[Physics] Lorentz transformation of vector field

coordinate systemsfield-theorylorentz-symmetrypoincare-symmetryVector Fields

Under a Lorentz transformation, a vector field transforms as: $$A'_{\mu}(x')=\Lambda^{\nu}_{\mu}A_{\nu}(x)$$

My question is, why is the Lorentz transformed vector field evaluated at $$x'=\Lambda x$$, rather than at $$x$$ only. One way I can make sense of it is if I take the example of a scalar field under a rotation, my transformed field evaluated at the rotated coordinates is same as my original field evaluated at the original coordinates, but somehow, I am not able to find a convincing argument for the vector field case.

Precisely because of the reason you have put forward,

the Lorentz transformation for a scalar field $$\phi(x)$$ is usually written as:

$$\phi(x) \rightarrow \phi(x)' = \phi(\Lambda^{-1}x),$$

and for a vector field $$V^{\mu}(x)$$ is usually written as:

$$V^\mu(x) \rightarrow V^\mu(x)' = \Lambda^{\mu}_{\nu}\,V^{\nu}(\Lambda^{-1}x).$$

So that you leave your coordinates $$x$$ fixed.

Hence, your expression is correct, but it’s in terms of the rotated coordinate system $$x’$$. You can just write $$x = \Lambda^{-1} \, x’$$ in the RHS and get to my expression.

EDIT

For clarity, the last expression may also be written as:

$$V^\mu(x) \rightarrow V^\mu(x')' = \Lambda^{\mu}_{\nu}\,V^{\nu}(x),$$ just by replacing $$x \rightarrow x'$$ and using $$x = \Lambda^{-1} \, x’$$.