# How does one include the magnetic scalar potential in the electromagnetic four-potential?

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?

• 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. – 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.

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