# Investigations into the infinitesimal Lorentz transformation

## Homework Statement

[/B]
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##.

## Homework Equations

3. The Attempt at a Solution [/B]

##\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?

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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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Gold Member
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
Homework Helper
Gold Member
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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Gold Member
Looks good.