A 3+1-dimensional Lorentz transformation can be written as(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\Lambda=\gamma\begin{pmatrix}1 & -v^T\beta \\ -v & \beta\end{pmatrix}[/tex]

where v is a 3×1 matrix representing the velocity difference, [itex]\gamma=1/\sqrt{1-v^2}[/itex], and [itex]\beta[/itex] is a 3×3 matrix that's orthogonal when v=0. When [itex]\beta=I[/itex], [itex]\Lambda[/itex] is a pure boost.

The product of two boosts with arbitrary velocities is

[tex]\Lambda(u)\Lambda(v)=\gamma_u\gamma_v(1+u^Tv)\begin{pmatrix}1 & -w^T \\ -w & \beta\end{pmatrix}[/tex]

where

[tex]w=\frac{u+v}{1+u^Tv}[/tex]

and

[tex]\beta=\frac{I+uv^T}{1+u^Tv}\neq I[/tex]

so this isn't another boost, but it's a Lorentz transformation with velocity w. (If you noticed that the upper right component of the matrix on the right-hand side is [tex]-w^T[/tex] even though the formula for a general Lorentz transformation says that we should have [tex]-w^T\beta[/tex] in that position, don't worry about it. It's not a mistake. This w happens to be an eigenvector with eigenvalue 1 of [tex]\beta^T[/tex]).

My question is this: Why doesn't this mean that the velocity addition rule in 3+1 dimensions is

[tex]u\oplus v=\frac{u+v}{1+u^Tv}[/tex]

The correct result is supposed to be

[tex]\vec u\oplus\vec v=\frac{1}{1+\vec u\cdot\vec v}\bigg(\vec u+\vec v+\frac{\gamma_{\vec u}}{1+\gamma_{\vec u}}\vec u\times(\vec u\times\vec v)\bigg)[/tex]

according to a book mentioned in another thread of this forum. (That book didn't include a derivation).

I'm pretty sure my algebra is correct. That's why I didn't bother posting all the details. If that last formula is the correct velocity addition rule, then there's something wrong with my whole approach to this.

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# Tell me what I'm doing wrong (velocity addition in 3+1 dimensions)

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