I have seen both equations, $A^{\alpha}=(\phi,\vec{A})$ is from Wikipedia and $A_{\alpha}=(\phi,\vec{A})$ is from my lecture. Which one is right?

My thoughts: As far as I know, $A$ is a 1-form, so $A(p)\in T_p^*M$ for all $p\in M\subset\mathbb R^4$. In addition, $(\text{d}x^1{}_p,...,\text{d}x^4{}_p)$ with $(x^1,...,x^4):=\text{id}_M$ is a basis of $T_p^*M$ and since we normally write the coefficients of dual vectors with the index below, I'd say $A=A_{\alpha}\text{d}x^{\alpha}$.


As the existing answer says, it's $A^\mu = (\phi, \mathbf{A})$. Here's a simple way to see that. In Lorenz gauge, the equation of motion is $$\partial^2 A^\mu = J^\mu.$$ We also know that $J^\mu = (\rho, \mathbf{J})$, and that the components of this equation are $$\partial^2 \phi = \rho, \quad \partial^2 \mathbf{A} = \mathbf{J}.$$ There are no minus signs anywhere, so we must have $A^\mu = (\phi, \mathbf{A})$.

  • $\begingroup$ First of all, thank you for your answer. Unfortunately, I don't know enough about electrodynamics yet to decide which answer to accept. Could you please give a reference (or even better: several references) for your equations? $\endgroup$ – Filippo Nov 2 '20 at 20:44
  • $\begingroup$ @Filippo You can find them in any standard electromagnetism textbook, such as Griffiths. Also, the other answer is equally correct. $\endgroup$ – knzhou Nov 2 '20 at 20:55
  • $\begingroup$ That's great, thank you very much :) $\endgroup$ – Filippo Nov 2 '20 at 21:34

It's $A^\mu = (\phi, \mathbf{A})$, no matter your metric convention.


I can never remember the signs in the correspondence between $F_{\mu\nu}$ and the electric and magnetic field (especially since they depend on the metric signature), so instead let's look at the Lorentz force per unit charge

$$f^\mu = F^\mu{}_\nu u^\nu.$$

The part of $f^i$ proportional to $u^0$ will then be the electric field $E^i$:

$$f^i = F^i{}_\mu u^\mu = F^i{}_0 u^0 + \dots = (\partial^i A_0 - \partial_0 A^i) u^0 + \dots.$$

Now, no matter the metric signature we have $\partial^i A_0 = -\partial_i A^0$, because we're flipping one time index and one space index, so we arrive at

$$E^i = -\partial_i A^0 - \partial_0 A^i.$$

Comparting with the known formula $\mathbf{E} = - \nabla \phi - \partial \mathbf{A}/\partial t$, we see that the contravariant components of $A^\mu$ are the potentials with no sign change.

  • 1
    $\begingroup$ Actually, the 4-force being the time derivative of the 4-momentum is also a co-vector, or a 1-form. So it should read $f_\mu = F_{\mu\nu} u^{\nu}$. $\endgroup$ – DanielC Nov 1 '20 at 21:57
  • $\begingroup$ It doesn't matter whether the force is "naturally" a vector or covector; at the end of the day, the equation of motion is $m\, d^2x^\mu / d\tau^2 = f^\mu$, so the contravariant components are the classical force (modulo some gamma factors and stuff like that). $\endgroup$ – Javier Nov 1 '20 at 22:29
  • $\begingroup$ The diff-geom. formulation of electromagnetism (in terms of connections on principal bundles) is usually done in -+++ metric, while it uses only forms (the E and B in nonrelativistic notation are 1-forms in R^3 and 2-forms, respectively. see Felsager, page 376 onwards. $\endgroup$ – DanielC Nov 1 '20 at 22:34
  • $\begingroup$ Do you only disagree on the proof, or on the answer to my question? $\endgroup$ – Filippo Nov 1 '20 at 22:40
  • $\begingroup$ I disagree that E and A are vectors in R^3, which is understood from boldface. $\endgroup$ – DanielC Nov 1 '20 at 22:51

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