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

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?

• the divergence of this $\mathbf V$ is not zero Commented Jul 16 at 13:52
• @hyportnex exactly, it's a magnetic monopole at the origin, and throwing a monopole in the middle of a magnetostatics problem will change the field.
– JEB
Commented Jul 17 at 1:52
• When I attempted to find a vector satisfying the conditions in the above problem, I didn't think about the validity of the vector or thinking whether it is possible to create such a field or not, because the validity of the field itself is given by the Maxwell's equations and boundary conditions. But yes, you are correct as the divergence of my V is indeed non-zero at the origin. Commented Jul 17 at 2:28

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.

• But why does the magnetic field have to be 𝐿² ? And why is such a condition not there for electric fields? Commented Jul 17 at 2:17
• @BipulKumar The squared integral of $B$ over some volume represents the energy of the field inside that volume. By restricting ourselves to $L^2$, we impose that the integral over all space, and thus the total energy, remains finite. There should be such a condition as well for electric fields? Commented Jul 17 at 13:10
• Yes, but why does the total energy have to be finite in the case of space containing magnetic field while it is not so in the case of electric fields, where the field due to a single point charge itself is not $L^2$ ? Commented Jul 17 at 13:36
• The $L^2$ condition is just to guarantee global regularity (tempered) but you can weaken it to be tempered and $L^2$ at infinity to guarantee the decay. In this case the Coulomb field is a valid solution. In general, you typically assume that your charge distribution is sufficiently regular as well.
– LPZ
Commented Jul 17 at 19:36
• Field being $L^2$ and Poynting energy being finite seems to be a red herring. There are examples of infinite sources with infinite Poynting energy, and unique field, like electric field of a charged line, or magnetic field of a straight wire. The sufficient condition for uniqueness seems to be that the source is not infinite in all directions, and in infinity, the field decays to zero. Commented Jul 17 at 23:00