Notes on Relativity and Cosmology - Physics Department, UCSB

4.3. MORE ON MINKOWSKIAN GEOMETRY 107
θ = tanh −1 (v/c).
Here, tanh −1 is the function such that tanh −1 (tanhθ) = tanh(tanh −1 θ) = 1.
This one is difficult to write in terms of more elementary functions (though it
can be done). However, we can draw a nice graph simply by ‘turning the above
picture on its side.’ The horizontal axis on the graph below is x and the vertical
axis is tanh −1 x. Note that two reference frames that differ by the speed of light
in fact differ by an infinite boost parameter.
boost parameter
5
0
−5
−1 0 1
velocity
Now for the magic: Let’s consider three inertial reference frames, Alice, Bob,
and Charlie. Let Bob have boost parameter θ BC = tanh −1 (v BC /c) relative to
Charlie, and let Alice have boost parameter θ AB = tanh −1 (v AB /c) relative to
Bob. Then the relative velocity of Alice and Charlie is
Let’s write this in terms of the boost parameter:
v AC /c = v AB/c + v BC /c
1 + v AB v BC /c2. (4.14)
v AC /c = tanh(θ AB) + tanh(θ BC )
1 + tanh(θ AB )tanh(θ BC /c 2 ) . (4.15)
After a little algebra (which I have saved for your next homework), one can
show that this is in fact:
v AC /c = tanh(θ AB + θ BC ). (4.16)
In other words, the boost parameter θ AC relating Alice to Charlie is just the
sum of the boost parameters θ AB and θ BC .
Boost parameters add:
θ AC = θ AB + θ BC !! (4.17)