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If a particle is moving in the $x$-direction with velocity $c/2$, then the Lorentz transformation $\Lambda = \begin{pmatrix}\gamma & -\beta \gamma & 0 & 0 \\ -\beta \gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix}\cosh\ \phi & -\sinh\ \phi & 0 & 0 \\ -\sinh\ \phi & \cosh\ \phi & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix}\frac{2}{\sqrt{3}} & -\frac{1}{\sqrt{3}} & 0 & 0 \\ -\frac{1}{\sqrt{3}} & \frac{2}{\sqrt{3}} & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$, where the rapidity $\phi$ is given by $\tanh\ \phi = \frac{v}{c}$.

Subsequently, if the particle is moving in the $y$-direction with velocity $c/2$, then the Lorentz transformation $\Lambda' = \begin{pmatrix}\gamma & 0 & -\beta \gamma & 0 \\ 0 & 1 & 0 & 0 \\ -\beta \gamma & 0 & \gamma & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix}\cosh\ \phi & 0 & -\sinh\ \phi & 0 \\ 0 & 1 & 0 & 0 \\ -\sinh\ \phi & 0 & \cosh\ \phi & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix}\frac{2}{\sqrt{3}} & 0 & -\frac{1}{\sqrt{3}} & 0 \\ 0 & 1 & 0 & 0 \\ -\frac{1}{\sqrt{3}} & 0 & \frac{2}{\sqrt{3}} & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$.

Therefore, the combined transformation $\Lambda''(w) = \Lambda' \Lambda = \begin{pmatrix}\cosh^{2}\ \phi & -\sinh\ \phi\ \cosh\ \phi & -\sinh\ \phi & 0 \\ -\sinh\ \phi & \cosh\ \phi & 0 & 0 \\ -\sinh\ \phi\ \cosh\ \phi & \sinh^{2}\ \phi & \cosh\ \phi & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$.

But now I'm having a bit of a trouble finding the boost velocity $w$ for $\Lambda''(w)$. Any suggestions?

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An alternative approach to using a matrix representation for the rotation operator is to use clifford algebra instead. You might know clifford algebra in the guise of Pauli and Dirac matrices, from quantum mechanics.

A boost in the $i$th direction is described fully by a "rotor", which takes the form $$q = \exp(-\gamma_0 \gamma_i \phi/2) = \cosh \frac{\phi}{2} - \gamma_0 \gamma_i \sinh \frac{\phi}{2}$$

If you're used to thinking of the Dirac matrices as, well, matrices, then feel free to write $I \cosh \frac{\phi}{2}$ instead. Here, however, I eschew treating these objects as matrices--I only need their multiplication law to use their clifford algebra properties--and as such, I consider such a term to be "scalar" compared to the vector space formed by the $\gamma_\mu$.

Now, if you're familiar with quaternions, you might recognize what we're doing: indeed, we can proceed exactly as in quaternions (and a good exercise is to derive quaternions using the Pauli algebra). A vector $v$ that is a linear combination $v = v^\mu \gamma_\mu$ can be boosted by

$$R(v) = q v q^{-1} = \exp(-\gamma_0 \gamma_i \phi/2) v \exp(\gamma_0 \gamma_i \phi/2)$$

Multiply all this out, and you get the usual formula for a boost.


Now let's take your example, and let's multiply two rotors: one boosting along $e_x = \gamma_1$ and another boosting along $e_y = \gamma_2$:

$$\left(\cosh \frac{\phi}{2} - \gamma_0 \gamma_1 \sinh \frac{\phi}{2} \right) \left(\cosh \frac{\phi}{2} - \gamma_0 \gamma_2 \sinh \frac{\phi}{2} \right)$$

Remember the clifford multiplication rules: $\gamma_\mu \gamma_\nu = -\gamma_\nu \gamma_\mu$ if $\mu \neq \nu$. and $\gamma_\mu \gamma_\mu = \eta_{\mu \mu}$ (no sum). The gammas are associative, so we can manipulate the expression to get

$$\cosh^2 \frac{\phi}{2} - \left(\sinh\frac{\phi}{2} \cosh \frac{\phi}{2}\right) [\gamma_0 \gamma_1 + \gamma_0 \gamma_2] - \gamma_1 \gamma_2 \sinh^2 \frac{\phi}{2}$$

using $\eta = (+,-,-,-)$ convention.

The presence of the $\gamma_1 \gamma_2$ term tells us that there is a rotation introduced when we compose boosts in this way (at least, when those boosts don't use the same plane).


Does it make sense to talk about the boost velocity for an operation that has both boost terms and pure rotation terms? Remember, this is not merely a boost and then a rotation, nor a rotation and then a boost--they're both happening together as you apply this operator.

If you have a definition of boost velocity that applies in this case, though, I'd be happy to turn the crank and compute it.

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  • $\begingroup$ Every Lorentz translation that keeps the origin fixed can be decomposed as a boost and then a rotation, or alternatively a rotation and then a boost. See Wikipedia. So yes, the boost velocity makes perfect sense here. $\endgroup$ – Peter Shor Jun 8 '16 at 16:03
  • $\begingroup$ Moreover, it's worth noting that the rapidities of the boosts $B$ and $B^\prime$ in the two different decompositions -$R\,B$ and $B^\prime\,R$ - @PeterShor speaks of are the same, so there is a unique, well defined rapidity / speed even though the boosts themselves are different operators ($B^\prime = R\,B\,R^{-1}$), and also that the rotation operators in the two decompositions are the same. $\endgroup$ – WetSavannaAnimal Jul 11 '16 at 1:47
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If you are doing what I think you are doing then, you are trying to get the Boost matrix for an arbitrary direction. Then the way to go about that will be to use the generalized boost matrix (see J D Jackson page547 or http://en.wikipedia.org/wiki/Lorentz_transformation). The the matrix against which you can compare to get the velocity from wikipedia is image.

In particular, look at the firs row and first column then $\Lambda_{11} = \gamma, \frac{c \Lambda_{12}}{\Lambda_{11}} = v_x, \frac{c \Lambda_{13}}{\Lambda_{11}} = v_y$ and $\frac{c \Lambda_{14}}{\Lambda_{11}} = v_z$

Aside: Are you sure your last matrix is correct? It isn't symmetric.

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  • $\begingroup$ I think so! I can't find any mistake in my calculations. But, then again, I think it's wrong because it's not symmetric. I'll have to take a closer look at my calculation again. $\endgroup$ – nightmarish Apr 7 '15 at 4:12
  • $\begingroup$ Why do you think multiplying two symmetric matrices gives a symmetric product? $\left(\begin{array}{cc}1&0\\0&0\end{array}\right) \cdot \left(\begin{array}{cc}0&1\\1&0\end{array}\right) =\left(\begin{array}{cc}0&1\\0&0\end{array}\right)$ $\endgroup$ – Peter Shor Jun 8 '16 at 15:56

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