What does the minus sign in Maxwell's third equation imply?

Question

If we write out Maxwell's equations with magnetic charges, we get

$$ \begin{align} \nabla \cdot \mathbf{E} &= 4 \pi \rho_e \tag{1}\\ \nabla \cdot \mathbf{B} &= 4 \pi \rho_m \tag{2}\\ -\nabla \times \mathbf{E} &= \frac{\partial \mathbf{B}}{\partial t} + 4 \pi \mathbf{J}_m \tag{3}\label{Eq:Faraday}\\ \nabla \times \mathbf{B} &= \frac{\partial \mathbf{E}}{\partial t} + 4 \pi \mathbf{J}_e \tag{4}\label{Eq:Ampere} \end{align} $$

In particular, Faraday's law \eqref{Eq:Faraday} contains a minus sign that Ampere's law \eqref{Eq:Ampere} does not. This always struck me as odd because it's often said the fields are dual to each other (i.e. you can replace E with B and "get the same result"), but that requires a bit of mental recalibration to accommodate that minus. So I'm curious what the origin of that negative is and what it means. Are there any intuitive explanations for how to think about it?

Answer 1

In plain English, it is just Lenz’s law :

Lenz's law, named after the physicist Emil Lenz who formulated it in 1834, states that the direction of the electric current which is induced in a conductor by a changing magnetic field is such that the magnetic field created by the induced current opposes changes in the initial magnetic field.

It is the basic principle behind all electric motors and dynamos, alternators, etc.

Wikipedia: Lenz's Law

Answer 2

The minus sign is what makes Maxwell's equations obey causality, so it's a good thing it's there! To see this, you can write out the source-free Maxwell's equations with the sign of $\nabla \times \mathbf{E}$ reversed in Ampère's Law. If you then to follow the standard construction to extract the wave equation from Maxwell's equations, you would obtain that $$ \frac{1}{c^2} \frac{\partial^2 \mathbf{E}}{\partial t^2} + \nabla^2 \mathbf{E} = 0, $$ and similarly for $\mathbf{B}$. In other words, we wouldn't get the wave equation (which would have a negative sign on the first term); we would get a 4-D version of Laplace's equation instead.

The solutions to Laplace's equation don't have the nice causality properties that solutions to the wave equation do. In mathematical terms, the PDEs are elliptic rather than hyperbolic, and elliptic PDEs behave rather differently than hyperbolic PDEs in some important ways. For example, suppose we specify the value of $\mathbf{E}$ at some moment $t = 0$ in some finite region of space. It can be shown that changes to this initial data cause changes in the solution for $t > 0$ arbitrarily far away from our initial "finite region". In other words, there is no upper limit for the speed of signal propagation in such a system.

Answer 3

Duality is actually not $\mathbf{E}\leftrightarrow \mathbf{B}$ (I've used $c=1$), i.e. $(\mathbf{E},\,\mathbf{B})\to(\mathbf{B},\,\mathbf{E})$. It's $(\mathbf{E},\,\mathbf{B})\to(-\mathbf{B},\,\mathbf{E})$. Defining $\mathbf{F}:=\mathbf{E}+i\mathbf{B}$ is a popular way to check this; the above duality is $\mathbf{F}\to i\mathbf{F}$. It's instructive to compute $\nabla\cdot\mathbf{F}$ and $\nabla\times\mathbf{F}-i\frac{\partial\mathbf{F}}{\partial t}$.

Answer 4

It really comes from relativity, where one uses the field strength tensor:

$$ F_{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}=\left(\begin{array}{cccc} 0 & E_x & E_y & E_z\\ -E_x & 0 & -B_z & B_x\\ -E_y & B_z & 0 & -B_y\\ -E_z & -B_y & B_x & 0\\ \end{array} \right)$$

When the indices are raised:

$$F^{\mu\nu}\equiv=\eta^{\mu\alpha}F_{\alpha\beta}\left(\begin{array}{cccc} 0 & -E_x & -E_y & -E_z\\ E_x & 0 & -B_z & -B_x\\ E_y & B_z & 0 & -B_y\\ E_z & -B_y & B_x & 0\\ \end{array} \right) $$

the time/space and space/time (E-field) terms flip sign, which the space/space (B-field) terms pick up a factor of $(-1)^2=1$ thanks to $ \eta={\rm diag}(+,-,-,-)$.

The four-current is

$$ j^{\mu}=(\rho, {\bf J})$$

Gauss and Ampère's laws are combined in:

$$\partial_{\mu}F^{\mu\nu}=\mu_0j^{\nu}$$

which is where the ${\dot{\bf E}}$ and $\nabla \times {\bf B}$ have opposite signs on the LHS.

Meanwhile, the Gauss-Farady law is:

$$\partial_{\mu}G^{\mu\nu}=0$$

(you can add a magnetic current on the RHS, as needed). Here $G$ is the Hodge dual tensor:

$$G^{\mu\nu}=(\frac 1 2 \epsilon^{\mu\nu\sigma\lambda}F_{\sigma\lambda}) =\left(\begin{array}{cccc} 0 & -B_x & -B_y & -B_z\\ B_x & 0 & E_z & -E_x\\ B_y & -E_z & 0 & E_y\\ B_z & E_y & -E_x & 0\\ \end{array} \right) $$

so that ${\dot{\bf B}}$ and $\nabla \times {\bf E}$ have the same sign on LHS.

So the blame falls on the metric $\eta_{\mu\nu}$ and on $\epsilon_{\alpha\beta\gamma\delta}$.

Answer 5

What should really bother you, is not the minus sign in $(3)$. Is it's absence of in $(4)$!.

The minus sign in $(3)$ actually prevents a run-away effect where an induced electric current would create a positive feedback on itself resulting in an unstable, ever-growing, electric current that: 1) would probably destroy your planet, and 2) violate energy conservation. The minus sign in $(3)$ is a consequence of Lenz’s law, which is a good thing, and since in our universe magnetic monopoles don't seem to exist, the absence of the minus sign in $(4)$ causes no harm.

In a universe with magnetic monopoles, it should be expected there to be a sort of Lenz’s law for magnetic current, to prevent all those bad things to happen or there would be no curious minds there, to check if your set of equations exists. In other words, in a universe with magnetic monopoles, Maxwell’s equations should probably looks like this:

$$\nabla \cdot \mathbf{E} = 4 \pi \rho_e \tag{1'}$$ $$\nabla \cdot \mathbf{B} = 4 \pi \rho_m \tag{2'}$$ $$-\nabla \times \mathbf{E} = \frac{\partial \mathbf{B}}{\partial t} + 4 \pi \mathbf{J}_m \tag{3'}$$ $$-\nabla \times \mathbf{B} = \frac{\partial \mathbf{E}}{\partial t} + 4 \pi \mathbf{J}_e \tag{4'}$$

Which is the perfect symmetry you seem to be looking for.

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EDIT:

Please notice that the introduction of magnetic monopoles does not just change the form of the equations, it profoundly changes the magnetic field itself. Look at (2), we now have a divergent magnetic field $\mathbf{B}$ which is a generalization of “our” magnetic field $\bar{\mathbf{B}}$. You now are in the position to make the discovery that Lenz’s law of magnetic current implies $$\lim_{\rho_m \to 0} (-\nabla \times \mathbf{B})=\nabla \times \bar{\mathbf{B}} \tag{5'}$$ and $(5')$ is the reason why in our universe, there is a minus sign in $(3)$ but not in $(4)$.

Answer 6

There's an approach from geometric algebra that considers the electromagnetic field as a field of bivectors, and with this geometric interpretation the electric and magnetic components can be rotated into each other. It's too long a discussion to attempt to explain this properly here; read the linked paper if you are interested in the details. I'll just give a quick hand-wavey sketch.

Where vectors represent oriented lines and lengths, bivectors represent oriented planes and areas. In 4d spacetime, a bivector has a 6D basis consisting of the pairs $xt$, $yt$, $zt$, $yz$, $zx$, and $xy$. If we select a particular observer reference frame, represented by the $t$ axis tangent to the observer's worldline, we can split the bivector field into two vector fields by multiplying by $t$. $(axt+byt+czt)t = (ax+by+cz)$, and this is the electric field. $(dyz+ezx+fxy)t = (dyzt+ezxt+fxyt) = xyzt(dx+ey+fz)$ and this is the magnetic field times a constant $I=xyzt$, which is a special quantity with the property that $I^2=-1$ and such that multiplying any geometric entity by it gives the 'orthogonal complement' subspace (or the Hodge-Star operation used in differential geometry).

This allows you to think of the electromagnetic field (in a particular observer's reference frame) as a complex vector $E+IB$ called the Riemann-Silberstein vector. The duality between them is that brought about by multiplying by $I$, but it's not an exact duality because of that -1. Taking $E+IB$ and multiplying by $-I$ gives $-I(E+IB)=B-IE$. That is to say, it's not a reflection, it's actually a $90^{\circ}$ rotation in the complex plane. It rotates the E axis and B axis into each other, which is why people identified an apparent duality between them, but with a sign-flip on one axis.

In fact, there is a continuous symmetry here. We can multiply the field by any unitary exponential $(E+IB)e^{I\theta}$ and get a mixed quantity that still obeys the extended Maxwell equations with magnetic charge. (And in a vacuum, where there are no charges, it satisfies the unmodified Maxwell equations.) We're just multiplying by a constant, and since the equations are linear, they are still satisfied.

Answer 7

In vacuum (all $\mathbf{J}$ and all $\rho$ zero) the Maxwell equations imply the wave equations $$ \frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2}=\Delta\mathbf{E}\,,\quad\quad\frac{1}{c^2}\frac{\partial^2\mathbf{B}}{\partial t^2}=\Delta\mathbf{B}\,. $$ See Vacuum equations, electromagnetic waves and speed of light on Wikipedia.

If you remove the minus sign from Faraday's law (2) you get instead $$ \frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2}=-\Delta\mathbf{E}\,,\quad\quad\frac{1}{c^2}\frac{\partial^2\mathbf{B}}{\partial t^2}=-\Delta\mathbf{B}\,. $$ But these are wave equations only in imaginary time.