# B The velocity-composition law when c is in the picture

1. Jun 27, 2017

### SiennaTheGr8

($c = 1$)

So:

$v = \dfrac{u + w}{1 + uw}$.

If the thing with speed $u$ happens to be light, then the equation still works:

$v = \dfrac{(1) + w}{1 + (1)(w)} = 1$.

And the inverse formula still works, too:

$u = \dfrac{v - w}{1 - vw} = \dfrac{(1) - w}{1 - (1)w} = 1$

But is this mere happenstance? To wit:

1. The often-preferred method of deriving the formula involves Lorentz-transforming the relevant spatial component of the four-velocity. But four-velocity is undefined for anything whose $v = 1$.

2. A more elegant way of writing the formula is:

$\tanh \phi_v = \tanh (\phi_u + \phi_w)$,

where $\phi_v = \tanh^{-1} v$ (etc.). But the argument of the $\tanh^{-1}$ function must have an absolute value less than $1$, so if $v = 1$ then $\phi_v = \tanh^{-1} v$ is no good.

Hm.

On the other hand, you can derive the formula in a more "basic" way by Lorentz-transforming the $x$ and $t$ coordinates (in standard configuration), and differentiating the former with respect to the latter. As far as I can tell, this derivation avoids the problem noted above.

I'm not really sure what my point is, but I found this interesting.

2. Jun 27, 2017

### Orodruin

Staff Emeritus
This is really just a special case. The only requirement necessary is that the vector you consider is the tangent vector of a world line. What you are really looking at is the ratio
$$\frac{dx^i}{dx^0} = \frac{dx^i/ds}{dx^0/ds}$$
and its transformation properties. It does not matter what the world-line parameter $s$ is. You can take it to be the proper time, which gives you the ratio of the 4-velocity components, but you don't need to do that.