##### Definition; Gauge Choices
ems.ms.vp.A

Since \({\boldsymbol \nabla} \cdot {\bf B} = 0\) in magnetostatics, following Helmholtz's theorem (see c_m_vf_helm) we can write

- Gr (5.61)

\[ {\bf B} = {\boldsymbol \nabla} \times {\bf A} \tag{BcurlA}\label{BcurlA} \]

Substituting this in Ampère's law and using curlcurl gives

\[ {\boldsymbol \nabla} \times {\bf B} = {\boldsymbol \nabla} \times ({\boldsymbol \nabla} \times {\bf A} ) = {\boldsymbol \nabla} ({\boldsymbol \nabla} \cdot {\bf A}) - {\boldsymbol \nabla}^2 {\bf A} = \mu_0 {\bf J} \]

In the case of electrostatics, we could add any constant to electrostatic potential without affecting the physics. Here, in magnetostatics, we can add any curlless function (said otherwise: the gradient of any scalar field) to the vector potential, without changing the magnetic field. Making a specific choice is called making a gauge choice. For example, we can always choose to eliminate the divergence of \({\bf A}\), giving us the example of the

Coulomb gauge

- Gr (5.63)

\[ {\boldsymbol \nabla} \cdot {\bf A} = 0. \tag{CoulG}\label{CoulG} \]

Proof: suppose our starting \({\bf A}_0\) is not divergenceless. We add \({\boldsymbol \nabla} \lambda\) to the vector potential, so \({\bf A} = {\bf A}_0 + {\boldsymbol \nabla} \lambda\). Then, \[ {\boldsymbol \nabla} \cdot {\bf A} = {\boldsymbol \nabla} \cdot {\bf A}_0 + {\boldsymbol \nabla}^2 \lambda. \] The scalar field then obeys a Poisson-like equation, \[ {\boldsymbol \nabla}^2 \lambda = -{\boldsymbol \nabla} \cdot {\bf A}_0, \] whose solution we know how to find, because it's mathematically exactly the same as solving the Poisson equation 🐟 with p_vcd. Therefore, provided \({\boldsymbol \nabla} \cdot {\bf A}_0\) goes to zero at infinity, \[ \lambda ({\bf r}) = \frac{1}{4\pi} \int d\tau' \frac{{\boldsymbol \nabla}' \cdot{\bf A}_0 ({\bf r}')}{|{\bf r} - {\bf r}'|}. \] (exercise: prove this in one line using the Laplacian and Lap1or).

Under this gauge choice, Ampère's law becomes

- Gr (5.64)

\[ {\boldsymbol \nabla}^2 {\bf A} = -\mu_0 {\bf J} \tag{Amp_CoulG}\label{Amp_CoulG} \]

Again this is precisely like the Poisson equation, with one separate equation for each component. For currents falling off sufficiently rapidly at infinity, we thus have

- Gr (5.65)

\[ {\bf A} ({\bf r}) = \frac{\mu_0}{4\pi} \int d\tau' \frac{J({\bf r}')}{|{\bf r} - {\bf r}'|} \tag{A_CoulG}\label{A_CoulG} \]

For line and surface currents,

\[ {\bf A}({\bf r}) = \frac{\mu_0}{4\pi} \int dl' \frac{{\bf I ({\bf r}')}}{|{\bf r} - {\bf r}'|}, \tag{A_CoulG_l}\label{A_CoulG_l} \]

\[ {\bf A}({\bf r}) = \frac{\mu_0}{4\pi} \int da' \frac{{\bf K ({\bf r}')}}{|{\bf r} - {\bf r}'|}. \tag{A_CoulG_s}\label{A_CoulG_s} \]

Example: infinite solenoid

Task: find the vector potential of an infinite solenoid with \(n\) turns pet unit length, radius \(R\) and current \(I\).

Solution: one cannot use Amp_CoulG since the current extends to infinity (comment: you are encouraged to check that the integral converges anyway, by combining the integrals for \(z > 0\) and \(z < 0\) into one).

Nice trick: notice that

\[ \oint d{\bf l} \cdot {\bf A} = \int d{\bf a} \cdot ({\boldsymbol \nabla} \times {\bf A}) = \int d{\bf a} \cdot {\bf B} = \Phi. \] This is reminiscent of Ampère's law in integral form Amp_int, \[ \oint d{\bf l} \cdot {\bf B} = \mu_0 I_{enc}. \] Mathematically, it's the same equation! Replacement: \({\bf B} \rightarrow {\bf A}\) and \(\mu_0 I_{enc} \rightarrow \Phi\). And to paraphrase Feynman's lectures: the same equations have the same solutions.

Use symmetry: vector potential can only be cicumferential. Using an amperian loop at a radius \(r\) inside the solenoid, and the fact that the field inside a solenoid is \(\mu_0 n I\) (as per Btor), we get \[ \oint d{\bf l} \cdot {\bf A} = A (2\pi r) = \int d{\bf a} \cdot {\bf B} = \mu_0 n I (\pi r^2), \] so

\[ {\bf A} = \frac{\mu_0 n I}{2} r ~\hat{\boldsymbol \varphi}, \hspace{1cm} r < R. \] For an amperian loop outside, the flux is always \(\mu_0 n I (\pi R^2)\), so

\[ {\bf A} = \frac{\mu_0 n I}{2} \frac{R^2}{r} \hat{\boldsymbol \varphi}, \hspace{1cm} r > R. \] Exercise: check that \({\boldsymbol \nabla} \times {\bf A} = {\bf B}\) and that \({\boldsymbol \nabla} \cdot {\bf A} = 0\).

Except where otherwise noted, all content is licensed under a Creative Commons Attribution 4.0 International License.

Author: Jean-Sébastien Caux

Created: 2022-03-24 Thu 08:42