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Gauss's law for magnectic field states $\nabla\cdot\mathbf{B}=0$, and thus calculating the divergence of a field and finding it is nonzero que can promptly prove the vector field cannot represent a magnetic field.

However, I don't find an analogous giveaway using Ampère-Maxwell's law. Which conditions does the curl of a vector field have to verify in order to represent a magnetic field?

Is it possible for a magnetic field to have both null divergence and null curl?

Can any vector field with null divergence represent a magnetic field, if no additional information about currents or electric fields is mentioned?

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    $\begingroup$ if it is a static field in vacuum then you must also have $\mathbf{curl}\mathbf{B} =\mathbf{0}$ in places where there is no macroscopic current, i.e., $\mathbf{J}=\mathbf{0}$ $\endgroup$ – hyportnex Apr 9 '17 at 16:04
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Any static field with zero divergence, i.e. which obeys the magnetic Gauss law $\nabla\cdot\mathbf B=0$, is a valid magnetic field. The curl of the field can be anything: if it is nonzero then it requires a current density $\mathbf J$ to sustain it, given by Ampère's law $$ \nabla\times\mathbf B = \mu_0\mathbf J. $$ This curl can in principle be any suitable vector field, though because it is a curl it is required to have zero divergence, i.e. $\nabla\cdot(\nabla \times \mathbf B )= \mu_0\nabla \cdot\mathbf J=0$, which needs to happen anyway in a static situation, because of conservation of charge.

If there are no currents, i.e. in vacuum, then yes, the magnetic field will have zero curl. Most of the usual examples of magnetic fields fall into this category, and it is plenty possible for a magnetic field to have zero divergence and zero curl (want a simple example? try a constant field).

It is important to note, though, that typically you want your magnetic field to have sources somewhere, and this means that the in-vacuum condition $\nabla\times\mathbf B=0$ will only hold for some restricted region in space. In this spirit, it is possible to have a vector field with no divergence and no curl defined over all of space, but this requires its magnetic energy $U=\frac{\mu_0}{2}\int|\mathbf B|^2\mathrm d\mathbf r$ to be infinite, which is as unphysical as a nonzero static magnetic field with no sources.


If you want a time-dependent field, though, the problem changes a good bit. As far as the magnetic Maxwell equations are concerned, $$ \nabla\cdot\mathbf B=0 \text{ and } \nabla\times\mathbf B = \mu_0\varepsilon_0 \frac{\partial \mathbf E}{\partial t}+\mu_0\mathbf J, $$ any vector field with no divergence can in principle be interpreted as a magnetic field. However, this requires us to find a set of sources, $\mathbf J$ and $\rho$, and more importantly an electric field $\mathbf E(\mathbf r,t)$, to go with it, or the whole thing is a bit moot. This means, therefore, that we need to turn Maxwell's equations on their head a little bit, and now they become \begin{align} \mu_0\varepsilon_0 \frac{\partial \mathbf E}{\partial t}+\mu_0\mathbf J & = \nabla\times\mathbf B \\ \nabla\times\mathbf E & = - \frac{\partial \mathbf B}{\partial t} \end{align} as a problem to be solved for $\mathbf E$ and $\mathbf J$, which is generally always solvable.

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  • $\begingroup$ Thank you for your thorough answer, it is exactly what I was looking for. $\endgroup$ – Mario Apr 10 '17 at 16:20
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Which conditions does the curl of a vector field have to verify in order to represent a magnetic field?

Probably none, except the Maxwell equation itself.

The equation

$$ \nabla\cdot \mathbf B = 0 $$

restricts the set of possible magnetic fields, because the right-hand side is constant in time and there is no other variable in the equation than $\mathbf B$. This kind of equation is sometimes called a constraint equation.

The equation

$$ \nabla\times\mathbf B = \mu_0\mathbf j + \epsilon_0\mu_0 \frac{\partial \mathbf E}{\partial t}, $$ on the other hand, has no constant part; both $\mathbf j$ and $\mathbf E$ are unknown functions of time and position. So this equation by itself does not define any constraint delimiting set of possible $\mathbf B$.

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As hyportnex states, the condition for a static magnetic field in vacuum (thank you Emilio for the correction) is that $\nabla \times {\bf B} = 0$.

It is possible for a magnetic field to have both null divergence and null curl. A basis of solutions has the form ${\bf B}_{l,m} \propto - \nabla \left( r^l Y_{l,m}(\theta,\phi) \right)$ where $Y_{l,m}$ are real spherical harmonics. I think Jackson calls these "interior solutions" to Laplace's equation. The corresponding "exterior solutions" are of the form ${\bf B}_{l,m} \propto -\nabla ( r^{-l-1} Y_{l,m} )$.

I think the simplest way to say whether a given field is plausibly a magnetic field is that it has zero divergence and zero curl, however this applies to static situations only.

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  • $\begingroup$ Your first sentence is incorrect; that condition is required for a static magnetic field in vacuum. If there is a current, the curl can be given by any arbitrary (divergenceless) vector field. $\endgroup$ – Emilio Pisanty Apr 9 '17 at 18:31

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