# Path integral to Schrödinger equation for a vector potential

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 8. Section 8.6, Exercise 8.6.4.

When we showed that the path integral approach is equivalent to the Schrödinger equation, we did so for a scalar potential ${V}$, so that the Lagrangian is the usual ${L=T-V}$, and we can use that to calculate the action over an infinitestimal time interval ${\varepsilon}$, during which time the particle moves from ${x^{\prime}}$ to ${x}$. In the calculation, we chose the value of ${V}$ at the midpoint of this interval, that is ${V\left(\frac{x+x^{\prime}}{2}\right)}$. In fact, in this derivation it didn’t matter where in the interval ${\left[x^{\prime},x\right]}$ we chose to evaluate ${V}$, since we took only terms up to first order in ${\varepsilon}$, and moving the point at which we evaluate ${V}$ introduced terms only of order ${\varepsilon^{2}}$ or higher.

Things get a bit more complicated if we consider a system such as the electromagnetic force, where the Lagrangian is no longer just ${T-V}$, but becomes

$\displaystyle L=\frac{1}{2}m\mathbf{v}\cdot\mathbf{v}-q\phi+\frac{q}{c}\mathbf{v}\cdot\mathbf{A} \ \ \ \ \ (1)$

To examine the effect this has on the demonstration that the path integral approach is equivalent to the Schrödinger equation, we’ll consider only one dimension, and leave out the electrostatic potential ${\phi}$ since it’s just a scalar potential and we already know that such potentials do indeed convert to the Schrödinger equation. Thus the Lagrangian we’ll consider is

$\displaystyle L=\frac{1}{2}mv^{2}+\frac{q}{c}vA \ \ \ \ \ (2)$

Over the infinitesimal time interval ${\varepsilon}$ we have

$\displaystyle v=\frac{x-x^{\prime}}{\varepsilon} \ \ \ \ \ (3)$

The propagator over this time interval is

 $\displaystyle U\left(x,\varepsilon;x^{\prime},0\right)$ $\displaystyle =$ $\displaystyle \sqrt{\frac{m}{2\pi\hbar\varepsilon i}}\exp\left[\frac{i}{\hbar}\left(\frac{1}{2}m\frac{\left(x-x^{\prime}\right)^{2}}{\varepsilon}+\varepsilon\frac{q}{c}\frac{x-x^{\prime}}{\varepsilon}A\left(x+\alpha\left(x-x^{\prime}\right)\right)\right)\right]\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{m}{2\pi\hbar\varepsilon i}}\exp\left[\frac{i}{\hbar}\left(\frac{1}{2}m\frac{\eta^{2}}{\varepsilon}-\frac{q}{c}\eta A\left(x+\alpha\eta\right)\right)\right]\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{m}{2\pi\hbar\varepsilon i}}\exp\left(\frac{im\eta^{2}}{2\hbar\varepsilon}\right)\exp\left[-\frac{iq}{\hbar c}\eta A\left(x+\alpha\eta\right)\right] \ \ \ \ \ (6)$

where ${\alpha}$ is a parameter that we can vary between 0 and 1 in order to vary the point along the path from ${x^{\prime}}$ to ${x}$ at which we evaluate the vector potential ${A}$. Also,

$\displaystyle \eta\equiv x^{\prime}-x \ \ \ \ \ (7)$

Using the same argument as before, we require

$\displaystyle \left|\eta\right|\lesssim\sqrt{\frac{2\hbar\varepsilon\pi}{m}} \ \ \ \ \ (8)$

so calculations to first order in ${\varepsilon}$ must include terms up to second order in ${\eta}$.

Once we have ${U\left(x,\varepsilon;x^{\prime},0\right)}$, we can find ${\psi\left(x,\varepsilon\right)}$ from

$\displaystyle \psi\left(x,\varepsilon\right)=\int_{-\infty}^{\infty}U\left(x,\varepsilon;x^{\prime},0\right)\psi\left(x^{\prime},0\right)dx^{\prime} \ \ \ \ \ (9)$

To find ${U}$ to first order in ${\varepsilon}$, we need to expand the second exponential in 6 out to terms in ${\eta^{2}}$, so we first look at the argument of the exponential:

$\displaystyle -\frac{iq}{\hbar c}\eta A\left(x+\alpha\eta\right)=-\frac{iq}{\hbar c}\left(\eta A\left(x\right)+\alpha\eta^{2}\frac{\partial A}{\partial x}+\ldots\right) \ \ \ \ \ (10)$

where the derivative is evaluated at the endpoint ${x}$ and is constant in the integral. The second exponential in 6 now becomes, to second order in ${\eta}$:

$\displaystyle \exp\left[-\frac{iq}{\hbar c}\eta A\left(x+\alpha\eta\right)\right]=1-\frac{iq}{\hbar c}\left(\eta A\left(x\right)+\alpha\eta^{2}\frac{\partial A}{\partial x}\right)-\left(\frac{q}{\hbar c}\right)^{2}\frac{\eta^{2}A^{2}\left(x\right)}{2} \ \ \ \ \ (11)$

We also need the expansion of the wave function in 9 up to second order in ${\eta}$:

$\displaystyle \psi\left(x+\eta,0\right)=\psi\left(x,0\right)+\eta\frac{\partial\psi}{\partial x}+\frac{\eta^{2}}{2}\frac{\partial^{2}\psi}{\partial x^{2}} \ \ \ \ \ (12)$

Again, both derivatives are evaluated at the endpoint ${x}$ and are constants in the integral.

We now need to do the integral 9, which consists of several standard Gaussian integrals. From 7, ${dx^{\prime}=d\eta}$, so

 $\displaystyle \int_{-\infty}^{\infty}U\left(x,\varepsilon;x^{\prime},0\right)\psi\left(x^{\prime},0\right)dx^{\prime}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{m}{2\pi\hbar\varepsilon i}}\psi\left(x,0\right)\int_{-\infty}^{\infty}\exp\left(\frac{im\eta^{2}}{2\hbar\varepsilon}\right)d\eta+\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle$ $\displaystyle \sqrt{\frac{m}{2\pi\hbar\varepsilon i}}\left(\frac{\partial\psi}{\partial x}-\frac{iq}{\hbar c}A\left(x\right)\psi\left(x,0\right)\right)\int_{-\infty}^{\infty}\exp\left(\frac{im\eta^{2}}{2\hbar\varepsilon}\right)\eta\;d\eta+\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle$ $\displaystyle \sqrt{\frac{m}{2\pi\hbar\varepsilon i}}\left(\frac{1}{2}\frac{\partial^{2}\psi}{\partial x^{2}}-\frac{iq}{\hbar c}A\left(x\right)\frac{\partial\psi}{\partial x}+\psi\left(x,0\right)\left(-\frac{iq\alpha}{\hbar c}\frac{\partial A}{\partial x}-\frac{1}{2}\left(\frac{qA\left(x\right)}{\hbar c}\right)^{2}\right)\right)\times\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle$ $\displaystyle \int_{-\infty}^{\infty}\exp\left(\frac{im\eta^{2}}{2\hbar\varepsilon}\right)\eta^{2}d\eta \ \ \ \ \ (16)$

We can now do the integrals:

 $\displaystyle \int_{-\infty}^{\infty}\exp\left(\frac{im\eta^{2}}{2\hbar\varepsilon}\right)d\eta$ $\displaystyle =$ $\displaystyle \sqrt{\frac{2\pi\hbar\varepsilon i}{m}}\ \ \ \ \ (17)$ $\displaystyle \int_{-\infty}^{\infty}\exp\left(\frac{im\eta^{2}}{2\hbar\varepsilon}\right)\eta\;d\eta$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (18)$ $\displaystyle \int_{-\infty}^{\infty}\exp\left(\frac{im\eta^{2}}{2\hbar\varepsilon}\right)\eta^{2}d\eta$ $\displaystyle =$ $\displaystyle -\frac{\hbar\varepsilon}{im}\sqrt{\frac{2\pi\hbar\varepsilon i}{m}} \ \ \ \ \ (19)$

Plugging these in we get

 $\displaystyle \psi\left(x,\varepsilon\right)$ $\displaystyle =$ $\displaystyle \psi\left(x,0\right)-\frac{\hbar\varepsilon}{im}\left[\frac{1}{2}\frac{\partial^{2}\psi}{\partial x^{2}}-\frac{iq}{\hbar c}A\left(x\right)\frac{\partial\psi}{\partial x}+\psi\left(x,0\right)\left(-\frac{iq\alpha}{\hbar c}\frac{\partial A}{\partial x}-\frac{1}{2}\left(\frac{qA\left(x\right)}{\hbar c}\right)^{2}\right)\right]\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \psi\left(x,0\right)+\frac{\varepsilon}{i\hbar}\left[-\frac{\hbar^{2}}{2m}\frac{\partial^{2}\psi}{\partial x^{2}}+\frac{i\hbar q}{mc}A\left(x\right)\frac{\partial\psi}{\partial x}+\psi\left(x,0\right)\left(\frac{i\hbar q\alpha}{mc}\frac{\partial A}{\partial x}+\frac{1}{2m}\left(\frac{qA\left(x\right)}{c}\right)^{2}\right)\right] \ \ \ \ \ (21)$

We can compare this with the quantum version of the Hamiltonian for the vector potential part of the electromagnetic force. The classical Hamiltonian is

$\displaystyle H=\frac{\left|\mathbf{p}-q\mathbf{A}/c\right|^{2}}{2m} \ \ \ \ \ (22)$

Because ${\mathbf{A}}$ depends on ${x}$, it doesn’t commute with ${\mathbf{p}}$ so to get the quantum version we need to symmetrize when we expand the square. The one dimensional version is

$\displaystyle H=\frac{P^{2}}{2m}-\frac{q}{2mc}PA-\frac{q}{2mc}AP+\frac{q^{2}A^{2}}{2mc^{2}} \ \ \ \ \ (23)$

In the coordinate basis, we have, using ${P=-i\hbar\partial/\partial x}$

 $\displaystyle H\psi$ $\displaystyle =$ $\displaystyle -\frac{\hbar^{2}}{2m}\frac{\partial^{2}\psi}{\partial x^{2}}+\frac{i\hbar q}{2mc}\left(\frac{\partial\left(A\psi\right)}{\partial x}+A\frac{\partial\psi}{\partial x}\right)+\frac{q^{2}A^{2}}{2mc^{2}}\psi\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\hbar^{2}}{2m}\frac{\partial^{2}\psi}{\partial x^{2}}+\frac{i\hbar q}{2mc}\left(2A\frac{\partial\psi}{\partial x}+\psi\frac{\partial A}{\partial x}\right)+\frac{q^{2}A^{2}}{2mc^{2}}\psi\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\hbar^{2}}{2m}\frac{\partial^{2}\psi}{\partial x^{2}}+\frac{i\hbar q}{mc}\left(A\frac{\partial\psi}{\partial x}+\frac{1}{2}\psi\frac{\partial A}{\partial x}\right)+\frac{q^{2}A^{2}}{2mc^{2}}\psi \ \ \ \ \ (26)$

Returning to the result we got from the path integral, upon rearranging 21 we get

 $\displaystyle i\hbar\frac{\psi\left(x,\varepsilon\right)-\psi\left(x,0\right)}{\varepsilon}$ $\displaystyle =$ $\displaystyle -\frac{\hbar^{2}}{2m}\frac{\partial^{2}\psi}{\partial x^{2}}+\frac{i\hbar q}{mc}\left(A\left(x\right)\frac{\partial\psi}{\partial x}+\alpha\psi\left(x,0\right)\frac{\partial A}{\partial x}\right)+\frac{\psi\left(x,0\right)}{2m}\left(\frac{qA\left(x\right)}{c}\right)^{2}\ \ \ \ \ (27)$ $\displaystyle i\hbar\frac{\partial\psi}{\partial t}$ $\displaystyle =$ $\displaystyle -\frac{\hbar^{2}}{2m}\frac{\partial^{2}\psi}{\partial x^{2}}+\frac{i\hbar q}{mc}\left(A\left(x\right)\frac{\partial\psi}{\partial x}+\alpha\psi\frac{\partial A}{\partial x}\right)+\frac{\psi}{2m}\left(\frac{qA\left(x\right)}{c}\right)^{2} \ \ \ \ \ (28)$

where in the last line we took the limit as ${\varepsilon\rightarrow0}$ on the LHS to get Schrödinger’s equation in the form

$\displaystyle i\hbar\frac{\partial\psi}{\partial t}=H\psi \ \ \ \ \ (29)$

Comparing the RHS of 28 with 26, we see that they are equal provided we take ${\alpha=\frac{1}{2}}$. Thus in this case, we really do need to evaluate the vector potential ${A}$ at the midpoint of the path.

# Faraday field and magnetic vector potential

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problems 5.50a, 7.47.

We’ve seen that in the magnetostatic case, the Biot-Savart law gives the magnetic field produced by steady currents:

$\displaystyle \mathbf{B}\left(\mathbf{r}\right)=\frac{\mu_{0}}{4\pi}\int_{V}\frac{\mathbf{J}\left(\mathbf{r}^{\prime}\right)\times\left(\mathbf{r}-\mathbf{r}^{\prime}\right)}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|^{3}}d^{3}\mathbf{r}^{\prime} \ \ \ \ \ (1)$

This law can be used to derive Ampère’s law which in differential form is

$\displaystyle \nabla\times\mathbf{B}=\mu_{0}\mathbf{J} \ \ \ \ \ (2)$

The Biot-Savart law can be seen as the solution of Ampère’s law, combined with the divergence equation ${\nabla\cdot\mathbf{B}=0}$.

In the case where the magnetic field is changing with time, Faraday’s law gives the electric field produced as a result of the changing magnetic field:

$\displaystyle \nabla\times\mathbf{E}=-\frac{\partial\mathbf{B}}{\partial t} \ \ \ \ \ (3)$

If there is no free charge, then ${\nabla\cdot\mathbf{E}=0}$, so this pair of equations for the electric field are formally equivalent to those above for the magnetic field. We can therefore write down an analogous solution by substituting ${-\frac{\partial\mathbf{B}}{\partial t}}$ for ${\mu_{0}\mathbf{J}}$ in the Biot-Savart law to get

$\displaystyle \mathbf{E}=-\frac{1}{4\pi}\frac{\partial}{\partial t}\int_{V}\frac{\mathbf{B}\left(\mathbf{r}^{\prime}\right)\times\left(\mathbf{r}-\mathbf{r}^{\prime}\right)}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|^{3}}d^{3}\mathbf{r}^{\prime} \ \ \ \ \ (4)$

We can extend the same argument to the magnetic vector potential ${\mathbf{A}}$. We have ${\mathbf{B}=\nabla\times\mathbf{A}}$ and we can choose ${\nabla\cdot\mathbf{A}=0}$, so again by analogy with the Biot-Savart law we can write

$\displaystyle \mathbf{A}=\frac{1}{4\pi}\int_{V}\frac{\mathbf{B}\left(\mathbf{r}^{\prime}\right)\times\left(\mathbf{r}-\mathbf{r}^{\prime}\right)}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|^{3}}d^{3}\mathbf{r}^{\prime} \ \ \ \ \ (5)$

Comparing the last two equations, we get

$\displaystyle \mathbf{E}=-\frac{\partial\mathbf{A}}{\partial t} \ \ \ \ \ (6)$

Taking the curl of both sides gives us back Faraday’s law: ${\nabla\times\mathbf{E}=-\frac{\partial\nabla\times\mathbf{A}}{\partial t}=-\frac{\partial\mathbf{B}}{\partial t}}$. Remember that this is true only for those electric fields produced as a result of changing magnetic fields; it does not apply to electric fields produced by free charge.

Suppose now we have a spherical shell of radius ${R}$ with a uniform surface charge density ${\sigma}$ which spins at a variable rate ${\omega\left(t\right)}$. Griffiths works out the vector potential of this sphere in his example 5.11 and gets

$\displaystyle \mathbf{A}=\begin{cases} \frac{\mu_{0}R\omega\sigma}{3}r\sin\theta\hat{\boldsymbol{\phi}} & r\le R\\ \frac{\mu_{0}R^{4}\omega\sigma}{3}\frac{\sin\theta}{r^{2}}\hat{\boldsymbol{\phi}} & r\ge R \end{cases} \ \ \ \ \ (7)$

Inside the sphere there is no Coulomb field, while outside, this field is

 $\displaystyle \mathbf{E}_{coulomb}$ $\displaystyle =$ $\displaystyle \frac{Q}{4\pi\epsilon_{0}r^{2}}\hat{\mathbf{r}}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{4\pi R^{2}\sigma}{4\pi\epsilon_{0}r^{2}}\hat{\mathbf{r}}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{R^{2}\sigma}{\epsilon_{0}r^{2}}\hat{\mathbf{r}} \ \ \ \ \ (10)$

Using 6 for the Faraday field, we get the total field:

$\displaystyle \mathbf{E}=\begin{cases} -\frac{\mu_{0}R\dot{\omega}\sigma}{3}r\sin\theta\hat{\boldsymbol{\phi}} & r\le R\\ \frac{R^{2}\sigma}{\epsilon_{0}r^{2}}\hat{\mathbf{r}}-\frac{\mu_{0}R^{4}\dot{\omega}\sigma}{3}\frac{\sin\theta}{r^{2}}\hat{\boldsymbol{\phi}} & r\ge R \end{cases} \ \ \ \ \ (11)$

# Magnetic potential and field of a rotating sphere of charge

Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Problem 5.58.

We return to the rotating solid sphere of charge. If the sphere has radius ${R}$, contains total charge ${Q}$ and is rotating with angular speed ${\omega}$ we can find its magnetic moment from the gyromagnetic ratio. The angular momentum of a sphere is ${L=\frac{2}{5}MR^{2}\omega}$, so

 $\displaystyle \gamma$ $\displaystyle =$ $\displaystyle \frac{m}{L}=\frac{Q}{2M}\ \ \ \ \ (1)$ $\displaystyle m$ $\displaystyle =$ $\displaystyle \frac{LQ}{2M}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{5}QR^{2}\omega \ \ \ \ \ (3)$

We can align the sphere so its magnetic moment is along the ${z}$ axis. From this we can find the average field within the sphere:

 $\displaystyle \mathbf{B}_{av}$ $\displaystyle =$ $\displaystyle \frac{2\mu_{0}}{4\pi R^{3}}\mathbf{m}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}}{4\pi}\frac{2Q\omega}{5R}\hat{\mathbf{z}} \ \ \ \ \ (5)$

We worked out the exact field earlier and found

 $\displaystyle \mathbf{B}$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}\omega Q}{4\pi R}\left[\left(1-\frac{3}{5}\frac{r^{2}}{R^{2}}\right)\cos\theta\hat{\mathbf{r}}+\left(\frac{6}{5}\frac{r^{2}}{R^{2}}-1\right)\sin\theta\hat{\boldsymbol{\theta}}\right]\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}\omega Q}{4\pi R}\left[\hat{\mathbf{z}}-\frac{3}{5}\frac{r^{2}}{R^{2}}\cos\theta\hat{\mathbf{r}}+\frac{6}{5}\frac{r^{2}}{R^{2}}\sin\theta\hat{\boldsymbol{\theta}}\right] \ \ \ \ \ (7)$

We can calculate the average field within the sphere from this formula by finding

$\displaystyle \mathbf{B}_{av}=\frac{3}{4\pi R^{3}}\int_{V}\mathbf{B}d^{3}\mathbf{r} \ \ \ \ \ (8)$

The average field has three contributions. The ${\hat{\mathbf{z}}}$ contribution is constant, so is just

$\displaystyle \mathbf{B}_{av;z}=\frac{\mu_{0}\omega Q}{4\pi R}\hat{\mathbf{z}} \ \ \ \ \ (9)$

The ${\hat{\mathbf{r}}}$ contribution, by symmetry, has a non-zero component only in the ${z}$ direction, and since the angle between ${\hat{\mathbf{r}}}$ and ${\hat{\mathbf{z}}}$ is ${\theta}$, this contribution will be ${\mathbf{B}_{r}=-\frac{\mu_{0}\omega Q}{4\pi R}\frac{3}{5}\frac{r^{2}}{R^{2}}\cos^{2}\theta\hat{\mathbf{z}}}$. If we take the average of this, we have

 $\displaystyle \mathbf{B}_{av;r}$ $\displaystyle =$ $\displaystyle \frac{3}{4\pi R^{3}}\int_{V}\mathbf{B}_{r}d^{3}\mathbf{r}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\hat{\mathbf{z}}\frac{\mu_{0}\omega Q}{4\pi R^{3}}\frac{3}{5}\frac{3}{4\pi R^{3}}\int_{0}^{R}\int_{0}^{\pi}\int_{0}^{2\pi}\left(r^{2}\cos^{2}\theta\right)\left(r^{2}\sin\theta\right)d\phi d\theta dr\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{3}{25}\frac{\mu_{0}\omega Q}{4\pi R}\hat{\mathbf{z}} \ \ \ \ \ (12)$

For the ${\hat{\boldsymbol{\theta}}}$ contribution, again the non-zero component is in the ${z}$ direction. This time, the angle between ${\hat{\boldsymbol{\theta}}}$ and ${\hat{\mathbf{z}}}$ is ${\theta+\frac{\pi}{2}}$ and ${\cos\left(\theta+\frac{\pi}{2}\right)=-\sin\theta}$, so the contribution is ${\mathbf{B}_{\theta}=-\frac{\mu_{0}\omega Q}{4\pi R}\frac{6}{5}\frac{r^{2}}{R^{2}}\sin^{2}\theta\hat{\mathbf{z}}}$ The average is

 $\displaystyle \mathbf{B}_{av;\theta}$ $\displaystyle =$ $\displaystyle \frac{3}{4\pi R^{3}}\int_{V}\mathbf{B}_{\theta}d^{3}\mathbf{r}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\hat{\mathbf{z}}\frac{\mu_{0}\omega Q}{4\pi R^{3}}\frac{6}{5}\frac{3}{4\pi R^{3}}\int_{0}^{R}\int_{0}^{\pi}\int_{0}^{2\pi}\left(r^{2}\sin^{2}\theta\right)\left(r^{2}\sin\theta\right)d\phi d\theta dr\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{12}{25}\frac{\mu_{0}\omega Q}{4\pi R}\hat{\mathbf{z}} \ \ \ \ \ (15)$

The total average field is then

 $\displaystyle \mathbf{B}_{av}$ $\displaystyle =$ $\displaystyle \mathbf{B}_{av;z}+\mathbf{B}_{av;r}+\mathbf{B}_{av;\theta}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(1-\frac{3}{25}-\frac{12}{25}\right)\frac{\mu_{0}\omega Q}{4\pi R}\hat{\mathbf{z}}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2}{5}\frac{\mu_{0}\omega Q}{4\pi R}\hat{\mathbf{z}} \ \ \ \ \ (18)$

This agrees with the calculation above using the magnetic moment.

The dipole term in the vector potential is

 $\displaystyle \mathbf{A}_{1}$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}}{4\pi r^{2}}\mathbf{m}\times\hat{\mathbf{r}}\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}}{4\pi r^{2}}\frac{R^{2}Q\omega}{5}\sin\theta\hat{\boldsymbol{\phi}} \ \ \ \ \ (20)$

so this is what we’d expect the potential to be for large distances from the sphere. However, we’ve seen that for a spherical shell, the potential outside the sphere is exactly equal to that of a perfect dipole. We might expect, therefore, that for a solid sphere, the dipole potential is also the exact potential. The potential for the shell of radius ${r^{\prime}}$ is

$\displaystyle \mathbf{A}=\frac{\mu_{0}r^{\prime4}\omega\sigma}{3}\frac{\sin\theta}{r^{2}}\hat{\boldsymbol{\phi}} \ \ \ \ \ (21)$

The surface charge density ${\sigma}$ for an infinitesimal shell of thickness ${dr^{\prime}}$ within the solid sphere is

$\displaystyle \sigma=\frac{3Q}{4\pi R^{3}}dr^{\prime} \ \ \ \ \ (22)$

so the total potential is (remember we’re holding the observation point constant, so ${r}$ and ${\theta}$ are both constants):

 $\displaystyle \mathbf{A}$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}\omega}{3}\frac{3Q}{4\pi R^{3}}\frac{\sin\theta}{r^{2}}\hat{\boldsymbol{\phi}}\int_{0}^{R}r^{\prime4}dr^{\prime}\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}}{4\pi r^{2}}\frac{R^{2}Q\omega}{5}\sin\theta\hat{\boldsymbol{\phi}} \ \ \ \ \ (24)$

Thus the exact potential is indeed equal to the dipole potential.

# Average magnetic field within a sphere

Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Problem 5.57.

We worked out the average electric field at the centre of a sphere earlier, so can we do something similar for the average magnetic field? Suppose we are given a steady current density within the sphere.

The starting point is the definition of the average field:

$\displaystyle \mathbf{B}_{av}=\frac{3}{4\pi R^{3}}\int_{V}\mathbf{B}d^{3}\mathbf{r} \ \ \ \ \ (1)$

We can write this in terms of the vector potential and then in terms of the current density:

 $\displaystyle \frac{3}{4\pi R^{2}}\int_{V}\mathbf{B}d^{3}\mathbf{r}$ $\displaystyle =$ $\displaystyle \frac{3}{4\pi R^{3}}\int_{V}\nabla\times\mathbf{A}d^{3}\mathbf{r}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{3}{4\pi R^{3}}\int_{A}\mathbf{A}\times d\mathbf{a}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{3\mu_{0}}{16\pi^{2}R^{3}}\int_{V}\int_{A}\frac{\mathbf{J}\left(\mathbf{r}\right)}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}\times d\mathbf{a}d^{3}\mathbf{r} \ \ \ \ \ (4)$

In the second line we used the vector identity ${\int_{V}\nabla\times\mathbf{A}d^{3}\mathbf{r}=-\int_{A}\mathbf{A}\times d\mathbf{a}}$ and in line 3 we used the definition of the vector potential.

We are now faced with a surface integral and a volume integral. We can do the surface integral first, but we need to be clear about which coordinates are which. If we do the surface integral first, then we are choosing a particular volume element ${d^{3}\mathbf{r}}$ and holding it constant while we integrate over the surface. That is, the observation point ${\mathbf{r}}$ points to the volume element and the source point ${\mathbf{r}^{\prime}}$ points to the surface element. This is why we’ve explicitly stated the dependence of ${\mathbf{J}\left(\mathbf{r}\right)}$ since it depends on the volume element, and not the surface element.

The only term containing ${\mathbf{r}^{\prime}}$ is therefore the ${\frac{1}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}}$ factor, so the integral we need to do is ${\int_{A}\frac{1}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}d\mathbf{a}}$. Since ${\mathbf{r}}$ is fixed, we can point it along the ${z}$ axis, making ${\theta}$ the angle between ${\mathbf{r}}$ and ${\mathbf{r}^{\prime}}$. Also, since ${\mathbf{r}^{\prime}}$ always points to a surface element, its magnitude is always ${R}$. We therefore have

$\displaystyle \frac{1}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}=\frac{1}{\sqrt{r^{2}+R^{2}-2rR\cos\theta}} \ \ \ \ \ (5)$

The surface element ${d\mathbf{a}}$ always points radially outward since it’s on the surface of a sphere, so by symmetry the integral over the ${x}$ and ${y}$ components of ${d\mathbf{a}}$ will cancel out and we’re left with only the ${z}$ component, which is ${\cos\theta\left|d\mathbf{a}\right|=R^{2}\sin\theta\cos\theta d\theta d\phi}$. The integral we must do is thus

$\displaystyle \int_{A}\frac{1}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}d\mathbf{a}=R^{2}\hat{\mathbf{z}}\int_{0}^{\pi}\int_{0}^{2\pi}\frac{\sin\theta\cos\theta d\theta d\phi}{\sqrt{r^{2}+R^{2}-2rR\cos\theta}} \ \ \ \ \ (6)$

Using software, we find that

$\displaystyle \int\frac{\sin\theta\cos\theta d\theta}{\sqrt{r^{2}+R^{2}-2rR\cos\theta}}=\frac{1}{3}{\frac{\sqrt{{r}^{2}+{R}^{2}-2\, rR\cos\left(\theta\right)}\left({r}^{2}+{R}^{2}+rR\cos\left(\theta\right)\right)}{{r}^{2}}} \ \ \ \ \ (7)$

Evaluating the limits requires specifying whether ${R}$ is larger or smaller than ${r}$. If we’re interested in currents within the sphere, then ${R>r>0}$ and we have

$\displaystyle R^{2}\hat{\mathbf{z}}\int_{0}^{\pi}\int_{0}^{2\pi}\frac{\sin\theta\cos\theta d\theta d\phi}{\sqrt{r^{2}+R^{2}-2rR\cos\theta}}=\frac{4}{3}\pi r\hat{\mathbf{z}} \ \ \ \ \ (8)$

Plugging this back into the average field formula above we get

$\displaystyle \mathbf{B}_{av}=-\frac{\mu_{0}}{4\pi R^{3}}\int_{V}\mathbf{J}\times\mathbf{r}d^{3}\mathbf{r} \ \ \ \ \ (9)$

We can write this in terms of the magnetic dipole moment as follows:

$\displaystyle \mathbf{B}_{av}=\frac{2\mu_{0}}{4\pi R^{3}}\mathbf{m} \ \ \ \ \ (10)$

If we look at currents outside the sphere, then ${R when we evaluate the limits on the integral above, giving

$\displaystyle R^{2}\hat{\mathbf{z}}\int_{0}^{\pi}\int_{0}^{2\pi}\frac{\sin\theta\cos\theta d\theta d\phi}{\sqrt{r^{2}+R^{2}-2rR\cos\theta}}=\frac{2R^{3}}{3r^{2}}\hat{\mathbf{z}} \ \ \ \ \ (11)$

Plugging this into the average field formula we get

$\displaystyle \mathbf{B}_{av}=-\frac{\mu_{0}}{4\pi}\int_{V}\frac{\mathbf{J}\times\mathbf{r}}{r^{3}}d^{3}\mathbf{r} \ \ \ \ \ (12)$

Remember that ${\mathbf{r}}$ points from the centre to the volume element, so it is the negative of the vector ${\mathbf{r}_{c}}$ from the volume element to the centre. In this form we get

$\displaystyle \mathbf{B}_{av}=\frac{\mu_{0}}{4\pi}\int_{V}\frac{\mathbf{J}\times\mathbf{r}_{c}}{r_{c}^{3}}d^{3}\mathbf{r} \ \ \ \ \ (13)$

This is just the volume form of the Biot-Savart law for calculating the field at the centre of the sphere thus for currents outside the sphere, the average of their field over the sphere is equal to the field produced at the centre of the sphere.

# Magnetic vector potential as the curl of another function

Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Problem 5.53.

The fact that ${\mathbf{B}}$ is divergenceless allows us to express it as the curl of a vector potential: ${\mathbf{B}=\nabla\times\mathbf{A}}$. By assuming ${\nabla\cdot\mathbf{A}=0}$ as well, can express it in turn as the curl of another function: ${\mathbf{A}=\nabla\times\mathbf{W}}$. Although this is rarely useful, we can try to work out ${\mathbf{W}}$ in a few cases.

First, we can express ${\mathbf{W}}$ in terms of ${\mathbf{B}}$ by noting

 $\displaystyle \mathbf{B}$ $\displaystyle =$ $\displaystyle \nabla\times\mathbf{A}\ \ \ \ \ (1)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \nabla\times\left(\nabla\times\mathbf{W}\right)\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \nabla\left(\nabla\cdot\mathbf{W}\right)-\nabla^{2}\mathbf{W} \ \ \ \ \ (3)$

If we assume that ${\mathbf{W}}$ is divergenceless as well, then we get

$\displaystyle \nabla^{2}\mathbf{W}=-\mathbf{B} \ \ \ \ \ (4)$

This is the same form as the equation for the vector potential in terms of the current density:

$\displaystyle \nabla^{2}\mathbf{A}=-\mu_{0}\mathbf{J} \ \ \ \ \ (5)$

This could be written in integral form as

$\displaystyle \mathbf{A}\left(\mathbf{r}\right)=\frac{\mu_{0}}{4\pi}\int_{V}\frac{\mathbf{J}\left(\mathbf{r}^{\prime}\right)}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}d^{3}\mathbf{r}^{\prime} \ \ \ \ \ (6)$

Therefore we can write

$\displaystyle \mathbf{W}\left(\mathbf{r}\right)=\frac{1}{4\pi}\int_{V}\frac{\mathbf{B}\left(\mathbf{r}^{\prime}\right)}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}d^{3}\mathbf{r}^{\prime} \ \ \ \ \ (7)$

It’s important to note that this equation relies on ${\mathbf{B}\rightarrow0}$ at infinite distance, just as the equation for ${\mathbf{A}}$ relied on the current density being finite in extent.

We can now find ${\mathbf{W}}$ in a couple of special cases. First, consider the case of a constant field. We might be tempted to try calculating ${\mathbf{W}}$ by simply taking ${\mathbf{B}}$ outside the integral above, but this integral formula doesn’t apply in this case since ${\mathbf{B}}$ extends to infinity. We can, however, use the result we got earlier for the vector potential for a constant field.

$\displaystyle \mathbf{A}=-\frac{1}{2}\mathbf{r}\times\mathbf{B} \ \ \ \ \ (8)$

We need to solve the equation

$\displaystyle \nabla\times\mathbf{W}=-\frac{1}{2}\mathbf{r}\times\mathbf{B} \ \ \ \ \ (9)$

One way of approaching this is to split the vector equation into its components. For the ${x}$ component we have

$\displaystyle \frac{\partial W_{z}}{\partial y}-\frac{\partial W_{y}}{\partial z}=-\frac{1}{2}\left(yB_{z}-zB_{y}\right) \ \ \ \ \ (10)$

We can try a solution of the form

 $\displaystyle \frac{\partial W_{z}}{\partial y}$ $\displaystyle =$ $\displaystyle -\frac{1}{2}yB_{z}\ \ \ \ \ (11)$ $\displaystyle W_{z}$ $\displaystyle =$ $\displaystyle -\frac{1}{4}y^{2}B_{z}+f\left(x,z\right)\ \ \ \ \ (12)$ $\displaystyle \frac{\partial W_{y}}{\partial z}$ $\displaystyle =$ $\displaystyle -\frac{1}{2}zB_{y}\ \ \ \ \ (13)$ $\displaystyle W_{y}$ $\displaystyle =$ $\displaystyle -\frac{1}{4}z^{2}B_{y}+g\left(x,y\right) \ \ \ \ \ (14)$

where ${f}$ and ${g}$ are functions of integration.

Working out the ${z}$ component, we get

$\displaystyle W_{y}=-\frac{1}{4}x^{2}B_{y}+h\left(y,z\right) \ \ \ \ \ (15)$

Comparing the two equations for ${W_{y}}$ we get

$\displaystyle W_{y}=-\frac{1}{4}B_{y}\left(x^{2}+z^{2}\right) \ \ \ \ \ (16)$

The other two components can be worked out the same way and we get

$\displaystyle \mathbf{W}=-\frac{1}{4}\left[B_{x}\left(y^{2}+z^{2}\right)\hat{\mathbf{x}}+B_{y}\left(x^{2}+z^{2}\right)\hat{\mathbf{y}}+B_{z}\left(x^{2}+y^{2}\right)\hat{\mathbf{z}}\right] \ \ \ \ \ (17)$

By direct calculation we can check that ${\nabla\cdot\mathbf{W}=0}$ so this is a valid solution. (There may be some fancy way of expressing this entirely in terms of vectors and their products, but if so, it eluded me.) The solution is unlikely to be unique.

For a second example, we can return to the infinite solenoid. Griffiths shows in his Example 5.12 that the vector potential inside a solenoid with ${n}$ turns per unit length and of radius ${R}$ is

$\displaystyle \mathbf{A}=\begin{cases} \frac{\mu_{0}nIr}{2}\hat{\boldsymbol{\phi}} & rR \end{cases} \ \ \ \ \ (18)$

Again, we seek components of ${\mathbf{W}}$ that satisfy the equation ${\mathbf{A}=\nabla\times\mathbf{W}}$. Outside the solenoid, we have, using the equations for the curl in cylindrical coordinates:

 $\displaystyle \frac{1}{r}\frac{\partial W_{z}}{\partial\phi}-\frac{\partial W_{\phi}}{\partial z}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (19)$ $\displaystyle \frac{\partial W_{r}}{\partial z}-\frac{\partial W_{z}}{\partial r}$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}nIR^{2}}{2r}\ \ \ \ \ (20)$ $\displaystyle \frac{1}{r}\left[\frac{\partial\left(rW_{\phi}\right)}{\partial r}-\frac{\partial W_{r}}{\partial\phi}\right]$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (21)$

We can try the solution

 $\displaystyle W_{r}$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}nIR^{2}z}{2r}\ \ \ \ \ (22)$ $\displaystyle W_{\phi}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (23)$ $\displaystyle W_{z}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (24)$

This satisfies all three curl equations, and also satisfies

 $\displaystyle \nabla\cdot\mathbf{W}$ $\displaystyle =$ $\displaystyle \frac{1}{r}\frac{\partial\left(rW_{r}\right)}{\partial r}+0+0\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (26)$

Inside the solenoid, we have

 $\displaystyle \frac{1}{r}\frac{\partial W_{z}}{\partial\phi}-\frac{\partial W_{\phi}}{\partial z}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (27)$ $\displaystyle \frac{\partial W_{r}}{\partial z}-\frac{\partial W_{z}}{\partial r}$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}nIr}{2}\ \ \ \ \ (28)$ $\displaystyle \frac{1}{r}\left[\frac{\partial\left(rW_{\phi}\right)}{\partial r}-\frac{\partial W_{r}}{\partial\phi}\right]$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (29)$

If we try a similar solution, we have

 $\displaystyle W_{r}$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}nIrz}{2}\ \ \ \ \ (30)$ $\displaystyle W_{\phi}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (31)$ $\displaystyle W_{z}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (32)$

This satisfies all of the curl equations but unfortunately the divergence isn’t zero:

 $\displaystyle \nabla\cdot\mathbf{W}$ $\displaystyle =$ $\displaystyle \frac{1}{r}\frac{\partial\left(rW_{r}\right)}{\partial r}+0+0\ \ \ \ \ (33)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \mu_{0}nIz \ \ \ \ \ (34)$

We can fix this by adding in a ${z}$ component, so we get

 $\displaystyle W_{r}$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}nIrz}{2}\ \ \ \ \ (35)$ $\displaystyle W_{\phi}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (36)$ $\displaystyle W_{z}$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}nIz^{2}}{2} \ \ \ \ \ (37)$

This doesn’t affect the curl, and makes the divergence zero. Again, this solution is unlikely to be unique.

# Magnetic vector potential from magnetic field

Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Problem 5.51.

The magnetic vector potential can be used to find the field through the relation ${\mathbf{B}=\nabla\times\mathbf{A}}$, but there is no common formula for calculating the potential from the field. The potential is usually calculated from the current density via the formula

$\displaystyle \mathbf{A}\left(\mathbf{r}\right)=\frac{\mu_{0}}{4\pi}\int_{V}\frac{\mathbf{J}\left(\mathbf{r}^{\prime}\right)}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}d^{3}\mathbf{r}^{\prime} \ \ \ \ \ (1)$

One attempt at a formula for deriving ${\mathbf{A}}$ from ${\mathbf{B}}$ is

$\displaystyle \mathbf{A}\left(\mathbf{r}\right)=\int_{\mathcal{O}}^{\mathbf{r}}\mathbf{B}\times d\mathbf{l} \ \ \ \ \ (2)$

where the integral is a line integral over some path from the origin (which must be defined independently) to the observation point ${\mathbf{r}}$. For a constant field, this gives

 $\displaystyle \mathbf{A}$ $\displaystyle =$ $\displaystyle \mathbf{B}\times\mathbf{r}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\mathbf{r}\times\mathbf{B} \ \ \ \ \ (4)$

This differs from the correct value by a factor of ${\frac{1}{2}}$, but there is a more serious problem in that the line integral isn’t always independent of the path. To see this, consider the case of an infinite wire with a current ${I}$. The magnetic field circles the wire so in cylindrical coordinates

$\displaystyle \mathbf{B}=\frac{\mu_{0}I}{2\pi r}\hat{\boldsymbol{\phi}} \ \ \ \ \ (5)$

If we integrate ${\mathbf{B}}$ around a closed square loop of side 1 that is perpendicular to the field, the contributions from the two edges perpendicular to the wire will cancel so we get

$\displaystyle \oint\mathbf{B}\times d\mathbf{l}=\frac{\mu_{0}I}{2\pi}\left(\frac{1}{a}-\frac{1}{a+1}\right) \ \ \ \ \ (6)$

where ${a}$ is the distance of the near edge from the wire. This is clearly not zero, which it would have to be for the integral to be path independent.

Another attempt at a formula is

$\displaystyle \mathbf{A}\left(\mathbf{r}\right)=-\mathbf{r}\times\int_{0}^{1}\lambda\mathbf{B}\left(\lambda\mathbf{r}\right)d\lambda \ \ \ \ \ (7)$

(This formula is stated in Griffiths’s problem.)

For a constant field, this gives

 $\displaystyle \mathbf{A}$ $\displaystyle =$ $\displaystyle -\mathbf{r}\times\mathbf{B}\int_{0}^{1}\lambda d\lambda\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{2}\mathbf{r}\times\mathbf{B} \ \ \ \ \ (9)$

which is correct.

In the case of the infinite wire, we have

$\displaystyle \mathbf{B}\left(\lambda\mathbf{r}\right)=\frac{\mu_{0}I}{2\pi\left(\lambda s\right)}\hat{\boldsymbol{\phi}} \ \ \ \ \ (10)$

where ${s}$ is the perpendicular distance from the wire to the point ${\mathbf{r}}$. Therefore

 $\displaystyle \int_{0}^{1}\lambda\mathbf{B}\left(\lambda\mathbf{r}\right)d\lambda$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}I}{2\pi s}\hat{\boldsymbol{\phi}}\int_{0}^{1}d\lambda\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}I}{2\pi s}\hat{\boldsymbol{\phi}} \ \ \ \ \ (12)$

In cylindrical coordinates, the position vector ${\mathbf{r}=s\hat{\mathbf{s}}+z\hat{\mathbf{z}}}$ so

 $\displaystyle \mathbf{A}$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}I}{2\pi s}\left(s\hat{\mathbf{s}}+z\hat{\mathbf{z}}\right)\times\hat{\boldsymbol{\phi}}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}I}{2\pi s}\left(z\hat{\mathbf{s}}-s\hat{\mathbf{z}}\right) \ \ \ \ \ (14)$

In this case, ${\nabla\cdot\mathbf{A}=0}$ and ${\nabla\times\mathbf{A}=\mathbf{B}}$ as can be checked by direct calculation, so this is a valid potential. However, in general, a potential calculated this way does not satisfy ${\nabla\cdot\mathbf{A}=0}$. We have

 $\displaystyle \nabla\cdot\mathbf{A}$ $\displaystyle =$ $\displaystyle -\nabla\cdot\left[\mathbf{r}\times\int_{0}^{1}\lambda\mathbf{B}\left(\lambda\mathbf{r}\right)d\lambda\right]\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \mathbf{r}\cdot\int_{0}^{1}\lambda\nabla\times\mathbf{B}\left(\lambda\mathbf{r}\right)d\lambda-\int_{0}^{1}\lambda\mathbf{B}\left(\lambda\mathbf{r}\right)\cdot\nabla\times\mathbf{r}d\lambda\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \mathbf{r}\cdot\int_{0}^{1}\lambda^{2}\mu_{0}\mathbf{J}\left(\lambda\mathbf{r}\right)d\lambda-0 \ \ \ \ \ (17)$

where we’ve used Ampère’s law and the chain rule in the first term, and the fact that ${\nabla\times\mathbf{r}=0}$ in the second. Thus the only way this can be zero for all observation points ${\mathbf{r}}$ is for the integral to be zero, which effectively means that there can be no current. So this formula for the potential is not in general divergence-free.

# Magnetostatic boundary conditions

Required math: calculus

Required physics: electrostatics & magnetostatics

Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Problem 5.32.

A while back, we worked out the behaviour of the electric field and potential at a layer of surface charge. We can do a similar analysis for the magnetic field and its behaviour as we cross a surface current.

Suppose we have a surface with a surface current ${\mathbf{K}}$. At a local point, let’s say this current flows in the ${+x}$ direction. Consider a small patch of area and build a little pillbox that straddles the surface at this area. If we make the thickness of the box infinitesimally thin, then a surface integral is essentially over the two sides of the box on either side of the current. Since ${\nabla\cdot\mathbf{B}=0}$, Gauss’s law says that

$\displaystyle \int_{V}\nabla\cdot\mathbf{B}d^{3}\mathbf{r}=\int_{A}\mathbf{B}\cdot d\mathbf{a}=0 \ \ \ \ \ (1)$

Now ${\mathbf{B}\cdot d\mathbf{a}=B_{\perp}da}$ so the second integral says that

$\displaystyle \int_{A}\mathbf{B}\cdot d\mathbf{a}=\left(B_{\perp}^{above}-B_{\perp}^{below}\right)A=0 \ \ \ \ \ (2)$

Therefore, the normal component of ${\mathbf{B}}$ is continuous across a surface current.

For the parallel component, consider first the component of ${\mathbf{B}}$ parallel to the surface but perpendicular to the current. We can define a little loop that straddles the surface, where the area enclosed by the loop is perpendicular to the current. If we make the vertical sides of the loop infinitesimal and the horizontal sides of length 1, then we can use Stokes’s theorem to say

$\displaystyle \oint\mathbf{B}\cdot d\mathbf{l}=B_{\parallel}^{above}-B_{\parallel}^{below}=\mu_{0}I=\mu_{0}K \ \ \ \ \ (3)$

Finally, if we take a loop perpendicular to the surface but parallel to the current, then the loop encloses zero current so ${\oint\mathbf{B}\cdot d\mathbf{l}=0}$ and this component of the field is continuous.

Thus the only component of ${\mathbf{B}}$ that has a discontinuity is the one parallel to the surface but perpendicular to the current. That is, the discontinuity is perpendicular both to the normal ${\hat{\mathbf{n}}}$ to the surface and the current ${\mathbf{K}}$, so the difference must be expressible as the cross product of these two vectors:

$\displaystyle \mathbf{B}^{above}-\mathbf{B}^{below}=\mu_{0}\mathbf{K}\times\hat{\mathbf{n}} \ \ \ \ \ (4)$

To get the direction of the discontinuity, suppose we look in the ${+x}$ direction (that is, in the direction of the current). Then the field above the current points to the right (that is, in the ${-y}$ direction) and below it points to the left (${+y}$ direction). Thus the difference ${\mathbf{B}^{above}-\mathbf{B}^{below}}$ points to the right. This works out properly if ${\mathbf{K}}$ points in the ${+x}$ direction and ${\hat{\mathbf{n}}}$ points in the ${+z}$ direction.

As for the vector potential, assuming ${\nabla\cdot\mathbf{A}=0}$ means that ${A_{\perp}}$ is continuous (using the same argument as above). Since ${\mathbf{B}=\nabla\times\mathbf{A}}$, we can again use Stokes’s theorem to integrate ${\mathbf{A}}$ around a loop straddling the surface:

$\displaystyle \oint\mathbf{A}\cdot d\mathbf{l}=\int\mathbf{B}\cdot d\mathbf{a} \ \ \ \ \ (5)$

where the second integral is over the area enclosed by the loop. Unlike the integral ${\oint\mathbf{B}\cdot d\mathbf{l}}$ above, though, the flux of ${\mathbf{B}}$ enclosed by the loop diminishes to zero as we make the loop thinner and thinner. There is no infinitesimally thin sheet of magnetic field that is always enclosed by the loop as there was in the case of enclosed current. Thus in the limit, ${\oint\mathbf{A}\cdot d\mathbf{l}=0}$ for any orientation of the loop across the surface, and all components of ${\mathbf{A}}$ are continuous across a surface current.

The derivative of ${\mathbf{A}}$ does have a discontinuity however. To see this, suppose we set up a coordinate system with ${\hat{\mathbf{z}}}$ the normal to the area patch and ${\hat{\mathbf{y}}}$ the direction of the current ${\mathbf{K}}$. Then

$\displaystyle \Delta\mathbf{B}=\mathbf{B}^{above}-\mathbf{B}^{below}=\mu_{0}\mathbf{K}\times\hat{\mathbf{n}}=\mu_{0}K\hat{\mathbf{x}} \ \ \ \ \ (6)$

Writing out ${\Delta\mathbf{B}=\nabla\times\left(\Delta\mathbf{A}\right)}$ we have

 $\displaystyle \Delta B_{x}$ $\displaystyle =$ $\displaystyle \partial_{y}\left(\Delta A_{z}\right)-\partial_{z}\left(\Delta A_{y}\right)=\mu_{0}K\ \ \ \ \ (7)$ $\displaystyle \Delta B_{y}$ $\displaystyle =$ $\displaystyle \partial_{z}\left(\Delta A_{x}\right)-\partial_{x}\left(\Delta A_{z}\right)=0\ \ \ \ \ (8)$ $\displaystyle \Delta B_{z}$ $\displaystyle =$ $\displaystyle \partial_{x}\left(\Delta A_{y}\right)-\partial_{y}\left(\Delta A_{x}\right)=0 \ \ \ \ \ (9)$

Since ${\mathbf{A}}$ is continuous across the surface, the derivatives in directions parallel to the surface (that is, ${x}$ and ${y}$) will be the same on both sides, so ${x}$ and ${y}$ derivatives of ${\Delta\mathbf{A}}$ will all be zero. This gives us

 $\displaystyle \Delta B_{x}$ $\displaystyle =$ $\displaystyle \partial_{z}\left(\Delta A_{y}\right)=-\mu_{0}K\ \ \ \ \ (10)$ $\displaystyle \Delta B_{y}$ $\displaystyle =$ $\displaystyle \partial_{z}\left(\Delta A_{x}\right)=0 \ \ \ \ \ (11)$

To get the third derivative we use ${\nabla\cdot\mathbf{A}=0}$, which gives

$\displaystyle \partial_{z}\left(\Delta A_{z}\right)=-\partial_{x}\left(\Delta A_{x}\right)-\partial_{y}\left(\Delta A_{y}\right)=0 \ \ \ \ \ (12)$

Therefore, the ${y}$ component of the normal derivative of ${\mathbf{A}}$ has a discontinuity:

$\displaystyle \partial_{z}\left(\Delta\mathbf{A}\right)=-\mu_{0}K\hat{\mathbf{y}} \ \ \ \ \ (13)$

# Divergenceless vector field as a curl

Required math: calculus

Required physics: electrostatics & magnetostatics

Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Problem 5.30.

This is more mathematics than physics, but it relates to the magnetic vector potential so here we go. The vector potential ${\mathbf{A}}$ is defined so that ${\mathbf{B}=\nabla\times\mathbf{A}}$ and ${\nabla\cdot\mathbf{B}=0}$. We’d like to prove that, in general, any divergenceless vector field ${\mathbf{F}}$ can be written as the curl of another vector field ${\mathbf{A}}$. The curl ${\mathbf{F}=\nabla\times\mathbf{A}}$ has three components:

 $\displaystyle \partial_{y}A_{z}-\partial_{z}A_{y}$ $\displaystyle =$ $\displaystyle F_{x}\ \ \ \ \ (1)$ $\displaystyle \partial_{x}A_{z}-\partial_{z}A_{x}$ $\displaystyle =$ $\displaystyle -F_{y}\ \ \ \ \ (2)$ $\displaystyle \partial_{x}A_{y}-\partial_{y}A_{x}$ $\displaystyle =$ $\displaystyle F_{z} \ \ \ \ \ (3)$

Thus in principle, we have 3 coupled PDEs to solve, but we can get a solution by starting with the assumption that ${A_{x}=0}$ (this isn’t the only solution, of course, but it does give a solution). Then

 $\displaystyle \partial_{x}A_{z}$ $\displaystyle =$ $\displaystyle -F_{y}\ \ \ \ \ (4)$ $\displaystyle A_{z}$ $\displaystyle =$ $\displaystyle -\int F_{y}\left(x^{\prime},y,z\right)dx^{\prime}+G\left(y,z\right)\ \ \ \ \ (5)$ $\displaystyle \partial_{x}A_{y}$ $\displaystyle =$ $\displaystyle F_{z}\ \ \ \ \ (6)$ $\displaystyle A_{y}$ $\displaystyle =$ $\displaystyle \int F_{z}\left(x^{\prime},y,z\right)dx^{\prime}+H\left(y,z\right) \ \ \ \ \ (7)$

where ${G}$ and ${H}$ are functions of integration; since the integrals are with respect to ${x}$, the ‘constants’ of integration can be functions of ${y}$ and ${z}$.

We can now plug these into the first component of the curl above:

 $\displaystyle \partial_{y}A_{z}-\partial_{z}A_{y}$ $\displaystyle =$ $\displaystyle -\int\partial_{z}F_{z}\left(x^{\prime},y,z\right)dx^{\prime}+\partial_{y}G\left(y,z\right)-\int\partial_{y}F_{z}\left(x^{\prime},y,z\right)dx^{\prime}-\partial_{z}H\left(y,z\right)\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int\left(-\nabla\cdot\mathbf{F}+\partial_{x^{\prime}}F_{x^{\prime}}\right)dx^{\prime}+\partial_{y}G\left(y,z\right)-\partial_{z}H\left(y,z\right) \ \ \ \ \ (9)$

If we now want the result at a particular point ${\left(x,y,z\right)}$, we can introduce limits on the integral, and also use the requirement that ${\nabla\cdot\mathbf{F}=0}$:

 $\displaystyle F_{x}\left(x,y,z\right)$ $\displaystyle =$ $\displaystyle \int_{0}^{x}\partial_{x^{\prime}}F_{x^{\prime}}dx^{\prime}+\partial_{y}G\left(y,z\right)-\partial_{z}H\left(y,z\right)\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle F_{x}\left(x,y,z\right)-F_{x}\left(0,y,z\right)+\partial_{y}G\left(y,z\right)-\partial_{z}H\left(y,z\right)\ \ \ \ \ (11)$ $\displaystyle 0$ $\displaystyle =$ $\displaystyle -F_{x}\left(0,y,z\right)+\partial_{y}G\left(y,z\right)-\partial_{z}H\left(y,z\right) \ \ \ \ \ (12)$

At this point we can proceed in various ways, since the functions ${G}$ and ${H}$ have between them only this one condition. One option is to choose

 $\displaystyle G\left(y,z\right)$ $\displaystyle =$ $\displaystyle \int_{0}^{y}F_{x}\left(0,y^{\prime},z\right)dy^{\prime}\ \ \ \ \ (13)$ $\displaystyle H\left(y,z\right)$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (14)$

With this option, we get

 $\displaystyle A_{z}$ $\displaystyle =$ $\displaystyle \int_{0}^{y}F_{x}\left(0,y^{\prime},z\right)dy^{\prime}-\int_{0}^{x}F_{y}\left(x^{\prime},y,z\right)dx^{\prime}\ \ \ \ \ (15)$ $\displaystyle A_{y}$ $\displaystyle =$ $\displaystyle \int_{0}^{x}F_{z}\left(x^{\prime},y,z\right)dx^{\prime}\ \ \ \ \ (16)$ $\displaystyle A_{x}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (17)$

We could equally as well have chosen

 $\displaystyle G\left(y,z\right)$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (18)$ $\displaystyle H\left(y,z\right)$ $\displaystyle =$ $\displaystyle -\int_{0}^{z}F_{x}\left(0,y,z^{\prime}\right)dz^{\prime} \ \ \ \ \ (19)$

or some combination of the two.

Using the first option, we can check the curl.

 $\displaystyle \left(\nabla\times\mathbf{A}\right)_{x}$ $\displaystyle =$ $\displaystyle \partial_{y}\left[\int_{0}^{y}F_{x}\left(0,y^{\prime},z\right)dy^{\prime}-\int_{0}^{x}F_{y}\left(x^{\prime},y,z\right)dx^{\prime}\right]-\partial_{z}\left[\int_{0}^{x}F_{z}\left(x^{\prime},y,z\right)dx^{\prime}\right]\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle F_{x}\left(0,y,z\right)+\int_{0}^{x}\left(-\nabla\cdot\mathbf{F}+\partial_{x^{\prime}}F_{x^{\prime}}\right)dx^{\prime}\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle F_{x}\left(0,y,z\right)+F_{x}\left(x,y,z\right)-F_{x}\left(0,y,z\right)\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle F_{x}\left(x,y,z\right)\ \ \ \ \ (23)$ $\displaystyle \left(\nabla\times\mathbf{A}\right)_{y}$ $\displaystyle =$ $\displaystyle -\partial_{x}\left[\int_{0}^{y}F_{x}\left(0,y^{\prime},z\right)dy^{\prime}-\int_{0}^{x}F_{y}\left(x^{\prime},y,z\right)dx^{\prime}\right]+\partial_{z}\left(0\right)\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle F_{y}\left(x,y,z\right)\ \ \ \ \ (25)$ $\displaystyle \left(\nabla\times\mathbf{A}\right)_{z}$ $\displaystyle =$ $\displaystyle \partial_{x}\int_{0}^{x}F_{z}\left(x^{\prime},y,z\right)dx^{\prime}-\partial_{y}\left(0\right)\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle F_{z}\left(x,y,z\right) \ \ \ \ \ (27)$

For the divergence

$\displaystyle \nabla\cdot\mathbf{A}=\partial_{x}\left(0\right)+\partial_{y}\int_{0}^{x}F_{z}\left(x^{\prime},y,z\right)dx^{\prime}+\partial_{z}\left[\int_{0}^{y}F_{x}\left(0,y^{\prime},z\right)dy^{\prime}-\int_{0}^{x}F_{y}\left(x^{\prime},y,z\right)dx^{\prime}\right] \ \ \ \ \ (28)$

This isn’t zero in general.

For a specific example, consider

$\displaystyle \mathbf{F}=y\hat{\mathbf{x}}+z\hat{\mathbf{y}}+x\hat{\mathbf{z}} \ \ \ \ \ (29)$

Then

 $\displaystyle A_{x}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (30)$ $\displaystyle A_{y}$ $\displaystyle =$ $\displaystyle \int_{0}^{x}F_{z}\left(x^{\prime},y,z\right)dx^{\prime}\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int_{0}^{x}x^{\prime}dx^{\prime}\ \ \ \ \ (32)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{x^{2}}{2}\ \ \ \ \ (33)$ $\displaystyle A_{z}$ $\displaystyle =$ $\displaystyle \int_{0}^{y}F_{x}\left(0,y^{\prime},z\right)dy^{\prime}-\int_{0}^{x}F_{y}\left(x^{\prime},y,z\right)dx^{\prime}\ \ \ \ \ (34)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int_{0}^{y}y^{\prime}dy^{\prime}-\int_{0}^{x}zdx^{\prime}\ \ \ \ \ (35)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{y^{2}}{2}-xz \ \ \ \ \ (36)$

By direct calculation

 $\displaystyle \left(\nabla\times\mathbf{A}\right)_{x}$ $\displaystyle =$ $\displaystyle \partial_{y}A_{z}-\partial_{z}A_{y}\ \ \ \ \ (37)$ $\displaystyle$ $\displaystyle =$ $\displaystyle y=F_{x}\ \ \ \ \ (38)$ $\displaystyle \left(\nabla\times\mathbf{A}\right)_{y}$ $\displaystyle =$ $\displaystyle \partial_{z}A_{x}-\partial_{x}A_{z}\ \ \ \ \ (39)$ $\displaystyle$ $\displaystyle =$ $\displaystyle z=F_{y}\ \ \ \ \ (40)$ $\displaystyle \left(\nabla\times\mathbf{A}\right)_{z}$ $\displaystyle =$ $\displaystyle \partial_{x}A_{y}-\partial_{y}A_{x}\ \ \ \ \ (41)$ $\displaystyle$ $\displaystyle =$ $\displaystyle x=F_{z} \ \ \ \ \ (42)$

# Magnetic vector potential: div, curl and Laplacian

Required math: calculus

Required physics: electrostatics & magnetostatics

Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Problem 5.27.

We’ve seen how the magnetic vector potential is derived from the fact that the magnetic field can be expressed as the curl of a vector field. The fact that ${\nabla\times\mathbf{A}}$ gives the Biot-Savart equation for the magnetic field can be obtained by just reversing the derivation of ${\mathbf{A}}$.

We can check a couple of other derivatives from the definition of ${\mathbf{A}}$, which is

$\displaystyle \mathbf{A}\left(\mathbf{r}\right)=\frac{\mu_{0}}{4\pi}\int_{V}\frac{\mathbf{J}\left(\mathbf{r}^{\prime}\right)}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}d^{3}\mathbf{r}^{\prime} \ \ \ \ \ (1)$

First, the divergence ${\nabla\cdot\mathbf{A}}$. Remember that the derivative is taken only with respect to the unprimed coordinates, so the divergence works out to

$\displaystyle \nabla\cdot\mathbf{A}=\frac{\mu_{0}}{4\pi}\int_{V}\mathbf{J}\left(\mathbf{r}^{\prime}\right)\cdot\nabla\frac{1}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}d^{3}\mathbf{r}^{\prime} \ \ \ \ \ (2)$

If we look at the derivative with respect to ${x}$ term, we have

 $\displaystyle \int_{V}J_{x}\left(\mathbf{r}^{\prime}\right)\cdot\frac{\partial}{\partial x}\left(\frac{1}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}\right)d^{3}\mathbf{r}^{\prime}$ $\displaystyle =$ $\displaystyle -\int_{V}J_{x}\left(\mathbf{r}^{\prime}\right)\frac{\partial}{\partial x^{\prime}}\left(\frac{1}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}\right)d^{3}\mathbf{r}^{\prime}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\left.J_{x}\left(\mathbf{r}^{\prime}\right)\frac{1}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}\right|_{A}+\int_{V}\frac{1}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}\frac{\partial}{\partial x^{\prime}}\left(J_{x}\left(\mathbf{r}^{\prime}\right)\right)d^{3}\mathbf{r}^{\prime} \ \ \ \ \ (4)$

In the first line, we used the fact that the derivative of ${\frac{1}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}}$ with respect to ${x}$ is the negative of the derivative with respect to ${x^{\prime}}$. We then integrated by parts. The integrated term is evaluated over the boundary surface ${A}$. We can take this surface to lie outside the region in which currents exist, so ${J_{x}=0}$ there. We will get similar results for the ${y}$ and ${z}$ terms in the original integral, so if we combine them, we get

$\displaystyle \int_{V}\mathbf{J}\left(\mathbf{r}^{\prime}\right)\cdot\nabla\frac{1}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}d^{3}\mathbf{r}^{\prime}=\int_{V}\frac{1}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}\nabla^{\prime}\cdot\mathbf{J}\left(\mathbf{r}^{\prime}\right)d^{3}\mathbf{r}^{\prime} \ \ \ \ \ (5)$

For steady currents, ${\nabla^{\prime}\cdot\mathbf{J}\left(\mathbf{r}^{\prime}\right)=0}$, so in the case of localized, steady currents, we have ${\nabla\cdot\mathbf{A}=0}$, which is what we assumed in our derivation of ${\mathbf{A}}$, so this is consistent.

We also had a version of Poisson’s equation for the potential:

$\displaystyle \nabla^{2}\mathbf{A}=-\mu_{0}\mathbf{J} \ \ \ \ \ (6)$

Taking the ${x}$ component, we have

 $\displaystyle \nabla^{2}A_{x}\left(\mathbf{r}\right)$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}}{4\pi}\int_{V}J_{x}\left(\mathbf{r}^{\prime}\right)\nabla^{2}\frac{1}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}d^{3}\mathbf{r}^{\prime}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\mu_{0}\int_{V}J_{x}\left(\mathbf{r}^{\prime}\right)\delta^{3}\left(\left|\mathbf{r}-\mathbf{r}^{\prime}\right|\right)d^{3}\mathbf{r}^{\prime}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\mu_{0}J_{x}\left(\mathbf{r}\right) \ \ \ \ \ (9)$

where we’ve used a result for the delta function.

Similar relations for ${y}$ and ${z}$ verify Poisson’s equation.

# Magnetic vector potential: sheet of current

Required math: calculus

Required physics: electrostatics & magnetostatics

Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Problem 5.26.

Another example of calculating the magnetic vector potential in a case where the current extends to infinity. We consider a uniform sheet of current in the ${xy}$ plane, carrying surface current density ${K\hat{\mathbf{x}}}$. Using the same argument as in the case of a slab of current, the magnetic field due to this current is

$\displaystyle \mathbf{B}=\begin{cases} -\frac{\mu_{0}}{2}K\hat{\mathbf{y}} & z>0\\ \frac{\mu_{0}}{2}K\hat{\mathbf{y}} & z<0 \end{cases} \ \ \ \ \ (1)$

Note that the field is independent of the distance from the sheet of current.

We can work out the potential by applying Stokes’s theorem. From the definition of ${\mathbf{A}}$, we know that ${\nabla\times\mathbf{A}=\mathbf{B}}$, so if we define a closed loop then we have

$\displaystyle \oint\mathbf{A}\cdot d\mathbf{l}=\int_{A}\mathbf{B}\cdot d\mathbf{a} \ \ \ \ \ (2)$

where the integral on the LHS is a line integral around the loop and the integral on the right is over the area ${A}$ enclosed by the loop. For a flat loop, the direction of integration on the LHS is such that the right hand rule gives a vector pointing in the same direction as ${d\mathbf{a}}$ on the RHS.

We need to be careful in defining directions for the area and line integrals to get the signs right. We’ll take as our area ${A}$ a rectangle above the ${xy}$ plane and parallel to the ${xz}$ plane. The sides of the rectangle parallel to the ${xy}$ plane are of length 1, with the lower side at a distance ${a}$ and the upper side at a distance ${b}$ from this plane. The normal to the area points in the ${\hat{\mathbf{y}}}$ direction Since ${\mathbf{B}}$ is uniform everywhere above the plane, we get

$\displaystyle \int_{A}\mathbf{B}\cdot d\mathbf{a}=-\frac{\mu_{0}}{2}K\left(b-a\right) \ \ \ \ \ (3)$

For the line integral, the path around the rectangle is clockwise when looking in the ${+y}$ direction, and by symmetry, the integral of ${\mathbf{A}\cdot d\mathbf{l}}$ along the vertical sides cancels out, so

$\displaystyle \oint\mathbf{A}\cdot d\mathbf{l}=A_{x}\left(b\right)-A_{x}\left(a\right) \ \ \ \ \ (4)$

Comparing these two, a reasonable candidate is

$\displaystyle \mathbf{A}=-\frac{\mu_{0}}{2}Kz\hat{\mathbf{x}} \ \ \ \ \ (5)$

We can check this by finding the div and curl, as usual:

 $\displaystyle \nabla\cdot\mathbf{A}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (6)$ $\displaystyle \nabla\times\mathbf{A}$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}}{2}K\hat{\mathbf{y}} \ \ \ \ \ (7)$

Below the sheet of current, the sign is reversed so we have

$\displaystyle \mathbf{A}=\frac{\mu_{0}}{2}Kz\hat{\mathbf{x}} \ \ \ \ \ (8)$