I'm trying to find the infinitesimal Lorentz transformation of a rank 2 antisymmetric tensor. Looking through Peskin, all I can see is the transformation of a vector, and even there it is simply given. I thought about developing it by writing it as a tensor product of two rank-1 tensors. This gives me: $$ C_{\mu\nu} = A_\mu B_\nu - A_\nu B_\mu\\ \delta C_{\mu\nu} = \delta A_\mu B_\nu + A_\mu\delta B_\nu - \delta A_\nu B_\mu - A_\nu\delta B_\mu $$ where $$ \delta A_\mu = \epsilon_\mu^{\phantom\mu\nu}A_\nu - \epsilon^\nu_{\phantom\nu\mu}A_\nu $$

Can this also be generalized for any rank-n tensor? Is this the way to find the spin tensor for every integer Lorentz representation?

  • $\begingroup$ Please note that questions whose potential answer is a simple "Yes." are not a good fit for the SE format. $\endgroup$
    – ACuriousMind
    Jun 14, 2016 at 12:48
  • $\begingroup$ I see what you mean. Is it better to just ask what is the transformation, without showing my idea on the matter? $\endgroup$
    – golanor
    Jun 14, 2016 at 12:51
  • 1
    $\begingroup$ To my mind your question is perfectly fine as is, in part because your approach is incorrect. It's helpful to see what kinds of mistakes you're making in order to give an effective answer. $\endgroup$
    – user_35
    Jun 14, 2016 at 12:54
  • $\begingroup$ No, showing your idea is definitely better than showing none. This type of questions is difficult to ask here because they quickly run afoul of our homework policy. I'm honestly not sure how to make this into a good question, although I think there should be a version of this that is on-topic... $\endgroup$
    – ACuriousMind
    Jun 14, 2016 at 12:57
  • $\begingroup$ What do you think about this version? $\endgroup$
    – golanor
    Jun 14, 2016 at 13:01

1 Answer 1


Sort of, except that you can't generally decompose a rank-2 tensor into a product of rank-1 tensors.

Let $\Lambda^{\mu}{}_{\nu}$ be an arbitrary Lorentz transformation. As you probably saw in Peskin, this transformation acts on vectors as

$$x^{\mu} \mapsto \Lambda^{\mu}{}_{\nu} x^{\nu}.$$

We can extend this principle to a tensor with an arbitrary number of up-indices. For example, for a rank-2 tensor $T^{\mu \nu},$ we have

$$T^{\mu \nu} \mapsto \Lambda^{\mu}{}_{\rho} \Lambda^{\nu}{}_{\sigma} T^{\rho \sigma}.$$

So, for example, since the statement that $\Lambda$ is a Lorentz transformation is equivalent to the statement that it leaves the Minkowski metric $\eta^{\mu \nu}$ invariant, $\Lambda$ must satisfy $\eta^{\mu \nu} \mapsto \eta^{\mu \nu},$ or

$$\Lambda^{\mu}{}_{\rho} \Lambda^{\nu}{}_{\sigma} \eta^{\rho \sigma} = \eta^{\mu \nu}.$$

Now, how should $\Lambda$ act on down-indices? Well, we can obtain a down-index from an up-index by lowering using the metric. So, starting with $x_{\mu} = \eta_{\mu \nu} x^{\nu}$ and using the fact that $\Lambda$ leaves the metric invariant, we have

$$x_{\mu} = \eta_{\mu \nu} x^{\nu} \mapsto \eta_{\mu \nu} \Lambda^{\nu}{}_{\rho} x^{\rho} = \Lambda_{\mu}{}^{\rho} x_{\rho}.$$

This tells us how $\Lambda$ should act on down-indices. However, in the particular case of Lorentz transformations, the tensor $\Lambda_{\mu}{}^{\rho}$ has a particular property. If we multiply it by the tensor $\Lambda^{\tau}{}_{\rho},$ we find

$$\Lambda^{\tau}{}_{\rho} \Lambda_{\mu}{}^{\rho} = \Lambda^{\tau}{}_{\rho} (\eta_{\mu \nu} \Lambda^{\nu}{}_{\sigma} \eta^{\sigma \rho}) = \eta_{\mu \nu} (\Lambda^{\nu}{}_{\sigma} \Lambda^{\tau}{}_{\rho} \eta^{\sigma \rho}) = \eta_{\mu \nu} \eta^{\nu \tau} = \delta_{\mu}{}^{\tau}.$$

So we have $\Lambda^{\tau}{}_{\rho} \Lambda_{\mu}{}^{\rho} = \delta_{\mu}{}^{\tau} = \Lambda^{\tau}{}_{\rho} (\Lambda^{-1})^{\rho}{}_{\mu}.$

So, generally, we conclude $\Lambda_{\mu}{}^{\rho} = (\Lambda^{-1})^{\rho}{}_{\mu}.$ So while up-indices transform naturally under $\Lambda$, down-indices transform naturally under $\Lambda^{-1}.$ That is,

$$x_{\mu} \mapsto \Lambda_{\mu}{}^{\rho} x_{\rho} = (\Lambda^{-1})^{\rho}{}_{\mu} x_{\rho}.$$

Now it's easy to answer your original question of "how does a rank-2 tensor $C_{\mu \nu}$ transform under a Lorentz transformation?" As in the case of multiple up-indices, we can just extend our principle to see

$$ C_{\mu \nu} \mapsto (\Lambda^{-1})^{\rho}{}_{\mu} (\Lambda^{-1})^{\sigma}{}_{\nu} C_{\rho \sigma}. $$

Or, equivalently,

$$ C_{\mu \nu} \mapsto \Lambda_{\mu}{}^{\rho} \Lambda_{\nu}{}^{\sigma} C_{\rho \sigma}. $$

  • $\begingroup$ Yes, but this is for a finite transformation. I'm trying to understand what is the spin tensor for an antisymmetric (or a general) rank-2 tensor, which is given by the infinitesimal transformation, not the finite one. $\endgroup$
    – golanor
    Jun 14, 2016 at 12:58
  • $\begingroup$ Oh I completely missed the "infinitesimal" part of your question! You can write an infinitesimal Lorentz transformation as the identity plus some antisymmetric tensor omega, and apply the same principle while keeping only terms up to first order in omega. $\endgroup$
    – user_35
    Jun 14, 2016 at 13:05
  • $\begingroup$ But perhaps it would help to have a better idea of exactly what you're trying to compute. Different techniques are more useful in different circumstances. $\endgroup$
    – user_35
    Jun 14, 2016 at 13:06
  • $\begingroup$ I want to find the generalisation of $$\left[S_{\alpha\beta}\right]^\mu_\nu = i(\delta^\mu_\alpha\eta_{\beta\nu}-\delta^\mu_\beta\eta_{\alpha\nu})$$ to rank-2 tensors. $\endgroup$
    – golanor
    Jun 14, 2016 at 13:08

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