How to deduce the general boost transformation matrix?

Question

The general matrix for a boost in an arbitrary direction, is given by:

$$\Lambda (\vec{v})= \begin{pmatrix} \gamma & -\gamma\beta_x & -\gamma\beta_y & -\gamma\beta_z\\ -\gamma\beta_x & 1+(1-\gamma)\frac{\beta_x^2}{\beta^2} & (\gamma-1)\frac{\beta_x\beta_y}{\beta^2} & (\gamma-1)\frac{\beta_x\beta_z}{\beta^2}\\ -\gamma\beta_y & (\gamma-1)\frac{\beta_x\beta_y}{\beta^2} & 1+(\gamma-1)\frac{\beta_y^2}{\beta^2} & (\gamma-1)\frac{\beta_y\beta_z}{\beta^2}\\ -\gamma\beta_z & (\gamma-1)\frac{\beta_x\beta_z}{\beta^2} & (\gamma-1)\frac{\beta_y\beta_z}{\beta^2} & 1+ (\gamma-1)\frac{\beta_z^2}{\beta^2} \end{pmatrix} $$

And the boosts in the cartesian directions, are given by

$$ \Lambda_x = \begin{pmatrix} \gamma_x & -\gamma_x\beta_x & 0 & 0\\ -\gamma_x & \gamma_x & 0 & 0 \\ 0 & 0 &1&0\\ 0 & 0& 0 & 1\\ \end{pmatrix}, $$ $$ \Lambda_y = \begin{pmatrix} \gamma_y & 0 & -\gamma_y\beta_y & 0 \\ 0 & 1 & 0 & 0\\ -\gamma_y\beta_y & 0 & \gamma_y & 0 \\ 0 & 0 & 0 &1 \end{pmatrix},$$ $$ \Lambda_z = \begin{pmatrix} \gamma_z & 0 & 0 & -\gamma_z\beta_z \\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ -\gamma_z\beta_z & 0 & 0 & \gamma_z \end{pmatrix}, $$

Where $\gamma_i = \sqrt{1-\beta_i^2}$, and $\beta_i = \frac{v_i}{c} $ for $i=x,y,z$.

From what I have searched, in order to have a continuous transformation, we can obtain the general matrix by applying successive boosts and rotations, but for me is not clear exactly how to combine this to get the general boost matrix. What I have tried is: $\Lambda_xR_x\Lambda_yR_z\Lambda_x$, this is; the first boost in the $x$ direction, then rotate around the $z$-axes counterclockwise by 90 degrees to position the $x$-axis onto the $y$-axis, boost in the $y$ direction, rotate about the $x$-axis by 90 degrees counterclockwise to position the actual $y$-axis on the $z$-axis and finally boost in the $z$-direction. However, this doesn't give the general boost. Since rotations do not commute, I think there must be a concrete order.

On the other hand; I have looked at the way the boost in cartesian $x$-direction is obtained through the Lie Theory of infinitesimal rotations, but I would like to obtain the general matrix with this procedure. If possible, I would appreciate any other method to obtain the general boost in a formal way.

Answer 1

Let us use bold letters to denote the spatial components of vectors. One convenient way of obtaining the general Lorentz boost with velocity $\boldsymbol{\beta}$ is to consider separately the transformations of the parallel and perpendicular components of the position vector, here denoted $\mathbf{x}_{\parallel}$ and $\mathbf{x}_{\perp}$. We can express these two vectors easily in terms of $\mathbf{x}$ and $\boldsymbol{\beta}$:

\begin{equation} \mathbf{x}_{\parallel} = \frac{\mathbf{x} \cdot \boldsymbol{\beta}}{\beta^2} \boldsymbol{\beta}, \quad\mathbf{x}_{\perp} = \mathbf{x} - \mathbf{x}_{\parallel} \end{equation}

Moreover, they also have simple Lorentz transformations:

\begin{equation} x^{\prime 0} = \gamma\left( x^0 - \boldsymbol{\beta} \cdot \mathbf{x}\right) \\ \mathbf{x}_{\parallel}^{\prime} = \gamma\left( \mathbf{x}_{\parallel} - \boldsymbol{\beta} x^0\right) \\ \mathbf{x}_{\perp}^{\prime} = \mathbf{x}_{\perp} \end{equation}

From this we obtain the transformation for $\mathbf{x}$ and put everything in index notation:

\begin{align} \mathbf{x}^{\prime} & = \mathbf{x}_{\parallel}^{\prime} + \mathbf{x}_{\perp}^{\prime} \\ & = \gamma\left( \mathbf{x}_{\parallel} - \boldsymbol{\beta} x^0\right) + \mathbf{x}_{\perp} \\ & = \gamma\left( \frac{\mathbf{x} \cdot \boldsymbol{\beta}}{\beta^2} \boldsymbol{\beta} - \boldsymbol{\beta} x^0\right) + \mathbf{x} - \frac{\mathbf{x} \cdot \boldsymbol{\beta}}{\beta^2} \boldsymbol{\beta} \\ & = \mathbf{x} + (\gamma - 1) \frac{\mathbf{x} \cdot \boldsymbol{\beta}}{\beta^2} \boldsymbol{\beta} - \gamma \boldsymbol{\beta} x^0 \\ \implies x^{\prime i} & = \left[\delta^i_k + (\gamma - 1) \frac{\beta^i \beta_k}{\beta^2}\right] x^k - \gamma \beta^i x^0 \end{align}

Together with $x^{\prime 0} = \gamma\left( x^0 - \beta_i x^i\right)$, we can then simply read off the expression for the boost matrix:

\begin{equation} \Lambda^{0}_{0} = \gamma \\ \Lambda^{0}_{i} = - \gamma \beta_i \\ \Lambda^{i}_{j} = \delta^{i}_j + (\gamma - 1) \frac{\beta^i \beta_j}{\beta^2} \end{equation}

Answer 2

I didn't try this but I'm pretty sure this is how it works:

Imagine you have a vector $\vec v$.

Now we rotate that vector such that it is parallel to an axis of our choice say $z$. Then we boost along the z-axis. Then we rotate the vector back to its original position:

$R_z(\phi)R_y(\theta)B_zR_y(-\theta)R_z(-\phi) v$

Then by group theoretical arguments, multiplying the first matrices should give you the wanted matrix.