📂Electrodynamics

Magnetic Vector Potential

Explanation¹

In electrostatics, the electric field is easily handled by using a property called $\nabla \times \mathbf{E} = \mathbf{0}$ to define the scalar potential $V$. Similarly, in magnetostatics, the vector potential $A$ is defined and used by utilizing a property called $\nabla \cdot \mathbf{B} = 0$. Let’s say the magnetic field $\mathbf{B}$ is the curl of some vector $\mathbf{A}$.

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

Since the divergence of a curl is 0, naturally, the following equation holds.

$$\nabla \cdot \mathbf{B} = \nabla \cdot (\nabla \times \mathbf{A}) = 0$$

Therefore, the vector $\mathbf{A}$ which becomes the magnetic field when the curl is taken is defined as the vector potential of the magnetic field. The key in dealing with the scalar potential of the electric field was that the value of the potential itself was not important, but the difference of potentials was. Therefore, a difference of constant $K$ did not affect handling the electric field. Similarly, the vector potential $\mathbf{A}$ can be chosen as a vector that makes its divergence $0$. It doesn’t matter if the vector’s divergence is not $0$, but the equation becomes cleanest when $\nabla \cdot \mathbf{A}=0$ is satisfied. Substituting the vector potential $\mathbf{A}$ into the differential form of Ampère’s law, we obtain the following equation.

$$\nabla \times \mathbf{B}=\nabla \times (\nabla \times \mathbf{A} ) = \nabla(\nabla \cdot \mathbf{A})-\nabla ^2 \mathbf{A} = \mu_{0} \mathbf{J}$$

(Refer to) If it’s $\nabla \cdot \mathbf{A}=0$, Ampère’s law neatly becomes as follows.

$$\begin{equation} \nabla ^2 \mathbf{A}=-\mu_{0} \mathbf{J} \label{1} \end{equation}$$

Let’s see why we can freely set $\mathbf{A}$ as a function whose divergence is $0$. Let’s denote a potential whose divergence is not $0$ as $\mathbf{A}_{0}$. Call $\mathbf{A}$, which is obtained by adding the gradient of some arbitrary scalar $\lambda$ to it.

$$\mathbf{A}=\mathbf{A}_{0} + \nabla \lambda$$

Taking the curl on both sides, since the curl of a gradient is $\mathbf{0}$,

$$\nabla \times \mathbf{A} = \nabla \times \mathbf{A}_{0} + \nabla \times (\nabla\lambda)=\nabla \times \mathbf{A}_{0}$$

Thus, the curl of the two vectors $\mathbf{A}, \mathbf{A}_{0}$ is the same, and the following holds.

$$\mathbf{B}=\nabla \times \mathbf{A} = \nabla \times \mathbf{A}_{0}$$

Therefore, adding the gradient of any arbitrary scalar to the vector potential does not affect the representation of the magnetic field. To determine the condition for scalar $\lambda$, taking the divergence of the two vector potentials,

$$\nabla \cdot \mathbf{A} = \nabla \cdot \mathbf{A}_{0} + \nabla^2 \lambda$$

Therefore, by choosing $\lambda$ that satisfies $\nabla ^2 \lambda=-\nabla \cdot \mathbf{A}_{0}$, we can make the divergence of the vector potential $\mathbf{A}$ into $0$. If it holds that $\nabla \cdot \mathbf{A}_{0}=0$ at a very far distance, the following equation is obtained.

$$\lambda=\dfrac{1}{4 \pi}\int \dfrac{\nabla \cdot \mathbf{A}_{0} } {\cR} d\tau^{\prime}$$

Solving $(1)$ to directly find $\mathbf{A}$ (at a very far distance when $\mathbf{J}=0$),

$$\mathbf{A}(\mathbf{r})=\dfrac{\mu_{0}}{4\pi} \int \dfrac{\mathbf{J} (\mathbf{r}^{\prime}) }{\cR} d\tau^{\prime}$$

As the equation suggests, if the direction of the current is constant, the direction of the vector potential and the current are the same. The vector potentials for line currents and surface currents are

$$\mathbf{A}=\dfrac{\mu_{0}}{4\pi} \int \dfrac{\mathbf{I} } {\cR} dl^{\prime}=\dfrac{\mu_{0} I}{4\pi} \int \dfrac{1}{\cR} d\mathbf{l}^{\prime}$$

$$\mathbf{A}=\dfrac{\mu_{0}}{4\pi}\int \dfrac{K}{\cR} da^{\prime}$$