6
$\begingroup$

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 $A=A_{\alpha}\text{d}x^{\alpha}$.

$\endgroup$
1
5
$\begingroup$

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 $A^\mu = (\phi, \mathbf{A})$.

$\endgroup$
3
  • $\begingroup$ 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
4
$\begingroup$

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

Proof:

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.

$\endgroup$
12
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.