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Does the vector potential $A_\mu$ transform when we merely relabel events in space-time (coordinate transformation), or does it transform with the basis vectors of a tangent space in which it lives?

If the latter, is there any difference between stating that we always have to transform the tangent space with the coordinates, and just saying that $A_\mu$ transforms with the coordinates?

Since it is part of the covariant derivative $\partial_\mu - i e A_\mu$ and the gauge transformation goes like $A_\mu \to A_\mu + \partial_\mu \lambda$, I would think it should transform with the coordinates, as does $\partial_\mu$. I don't see how the gauge transformation can be consistent if $A_\mu$ and $\partial_\mu \lambda$ don't transform in the same way.

Edit: I'm sorry, I should have given more context to the question. Thank you for the well reasoned answers so far. I'm thinking of the action for a Dirac field: $$ \int \mathrm d^4 x \, i\bar\psi \gamma^\mu (\partial_\mu - i e A_\mu) \psi $$ I'm imagining breaking up the transformation into two steps: first transforming the coordinates and then transforming the basis/components of the fields. Let's assume that the volume element is invariant. After the coordinate transformation, we have $$ \int \mathrm d^4 x \, i\bar\psi(\Lambda^{-1}x) \gamma^\mu (\partial_\mu - i e A_\mu(\Lambda^{-1}x)) \psi(\Lambda^{-1}x) $$ Since we're integrating over all of space, this is equivalent to $$ \int \mathrm d^4 x \, i\bar\psi(x) \gamma^\mu (\Lambda_\mu^{\phantom\mu\nu}\partial_\nu - i e A_\mu(x)) \psi(x) $$ Then we complete the transformation by changing the basis/components: $$ \int \mathrm d^4 x \, i\bar\psi(x) S^{-1} \gamma^\mu \Lambda_\mu^{\phantom\mu\nu}(\partial_\nu - i e A_\nu(x)) (S \psi(x)) $$ Alternatively, it could be that changing the coordinates (first step) would have to also change the components of $A_\mu$ in the third equation, which would be invariant in the second step.

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    $\begingroup$ I do not understand where the difference between "transforming with the coordinates" and "transforming as a tangent vector" is supposed to be. Can you give an example for objects that transform as the former, but not as the latter, and vice versa? $\endgroup$ – ACuriousMind Mar 9 '19 at 3:19
  • $\begingroup$ The components of a vector are defined with respect to a basis. The basis is usually set up so that it is oriented along the directions in which the coordinates increase, but it doesn't have to be. So we could change the coordinates without changing the basis vectors and components of the four-vector in question – unless the components are defined in terms of the coordinates, like $\partial_\mu \lambda$. I'll try to think of an example. $\endgroup$ – rossng Mar 9 '19 at 3:33
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Please Notice that $A_\mu$ was $\textbf{coefficients}$ of the tensor. Your tensor was invariant and written $\sum A_{\mu_i} d\mu_i$ where $d\mu_i$ were basis. The way people treated it was to use $A_\mu$ the coefficients and doing the algebra there. The tensor itself was always invariant, but $A_\mu$'s representation may differ in different coordinates. The derivation of the transformation formula actually needed to include the basis.

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I'm not entirely clear that I understand the question, but the complete transformation properties of spin-1 fields transforms under the Lorentz group as

$$ A_\mu \to \Lambda_{\mu}^{\ \nu} A_\nu(\Lambda^{-1} x). $$

I have taken the "passive" transformation convention where I have chosen to transform the field $A_\mu$ and keep the point of observation fixed.

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  • $\begingroup$ Yes, I'm having a bit of trouble communicating what I mean. How about this: If I multiply the x coordinate of every point in space by 2 and consider that my new x coordinate, then the derivative of a scalar field with respect to x will double. If that's all I do, will $A_\mu$ transform in the way you wrote? Or will it only become $A_\mu(\Lambda^{-1}x)$, and some extra step is needed to make the Greek index transform? Thanks- $\endgroup$ – rossng Mar 9 '19 at 1:46
  • $\begingroup$ I think you may be confused on the meaning of "transforming" a field. The entire goal of transforming a field is to see what it looks like in a new coordinate system: If I just rotate the coordinates but leave the field the same, that's equivalent to just changing the field but leaving the coordinates fixed (think temperature map). However, if I transform both nothing will change at all. $\endgroup$ – InertialObserver Mar 9 '19 at 1:55
  • $\begingroup$ But to answer your question, under an arbitrary Lorentz transformation the coordinates transform in the way in my answer. $\endgroup$ – InertialObserver Mar 9 '19 at 1:57
  • $\begingroup$ I guess what I'm asking is whether it's mathematically possible to rotate the coordinates without mixing $A_1,A_2,A_3$, in the same sense that it is not mathematically possible to rotate the coordinates without mixing $\partial_1 \lambda, \partial_2 \lambda, \partial_3 \lambda$. $\endgroup$ – rossng Mar 9 '19 at 3:02
  • $\begingroup$ What you’re describing is just the field evaluated at x’=Rx $\endgroup$ – InertialObserver Mar 9 '19 at 3:05
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rossng asked: Does the vector potential Aμ transform when we merely relabel events in space-time (coordinate transformation), or does it transform with the basis vectors of a tangent space in which it lives?

The rule to transform from the coordiantes $x$ to the coordinates $\bar{x}$ is

$$ \bar{A}_{\beta}(\bar{x}) = \frac{\partial x^{\gamma}}{\partial \bar{x}^{\beta}} A_{\gamma}(x) $$

for more details see here.

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Does the vector potential $A_\mu$ transform when we merely relabel events in space-time (coordinate transformation), or does it transform with the basis vectors of a tangent space in which it lives?

transformations in spacetime like lorentz boosts ($\Lambda^{\beta}{}_{\alpha} x^\alpha$) transforms $A_\alpha$ into $\Lambda^{\beta}{}_{\alpha}A_{\beta}({\Lambda}^{-1}x)$. Note that $A_\mu$ is an covariant tensor, and transforms under changes in basis, See

is there any difference between stating that we always have to transform the tangent space with the coordinates, and just saying that $A_\mu$ transforms with the coordinates?

in curved coordinates which tangent spaces varies, $A=A^{\alpha}e_{\alpha}$ has a curvature form $F$ defined by $F=dA$, in indice notation $F=dA$ can be rewrite as $F_{\alpha\beta}={\partial}_{\alpha}A_{\beta}-{\partial}_{\beta}A_{\alpha}$, just turns $\partial_\mu$ into Levi-civita's $\nabla_\mu$.

Since it is part of the covariant derivative $\partial_\mu−ieA_\mu$ and the gauge transformation goes like $A_\mu \to A_\mu + \partial_\mu \lambda$, I would think it should transform with the coordinates, as does $\partial_\mu$. I don't see how the gauge transformation can be consistent if $A_\mu$ and $\partial_\mu \lambda$ don't transform in the same way.

the covariant derivative is an object in QFT's constructed to make field lagrangians invariant under some groups transformations. $U(1)$ generates a local phase change in matter fields, to avoid "unphysical" local terms in the lagrangian, is introduced the covariant derivative. Also note that $A_\mu$ satisfies $\partial^\mu F_{\mu\nu}=0$, therefore arbitrary transformations like $\partial_\nu h(x)$ does not change physics of the system

edit: I don't know if this will answner your question, but fields are maps from the Manifold to his "space" i.e. $F:M\to W$; an arbitrary scalar field $\phi$ can be denoted as $\phi:M \to \mathbb{R}$, vectorial field can be represented as $A:M \to V$, and an spinor field as $\psi:M \to S$, what you're doing is transforming $M$, and later $W$.

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    $\begingroup$ Lorentz transformations do change $A_\mu$. Covariant does not mean invariant. $\endgroup$ – G. Smith Mar 9 '19 at 4:13
  • $\begingroup$ Yes, what i mean't with "invariant" was that $A_\mu$ changes "invariantly" with respect to the local basis; i made some edits on the post. $\endgroup$ – user164843 Mar 11 '19 at 1:21

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