7.3 Magnetic vector potential

The magnetic vector potential $\vec{A}$ provides an alternative way to describe magnetic fields that often makes calculations more tractable. Instead of working directly with $\vec{B}$, you define a vector field whose curl gives you the magnetic field. This automatically satisfies $\nabla \cdot \vec{B} = 0$ and opens the door to powerful techniques like gauge freedom and multipole expansions.

Definition of magnetic vector potential

The magnetic vector potential $\vec{A}$ is a vector field defined so that the magnetic field can be recovered from it via a curl operation. It doesn't correspond to a directly measurable quantity the way $\vec{B}$ does, but it streamlines many calculations, especially for current distributions and time-varying fields.

A key feature of $\vec{A}$ is that it is not unique. You can add the gradient of any scalar function to $\vec{A}$ without changing the physical magnetic field. This freedom is called gauge invariance, and choosing a particular gauge amounts to imposing an extra condition on $\vec{A}$ to pin down a unique solution.

Relationship to magnetic field

The defining relation is:

$\vec{B} = \nabla \times \vec{A}$

This single equation does two things at once. First, it gives you a recipe for computing $\vec{B}$ from $\vec{A}$. Second, it automatically guarantees $\nabla \cdot \vec{B} = 0$, because the divergence of any curl is identically zero. That's the mathematical statement that magnetic monopoles don't exist.

Different choices of $\vec{A}$ can produce the same $\vec{B}$. Specifically, if $\vec{A}$ is a valid vector potential, then so is $\vec{A}' = \vec{A} + \nabla \lambda$ for any scalar function $\lambda$, since $\nabla \times (\nabla \lambda) = 0$.

Gauge transformations

A gauge transformation replaces $\vec{A}$ with $\vec{A}' = \vec{A} + \nabla \lambda$ (and correspondingly $\phi' = \phi - \partial \lambda / \partial t$ for the scalar potential). The fields $\vec{B}$ and $\vec{E}$ are unchanged. This freedom lets you choose whichever $\vec{A}$ makes your problem easiest to solve.

Coulomb gauge

The Coulomb gauge imposes:

$\nabla \cdot \vec{A} = 0$

This is the natural choice in magnetostatics. With this condition, the vector identity that appears when you substitute $\vec{B} = \nabla \times \vec{A}$ into Ampère's law simplifies directly to Poisson's equation:

$\nabla^2 \vec{A} = -\mu_0 \vec{J}$

Each Cartesian component of $\vec{A}$ independently satisfies a scalar Poisson equation, which is a huge computational advantage. The Coulomb gauge also decouples $\vec{A}$ from the scalar potential $\phi$ in the static case.

Lorenz gauge

The Lorenz gauge imposes:

$\nabla \cdot \vec{A} + \frac{1}{c^2} \frac{\partial \phi}{\partial t} = 0$

This is the preferred gauge in electrodynamics because it treats space and time symmetrically. Both potentials then satisfy inhomogeneous wave equations:

$\nabla^2 \vec{A} - \frac{1}{c^2}\frac{\partial^2 \vec{A}}{\partial t^2} = -\mu_0 \vec{J}$

$\nabla^2 \phi - \frac{1}{c^2}\frac{\partial^2 \phi}{\partial t^2} = -\frac{\rho}{\epsilon_0}$

The Lorenz gauge is also manifestly Lorentz covariant, which matters when you move to relativistic formulations.

Calculation from current distribution

Biot-Savart law for vector potential

For a volume current density $\vec{J}(\vec{r}')$, the vector potential in the Coulomb gauge is:

$\vec{A}(\vec{r}) = \frac{\mu_0}{4\pi} \int \frac{\vec{J}(\vec{r}')}{|\vec{r} - \vec{r}'|} \, d^3\vec{r}'$

This is the direct analog of the Coulomb integral for the scalar potential $\phi$ in electrostatics, with $\rho/\epsilon_0$ replaced by $\mu_0 \vec{J}$. For a line current $I$ along a path, the integral reduces to:

$\vec{A}(\vec{r}) = \frac{\mu_0 I}{4\pi} \int \frac{d\vec{l}'}{|\vec{r} - \vec{r}'|}$

Notice that $\vec{A}$ points in the same direction as the current, which is often a useful intuitive check.

Examples of simple current distributions

Infinite straight wire carrying current $I$ along $\hat{z}$: The vector potential has only a $z$-component and, up to a constant that depends on your reference point, goes as:

$\vec{A}(\vec{r}) = -\frac{\mu_0 I}{2\pi} \ln(s) \, \hat{z}$

where $s$ is the perpendicular distance from the wire. Taking the curl recovers the familiar $\vec{B} = \frac{\mu_0 I}{2\pi s}\hat{\phi}$.

Circular current loop of radius $R$ carrying current $I$: On the axis (distance $z$ from the center), symmetry forces $\vec{A} = 0$ because the azimuthal components cancel by symmetry at axial points. Off-axis, $\vec{A}$ has only a $\hat{\phi}$ component and generally requires elliptic integrals to evaluate. The far-field form is captured by the dipole term of the multipole expansion.

Boundary conditions

At an interface between two regions, the vector potential must satisfy conditions that enforce continuity of the physical field $\vec{B}$.

Continuity of the tangential component of $\vec{A}$

The tangential components of $\vec{A}$ are continuous across any boundary (assuming no delta-function singularities in the current):

$\vec{A}_1^{\text{tan}} = \vec{A}_2^{\text{tan}}$

This follows from requiring that the normal component of $\vec{B} = \nabla \times \vec{A}$ be continuous, which is itself a consequence of $\nabla \cdot \vec{B} = 0$.

Discontinuity related to surface currents

When a surface current density $\vec{K}$ exists at the boundary, the standard boundary condition on $\vec{B}$ is:

$\hat{n} \times (\vec{B}_1 - \vec{B}_2) = \mu_0 \vec{K}$

In terms of $\vec{A}$, this translates into a condition on the normal derivatives of $\vec{A}$ across the surface rather than a simple jump in $\vec{A}$ itself. The vector potential is typically taken to be continuous across the boundary, with the discontinuity appearing in $\partial \vec{A}/\partial n$.

Poisson's equation for vector potential

Derivation from Maxwell's equations

Starting from Ampère's law (magnetostatic form):

1. Write $\nabla \times \vec{B} = \mu_0 \vec{J}$.

2. Substitute $\vec{B} = \nabla \times \vec{A}$ to get $\nabla \times (\nabla \times \vec{A}) = \mu_0 \vec{J}$.

3. Apply the vector identity: $\nabla \times (\nabla \times \vec{A}) = \nabla(\nabla \cdot \vec{A}) - \nabla^2 \vec{A}$.

4. Choose the Coulomb gauge $\nabla \cdot \vec{A} = 0$.

5. Arrive at: $\nabla^2 \vec{A} = -\mu_0 \vec{J}$

This is three decoupled scalar Poisson equations, one for each Cartesian component of $\vec{A}$.

Solution methods

• Direct integration: The Green's function for the Laplacian gives the Biot-Savart integral for $\vec{A}$ shown above.
• Separation of variables: Works well when the current distribution and boundaries respect a coordinate system (cylindrical, spherical, etc.).
• Numerical methods: Finite element or finite difference methods handle arbitrary geometries where analytic solutions aren't available.

Once you have $\vec{A}$, take $\vec{B} = \nabla \times \vec{A}$ to get the magnetic field.

Magnetic vector potential in magnetostatics

Current-carrying wires

For the infinite straight wire along $\hat{z}$, the vector potential is:

$\vec{A} = -\frac{\mu_0 I}{2\pi} \ln s \, \hat{z}$

Taking the curl in cylindrical coordinates:

$\vec{B} = \nabla \times \vec{A} = -\frac{\partial A_z}{\partial s}\hat{\phi} = \frac{\mu_0 I}{2\pi s}\hat{\phi}$

This confirms the result you already know from Ampère's law. The vector potential approach is more useful for configurations without enough symmetry to apply Ampère's law directly.

Solenoids and toroids

Ideal solenoid ($n = N/L$ turns per unit length, current $I$, radius $R$): Inside the solenoid, $\vec{B} = \mu_0 n I \, \hat{z}$ is uniform. The vector potential consistent with this field and with azimuthal symmetry is:

$\vec{A} = \frac{1}{2}\mu_0 n I \, s \, \hat{\phi} \quad (s < R)$

Outside ($s > R$), the field vanishes but $\vec{A}$ does not:

$\vec{A} = \frac{\mu_0 n I R^2}{2s} \hat{\phi} \quad (s > R)$

This is a striking result: $\vec{B} = 0$ outside the solenoid, yet $\vec{A} \neq 0$. The Aharonov-Bohm effect in quantum mechanics shows that this nonzero $\vec{A}$ has physical consequences even where $\vec{B}$ vanishes.

Toroid with $N$ turns and rectangular or circular cross-section: The calculation is more involved, but the vector potential again has only a $\hat{\phi}$ component (in the toroidal coordinate sense) and can be found by integrating the Biot-Savart expression or by matching boundary conditions.

Faraday's law in terms of vector potential

Induced electric field

In the most general case, the electric field is related to both potentials by:

$\vec{E} = -\nabla \phi - \frac{\partial \vec{A}}{\partial t}$

The term $-\partial \vec{A}/\partial t$ is the piece of the electric field generated by changing magnetic fields. The term $-\nabla \phi$ is the conservative (Coulomb) part. Together they give the full electric field.

Taking the curl of both sides recovers the differential form of Faraday's law:

$\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}$

since $\nabla \times (\nabla \phi) = 0$ and $\nabla \times \vec{A} = \vec{B}$.

Gauge invariance of induced EMF

The EMF around a closed loop is:

$\mathcal{E} = \oint \vec{E} \cdot d\vec{l} = -\oint \frac{\partial \vec{A}}{\partial t} \cdot d\vec{l} - \oint \nabla \phi \cdot d\vec{l}$

The second integral vanishes identically for any closed loop (gradient fields are conservative). So:

$\mathcal{E} = -\oint \frac{\partial \vec{A}}{\partial t} \cdot d\vec{l} = -\frac{d}{dt}\oint \vec{A} \cdot d\vec{l} = -\frac{d\Phi_B}{dt}$

The last step uses Stokes' theorem: $\oint \vec{A} \cdot d\vec{l} = \int (\nabla \times \vec{A}) \cdot d\vec{a} = \int \vec{B} \cdot d\vec{a} = \Phi_B$. This shows that the magnetic flux through a loop equals the circulation of $\vec{A}$ around it, a result that's both elegant and practically useful.

Magnetic energy in terms of vector potential

Energy density

The standard expression for magnetic energy density is $u_m = \frac{B^2}{2\mu_0}$. An equivalent form, valid when the fields are produced by a current distribution $\vec{J}$, is:

$u_m = \frac{1}{2}\vec{J} \cdot \vec{A}$

This expression holds within the region where currents flow. Be careful: it doesn't mean the energy is localized only where $\vec{J} \neq 0$. The $B^2/2\mu_0$ form gives the energy density everywhere in space, while the $\vec{J} \cdot \vec{A}$ form is an alternative that, when integrated over all currents, gives the same total energy.

Total stored magnetic energy

Integrating over the current distribution:

$U_m = \frac{1}{2}\int \vec{J} \cdot \vec{A} \, d^3\vec{r}$

For a circuit carrying current $I$, this reduces to:

$U_m = \frac{1}{2}I \oint \vec{A} \cdot d\vec{l} = \frac{1}{2}I\Phi_B = \frac{1}{2}LI^2$

where $L$ is the self-inductance. This connects the vector potential formalism directly to the circuit concept of inductance.

Multipole expansion of vector potential

When you're far from a localized current distribution ($r \gg$ the size of the source), you can expand $1/|\vec{r} - \vec{r}'|$ in powers of $r'/r$. The resulting series for $\vec{A}$ is the multipole expansion.

Dipole term

The monopole term (zeroth order) vanishes identically because $\int \vec{J} \, d^3\vec{r}' = 0$ for any localized steady current distribution. (This is a consequence of $\nabla \cdot \vec{J} = 0$.)

The leading surviving term is the magnetic dipole:

$\vec{A}_{\text{dipole}}(\vec{r}) = \frac{\mu_0}{4\pi}\frac{\vec{m} \times \hat{r}}{r^2}$

where the magnetic dipole moment is:

$\vec{m} = \frac{1}{2}\int \vec{r}' \times \vec{J}(\vec{r}') \, d^3\vec{r}'$

For a planar current loop, this simplifies to $\vec{m} = I\vec{a}$, where $\vec{a}$ is the area vector of the loop. Taking the curl of $\vec{A}_{\text{dipole}}$ gives the dipole magnetic field, which falls off as $1/r^3$.

Quadrupole and higher-order terms

The magnetic quadrupole term falls off as $1/r^3$ in $\vec{A}$ (and $1/r^4$ in $\vec{B}$). It becomes relevant when the dipole moment vanishes or when you need more accuracy at intermediate distances.

Higher-order terms (octupole, etc.) decrease as successively higher powers of $1/r$. In practice, the dipole approximation is sufficient for most far-field calculations. The quadrupole and beyond matter primarily in precision work or for current distributions with special symmetry that causes lower-order terms to vanish.

Magnetic vector potential in electrodynamics

Retarded vector potential

When currents vary in time, the changes in $\vec{A}$ propagate outward at the speed of light. The vector potential at position $\vec{r}$ and time $t$ depends on what the currents were doing at the retarded time $t_r = t - |\vec{r} - \vec{r}'|/c$:

$\vec{A}(\vec{r}, t) = \frac{\mu_0}{4\pi}\int \frac{\vec{J}(\vec{r}', t_r)}{|\vec{r} - \vec{r}'|} \, d^3\vec{r}'$

This is the retarded solution to the inhomogeneous wave equation in the Lorenz gauge. It reduces to the magnetostatic Biot-Savart integral when the currents are steady. The retarded potential is the physically correct solution because it respects causality: effects propagate forward in time from source to field point.

Jefimenko's equations

Jefimenko's equations express $\vec{E}$ and $\vec{B}$ directly in terms of the charge and current distributions (evaluated at retarded times), without going through the potentials. They are the time-dependent generalizations of Coulomb's law and the Biot-Savart law. While they're less commonly used for direct calculation than the retarded potentials, they make the causal structure of electrodynamics explicit and confirm that the fields depend on the retarded-time values of $\rho$, $\vec{J}$, and their time derivatives.