# Why aren't Faraday's law of induction and Maxwell-Ampere's law (without sources) symmetric?

I was wondering why Faraday's law of induction and Maxwell-Ampere's law (without sources) are not totally symmetric in the sense that Maxwell-Ampere's law has a $\epsilon_0 \mu_0$ term on the right (in SI units) while Faraday's law doesn't, as symmetry is an important feature in most physical laws.

\begin{align} \nabla\times\mathbf E&=-\frac{\partial\mathbf B}{\partial t} \\ \nabla\times\mathbf B&=\color{blue}{\mu_0\varepsilon_0}\frac{\partial\mathbf E}{\partial t} \end{align}

A popular reference book states the reason being "that we use SI units". Can anyone tell me how using a particular unit can affect the symmetry of physical laws written in their mathematical form?

• What do you mean by Maxwell's law? I thought there was Maxwell's equations and each one was associated with some past person (e.g., Ampere's law). Are you referring to one specific Maxwell equation? – honeste_vivere Oct 29 '14 at 15:43
• Specifically, Maxwell's law of induction. – Gaurav Oct 29 '14 at 15:47
• I've never heard the term "Maxwell's law of induction." Are you referring to the $\partial E/\partial t$ term in Maxwell's equations? – Ben Crowell Oct 29 '14 at 15:54
• Exactly. (if 'E' means electric flux). Please edit the question if you think it is not sufficiently stated. – Gaurav Oct 29 '14 at 15:57
• These parameters can be absorbed into the equations by a mere choice of dimensions or a re-definition of $\vec{E}$ and $\vec{B}$ (thus making the equations symmetric). However there is another asymmetry which is the most important, lack of any magnetic monopoles (in contrast to electric monopoles , e.g electrons) – Nikos M. Oct 29 '14 at 21:14

Maxwell's equations in vacuum are symmetric bar the problem with units that you have identified. In SI units $$\nabla \cdot {\bf E} = 0\ \ \ \ \ \ \nabla \cdot {\bf B} =0$$ $$\nabla \times {\bf E} = -\frac{\partial {\bf B}}{\partial t}\ \ \ \ \ \ \nabla \times {\bf B} = \mu_0 \epsilon_0 \frac{\partial {\bf E}}{\partial t}$$

If we let $\mu_0=1$, $\epsilon_0 =1$ (effectively saying we are adopting a system of units where $c=1$, then these equations become completely symmetric to the exchange of ${\bf E}$ and ${\bf B}$ except for the minus sign in Faraday's law. They are symmetric to a rotation (see below).

If the source terms are introduced then this breaks the symmetry, but only because we apparently inhabit a universe where magnetic monopoles do not exist. If they did, then Maxwell's equations can be written symmetrically. We suppose a magnetic charge density $\rho_m$ and a magnetic current density ${\bf J_{m}}$, then we write $$\nabla \cdot {\bf E} = \rho\ \ \ \ \ \ \nabla \cdot {\bf B} = \rho_m$$ $$\nabla \times {\bf E} = -\frac{\partial {\bf B}}{\partial t} - {\bf J_m}\ \ \ \ \ \ \nabla \times {\bf B} = \frac{\partial {\bf E}}{\partial t} + {\bf J}$$

With these definitions, Maxwell's equations acquire symmetry to duality transformations. If you put $\rho$ and $\rho_m$; ${\bf J}$ and ${\bf J_m}$; ${\bf E}$ and ${\bf H}$; ${\bf D}$ and ${\bf B}$ into column matrices and operate on them all with a rotation matrix of the form $$\left( \begin{array}{cc} \cos \phi & -\sin \phi \\ \sin \phi & \cos \phi \end{array} \right),$$ where $\phi$ is some rotation angle, then the resulting transformed sources and fields also obey the same Maxwell's equations. For instance if $\phi=\pi/2$ then the E- and B-fields swap identities; electrons would have a magnetic charge, not an electric charge and so on.

Whilst one can argue then about what we define as electric and magnetic charges, it is an empirical fact at present that whatever the ratio of electric to magnetic charge (because any ratio can be made to satisfy the symmetric Maxwell's equations) all particles appear to have the same ratio, so we choose to fix it that one of the charge types is always zero - i.e. no magnetic monopoles.

I mention all this really as a curiosity. It seems to me that the real symmetries of Maxwell's equations only emerge when one considers the electromagnetic potentials.

e.g. if we insert $B = \nabla \times {\bf A}$ and $E= -{\bf \nabla V} - \partial {\bf A}/\partial t$ into our Ampere's law $$\nabla \times (\nabla \times {\bf A}) = \frac{\partial}{\partial t} \left({\bf -\nabla V} - \frac{\partial {\bf A}}{\partial t}\right) +{\bf J},$$ $$-\nabla^2 {\bf A} +\nabla(\nabla \cdot {\bf A}) = -\nabla \frac{\partial V}{\partial t} - \frac{\partial^2 {\bf A}}{\partial t^2} + {\bf J}.$$ Then using the Lorenz gauge $$\nabla \cdot {\bf A} + \frac{\partial V}{\partial t} = 0$$ we can get $$\nabla^2 {\bf A} - \frac{\partial^2 {\bf A}}{\partial t^2} + {\bf J} = 0$$ A so-called inhomogeneous wave equation. A similar set of operations on Gauss's law yields $$\nabla^2 V - \frac{\partial^2 V}{\partial t^2} + \rho= 0$$

These remarkably symmetric equations betray the close connection between relativity and electromagnetism and that electric and magnetic fields are actually part of the electromagnetic field. Whether one observes $\rho$ or ${\bf J}$; ${\bf E}$ or ${\bf B}$, is entirely dependent on frame of reference.

• Thanks for the time and the comprehensive answer, Mr.Rob! Since you say that the "remarkably symmetric equations betray the close connection between relativity", is it safe to conclude the non-existence of magnetic monopoles ? If yes, then why is there still an uncertainty in the scientific community as to their non-existence ? – Gaurav Oct 30 '14 at 4:34
• Thanks, but I think others on the site could possibly answer even more comprehensively - I am a student of these issues. I believe if you add monopoles, the inhomogeneous wave equations retain their symmetry. – Rob Jeffries Oct 30 '14 at 8:24
• Sir, I would like to let you know that my physics course has not yet reached the stage where I could comprehend the mathematical terms you have used from the 'Lorenz gauge' onwards. However, I am familiar with del operators, differential equations and wave equations. I will revisit this question once I reach that stage. For now, what interests me is the last paragraph of your answer. Since you say the existence of magnetic monopoles contradicts the existing close link between relativity and electromagnetism, does it prove their non-existence ? – Gaurav Oct 30 '14 at 10:04
• I don't think it does, at least not at the level presented here. And I didn't say it did. I just argued you don't need monopoles to write the laws of electromagnetism in a symmetric way, even with charges and current sources. – Rob Jeffries Oct 30 '14 at 12:12
• Doubt clarified ! – Gaurav Oct 30 '14 at 12:38

In Gaussian units, we set $\epsilon_0 = \frac1{4\pi}$ (and so $\mu_0 = \frac{4\pi}{c^2}$) and change the units of $B$ so both electric and magnetic fields have the same dimension. In these units, Maxwell's equations are as follows:

\begin{align} \nabla \cdot \mathbf{E} &= 4\pi \rho \\ \nabla \times \mathbf{E} &= - \frac1{c} \frac{\partial \mathbf{B}}{\partial t} \\ \nabla \cdot \mathbf{B} &= 0 \\ \nabla \times \mathbf{B} &= \frac{4\pi}{c} \mathbf{J} + \frac1{c} \frac{\partial \mathbf{E}}{\partial t} \end{align}

The symmetry you're looking for is there, I guess. The important bit, as far as I can tell, is making things so $E$ and $B$ have the same units (and using the fact that $\epsilon_0 \mu_0 = \frac1{c^2}$). You won't be able to get rid of the minus sign, but then again without that minus sign you wouldn't get waves, so it's pretty important.

Actually, the laws of electromagnetism are symmetrical. Any event the laws of electromagnetism allow, the laws allow it's mirror image. Let's consider the situation of a magnet moving through a coil. The electrons will move a certain way. Let's see what happens if you do the mirror image of that experiment, since it's the spinning electrons that give make it a magnetic field, in the mirror image, the electrons will be moving the opposite way so the north end will be replaces with a south end so the magnetic field will point in the opposite direction so the magnetic field will go in the opposite way. If you invert an object with an permanent electric dipole, the positive end does not turn into a negative end but an object will an electric dipole will not induce a current in a coil either. Therefore, the laws of electromagnetism are symmetrical.