(##c = 1##)(adsbygoogle = window.adsbygoogle || []).push({});

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.

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

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