# Is the Lorentz Boost Generator Commutator Zero?

• han
han
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.

Last edited:
han said:
[....]

Using above formula, I could calculate the given commutator.
$$[\epsilon^{\mu\nu\rho\sigma} M_{\mu \nu}M_{\rho\sigma},M_{\alpha\beta}]=i\epsilon^{\mu\nu\rho\sigma}(M_{\mu \nu}[M_{\rho\sigma},M_{\alpha\beta}]+[M_{\rho\sigma},M_{\alpha\beta}]M_{\mu \nu})$$
I believe this should read...
$$[\epsilon^{\mu\nu\rho\sigma} M_{\mu \nu}M_{\rho\sigma},M_{\alpha\beta}]=i\epsilon^{\mu\nu\rho\sigma}(M_{\mu \nu}[M_{\rho\sigma},M_{\alpha\beta}]+[M_{\mu\nu},M_{\alpha\beta}]M_{\rho \sigma})$$
Lie brackets obey a Leibniz rule: [AB,C] = A[B,C]+[A,C]B.
In detail: [AB,C]=ABC-CAB = ABC-ACB+ACB-CAB.

vanhees71
jambaugh said:
I believe this should read...
$$[\epsilon^{\mu\nu\rho\sigma} M_{\mu \nu}M_{\rho\sigma},M_{\alpha\beta}]=i\epsilon^{\mu\nu\rho\sigma}(M_{\mu \nu}[M_{\rho\sigma},M_{\alpha\beta}]+[M_{\mu\nu},M_{\alpha\beta}]M_{\rho \sigma})$$
Lie brackets obey a Leibniz rule: [AB,C] = A[B,C]+[A,C]B.
In detail: [AB,C]=ABC-CAB = ABC-ACB+ACB-CAB.
You can check that
$$[\epsilon^{\mu\nu\rho\sigma} M_{\mu \nu}M_{\rho\sigma},M_{\alpha\beta}]=\epsilon^{\mu\nu\rho\sigma}(M_{\mu \nu}[M_{\rho\sigma},M_{\alpha\beta}]+[M_{\mu\nu},M_{\alpha\beta}]M_{\rho \sigma})=\epsilon^{\mu\nu\rho\sigma}(M_{\mu \nu}[M_{\rho\sigma},M_{\alpha\beta}]+[M_{\rho\sigma},M_{\alpha\beta}]M_{\mu \nu})$$

Because
$$\epsilon^{\mu\nu\rho\sigma}[M_{\mu\nu},M_{\alpha\beta}]M_{\rho \sigma})=\epsilon^{\rho\sigma\mu\nu}[M_{\rho \sigma},M_{\alpha\beta}]M_{\mu\nu})=\epsilon^{\mu\nu\rho\sigma}[M_{\rho \sigma},M_{\alpha\beta}]M_{\mu\nu})$$
##\mu\nu\rho\sigma## and ##\rho\sigma\mu\nu## both are even permutations.

## What is a Lorentz Boost Generator?

A Lorentz Boost Generator is an operator in the context of special relativity and quantum field theory that generates Lorentz boosts, which are transformations between different inertial frames moving at constant velocities relative to each other. These generators form part of the Lorentz algebra, which describes the symmetries of spacetime.

## What does it mean for the commutator of Lorentz Boost Generators to be zero?

If the commutator of Lorentz Boost Generators were zero, it would imply that the operations of applying two different boosts in sequence would be independent of the order in which they are applied. In mathematical terms, for two generators $$K_i$$ and $$K_j$$, this would mean $$[K_i, K_j] = 0$$.

## Is the commutator of Lorentz Boost Generators actually zero?

No, the commutator of Lorentz Boost Generators is not zero. In fact, the commutator of two Lorentz Boost Generators is related to the generators of rotations. Specifically, $$[K_i, K_j] = -i \epsilon_{ijk} J_k$$, where $$\epsilon_{ijk}$$ is the Levi-Civita symbol and $$J_k$$ are the rotation generators. This non-zero commutator reflects the non-commutative nature of Lorentz transformations.

## Why is the commutator of Lorentz Boost Generators not zero?

The commutator of Lorentz Boost Generators is not zero because Lorentz boosts and rotations form a non-abelian Lie algebra, specifically the Lorentz algebra. This algebra encapsulates the fact that performing boosts in different directions does not commute and can result in a rotation, reflecting the complex structure of spacetime symmetries.

## What are the implications of a non-zero commutator for physical theories?

A non-zero commutator of Lorentz Boost Generators has significant implications for physical theories, particularly in special relativity and quantum field theory. It means that the order in which transformations are applied affects the outcome, leading to richer and more complex behavior of particles and fields under Lorentz transformations. This non-commutativity is essential for the consistency of relativistic theories and has profound effects on the structure of spacetime and the behavior of fundamental particles.

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