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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 $W$. Btw, the extra term in Dirac-action $\int \mathrm d^4x \bar\psi(x) ig\gamma^{\mu}A_\mu(x)\psi(x)$ is also lorentz invariant, to see that, just transform $\psi(x) \to S[\Lambda]\psi(\Lambda^{-1}x)$ and use $S^{-1}[\Lambda]\gamma^{\mu}S[\Lambda]=\Lambda^{\mu}{}_{\nu}\gamma^{\nu}$ and $\Lambda^{-1}x=y$.

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