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

  1. Jun 27, 2017 #1
    (##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. jcsd
  3. Jun 27, 2017 #2

    Orodruin

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    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.
     
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