Finding the vector potential of magnetic field via line integration

Question

If we look at the relationship between the scalar electric potential and electric field in electrostatics, $\vec{E} = - \vec{\nabla} \phi$, we can easily invert this relationship by $$ V (\vec{r}) = -\int \limits_{\vec{r}_0}^{\vec{r}} \mathrm{d} \vec{\ell} \cdot \vec{E} $$ where $\vec{r}_0$ is arbitrary.

This got me thinking; is the same possible for magnetic field $\vec{B} = \vec{\nabla} \times \vec{A}$? Can we find $\vec{A}$ (in some specific gauge) from $\vec{B}$ by inverting this relationship in terms of a line integral (I'm aware of finding $\vec{A}$ from current distribution, but that's a volume integral and it involves sources, which I'd like to avoid).

So I made a guess: $$ \vec{A} (\vec{r}) \overset{?}{=} \int \limits_{\vec{r}_0}^{\vec{r}} \mathrm{d} \vec{\ell} \times \vec{B} $$

Of course, we need to verify that this is correct (it isn't, but it's only a constant factor), which we can do $$ A_j \overset{?}{=} \varepsilon_{jab} \int \limits_{\vec{r}_0}^{\vec{r}} \mathrm{d} \ell_a B_b \quad \to \quad \left( \vec{\nabla} \times \vec{A} \right)_i = \varepsilon_{ikj} \partial_k A_j = \varepsilon_{ikj} \varepsilon_{jab} \partial_k \int \limits_{\vec{r}_0}^{\vec{r}} \mathrm{d} \ell_a B_b $$

I'm not so sure about this, but from what I understand, a derivative acting on a line integral like this will pluck out the index of the line element, $\mathrm{d} \ell$, in this case, $a$ becomes $k$ $$ \left( \vec{\nabla} \times \vec{A} \right)_i = \varepsilon_{ikj} \varepsilon_{jkb} B_b = \left( \delta_{ik} \delta_{kb} - \delta_{ib} \delta_{kk} \right) B_b = B_i - 3 B_i = - 2 B_i $$

So the correct formula would seem to be $$ \vec{A} (\vec{r}) \overset{\checkmark}{=} - \frac{1}{2} \int \limits_{\vec{r}_0}^{\vec{r}} \mathrm{d} \vec{\ell} \times \vec{B} $$

I verified that this should be the case on a simple magnetic field $\vec{B} = B_0 \hat{z}$ and for path that is a straight line between $\vec{r}_0 = \vec{0}$ and $\vec{r}$. In that case $\mathrm{d} \vec{\ell} = \hat{r} \mathrm{d} \ell$. We also need $\hat{r} \times \hat{z} = - \hat{\varphi}$ and the integral becomes $$ \vec{A} (\vec{r}) = \frac{1}{2} B_0 \hat{\varphi} \int \limits_0^r \mathrm{d} \ell = \frac{1}{2} B_0 r \hat{\varphi} = \frac{1}{2} B_0 \left( -y, x, 0 \right) $$

Taking a curl of this gives the original magnetic field.

An alternative way is to plug in $\vec{B} = \vec{\nabla} \times \vec{A}$ under the integral $$ \mathrm{d} \vec{\ell} \times \vec{B} = \mathrm{d} \vec{\ell} \times \left( \vec{\nabla} \times \vec{A} \right) = \vec{\nabla} \left( \mathrm{d} \vec{\ell} \cdot \vec{A} \right) - \left( \mathrm{d} \vec{\ell} \cdot \vec{\nabla} \right) \vec{A} $$

The problem is I am not sure what to do with the first term $\vec{\nabla} \left( \mathrm{d} \vec{\ell} \cdot \vec{A} \right)$, it doesn't seem to simplify to anything reasonable, whereas the second term $\left( \mathrm{d} \vec{\ell} \cdot \vec{\nabla} \right) \vec{A}$, when integrated, yields $\vec{A} (\vec{r}) - \vec{A} (\vec{r}_0)$. If my thoughts are correct though, the term $\mathrm{d} \ell_j \partial_i A_j$ should be equal to $- \mathrm{d} \ell_j \partial_j A_i$, i.e. the combination $\partial_i A_j + \partial_j A_i$ should be zero. I feel like this is rather a gauge choice than something that should always hold (although, it brings 6 equations, which seems like a lot for a gauge).

Can someone point me to the right direction? Is my formula correct? What about the $\partial_i A_j + \partial_j A_i = 0$?

Answer 1

Firstly, you do not invert $\mathbf{E}=-\boldsymbol{\nabla}V$ the way you wrote it since:

$$ \int_{\mathbf{r}_0}^{\mathbf{r}}\boldsymbol{\nabla}'V\left(\mathbf{r}'\right).d\mathbf{r}'=V\left(\mathbf{r}\right)-V\left(\mathbf{r}_0\right) $$

i.e. you only get back the potential difference between two points, not the actual value of the potential.

Secondly, this is in fact a special case of the Generalized Stokes Theorem.

$$ \int_\Omega d\omega = \oint_{\partial\Omega}\omega $$

Where $\omega$ is a $k$-form and $\Omega$ is a $k+1$-chain. $d$ here is the exterior derivative.

I will now proceed to briefly explain how gradient and curl cases map onto the above expression.

In case of gradient: In this case $V\to \omega$ is a 0-form, a scalar, whilst $\Omega$ the curve from $\mathbf{r}_0$ to $\mathbf{r}$ is a 1-chain - essentially a set of straight-line segments that together make up the curve. $\partial \Omega$ is the boundary to the chain $\Omega$, and is itself a $k$-chain. Here it is simply two points.

In case of curl: $\mathbf{A}\to \omega$ is a 1-form (see musical isomorphisms to understand how vectors map to 1-forms). $\Omega$ is then a 2-chain, a set of, essentially, triangles, that mesh the two-dimensional space. Essentially it encodes normal Stokes theorem.

So a proper generalization of your starting expression with scalar potential to the case of curl, is actually Stokes theorem.

Thirdly, consider vector fields:

$$ \begin{align} \mathbf{A}_1 &= -\mathbf{\hat{x}} \,y \\ \boldsymbol{\nabla}\times\mathbf{A}_1 &= \mathbf{\hat{z}} \\ \boldsymbol{\nabla}.\mathbf{A}_1 &= 0 \\ \mathbf{A}_2 &= \mathbf{\hat{y}} \,x \\ \boldsymbol{\nabla}\times\mathbf{A}_2 &= \mathbf{\hat{z}} \\ \boldsymbol{\nabla}.\mathbf{A}_2 &= 0 \end{align} $$

Both vector fields have identical curls and satisfy Coulomb gauge, how can your procedure distinguish them? I think it cannot. This is sufficient to show that your approach does not generalize

Answer 2

The answer is found in this Wikipedia article: https://en.wikipedia.org/wiki/Gauge_fixing#Multipolar_gauge

In specific, the relationship between the electric field and potential can be reversed as $$ \varphi (\vec{r}, t) = - \vec{r} \cdot \int \limits_0^1 \vec{E} (u \vec{r}, t) \mathrm{d} u $$ and the relationship between the magnetic field and vector potential can be reversed, too $$ \vec{A} (\vec{r}, t) = - \vec{r} \times \int \limits_0^1 \vec{B} (u \vec{r}, t) \, u \, \mathrm{d} u $$

To see how this works, we take the first equation and apply $\partial_i$ to it. We get $$ \partial_i \varphi (\vec{r}, t) = - \partial_i \left[ x_j \cdot \int \limits_0^1 E_j (u \vec{r}, t) \, \mathrm{d} u \right] = - \int \limits_0^1 \left[ E_i (u \vec{r}, t) + u \, x_j \, E^\prime_{j,i} (u \vec{r}, t) \, \right] \mathrm{d} u $$

Here the chain rule when derivative is applied to $\vec{E}$ affects all spatial components the same way, since $u$ appears in all of them, which gives us a factor of $u$ in the second term. Now we recognize the total $u$ derivative $$ E_i (u \vec{r}, t) + u \, x_j E^\prime_{j,i} (u \vec{r}, t) = \frac{\mathrm{d}}{\mathrm{d} u} \left[ u \, E_i (u \vec{r}, t) \right] $$ so, since the total derivative appears under the integral, this gives us $$ \partial_i \varphi = - \left[ u \, E_i (u \vec{r}, t) \right]_{u = 0}^1 = - E_i (\vec{r}, t) $$

It works very similarly for $\vec{B}$ and $\vec{A}$. Let's apply curl to the equation for $\vec{A}$ (note: $\vec{\nabla}$ acts on everything to the right of it) $$ \vec{\nabla} \times \vec{A} (\vec{r}, t) = - \vec{\nabla} \times \left[ \vec{r} \times \int \limits_0^1 \vec{B} (u \vec{r}, t) \, u \, \mathrm{d} u \right] = - \int \limits_0^1 \left[ \left( \vec{\nabla} \cdot \vec{B} (u \vec{r}, t) \right) \vec{r} - \left( \vec{\nabla} \cdot \vec{r} \right) \vec{B} \right] u \, \mathrm{d} u $$

Using $\vec{\nabla} \cdot \vec{B} = 0$ and index notation, we get $$ \left( \vec{\nabla} \times \vec{A} (\vec{r}, t) \right)_i = - \int \limits_0^1 \left( B_j (u \vec{r}, t) \partial_j x_i - B_i (u \vec{r}, t) \partial_j x_j - x_j \partial_j B_i (u \vec{r}, t) \right) u \, \mathrm{d} u = \\ - \int \limits_0^1 \left( B_j (u \vec{r}, t) \, \delta_{ij} - 3 B_i (u \vec{r}, t) - u \, x_j B^\prime_{i, j} (u \vec{r}, t) \right) u \, \mathrm{d} u = \\ \int \limits_0^1 \left( 2 u \, B_i (u \vec{r}, t) + u^2 \, x_j B^\prime_{i, j} (u \vec{r}, t) \right) \, \mathrm{d} u $$

Again, we recognize a total $u$ derivative $$ 2 u \, B_i (u \vec{r}, t) + u^2 \, x_j B^\prime_{i, j} (u \vec{r}, t) = \frac{\mathrm{d}}{\mathrm{d} u} \left[ u^2 B_i (u \vec{r}, t) \right] $$ and so the integral yields $$ \left( \vec{\nabla} \times \vec{A} (\vec{r}, t) \right)_i = \left[ u^2 B_i (u \vec{r}, t) \right]_{u = 0}^1 = B_i (\vec{r}, t) $$

The Wikipedia article also answers my question about the gauge: this works in a particular gauge in which $\vec{r} \cdot \vec{A} = 0$, which is trivial to see (dot product combined with a cross product). This is called "Poincaré gauge". You learn something new every day :)