- #1

han

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- Homework Statement
- Show that ##[\epsilon^{\mu\nu\rho\sigma} M_{\mu \nu}M_{\rho\sigma},M_{\alpha\beta}]=0## where ##M_{\mu\nu}## is a Lorentz boost generator

- Relevant Equations
- The commutation relation of ##M_{\mu\nu}## is given: $$[M_{\rho \sigma},M_{\alpha\beta}]=i(g_{\rho\beta}M_{\sigma\alpha}+g_{\sigma\alpha}M_{\rho\beta}-g_{\rho\alpha}M_{\sigma\beta}-g_{\sigma\beta}M_{\rho\alpha}).$$

Using above formula, I could calculate the given commutator.

$$

[\epsilon^{\mu\nu\rho\sigma} M_{\mu \nu}M_{\rho\sigma},M_{\alpha\beta}]=i\epsilon^{\mu\nu\rho\sigma}(M_{\mu \nu}[M_{\rho\sigma},M_{\alpha\beta}]+[M_{\rho\sigma},M_{\alpha\beta}]M_{\mu \nu})

$$

(because ##\epsilon^{\mu\nu\rho\sigma}=\epsilon^{\rho\sigma\mu\nu}##, ##(\mu\nu)\leftrightarrow(\rho\sigma)## preserves the result in the 2nd term)

$$

=i\epsilon^{\mu\nu\rho\sigma}(g_{\rho\beta}(M_{\mu\nu}M_{\sigma\alpha}+M_{\sigma\alpha}M_{\mu\nu})+g_{\sigma\alpha}(M_{\mu\nu}M_{\rho\beta}+M_{\rho\beta}M_{\mu\nu})-g_{\rho\alpha}(M_{\mu\nu}M_{\sigma\beta}+M_{\sigma\beta}M_{\mu\nu})-g_{\sigma\beta}(M_{\mu\nu}M_{\rho\alpha}+M_{\rho\alpha}M_{\mu\nu}))

$$

And my calculation stuck here. I could not find any clue that the terms in above formula cancel each other.

I personally checked that for a specific example like taking ##\alpha=1, \beta=2##, the commutator is indeed zero.

It feels like any sign in the 3rd or 4th term is miscalculated and symmetricity in ##\rho## and ##\sigma## combines with the antisymmetric tensor and give the result zero, but I could not find where did I make a mistake on the signs.

Additionally, the term ##\epsilon^{\mu\nu\rho\sigma} M_{\mu \nu}M_{\rho\sigma}## is not explicity zero for example on spinor, where ##M_{\mu \nu}=\frac{i}{4}[\gamma^{\mu},\gamma^{\nu}]##, you can check that the given expression is proportional to ##\gamma^5##.

Edit: I found out by directly calculating that $$

\epsilon^{\mu\nu\rho\sigma}g_{\rho\beta}(M_{\mu\nu}M_{\sigma\alpha}+M_{\sigma\alpha}M_{\mu\nu})$$

term itself is already zero. Again for example when ##\alpha=1,\beta=2## case,

$$

\begin{align}

\epsilon^{\mu\nu\rho\sigma}g_{\rho 2}(M_{\mu\nu}M_{\sigma 1}+M_{\sigma 1}M_{\mu\nu})\\ \nonumber

&=\epsilon^{\mu\nu 23}g_{22}(M_{\mu\nu}M_{31}+M_{31}M_{\mu\nu})+\epsilon^{\mu\nu 20}g_{22}(M_{\mu\nu}M_{01}+M_{01}M_{\mu\nu})\\ \nonumber

&=\epsilon^{0123}g_{22}(M_{01}M_{31}+M_{31}M_{01})+\epsilon^{1320}g_{22}(M_{13}M_{01}+M_{01}M_{13})\\ \nonumber

&=-(M_{01}M_{31}+M_{31}M_{01})+(M_{31}M_{01}+M_{01}M_{31})=0 \nonumber

\end{align}

$$

(Using ##g_{00}=+1, g_{11}=g_{22}=g_{33}=-1## convention)

So it's enough to show that the form ##\epsilon^{\mu\nu\rho\sigma}g_{\rho\beta}(M_{\mu\nu}M_{\sigma\alpha}+M_{\sigma\alpha}M_{\mu\nu})## is zero. But I have no clue how to show this formula is zero with algebraic steps, like switching indicies and cancel the terms out.

$$

[\epsilon^{\mu\nu\rho\sigma} M_{\mu \nu}M_{\rho\sigma},M_{\alpha\beta}]=i\epsilon^{\mu\nu\rho\sigma}(M_{\mu \nu}[M_{\rho\sigma},M_{\alpha\beta}]+[M_{\rho\sigma},M_{\alpha\beta}]M_{\mu \nu})

$$

(because ##\epsilon^{\mu\nu\rho\sigma}=\epsilon^{\rho\sigma\mu\nu}##, ##(\mu\nu)\leftrightarrow(\rho\sigma)## preserves the result in the 2nd term)

$$

=i\epsilon^{\mu\nu\rho\sigma}(g_{\rho\beta}(M_{\mu\nu}M_{\sigma\alpha}+M_{\sigma\alpha}M_{\mu\nu})+g_{\sigma\alpha}(M_{\mu\nu}M_{\rho\beta}+M_{\rho\beta}M_{\mu\nu})-g_{\rho\alpha}(M_{\mu\nu}M_{\sigma\beta}+M_{\sigma\beta}M_{\mu\nu})-g_{\sigma\beta}(M_{\mu\nu}M_{\rho\alpha}+M_{\rho\alpha}M_{\mu\nu}))

$$

And my calculation stuck here. I could not find any clue that the terms in above formula cancel each other.

I personally checked that for a specific example like taking ##\alpha=1, \beta=2##, the commutator is indeed zero.

It feels like any sign in the 3rd or 4th term is miscalculated and symmetricity in ##\rho## and ##\sigma## combines with the antisymmetric tensor and give the result zero, but I could not find where did I make a mistake on the signs.

Additionally, the term ##\epsilon^{\mu\nu\rho\sigma} M_{\mu \nu}M_{\rho\sigma}## is not explicity zero for example on spinor, where ##M_{\mu \nu}=\frac{i}{4}[\gamma^{\mu},\gamma^{\nu}]##, you can check that the given expression is proportional to ##\gamma^5##.

Edit: I found out by directly calculating that $$

\epsilon^{\mu\nu\rho\sigma}g_{\rho\beta}(M_{\mu\nu}M_{\sigma\alpha}+M_{\sigma\alpha}M_{\mu\nu})$$

term itself is already zero. Again for example when ##\alpha=1,\beta=2## case,

$$

\begin{align}

\epsilon^{\mu\nu\rho\sigma}g_{\rho 2}(M_{\mu\nu}M_{\sigma 1}+M_{\sigma 1}M_{\mu\nu})\\ \nonumber

&=\epsilon^{\mu\nu 23}g_{22}(M_{\mu\nu}M_{31}+M_{31}M_{\mu\nu})+\epsilon^{\mu\nu 20}g_{22}(M_{\mu\nu}M_{01}+M_{01}M_{\mu\nu})\\ \nonumber

&=\epsilon^{0123}g_{22}(M_{01}M_{31}+M_{31}M_{01})+\epsilon^{1320}g_{22}(M_{13}M_{01}+M_{01}M_{13})\\ \nonumber

&=-(M_{01}M_{31}+M_{31}M_{01})+(M_{31}M_{01}+M_{01}M_{31})=0 \nonumber

\end{align}

$$

(Using ##g_{00}=+1, g_{11}=g_{22}=g_{33}=-1## convention)

So it's enough to show that the form ##\epsilon^{\mu\nu\rho\sigma}g_{\rho\beta}(M_{\mu\nu}M_{\sigma\alpha}+M_{\sigma\alpha}M_{\mu\nu})## is zero. But I have no clue how to show this formula is zero with algebraic steps, like switching indicies and cancel the terms out.

Last edited: