The electromagnetic four-potential is typically written as $$A_{\mu}=(\phi,\vec{A}),$$ where $\vec{A}$ is the magnetic vector potential, and $\phi$ is the electric scalar potential.

Should I want to use the magnetic scalar potential, $\phi_M$, instead of $\vec{A}$, what would the expression of $A_\mu$ look like?

  • $\begingroup$ It doesn't work. The magnetic scalar potential is an aid to simplifying some calculations, but it is not part of any 4-vector as far as I know. $\endgroup$ – Andrew Steane Oct 15 at 22:11

From the Helmholtz decomposition theorem, for any solenoidal field $$\mathbf B = \nabla \times \mathbf A$$ on some bounded domain $V\subseteq \mathbb R^3$, then

$$ \mathbf A(\mathbf r) = \frac{1}{4\pi} \int_V d^3 r' \frac{\nabla' \times \mathbf B(\mathbf r')}{|\mathbf r - \mathbf r'|} - \frac{1}{4\pi} \oint_{S} dS' \ \left(\hat n' \times \frac{\mathbf B(\mathbf r')}{|\mathbf r - \mathbf r'|}\right)$$

where $S$ is the surface enclosing $V$ and $\hat n'$ is the outward-facing normal vector.

For magnetostatic situations in the absence of any current density (in which $\nabla \times \mathbf B=0$, and so $\mathbf B = \nabla \phi_M$) it follows that

$$ \mathbf A = \frac{1}{4\pi} \oint dS' \left(\hat n' \times \frac{\nabla' \phi_M(\mathbf r')}{|\mathbf r - \mathbf r'|}\right)$$

Explicitly writing $A_\mu$ in terms of $\phi_M$ is unpleasant, but it is possible, if you are so inclined.

  • $\begingroup$ Thank you! This sounds convincing... although, as you say, writing it might be indeed unpleasant... $\endgroup$ – Hans Castrop Oct 16 at 8:08

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