In a paper I am reading, it is stated that under a Lorentz transformation, the coordinates transform as $x^{\mu} \to \Lambda^{\mu}_{\nu}x^{\nu}$, and so the change in the (vector) field at the same value of the co-ordinate is $\delta A^{\mu}=\omega^{\mu}_{\nu}A^{\nu}+\partial^{\mu}\alpha$, where $\alpha$ is a gauge transformation, which we use to preserve the gauge choice.

When I try to find the variation, I am getting $\delta A^{\mu}=\omega^{\mu}_{\nu}A^{\nu}(x)-\omega^{\nu}_ix^i\partial_{\nu}{A'}^{\mu}(x)$. Now I understand that in order to preserve the gauge choice, we would have to include an extra parameter $\partial^{\mu}\alpha$, however I can't seem to get rid of the extra term $-\omega^{\nu}_ix^i\partial_{\nu}{A'}^{\mu}(x)$. A similar question was also asked on this website here, but it is also unanswered.

  • $\begingroup$ What paper are you talking about? $\endgroup$ – DanielC Jan 6 '19 at 21:27
  • $\begingroup$ Can you write out the lagrangian your are transforming? that makes all the difference $\endgroup$ – InertialObserver Jan 6 '19 at 22:24
  • $\begingroup$ Where does $\alpha$ come from? Why consider gauge and Lorentz transformation at the same time? $\endgroup$ – my2cts Jan 7 '19 at 0:11
  • $\begingroup$ @my2cts If we are working in a gauge, say Light cone gauge, then after a lorentz transformation, to remain in the gauge, the field should transform like that. $\endgroup$ – Chetan Pandey Jan 7 '19 at 3:40
  • $\begingroup$ @InertialObserver we are just looking at the transformation of the four vector field; there is no Lagrangian. $\endgroup$ – Chetan Pandey Jan 7 '19 at 3:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.