In special relativity, when working with coordinates $(x^0, x^1, x^2, x^3)$ , the matrix corresponding to a Lorentz boost in the $x^2$ direction is $$ \begin{pmatrix} \gamma & -\gamma \beta & 0 & 0 \\ 0 & 1 & 0 & 0 \\ – \gamma \beta & 0 & \gamma & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}. $$ Similarly, the one corresponding to the $x^3$ direction is $$ \begin{pmatrix} \gamma & -\gamma \beta & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ – \gamma \beta & 0 & 0 & \gamma \end{pmatrix}. $$ I now want to compute the matrix corresponding to a boost in the $x^2 + x^3$ direction. I looked around on the internet and it seems that the product of two non-parellel boosts is not a boost. So my idea was this: I first boost in the $x^2$ direction (the $y$-axis), then I rotate my coordinate axes so that the $x^3$ axis (i.e. $z$-axis) gets mapped to the $x^2$-axis by rotating clockwise about the $x^1$-axis, that is, by applying the matrix $$ R = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & \cos \theta & – \sin \theta \\ 0 & 0 & \sin \theta & \cos \theta \end{pmatrix} $$ with $\theta = \pi/2$. Then I apply a boost again in the $x^2$ direction. Finally I rotate back by applying $R^{-1}$. So that would give me $$ R^{-1} \Lambda R \Lambda. $$
Can someone tell me if this is the correct reasoning, and that this is the way one finds out a general boost?
Best Answer
The Wikipedia General Lorentz Boosts article, from which the equations below are taken, stresses at least 3 times that, due to the non-commutation of boosts and rotations, that calculating what I think you are doing, is a non-trivial exercise, possibly involving computations as detailed below.
I don't really consider this an answer, I just wanted to draw your attention to potential problems in calculation. My apologies if it's familiar to you already.
The general boost matrix is
$${\displaystyle B({\boldsymbol {\zeta }})=e^{-{\boldsymbol {\zeta }}\cdot \mathbf {K} }}$$
where $K = (K_x, K_y, K_z)$ is a vector of matrices with components
$${\displaystyle K_{x}={\begin{bmatrix}0&1&0&0\\1&0&0&0\\0&0&0&0\\0&0&0&0\\\end{bmatrix}}\,,\quad K_{y}={\begin{bmatrix}0&0&1&0\\0&0&0&0\\1&0&0&0\\0&0&0&0\end{bmatrix}}\,,\quad K_{z}={\begin{bmatrix}0&0&0&1\\0&0&0&0\\0&0&0&0\\1&0&0&0\end{bmatrix}}}$$
and expanding the exponential in a Taylor series reveals a hyperbolic analogue of Rodrigues' rotation formula,
$${\displaystyle B({\boldsymbol {\zeta }})=I-\sinh \zeta (\mathbf {n} \cdot \mathbf {K} )+(\cosh \zeta -1)(\mathbf {n} \cdot \mathbf {K} )^{2}}$$
which reproduces the boost matrix after converting the rapidity back to relative velocity. Similarly the general rotation matrix is
$${\displaystyle R({\boldsymbol {\theta }})=e^{{\boldsymbol {\theta }}\cdot \mathbf {J} }}$$
where $J = (J_x, J_y, J_z)$ and
$${\displaystyle J_{x}={\begin{bmatrix}0&0&0&0\\0&0&0&0\\0&0&0&-1\\0&0&1&0\\\end{bmatrix}}\,,\quad J_{y}={\begin{bmatrix}0&0&0&0\\0&0&0&1\\0&0&0&0\\0&-1&0&0\end{bmatrix}}\,,\quad J_{z}={\begin{bmatrix}0&0&0&0\\0&0&-1&0\\0&1&0&0\\0&0&0&0\end{bmatrix}}}$$
and Taylor expanding the exponential reveals the actual Rodrigues' rotation formula
$${\displaystyle R({\boldsymbol {\theta }})=I+\sin \theta (\mathbf {e} \cdot \mathbf {J} )+(1-\cos \theta )(\mathbf {e} \cdot \mathbf {J} )^{2}\,.}$$
The axis-angle vector θ and rapidity vector ζ are altogether six continuous variables which make up the group parameters (in this particular representation), and J and K are the corresponding six generators of the group. Physically, the generators of the Lorentz group correspond to important symmetries in spacetime: J are the rotation generators which correspond to angular momentum, and K are the boost generators which correspond to the motion of the system in spacetime.
The derivative of any smooth curve $A(t)$ with $A(0) = I$ in the group depending on some group parameter t with respect to that group parameter, evaluated at t = 0, serves as a definition of a corresponding group generator $X$, and this reflects an infinitesimal transformation away from the identity. The smooth curve can always be taken as an exponential as the exponential will always map $X$ smoothly back into the group via t → exp(tX) for all t; this curve will yield $X$ again when differentiated at t = 0.
Three of the commutation relations of the Lorentz generators are
$${\displaystyle {\begin{aligned}\lbrack J_{x},J_{y}\rbrack &=J_{z}\\\lbrack K_{x},K_{y}\rbrack &=-J_{z}\\\lbrack J_{x},K_{y}\rbrack &=K_{z}\end{aligned}}}$$
where the bracket $[A, B] = AB − BA$ is a binary operation known as the commutator, and the other relations can be found by taking cyclic permutations of $x, y, z$ components (i.e. change x to y, y to z, and z to x, repeat).