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.

The easiest way is to use four-vectors. If an object has three-velocity $\mathbf{u} =\begin{bmatrix} u_x \\ u_y \\ u_z \end{bmatrix}$ in one frame, its four-velocity is
$$\gamma_u\begin{bmatrix} c \\ u_x \\ u_y \\ u_z\end{bmatrix}$$

The general form of a Lorentz transformation from frame $S$ to frame $S'$ is
$$\begin{bmatrix} \gamma&-\gamma\beta_x&-\gamma\beta_y&-\gamma\beta_z\\ -\gamma\beta_x&1+(\gamma-1)\frac{\beta_x^2}{\beta^2}&(\gamma-1)\frac{\beta_x \beta_y}{\beta^2}&(\gamma-1)\frac{\beta_x \beta_z}{\beta^2}\\ -\gamma\beta_y&(\gamma-1)\frac{\beta_x \beta_y}{\beta^2}&1+(\gamma-1)\frac{\beta_y^2}{\beta^2}&(\gamma-1)\frac{\beta_y \beta_z}{\beta^2}\\ -\gamma\beta_z&(\gamma-1)\frac{\beta_x \beta_z}{\beta^2}&(\gamma-1)\frac{\beta_y \beta_z}{\beta^2}&1+(\gamma-1)\frac{\beta_z^2}{\beta^2}\\ \end{bmatrix}$$
where $\begin{bmatrix} \beta_x \\ \beta_y \\ \beta_z \end{bmatrix}$ is the velocity of $S'$ relative to $S$. Since the four-vector transforms by matrix multiplication, and both four-vectors are given, it is then a straightforward task to solve for the relative velocity in the matrix.

Since the velocity of the particle in the y direction is the same with respect to both frames, can we say that the relative velocity of the frames in the y direction at least is 0?

Yes, this implies that all relative velocity must either be zero or be in the perpendicular direction.

## Best Answer

Look at your last equation:

$$\gamma (w)\begin{pmatrix} 1 &\frac{w}{c} \\ \frac{w}{c} & 1 \end{pmatrix} = \gamma(u) \gamma(v)\begin{pmatrix} 1 & \frac{v}{c} \\ \frac{v}{c} & 1 \end{pmatrix}\begin{pmatrix} 1 & \frac{u}{c} \\ \frac{u}{c} & 1 \end{pmatrix}$$

It implies, on equating matrix elements, two equations:

$$\gamma(w)=\gamma(u)\,\gamma(v)\,\left(1+\frac{u\,v}{c^2}\right)\quad\text{(equate elements (1,1) and (2,2))}$$ $$\gamma(w)\,w=\gamma(u)\,\gamma(v)\,(u+v)\quad\text{(equate elements (1,2) )}$$

Now eliminate the $\gamma$s (

e.g.divide one by the other) and you should make some headway.My personal favorite way for this problem is to reason with rapidities. Rapidities, defined by $\eta = \operatorname{artanh}\left(\frac{v}{c}\right)$, combine linearly. This assertion is readily proven by writing the boost matrix as $\exp\left(\eta\,\left(\begin{array}{cc}0&1\\1&0\end{array}\right)\right)$. Then the velocity addition law simply follows from the addition identity for hyperbolic tangents $\tanh(\eta_1+\eta_2) = \frac{\tanh \eta_1+\tanh\eta_2}{1+\tanh\eta_1\,\tanh\eta_2}$.