I'm supposed to consider a Lorentz transformation of the form $$\Lambda^{\mu}_{\ \nu} = \delta^{\mu}_{\ \nu} + \omega^{\mu}_{\ \nu},$$ where $\omega$ is some tensor.

Since Lorentz tranformations satisfy $$\Lambda^{\mu}_{\ \sigma} \eta^{\sigma \rho} \Lambda^{\nu}_{\ \rho}= \eta^{\mu \nu}$$ (where $\eta$ is the Minkowski metric), I was able to find that $\omega$ satisfies the condition;

$$\eta^{\mu \rho} \omega^{\nu}_{\ \rho} + \eta^{\nu \rho} \omega^{\mu}_{\ \rho} + \omega^{\mu}_{\ \sigma} \eta^{\sigma \rho} \omega^{\nu}_{\ \rho} = 0$$

$\ $

Now my question is asking me to consider a Klein-Gordon field $\phi$, and Taylor expand it, so that I can find the variation $\delta \phi$ in terms of the $\omega$.

Where do I begin with this?

My attempt:

I'm assuming that I'm taking a tranformation $\phi(x) \mapsto \phi( x + \delta x ) = \phi(x) + \delta \phi$.

A guess I have, using a Taylor-type expansion, is:

$$\phi( x + \delta x ) = \phi(x) + (\delta x)^{\mu} \partial_{\mu} \phi(x) + ....,$$ so that $$\delta\phi \approx (\delta x)^{\mu} \partial_{\mu} \phi(x)$$

But I am really quite lost from here...how do I incorporate $\omega$ into the above?

EDIT: I made a mistake; the tensor $\omega$ is infinitesimal, and this condition results in $\omega^{\mu \nu} = - \omega^{\nu \mu}$, so that $\omega$ is anti-symmetric.

  • 2
    $\begingroup$ Well, what is $\delta x$? It comes from the Lorentz transformation. $\endgroup$
    – Javier
    Oct 17, 2016 at 23:38

1 Answer 1


Thanks @Javier , I now understand:

If I take $x \mapsto x^{\prime} = \tilde{x} + \delta x$ to be a Lorentz transformation, then I write:

$\tilde{x}^{\mu} = \Lambda^{\mu}_{\ \nu} x^{\nu} = \delta^{\mu}_{\ \nu} x^{\nu} + \omega^{\mu}_{\ \nu} x^{\nu} = x^{\mu} + \omega^{\mu}_{\ \nu} x^{\nu}$

Then obviously, we have $(\delta x)^{\mu}=\omega^{\mu}_{\ \nu} x^{\nu}$.

And so: $\delta \phi = (\delta x)^{\mu} \partial_{\mu} \phi(x) = \omega^{\mu}_{\ \nu} x^{\nu} \partial_{\mu} \phi(x)$

I think this is right


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