In the lab frame, particle $B$ moves to the right with speed $u$, and particle $C$ moves to the left with speed $v$. In the frame of $C$, particle $B$ is seen to move to the right with speed $w$, while particle $C$ itself is of course at rest. Of course, $w$ can be written down in terms of $u$ and $v$ using the velocity addition formula, but we will re-derive this formula below using $4$-velocities.

I am trying to show invariance of the inner product of both $4$-velocities.

My Attempt

In Lab Frame The $4$-velocity of $B$ is $\gamma_u(c,u,0,0)$ and the $4$-velocity of $C$ is $\gamma_v(c,-v,0,0)$

Moving to the C-Frame the $4$ velocity of $B$ becomes $\gamma_w(c,w,0,0)$ and the $4$-velocity of $C$ becomes $\gamma_v(c,0,0,0)$ because it is stationary.

I have defined the inner product of $4$-velocity as $A*B=A_0B_0-A_1B_1-A_2B_2-A_3B_3$ where $A=(A_0,A_1,A_2,A_3)$ and $B=(B_0,B_1,B_2,B_3)$.....(Further review seems like this might be incorrect as it applys to Lorentz boost and rotation only?)

This calculation doesn't indicate invariance so I must be making an error somewhere. If someone could provide me with help on this it would be greatly appreciated.

  • $\begingroup$ I believe the question requires you to just equate the inner products and from this equation a 'known' expression for $\gamma_w$ (from velocity addition formula) should be easily derived $\endgroup$ – Student146 Apr 14 at 16:08
  • 2
    $\begingroup$ Does the stationary particle require a relativistic factor? I don't think so. That would change the calculation of your inner product.... After calculation the invariance equation should yield the following $\gamma_w c^2 = \gamma_u \gamma_v (c^2 + uv)$ Removing common terms this reduces to $\gamma_w c^2 = \gamma_u \gamma_v (c^2 + uv)$ Divide across by $c^2$ yields: $\gamma_w = \gamma_u \gamma_v (1+uv)$ $\endgroup$ – Student146 Apr 14 at 16:14

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