# Investigations into the infinitesimal Lorentz transformation

1. Homework Statement

A Lorentz transformation $x^{\mu} \rightarrow x'^{\mu} = {\Lambda^{\mu}}_{\nu}x^{\nu}$ is such that it preserves the Minkowski metric $\eta_{\mu\nu}$, meaning that $\eta_{\mu\nu}x^{\mu}x^{\nu}=\eta_{\mu\nu}x'^{\mu}x'^{\nu}$ for all $x$. Show that this implies that $\eta_{\mu\nu} = \eta_{\sigma\tau}{\Lambda^{\sigma}}_{\mu}{\Lambda^{\tau}}_{\nu}$.

Use this result to show that an infinitesimal transformation of the form ${\Lambda^{\mu}}_{\nu}={\delta^{\mu}}_{\nu}+{\omega^{\mu}}_{\nu}$ is a Lorentz transformation when $\omega^{\mu\nu}$ is antisymmetric: i.e. $\omega^{\mu\nu}=-\omega^{\nu\mu}$.

Write down the matrix form for ${\omega^{\mu}}_{\nu}$ that corresponds to a rotation through an infinitesimal angle $\theta$ about the $x^{3}$-axis.

Do the same for a boost along the $x^{1}$-axis by an infinitesimal velocity $v$.

2. Homework Equations

3. The Attempt at a Solution

$\eta_{\mu\nu}x^{\mu}x^{\nu}=\eta_{\mu\nu}x'^{\mu}x'^{\nu}$

$\implies \eta_{\mu\nu}x^{\mu}x^{\nu}=\eta_{\mu\nu}({\Lambda^{\mu}}_{\rho}x^{\rho})({\Lambda^{\mu}}_{\sigma}x^{\sigma})$

$\implies \eta_{\mu\nu}x^{\mu}x^{\nu}=\eta_{\rho\sigma}({\Lambda^{\rho}}_{\mu}x^{\mu})({\Lambda^{\sigma}}_{\nu}x^{\nu})$

$\implies \eta_{\mu\nu}x^{\mu}x^{\nu}=\eta_{\rho\sigma}{\Lambda^{\rho}}_{\mu}{\Lambda^{\sigma}}_{\nu}x^{\mu}x^{\nu}$

$\implies \eta_{\mu\nu}=\eta_{\rho\sigma}{\Lambda^{\rho}}_{\mu}{\Lambda^{\sigma}}_{\nu}$

Am I correct so far?

## Answers and Replies

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TSny
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$\implies \eta_{\mu\nu}x^{\mu}x^{\nu}=\eta_{\rho\sigma}{\Lambda^{\rho}}_{\mu}{\Lambda^{\sigma}}_{\nu}x^{\mu}x^{\nu}$

$\implies \eta_{\mu\nu}=\eta_{\rho\sigma}{\Lambda^{\rho}}_{\mu}{\Lambda^{\sigma}}_{\nu}$

Am I correct so far?
Yes, I think so. But someone might want to see explicitly how you get the last line from the next to last line.

Well, the next to last line consists of a sum of terms $\eta_{\mu\nu}x^{\mu}x^{\nu}$ and $\eta_{\rho\sigma}{\Lambda^{\rho}}_{\mu}{\Lambda^{\sigma}}_{\nu}x^{\mu}x^{\nu}$ with all possible combinations of values $0,1,2,3$ for the indices $\mu$ and $\nu$.

However, the last line only contains the terms $\eta_{\mu\nu}$ and $\eta_{\rho\sigma}{\Lambda^{\rho}}_{\mu}{\Lambda^{\sigma}}_{\nu}$ for a specific value of $\mu$ and $\nu$.

I thought that this was obvious, so I decided to skip the explanation. Isn't my reasoning sound, though?

TSny
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If I have two matrices $A_{\mu \nu}$ and $B_{\mu \nu}$ that satisfy $A_{\mu \nu}x^{\mu}x^{\nu} = B_{\mu \nu}x^{\mu}x^{\nu}$ for all possible $x^{\mu}$, can I conclude that $A_{\alpha \beta} = B_{\alpha \beta}$ for all $\alpha$ and $\beta$?

I think so, yes.

TSny
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Try to show explicitly that $A_{12} = B_{12}$.

Oh wait. The correct relation is $A_{12}+A_{21}=B_{12}+B_{21}$.

TSny
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Oh wait. The correct relation is $A_{12}+A_{21}=B_{12}+B_{21}$.
Right.

Well! In that case, I need to rewrite my solution:

$\eta_{\mu\nu}x^{\mu}x^{\nu}=\eta_{\mu\nu}x'^{\mu}x'^{\nu}$

$\implies \eta_{\mu\nu}x^{\mu}x^{\nu}=\eta_{\mu\nu}({\Lambda^{\mu}}_{\sigma}x^{\sigma})({\Lambda^{\nu}}_{\tau}x^{\tau})$

$\implies \eta_{\mu\nu}x^{\mu}x^{\nu}=\eta_{\sigma\tau}({\Lambda^{\sigma}}_{\mu}x^{\mu})({\Lambda^{\tau}}_{\nu}x^{\nu})$

$\implies \eta_{\mu\nu}x^{\mu}x^{\nu}=\eta_{\sigma\tau}{\Lambda^{\sigma}}_{\mu}{\Lambda^{\tau}}_{\nu}x^{\mu}x^{\nu}$

$\implies \eta_{\mu\nu}+\eta_{\nu\mu}=\eta_{\sigma\tau}{\Lambda^{\sigma}}_{\mu}{\Lambda^{\tau}}_{\nu}+\eta_{\sigma\tau}{\Lambda^{\sigma}}_{\nu}{\Lambda^{\tau}}_{\mu}$, since $x^{\mu}x^{\nu}=x^{\nu}x^{\mu}$

$\implies \eta_{\mu\nu}+\eta_{\mu\nu}=\eta_{\sigma\tau}{\Lambda^{\sigma}}_{\mu}{\Lambda^{\tau}}_{\nu}+\eta_{\tau\sigma}{\Lambda^{\tau}}_{\mu}{\Lambda^{\sigma}}_{\nu}$, since the metric tensor is symmetric, i.e. $\eta^{\mu\nu}=\eta^{\nu\mu}$

$\implies \eta_{\mu\nu}+\eta_{\mu\nu}=\eta_{\sigma\tau}{\Lambda^{\sigma}}_{\mu}{\Lambda^{\tau}}_{\nu}+\eta_{\sigma\tau}{\Lambda^{\sigma}}_{\mu}{\Lambda^{\tau}}_{\nu}$

$\implies 2 \eta_{\mu\nu}= 2 \eta_{\sigma\tau}{\Lambda^{\sigma}}_{\mu}{\Lambda^{\tau}}_{\nu}$

$\implies \eta_{\mu\nu}=\eta_{\sigma\tau}{\Lambda^{\sigma}}_{\mu}{\Lambda^{\tau}}_{\nu}$

Is my solution correct?

Last edited:
TSny
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Yes. The symmetry of the metric tensor is used here.

Next, I need to use the result $\eta_{\mu\nu}=\eta_{\sigma\tau}{\Lambda^{\sigma}}_{\mu}{\Lambda^{\tau}}_{\nu}$ to show that an infinitesimal transformation of the form ${\Lambda^{\mu}}_{\nu}={\delta^{\mu}}_{\nu}+{\omega^{\mu}}_{\nu}$ is a Lorentz transformation when $\omega^{\mu\nu}$ is antisymmetric: i.e. $\omega^{\mu\nu}=-\omega^{\nu\mu}$:

$\eta_{\mu\nu}=\eta_{\sigma\tau}{\Lambda^{\sigma}}_{\mu}{\Lambda^{\tau}}_{\nu}$

$\implies \eta_{\mu\nu}=\eta_{\sigma\tau}({\delta^{\sigma}}_{\mu}+{\omega^{\sigma}}_{\mu})({\delta^{\tau}}_{\nu}+{\omega^{\tau}}_{\nu})$

$\implies \eta_{\mu\nu}=\eta_{\sigma\tau}({\delta^{\sigma}}_{\mu}{\delta^{\tau}}_{\nu}+{\delta^{\sigma}}_{\mu}{\omega^{\tau}}_{\nu}+{\omega^{\sigma}}_{\mu}{\delta^{\tau}}_{\nu}+{\omega^{\sigma}}_{\mu}{\omega^{\tau}}_{\nu})$

$\implies \eta_{\mu\nu}=\eta_{\mu\nu}+\eta_{\mu\tau}{\omega^{\tau}}_{\nu}+\eta_{\nu\sigma}{\omega^{\sigma}}_{\mu}+\eta_{\sigma\tau}{\omega^{\sigma}}_{\mu}{\omega^{\tau}}_{\nu}$

$\implies \eta_{\mu\tau}{\omega^{\tau}}_{\nu}+\eta_{\nu\sigma}{\omega^{\sigma}}_{\mu} = 0$, where we neglect the term $\eta_{\sigma\tau}{\omega^{\sigma}}_{\mu}{\omega^{\tau}}_{\nu}$ because it is of second order in the infinitesimal ${\omega^{\mu}}_{\nu}$.

$\implies \omega_{\mu\nu}+\omega_{\nu\mu}=0$

$\implies \omega_{\mu\nu}=-\omega_{\nu\mu}$

$\implies \omega^{\mu\nu}=-\omega^{\nu\mu}$

so that $\omega^{\mu\nu}$ is antisymmetric.

Is my solution correct?

TSny
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Yes.

Finally, the remaining parts of the problem:

The matrix form for ${\Lambda^{\mu}}_{\nu}$ that corresponds to a rotation through a finite angle $\theta$ about the $x^{3}$-axis is given by

${\Lambda^{\mu}}_{\nu} = \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & \text{cos}\ \theta & -\text{sin}\ \theta & 0 \\ 0 & \text{sin}\ \theta & \text{cos}\ \theta & 0 \\ 0 & 0 & 0 & 1 \end{array} \right) = \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right) + \left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & -\theta & 0 \\ 0 & \theta & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right) + \cdots ={\delta^{\mu}}_{\nu}+{\omega^{\mu}}_{\nu}+\cdots ,$

where we used $\text{sin}\ \theta = \theta - \frac{\theta^{3}}{3!}+\cdots$ and $\text{cos}\ \theta = 1 + \frac{\theta^{2}}{2!}+\cdots$ and we only kept terms up to linear order in $\theta$ in the expansion of ${\Lambda^{\mu}}_{\nu}$,

so the matrix form for ${\omega^{\mu}}_{\nu}$ that corresponds to a rotation through an infinitesimal angle $\theta$ about the $x^{3}$-axis is given by

${\omega^{\mu}}_{\nu}= \left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & -\theta & 0 \\ 0 & \theta & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right).$

The matrix form for ${\Lambda^{\mu}}_{\nu}$ that corresponds to a boost along the $x^{1}$-axis by a finite velocity $v$ is given by

${\Lambda^{\mu}}_{\nu} = \left( \begin{array}{cccc} \gamma & -\gamma v & 0 & 0 \\ -\gamma v & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right) = \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right) + \left( \begin{array}{cccc} 0 & -v & 0 & 0 \\ v & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right) + \cdots ={\delta^{\mu}}_{\nu}+{\omega^{\mu}}_{\nu}+\cdots ,$

where we used $\gamma =\frac{1}{\sqrt{1-v^{2}}}=(1-v^{2})^{-1/2}=1+\frac{v^{2}}{2}+\dots$ and $\gamma v =v\frac{1}{\sqrt{1-v^{2}}}=v(1-v^{2})^{-1/2}=v+\frac{v^{3}}{2}+\dots$ and we only kept terms up to linear order in $v$ in the expansion of ${\Lambda^{\mu}}_{\nu}$,

so the matrix form for ${\omega^{\mu}}_{\nu}$ that corresponds to a boost along the $x^{1}$-axis by an infinitesimal velocity $v$ is given by

${\omega^{\mu}}_{\nu}= \left( \begin{array}{cccc} 0 & -v & 0 & 0 \\ v & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right).$

Is this solution correct?

TSny
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The matrix form for ${\Lambda^{\mu}}_{\nu}$ that corresponds to a boost along the $x^{1}$-axis by a finite velocity $v$ is given by

${\Lambda^{\mu}}_{\nu} = \left( \begin{array}{cccc} \gamma & -\gamma v & 0 & 0 \\ -\gamma v & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right) = \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right) + \left( \begin{array}{cccc} 0 & -v & 0 & 0 \\ v & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right) + \cdots$
Check the sign in the first column of the last matrix on the right side.

It is a typo. We should instead have the following:

${\Lambda^{\mu}}_{\nu} = \left( \begin{array}{cccc} \gamma & -\gamma v & 0 & 0 \\ -\gamma v & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right) = \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right) + \left( \begin{array}{cccc} 0 & -v & 0 & 0 \\ -v & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right) + \cdots ={\delta^{\mu}}_{\nu}+{\omega^{\mu}}_{\nu}+\cdots$

so that

${\omega^{\mu}}_{\nu}= \left( \begin{array}{cccc} 0 & -v & 0 & 0 \\ -v & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right)$.

I believe everything else is correct, isnt it?

TSny
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Looks good.