canalwaysfind a frame F' that satisfies your condition 3. It is interesting to note that the speed v between F and F' depends only on the spacetime coordinates of A.

The velocity of the frame in which [itex]B[/itex] and [itex]C[/itex] are simultaneous is, I believe,

[itex]v = c^2 \frac{t_C - t_B}{x_C - x_B}[/itex].

Of course, if [itex]v \ge c[/itex], then the frame is not physically realistic.