From what equations is magnetic field uniquely determined for a given current distribution?

Question

The Maxwell equations for magnetostatics in the absence of time varying electric field state that - $$ \mathbf{\overrightarrow{\nabla}} \cdot \mathbf{\overrightarrow{B}} = 0 $$ $$ \mathbf{\overrightarrow{\nabla}} \times \mathbf{\overrightarrow{B}} = \mu_0\mathbf{\overrightarrow{J}} $$

It is stated in the book 'Introduction to Electrodynamics' by David J. Griffiths that the above equations along with the boundary condition that $\mathbf{\overrightarrow{B}} \rightarrow 0$ far from all currents, determine the magnetic field which is equivalent to the magnetic field that we get by applying the Biot-Savart's law i.e., $$ \mathbf{\overrightarrow{B}}(\mathbf{\overrightarrow{r}}) = \frac{\mu_0}{4\pi} \int \frac{\mathbf{\overrightarrow{J}}(\mathbf{\overrightarrow{r'}}) \times {(\mathbf{\overrightarrow{r}}}-{\mathbf{\overrightarrow{r'}}})}{{|\mathbf{\overrightarrow{r}}-{\mathbf{\overrightarrow{r'}}}|}^3} {d{\tau}'} $$

Summarizing the above, it states that is there a unique solution for the set of equations $\mathbf{\overrightarrow{\nabla}}\cdot\mathbf{\overrightarrow{B}} = 0 $, $\mathbf{\overrightarrow{\nabla}}\times\mathbf{\overrightarrow{B}}= \mu_0\mathbf{\overrightarrow{J}}$ and $\mathbf{\overrightarrow{B}} \rightarrow 0$ at infinity.

But with the given information, I was able to conclude that it is possible to have more than one vector field which satisfies the above conditions for a given $\mathbf{\overrightarrow{J}}$.

My argument is simple, if we can find a vector $\mathbf{\overrightarrow{V}}$ such that $\mathbf{\overrightarrow{\nabla}} \cdot \mathbf{\overrightarrow{V}} = 0$ and $\mathbf{\overrightarrow{\nabla}} \times \mathbf{\overrightarrow{V}} = 0$, then $\mathbf{\overrightarrow{V}}$ can be added to an existing solution, say $\mathbf{\overrightarrow{B_0}}$ corresponding to a current density $\mathbf{\overrightarrow{J_0}}$, to give another vector which atleast satisfies the first two conditions.

And there are a plenty of such vectors, as we can take any constant vector with the dimensions of magnetic field. The problem arises for the boundary condition as we need to find a vector which vanishes at infinity, I think I was able to find such a vector -

$ \mathbf{\overrightarrow{V}} = k \cdot \dfrac{\mathbf{\hat{r}}}{r^2}$

Where $k$ is a scalar constant with appropriate dimensions.

My question is that if we add this $\mathbf{\overrightarrow{V}}$ to $\mathbf{\overrightarrow{B_0}}$ then the resulting field, $\mathbf{\overrightarrow{B_0}} + \mathbf{\overrightarrow{V}}$ must also be a magnetic field corresponding to the same current density $\mathbf{\overrightarrow{J_0}}$, but this statement contradicts the Biot-Savart's law which suggests that there must be a unique field for a given current distribution.

So, is there some extra condition for the validity of magnetic field? Or is there a fundamental flaw in my argument?

Answer 1

The domain is important. The solution is given by Biot-Savart if it is the entire $\mathbb R^3$ space, with some assumptions on the regularity and decay of $B$ (e.g. $L^2$). In this case, you can prove it mathematically, no discussion there.

Indeed, the key result is that under the previous assumptions ($L^2$ vector field on $\mathbb R^3$), if $V$ is curl and divergence free: $$ \nabla\times V = 0 \quad \nabla\cdot V = 0 $$ then $V=0$.

You example does not work since it is curl free but not divergence free: $$ \nabla\cdot V = 4\pi k\delta $$ Indeed, it is the field of monopole at the origin, so you have a punctual source there. Alternatively, it is a valid candidate if your domain is $\mathbb R^3-\{0\}$, but then the previous assumptions fail.

Actually, you could have though of another example, like: $$ V = xe_x-ye_y $$ which is curl and divergence free on the entire domain. However, this time, the issue is the divergence at infinity. Physically, the energy is infinite, mathematically it is not $L^2$.

Intuitively, the reason why it works is more obvious when you go in Fourier space: $$ ik\times \hat V=0 \quad ik\cdot \hat V = 0 $$ Curl free means that the field is longitudinal, while divergence free means that it is transversal. This means that away from the origin, $V\neq 0$. At the origin though, you could potentially have a singular part involving derivatives of Dirac deltas. This is why you can have harmonic solutions like the one I mentioned. This is where the sufficiently fast decay at infinity of $V$ (like $L^2$) gets involved to check that even those terms are zero.