Pre-Quantum Electrodynamics

The Vector Potential ems.ms.vp

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

\[ {\bf B} = {\boldsymbol \nabla} \times {\bf A} \label{Gr(5.59)} \]

Connection with Ampère's law: \[ {\boldsymbol \nabla} \times {\bf B} = {\boldsymbol \nabla} \times ({\boldsymbol \nabla} \times {\bf A} ) = {\boldsymbol \nabla} ({\boldsymbol \nabla} \cdot {\bf A}) - {\boldsymbol \nabla}^2 {\bf A} = \mu_0 {\bf J} \label{Gr(5.60)} \] Electrostatics: you could add any constant to electrostatic potential. Here: you can add any curlless function (so gradient of a scalar field) to the vector potential, without changing the magnetic field. This is called a {\bf gauge choice} in electrodynamics. For example, we can {\bf always} eliminate the divergence of \({\bf A}\),

{\bf Example gauge choice:} \[ {\boldsymbol \nabla} \cdot {\bf A} = 0. \label{Gr(5.61)} \]

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. Provided \({\boldsymbol \nabla} \cdot {\bf A}_0\) goes to zero at infinity, \[ \lambda ({\bf r}) = \frac{1}{4\pi} \int d\tau' \frac{{\boldsymbol \nabla}' \cdot{\bf A}_0 ({\bf r}')}{|{\bf r} - {\bf r}'|}. \] {\it Provide proof in one line using Laplacian leading to delta function.}

Under this gauge choice, Ampère's law becomes

\[ {\boldsymbol \nabla}^2 {\bf A} = -\mu_0 {\bf J} \label{Gr(5.62)} \]

Note: this is a Poisson equation for each component. For currents falling off sufficiently rapidly at infinity,

\[ {\bf A} ({\bf r}) = \frac{\mu_0}{4\pi} \int d\tau' \frac{J({\bf r}')}{|{\bf r} - {\bf r}'|} \label{Gr(5.63)} \]

For line and surface currents, (beware Griffiths' horrendous notation)

\[ {\bf A}({\bf r}) = \frac{\mu_0}{4\pi} \int dl' \frac{{\bf I ({\bf r}')}}{|{\bf r} - {\bf r}'|}, \hspace{2cm} {\bf A}({\bf r}) = \frac{\mu_0}{4\pi} \int da' \frac{{\bf K ({\bf r}')}}{|{\bf r} - {\bf r}'|}. \label{Gr(5.64)} \]

\paragraph{Example 5.11:} a spherical shell of radius \(R\), carrying a uniform surface charge \(\sigma\), is set spinning at angular velocity \(\omega\). Find the vector potential at \({\bf r}\). \paragraph{Solution:} do it by yourselves. Fun conclusion: the field inside the sphere is uniform ! \[ {\bf B} = \frac{2}{3} \mu_0 \sigma R {\boldsymbol \omega}. \label{Gr(5.68)} \]

\paragraph{Example 5.12:} find the vector potential of an infinite solenoid with \(n\) turns pet unit length, radius \(R\) and current \(I\). \paragraph{Solution:} cannot use \ref{Gr(5.64)} since the current extends to infinity ({\bf Comment:} check that the integral converges anyway, by combining the integrals for \(z > 0\) and \(z < 0\) into one).

{\bf Nice trick:} notice that \[ \oint {\bf A} \cdot d{\bf l} = \int ({\boldsymbol \nabla} \times {\bf A}) \cdot d{\bf a} = \int {\bf B} \cdot d{\bf a} = \Phi. \label{Gr(5.69)} \] This is reminiscent of Ampère's law in integral form, \ref{Gr(5.55)}, \[ \oint {\bf B} \cdot d{\bf l} = \mu_0 I_{enc}. \] It's the same equation ! Replacement: \({\bf B} \rightarrow {\bf A}\) and \(\mu_0 I_{enc} \rightarrow \Phi\). And to paraphrase Feynman's lectures: {\it the same equations have the same solutions.}

Use symmetry: vector potential can only be cicumferential. Using an 'amperian' loop at a radius \(s\) {\it inside} the solenoid, and the fact that the field inside a solenoid is \(\mu_0 n I\) (\ref{Gr(5.57)}), we get \[ \oint {\bf A} \cdot d{\bf l} = A (2\pi s) = \int {\bf B} \cdot d{\bf a} = \mu_0 n I (\pi s^2), \] so \[ {\bf A} = \frac{\mu_0 n I}{2} s \hat{\boldsymbol \varphi}, \hspace{1cm} s < R. \label{Gr(5.70)} \] For an 'amperian' loop outside, the flux is always \(\mu_0 n I (\pi R^2)\), so \[ {\bf A} = \frac{\mu_0 n I}{2} \frac{R^2}{s} \hat{\boldsymbol \varphi}, \hspace{1cm} s > R. \label{Gr(5.71)} \] \paragraph{Exercise:} check that \({\boldsymbol \nabla} \times {\bf A} = {\bf B}\) and that \({\boldsymbol \nabla} \cdot {\bf A} = 0\).

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Author: Jean-Sébastien Caux

Created: 2022-02-17 Thu 08:42