# Special Relativity: Verifying a general boost matrix is in the Lorentz group

I'm attempting the problem shown below. Using the hint, I have so far found:

\begin{align}B^T \eta B &= \begin{pmatrix}\gamma & -\gamma\beta^j \\ -\gamma \beta_k & \delta_k^j+\frac{(\gamma -1)\beta^j \beta_k}{\beta^2}\end{pmatrix}. \begin{pmatrix}-1 & 0 \\ 0 & 1\end{pmatrix}. \begin{pmatrix}\gamma & -\gamma \beta_n\\ -\gamma\beta^m & \delta_n^m+\frac{(\gamma -1)\beta^m \beta_n}{\beta^2}\end{pmatrix}\\&= \begin{pmatrix}-\gamma^2(1-\beta^j \beta^m) & \gamma^2 \beta_n-\gamma \beta^j\Big(\delta_n^m + \frac{(\gamma -1)\beta^m \beta_n}{\beta ^2}\Big) \\ \gamma^2 \beta_k -\gamma \beta^m \Big(\delta_k^j + \frac{(\gamma -1)\beta^j \beta_k}{\beta^2}\Big) & -\gamma^2 \beta_k \beta_n + \Big(\delta_k^j + \frac{(\gamma -1)\beta^j \beta_k}{\beta^2}\Big)\Big(\delta_n^m + \frac{(\gamma -1)\beta^m \beta_n}{\beta^2}\Big)\end{pmatrix}\end{align}

However, I don't understand what is meant by time-time, time-space, space-time and space-space components. Some elaboration on what this means would be appreciated. Am I supposed to equate the above matrix components with $$\begin{pmatrix}-1 & 0 \\ 0 & 1\end{pmatrix}.$$

If so, what am I trying to solve for to truly show this is in the Lorentz group?

Please note, I'm only a novice with the summation notation, so I apologize if I've written anything out incorrectly.

## Problem I'm Attempting

Hints :

$$\text{your matrix :}\:\: \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \text{is this matrix :}\:\: \eta= \begin{bmatrix} -1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} -1 & \boldsymbol{0}^{\mathsf{T}} \\ \boldsymbol{0} & \mathrm{I} \end{bmatrix} \tag{01}$$ where $$\boldsymbol{0}\equiv \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix} =\text{the null column vector}, \qquad \boldsymbol{0}^{\mathsf{T}}\equiv \begin{bmatrix} 0 & 0 & 0 \end{bmatrix} =\text{the null row vector} \tag{02}$$

$$B= \begin{bmatrix} \gamma & -\gamma\beta^j \\ -\gamma \beta_k & \delta_k^j+\dfrac{(\gamma -1)\beta^j \beta_k}{\beta^2} \end{bmatrix} = \begin{bmatrix} \gamma & -\gamma\boldsymbol{\beta}^{\mathsf{T}} \\ -\gamma \boldsymbol{\beta} & \mathrm{I}+\dfrac{(\gamma -1)\boldsymbol{\beta} \boldsymbol{\beta}^{\mathsf{T}}}{\beta^2} \end{bmatrix} \tag{03}$$

$$\boldsymbol{\beta}= \begin{bmatrix} \beta_{1}\\ \beta_{2}\\ \beta_{3} \end{bmatrix}, \qquad \boldsymbol{\beta}^{\mathsf{T}}= \begin{bmatrix} \beta^{1} & \beta^{2} & \beta^{3} \end{bmatrix}, \qquad \mathrm{I}= \begin{bmatrix} 1&0&0\\ 0&1&0\\ 0&0&1 \end{bmatrix} \tag{04}$$

$$\boldsymbol{\beta}^{\mathsf{T}}\boldsymbol{\beta}= \begin{bmatrix} \beta^{1} & \beta^{2} & \beta^{3} \end{bmatrix} \begin{bmatrix} \beta_{1}\\ \beta_{2}\\ \beta_{3} \end{bmatrix} =\beta^{m}\beta_{m} =\boldsymbol{\beta}\boldsymbol{\cdot}\boldsymbol{\beta}=\Vert \boldsymbol{\beta} \Vert ^{2}=\beta^{2}=\dfrac{v^2}{c^2}=\dfrac{\gamma^2-1}{\gamma^2} \tag{05}$$

$$\boldsymbol{\beta}\boldsymbol{\beta}^{\mathsf{T}}= \begin{bmatrix} \beta^{1}\\ \beta^{2}\\ \beta^{3} \end{bmatrix} \begin{bmatrix} \beta_{1} & \beta_{2} & \beta_{3} \end{bmatrix} =\beta_{m}\beta^{n}= \begin{bmatrix} \beta_{1}^{2} & \beta_{1}\beta_{2} & \beta_{1}\beta_{3}\\ \beta_{2}\beta_{1} & \beta_{2}^{2} & \beta_{2}\beta_{3}\\ \beta_{3}\beta_{1} & \beta_{3}\beta_{2} & \beta_{3}^{2} \end{bmatrix} \tag{06}$$

$$\left(\boldsymbol{\beta}\boldsymbol{\beta}^{\mathsf{T}}\right)^{2}= \left(\boldsymbol{\beta}\boldsymbol{\beta}^{\mathsf{T}}\right)\left(\boldsymbol{\beta}\boldsymbol{\beta}^{\mathsf{T}}\right)= \boldsymbol{\beta}\underbrace{ \left (\boldsymbol{\beta}^{\mathsf{T}} \boldsymbol{\beta}\right)}_{\Vert \boldsymbol{\beta} \Vert ^{2}}\boldsymbol{\beta}^{\mathsf{T}}= \Vert \boldsymbol{\beta} \Vert ^{2}\boldsymbol{\beta}\boldsymbol{\beta}^{\mathsf{T}}=\left(\dfrac{{\gamma}^2-1}{\gamma^2}\right)\boldsymbol{\beta}\boldsymbol{\beta}^{\mathsf{T}} \tag{07}$$

Note that if $\:\mathbf{n}\:$ is a unit 3-vector then $\:\mathrm{P}_{\mathbf{n}}=\mathbf{n}\mathbf{n}^{\mathsf{T}}\:$ is the projection on its direction, since for every $\:\mathbf{x} \in \mathbb{R}^{3}\:$ $$\mathrm{P}_{\mathbf{n}}\mathbf{x}=\mathbf{n}\mathbf{n}^{\mathsf{T}} \mathbf{x}= \begin{bmatrix} n_{1}\\ n_{2}\\ n_{3} \end{bmatrix} \underbrace{ \begin{bmatrix} n_{1} & n_{2} &n_{3} \end{bmatrix} \begin{bmatrix} x_{1}\\ x_{2}\\ x_{3} \end{bmatrix}}_{\left(\mathbf{x} \boldsymbol{\cdot}\mathbf{n} \right)} =\left(\mathbf{n} \boldsymbol{\cdot}\mathbf{x} \right)\mathbf{n} \tag{08}$$ with the well known property of projections $$\mathrm{P}_{\mathbf{n}}^{2}=\mathrm{P}_{\mathbf{n}} \tag{09}$$ Defining $$\mathbf{n} \equiv \dfrac{\boldsymbol{\beta}}{\Vert \boldsymbol{\beta} \Vert}=\dfrac{\boldsymbol{\beta}}{\beta} \tag{10}$$ then $$\mathrm{P}_{\mathbf{n}}=\mathbf{n}\mathbf{n}^{\mathsf{T}}=\left( \dfrac{ \boldsymbol{\beta}}{\beta} \right) \left(\dfrac{\boldsymbol{\beta}}{\beta}\right)^{\mathsf{T}}=\dfrac{\boldsymbol{\beta}\boldsymbol{\beta}^{\mathsf{T}} }{\beta^{2}} \tag{11}$$ and (07) is its property as projection.

EDIT :

\begin{align} & B^{\mathsf{T}}\eta B =\\ &\begin{bmatrix} \gamma & -\gamma\boldsymbol{\beta}^{\mathsf{T}} \\ -\gamma \boldsymbol{\beta} & \mathrm{I}+\dfrac{(\gamma -1)\boldsymbol{\beta} \boldsymbol{\beta}^{\mathsf{T}}}{\beta^2} \end{bmatrix} \begin{bmatrix} -1 & \boldsymbol{0}^{\mathsf{T}}\\ &\\ \boldsymbol{0} & \mathrm{I} \end{bmatrix} \begin{bmatrix} \gamma & -\gamma\boldsymbol{\beta}^{\mathsf{T}} \\ -\gamma \boldsymbol{\beta} & \mathrm{I}+\dfrac{(\gamma -1)\boldsymbol{\beta} \boldsymbol{\beta}^{\mathsf{T}}}{\beta^2} \end{bmatrix}=\\ &\begin{bmatrix} \gamma & -\gamma\boldsymbol{\beta}^{\mathsf{T}} \\ -\gamma \boldsymbol{\beta} & \mathrm{I}+\dfrac{(\gamma -1)\boldsymbol{\beta} \boldsymbol{\beta}^{\mathsf{T}}}{\beta^2} \end{bmatrix} \begin{bmatrix} -\gamma & +\gamma\boldsymbol{\beta}^{\mathsf{T}} \\ -\gamma \boldsymbol{\beta} & \mathrm{I}+\dfrac{(\gamma -1)\boldsymbol{\beta} \boldsymbol{\beta}^{\mathsf{T}}}{\beta^2} \end{bmatrix}= \begin{bmatrix} \sigma & \boldsymbol{\rho}^{\mathsf{T}} \\ &\\ \boldsymbol{\rho} & \mathrm{Z} \end{bmatrix}\equiv \xi \tag{12} \end{align}

where $\:\xi\:$ a real symmetric $\:4\times 4\:$ matrix with elements $\:\sigma,\boldsymbol{\rho},\mathrm{Z}\:$ a real scalar, a real 3-vector and a real symmetric $\:3\times 3\:$ matrix respectively, all to be determined. Now, \begin{align} \sigma=-\gamma^{2}\underbrace{\left(1-\boldsymbol{\beta}^{\mathsf{T}}\boldsymbol{\beta}\right)}_{1-\tfrac{v^2}{c^2}=\gamma^{-2}} \quad \Longrightarrow \quad \sigma=-1 \tag{13a} \end{align}

\begin{align} \boldsymbol{\rho} & =\gamma^{2}\boldsymbol{\beta}-\gamma\boldsymbol{\beta}-\dfrac{\gamma(\gamma -1)\boldsymbol{\beta}\boldsymbol{\beta}^{\mathsf{T}} \boldsymbol{\beta}}{\beta^2}\\ & =\gamma\left(\gamma-1\right)\boldsymbol{\beta}-\gamma\left(\gamma-1\right)\boldsymbol{\beta}\underbrace{\left(\dfrac{\boldsymbol{\beta}^{\mathsf{T}}\boldsymbol{\beta}}{\beta^2}\right)}_{=1} \quad \Longrightarrow \quad \boldsymbol{\rho}=\boldsymbol{0} \tag{13b} \end{align}

\begin{align} \mathrm{Z} & =-\gamma^{2}\boldsymbol{\beta}\boldsymbol{\beta}^{\mathsf{T}}+\biggl(\mathrm{I}+\dfrac{(\gamma -1)\boldsymbol{\beta} \boldsymbol{\beta}^{\mathsf{T}}}{\beta^2}\biggr)^{2}\\ & = -\gamma^{2}\boldsymbol{\beta}\boldsymbol{\beta}^{\mathsf{T}}+\mathrm{I}+\dfrac{2(\gamma -1)\boldsymbol{\beta} \boldsymbol{\beta}^{\mathsf{T}}}{\beta^2}+\left(\gamma -1\right)^{2}\overbrace{\biggl(\dfrac{\boldsymbol{\beta} \boldsymbol{\beta}^{\mathsf{T}}}{\beta^2}\biggr)^{2}}^{=\boldsymbol{\beta}\boldsymbol{\beta}^{\mathsf{T}}/\beta^2}\\ & = \mathrm{I}+ \underbrace{\left[-\gamma^{2}+\dfrac{2(\gamma -1)}{\beta^2}+\dfrac{(\gamma -1)^{2}}{\beta^2} \right]}_{=0}\boldsymbol{\beta}\boldsymbol{\beta}^{\mathsf{T}}\quad \Longrightarrow \quad \mathrm{Z}=\mathrm{I} \tag{13c} \end{align}

So, $$B^{\mathsf{T}}\eta B = \xi = \begin{bmatrix} \sigma & \boldsymbol{\rho}^{\mathsf{T}} \\ &\\ \boldsymbol{\rho} & \mathrm{Z} \end{bmatrix}= \begin{bmatrix} -1 & \boldsymbol{0}^{\mathsf{T}}\\ &\\ \boldsymbol{0} & \mathrm{I} \end{bmatrix} \equiv \eta \tag{14}$$

QED.

• Thanks. I ended up solving the problem, although I did stick with the summation notation. Your hints did help. – Akyidrian Jun 7 '16 at 6:21
• @Akyidrian : As a rule of this site we must not post complete answers to homework-like exercises but hints only. Since you did solve the problem by yoursefl using summation notation, I complete my answer with vectors. I think you'll find it useful for comparison with your solution and for future use. – Frobenius Jun 7 '16 at 8:12

Another, less messy way to do this is as follows. Link the said matrix to the identity matrix by a path defined by:

$$\Lambda:\mathbb{R}\to \mathscr{M}_{4\times4};\;\Lambda(\zeta) = \left(\begin{array}{c|c}\cosh\zeta & -\hat{B}^T \,\sinh\zeta\\\hline -\hat{B}\,\sinh\zeta &\mathrm{id} +\hat{B}\,\hat{B}^T\, (\cosh\zeta-1) \end{array}\right)\tag{1}$$

where $\zeta = \operatorname{artanh}\frac{v}{c}$ is the rapidity of the putative boost. Here $\hat{B} =\frac{1}{v} \begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix}$ is the unit vector of direction cosines pointing along the direction of the boost.

Exercise:Check that all matrices of the stated form can be written in the form of (1), so that they all form a smooth path through the identity (where they pass when $\zeta=0$).

Note the useful little formula $\hat{B}^T\,\hat{B} = 1$, so that you can almost manipulate matrices of the form in (1) as though their elements were scalar, aside from that $\hat{B}\,\hat{B}^T$ is left unsimplified. You get things like $(\hat{B}\,\hat{B}^T)^N = \hat{B}\,\hat{B}^T;\,N\geq 1$ (this means that $(\hat{B}\,\hat{B}^T)^N$ is the idempotent projector onto the direction of the boost) which you can use in the following.

Excercise: Prove that

$$\Lambda(\zeta)\Lambda(-\zeta) = \mathrm{id}\tag{2}$$

whence:

$$\frac{\mathrm{d}\Lambda}{\mathrm{d}\zeta} \Lambda^{-1} = -\left(\begin{array}{c|c}0&\hat{B}^T\\\hline\hat{B}&0\end{array}\right)\tag{3}$$

Now, given (3), we have very simply:

$$\frac{\mathrm{d}}{\mathrm{d}\zeta} \left(\Lambda^T\,\eta\,\Lambda\right) = \Lambda^T\left(-\left(\begin{array}{c|c}0&\hat{B}^T\\\hline\hat{B}&0\end{array}\right)^T\,\eta - \eta \,\left(\begin{array}{c|c}0&\hat{B}^T\\\hline\hat{B}&0\end{array}\right)\right)\,\Lambda=0\tag{4}$$

and, since the sought identity follows trivially for $\zeta=0$, we have, through (4) a Cauchy initial value problem wherein the the derivative is a Lipschitz-continuous function of $\Lambda^T\,\eta\,\Lambda$, therefore $\Lambda^T\,\eta\,\Lambda=\eta$, true for all $\zeta$, is the unique solution to this CIVP and the identity is proven.

You've therefore got yourself a rather useful and compact expression for a general boost in (1), and you can see, in the light of (3), that it is the matrix exponential of $\zeta$ times the simple matrix on the right hand side of (3). The matrix on the right hand side of (3) is sometimes called an infinitessimal boost; all infinitessimal boosts are linear combinations, with the direction cosines as weights, of the three infinitessimal boosts for the three spatial co-ordinate directions.