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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?

jacob1729
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rayhem
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    https://physics.stackexchange.com/a/639811/195949 – Claudio Saspinski Jun 22 '21 at 13:56
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    "you can replace E with B and 'get the same result'" Well, not quite. You have to replace $\mathbf{E}$ with $\mathbf{B}$ and $\mathbf{B}$ with $-\mathbf{E}$. Actually, there's a more general rule than this that you might find in an advanced text book on electrodynamics. But replacing E,B with B,E doesn't work. – bob.sacamento Jun 24 '21 at 20:23

7 Answers7

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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

Urb
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Serge Hulne
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    yes. Lenz's law is conservation of energy. +1 – joseph h Jun 23 '21 at 00:50
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    @josephh That never occurred to me but true: If the effect of the induced current amplified the original magnetic field it would create a runaway effect (and the world would end! ;-) ): – Peter - Reinstate Monica Jun 23 '21 at 14:35
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    this is the best answer. it provides the physical foundation for all the math that everyone else is dumping out. I think people here forget that the math is a consequence of the physics not the other way around. – user7257 Jun 23 '21 at 15:24
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    @josephh I'm pretty sure that you would still have a notion of energy conservation in the case where the minus sign was a plus sign (time-independence of the equations of motion & Noether's theorem & all that.) It's just that the electromagnetic field energy would not be bounded below and so you would end up with runaway solutions rather than a stable situation where $\mathbf{E} = \mathbf{B} = 0$. – Michael Seifert Jun 23 '21 at 17:09
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    This is answering what it is (the minus sign), but the question is why it doesn't appear on $(4)$. Therefore it doesn't answer the question posed by the OP. – J. Manuel Jun 23 '21 at 20:42
  • @MichaelSeifert Are you saying that if flipped signs from the two curl maxwell equations like $$\operatorname{curl}\textbf{E} = \partial\textbf{B}/\partial t$$ and $$\operatorname{curl}\textbf{B} = -\partial\textbf{E}/\partial t$$ we'd have physical solutions or are you saying that both equations with a positive sign will be physical?! We need opposite signs on these equations or as Peter stated, we would indeed have exponential solutions and not oscillating waves as solutions to the maxwell equations. – joseph h Jun 23 '21 at 22:52
  • @josephh: The point I was making is that in this hypothetical Universe, those exponentially growing solutions would still conserve energy, because the energy density of the field would be something like $\mathbf{E}^2 - \mathbf{B}^2$. So it's not quite correct to say that the sign is there to ensure energy conservation; it's more that it ensures stability. – Michael Seifert Jun 23 '21 at 23:35
  • @MichaelSeifert What exactly do you mean by hypothetical universe? You mean a universe where the Maxwell equations do not have opposite signs on the curl E and curl B equations? So you agree that they are not physical since in this universe energy is conserved? That is $E^2 + B^2$? I'm not sure what you are getting at in this last comment. What are you saying is the implications if the energy density of the EM field was $E^2 - B^2$? – joseph h Jun 24 '21 at 00:11
  • @josephh: If you wrote down Maxwell's equations with the sign on Faraday's Law reversed, you could show (via the standard techniques) that there exists a conserved quantity associated with time translations. This quantity could reasonably be called "the energy" in this model, per Noether's theorem. I am fairly confident that this quantity would be $\int (E^2 - B^2) d^3x$. However, this energy is not bounded below, which means that the fields could get arbitrarily large while still conserving the total energy of the system. These would be the runaway solutions mentioned by Peter. – Michael Seifert Jun 24 '21 at 00:35
  • This answers a different question, namely why are the signs in equations (3) and (4) opposite (or equivalently, why is their product negative). If we accept that, the negative sign could still be chosen to occur either in (3) or in (4), and the only difference would be the relative orientation that by convention the electric and magnetic field vectors would have (exactly one of them would have to change direction when we displace the minus sign). – Marc van Leeuwen Jun 25 '21 at 16:40
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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.

Michael Seifert
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  • Causality is not the same as finiteness speed of signal propagation. – Marc van Leeuwen Jun 25 '21 at 16:43
  • @MarcvanLeeuwen: True, but they're pretty closely connected. Really what I'm talking about here is a well-posed initial-value formulation of the equations, as described here. But I wasn't sure that was the level that the OP was at, and so I didn't want to introduce too much jargon. – Michael Seifert Jun 25 '21 at 16:48
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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 $\nabla\times\mathbf{F}-i\frac{\partial\mathbf{F}}{\partial t}$.

J.G.
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8

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 $\epsilon_{\alpha\beta\gamma\delta}$.

JEB
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  • Note that in general, $\star(\star \mathbf{F}) = \pm \mathbf{F}$, where the plus corresponds to Riemannian metrics and the minus corresponds to Lorentzian metrics. Presumably Riemannian electrodynamics wouldn't have this minus sign in the duality, which ties in to my answer above about the hyperbolic vs. elliptic character of the ODEs. – Michael Seifert Jun 23 '21 at 01:21
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    While the physics in this answer is true, it's quite an anachroism to say it "comes" from GR, since Maxwell's law were published in 1865, almost exactly 50 years prior to GR. In fact, as these equations were already known to be true, GR had to be construed in a way that it gives them back in flat spacetime - so it's no surprise it does, and I don't find it a really strong reason behind that minus sign. – Neinstein Jun 23 '21 at 06:34
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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.

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 $(4)$.

J. Manuel
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  • Aha, this is exactly the kind of analysis I was hoping for! Thank you! I'm still a little puzzled why we don't have the minus sign in (4), though. Whether our universe does not have magnetic monopoles or it does and all of their charges just happen to be zero is something better left to philosophers, but the interpretation shouldn't matter so these "symmetric" equations should be equally valid here, too, right? If that's the case, why don't we write them as such? – rayhem Jun 24 '21 at 14:31
  • @ConnorGlosser. Look at the EDIT I did to the answer. See if it helps. Hugs! – J. Manuel Jun 24 '21 at 21:31
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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.

Glorfindel
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Kurt G.
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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.

Nullius in Verba
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