# The middle frame lemma

1. Feb 10, 2009

### kron

Hi,

I'm currently reading "Introduction to SR", by Wolfgang Rindler (2nd Ed, 1991 Oxford Science Publications).
Now I have troubles with Problem Ex I No. 7.
Since this problem is not homework and also a general question on SR, I post it here.
(If I should post other problems from that book in the homework section anyway, I will do that.)

The middle frame lemma is given as follows:

Given two inertial frames (IF) S and S', there exists a third IF S''' relative to which S and S''
have opposite velocities of the same magnitude. We can picture it 'in the middle' between S and S'.

Now the question:

Prove that at any instant there is just one plane in S on which the clocks
of S agree with the clocks of S', and that this plane moves with velocity

$$\frac{c^2}{v}(1-\frac{1}{\gamma})$$

How is this plane related to the frame S''
of the 'middle frame' lemma ?

My idea:

From the Lorentz transformation (LT) of

$$t^{'}=\gamma(t-\frac{vx}{c^2})$$

I get,

$$t=t^{'}$$:

$$t=\frac{v}{c^2}( 1-\frac{1}{\gamma} )^{-1} x$$

in S and hence this is a plane through

$$x=\frac{c^2}{v}(1-\frac{1}{\gamma})t$$

. One now finds

$$v_x=\frac{c^2}{v}(1-\frac{1}{\gamma})$$

for the velocity of such a plane.

I now see that this gives

$$t^{'}=\frac{v}{c^2} x \frac{\gamma}{\gamma-1}$$

in S', so

$$v_x^{'}=-\frac{c^2}{v}(1-\frac{1}{\gamma})$$.

So this planes move in the opposite direction in S'.

But how are they connected with S'''. I would say, that this are the x'''-axes ?
But what is then the t'''-axis ?

Any comments are welcome, thanks.

Last edited: Feb 10, 2009