# Variation of vector field under Lorentz transformation and gauge transformation

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.

• What paper are you talking about? – DanielC Jan 6 '19 at 21:27
• Can you write out the lagrangian your are transforming? that makes all the difference – InertialObserver Jan 6 '19 at 22:24
• Where does $\alpha$ come from? Why consider gauge and Lorentz transformation at the same time? – my2cts Jan 7 '19 at 0:11
• @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. – Chetan Pandey Jan 7 '19 at 3:40
• @InertialObserver we are just looking at the transformation of the four vector field; there is no Lagrangian. – Chetan Pandey Jan 7 '19 at 3:46