# Show invariance of the inner product of $4$-velocities in different frames

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.

• 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 – Student146 Apr 14 at 16:08
• 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: $\gamma_w = \gamma_u \gamma_v (1+uv)$ – Student146 Apr 14 at 16:14