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I am kind of stuck with a problem mentioned in my current reading about special relativity. Given the Lorentz transformation

$$x^{\bar{i}} = L^i{}_k \, x^k \quad ,$$

one has to find the transformation law for the "ordinary" velocity ( =: 3-velocity) of a particle described by its spatial coordinates$\, \textbf{x}(t) \,$. The 3-velocity is of course

$$ \textbf{v} = \frac{\mathrm{d}\textbf{x}}{\mathrm{d}t} \quad .$$

Now, if I only consider e.g. boosts along the x-axis, the problem is easy to solve. However, for the general case I don't seem to find a satisfying solution.

First, lets define the range for latin letters as $\, i,j,k = 0,1,2,3 \,$ and the range for greek letters as $\, \mu, \nu, \varrho = 1,2,3\, $. Then I insert the definitions as

$$ v^{\, \overline{\mu}} \enspace = \enspace \dfrac{\mathrm{d} \left( L^{\, \mu}{}_{k} \, \, x^k \right)}{\mathrm{d} \left( L^{\,0}{}_j \, \, x^j \right)} \enspace = \enspace \dfrac{L^{\,\mu}{}_{k} \, \,\mathrm{d} x^k}{L^{\, 0}{}_j \, \, \mathrm{d} x^j}$$

But what are the next steps?

Urb
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Octavius
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  • Related : Transformation of 4− velocity. For the transformation of the 3-velocity vector see equation (08) and Figure-01 in my answer therein. – Frobenius Aug 12 '20 at 05:47
  • This only covers boosts in an arbitrary direction. What about Lorentz transformations that include rotations? – Octavius Aug 12 '20 at 23:29
  • You transform the 4-velocity and then take the spatial part of the result. Are you asking for something different than that? – Brick Aug 14 '20 at 15:24
  • I think that this would not work, since the spatial part of the 4-velocity does not coincide with the "ordinary" velocity. (difference: differentiating with respect to time coordinate as opposed to proper time) – Octavius Aug 14 '20 at 22:11

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