On Physical Lines of Force/Modern derivations/Gauss's law for electricity

On Physical Lines of Force → Modern derivations → Gauss's law

What Maxwell had

In Part III, Prop. XIV — the proposition that introduces the displacement current — Maxwell takes the three corrected equations of currents, equation (112), differentiates them with respect to $x$, $y$ and $z$ respectively, and adds. What comes out is

$$e+\frac{df}{dx}+\frac{dg}{dy}+\frac{dh}{dz}=0\mathrm{(}\mathrm{113}\mathrm{)}$$

Provenance: the equation number is attested; the body is a reconstruction in Maxwell's notation. See Sources.

Here $e$ is the density of free electricity — what we call the charge density $\rho$ — and $f,g,h$ are the three components of the electric displacement. The three-term sum is a divergence, as the notation page describes, so (113) says that the divergence of the displacement is the negative of the charge density.

That is Gauss's law for electricity: the last of the four to appear in this paper, and the fourth and final one to be assembled.

Displacement in the cells of the medium

Why the sign looks wrong

A modern reader meets (113) and suspects an error, because the textbook statement has a plus where Maxwell has a minus. It is not an error. It is a choice about which way the displacement points.

Maxwell's displacement is defined by equation (105) — reconstructed there, and on this wiki, as $R=-4\pi E^{2}h$ — in which the minus sign records that the medium pushes back against being displaced, as a spring does. With that definition, a displacement lying along the electromotive force counts as negative. The modern $𝐃$ is defined the other way round, pointing with the field.

Track the two conventions through and they agree exactly. Taking (105) and (113) as reconstructed here, Maxwell's triple $(f,g,h)$ corresponds to $-𝐃/4\pi$ in Gaussian units. Substituting into (113):

$$\rho +\mathrm{\nabla }\cdot (-\frac{𝐃}{4\pi })=0⟹\mathrm{\nabla }\cdot 𝐃=4\pi \rho$$

which is the standard Gaussian form. The two signs are internally consistent with one another, and neither is a claim about physics. Every prediction is unchanged. The concordance on the notation page lists the correspondence as $h\leftrightarrow 𝐃/4\pi$, matching magnitudes; the sign is the refinement this page adds.

What it says today

$$\mathrm{\nabla }\cdot 𝐃=\rho \text{(SI)}$$

$$\mathrm{\nabla }\cdot 𝐃=4\pi \rho \text{(Gaussian)}$$

and in vacuum, where $𝐃={\epsilon }_0𝐄$,

$$\mathrm{\nabla }\cdot 𝐄=\frac{\rho }{{\epsilon }_0}\text{(SI)}$$

In words: electric charge is a source of the electric displacement. Field lines begin on positive charges and end on negative ones, and the amount of field emerging from a region is set by how much charge is inside it. This is precisely what the magnetic equation denies for magnetism, and the contrast between the two is the whole difference between electricity and magnetism at this level.

The integral form

Integrate over a volume $V$ with boundary surface $S$, and use the divergence theorem to turn the left side into a flux:

$${\oint }_S𝐃\cdot d𝐒=Q_{\text{enc}}\text{(SI)}$$

The flux of the displacement through any closed surface equals the charge enclosed. Nothing about the shape of the surface enters, nor where the charge sits inside it; and charge outside contributes nothing at all, since whatever flux it sends in must come out again.

This is the form that does real work, because for a symmetric problem you can choose a surface on which the field is constant, and then the integral becomes a multiplication.

Coulomb's law, derived

Here is the classic example, worth doing carefully because it shows where the inverse square comes from.

Put a point charge $q$ at the origin and ask for the field.

1. Use the symmetry. Rotate the whole arrangement about any axis through the charge and nothing changes — there is no other object to spoil the symmetry, no preferred direction anywhere — so the field cannot change either. The only vector field with that property points radially and has a magnitude $E(r)$ depending on the distance alone. Everything else has been excluded before any calculation begins.
2. Choose the surface. Take $S$ to be a sphere of radius $r$ centred on the charge. On it $𝐄$ is everywhere perpendicular to the surface and everywhere of the same magnitude — which is exactly why the sphere is the right choice.
3. Evaluate the flux. Since $E$ is constant over the sphere and normal to it, the integral is just the magnitude times the area: ${\oint }_S𝐄\cdot d𝐒=E(r)\times 4\pi r^2$
4. Set it equal to the enclosed charge. In vacuum, $E(r)4\pi r^2=\frac{q}{{\epsilon }_0}⟹E(r)=\frac{q}{4\pi {\epsilon }_0{r}^2}$

That is Coulomb's law, and the derivation makes plain what the inverse square is doing there. It is the surface area of a sphere. The total flux out of any surface surrounding the charge is fixed — it depends only on $q$. As the sphere grows, that fixed total is spread over an area growing as $r^2$, so the amount per unit area must fall as $1/r^2$. The exponent is not a fact about electricity. It is a fact about three-dimensional space. In a world with four spatial dimensions the same argument would give an inverse cube.

Where the 4π went

The $4\pi$ in the denominator of Coulomb's law is the same $4\pi$ — the area of the unit sphere — and this is the clearest illustration of what the two unit systems are doing.

• SI puts no $4\pi$ in the field equation, $\mathrm{\nabla }\cdot 𝐃=\rho$, and so the $4\pi$ surfaces in Coulomb's law, $E=q/4\pi {\epsilon }_0{r}^2$.
• Gaussian, Maxwell's system, puts the $4\pi$ in the field equation, $\mathrm{\nabla }\cdot 𝐃=4\pi \rho$. Run the same derivation and the two factors cancel: $E4\pi r^2=4\pi q$, giving simply $E=q/r^2$.

Neither system removes the $4\pi$; they choose where it appears. Gaussian makes the force law clean and the field equations cluttered, SI the reverse. That is the whole of the dispute, and there is nothing in it to resolve. See A note on units.

The historical oddity

It is worth stopping to notice the shape of this paper's argument, because it is genuinely strange.

Coulomb's inverse-square law was the oldest and best-established result in the entire subject — measured with a torsion balance three-quarters of a century before Maxwell wrote, confirmed repeatedly, and the foundation on which Poisson and Green had built the whole mathematical theory of electrostatics. If you had asked any physicist in 1861 to name the one thing about electricity that was certain, this is what they would have said.

And Maxwell reaches it last. Not as a starting point, not as an assumption, but as a by-product of a correction to a different equation, arriving at number (113) in a paper whose equation numbering runs past a hundred and twenty. He does not even announce it. It falls out of differentiating the displacement-current equations and adding them up.

This inversion is the signature of the whole enterprise. A theory taking action at a distance as fundamental would have had to start from Coulomb, there being nothing else to start from. Maxwell starts from a medium, gives it mechanical properties, and finds the inverse-square law is one of the things such a medium does. What was the axiom becomes a theorem.

He does it twice over, in fact. Prop. XV recovers the same force by a different route — computing the energy stored in the strained cells and asking how it changes when two charged bodies move. That second derivation carries the physical claim: the energy is in the space between the bodies, not a property of the pair. It is a different statement from Coulomb's, and it makes different predictions.

Where the conventions differ

	Maxwell 1861	Modern
the statement	eq. (113), $e+\frac{df}{dx}+\frac{dg}{dy}+\frac{dh}{dz}=0$	$\mathrm{\nabla }\cdot 𝐃=\rho$
charge	$e$, "free electricity"	$\rho$
displacement	$f,g,h$; corresponds to $-𝐃/4\pi$	$𝐃$, pointing with the field
sign	charge and divergence on the same side, summing to zero	opposite sides, hence a plus
factor $4\pi$	explicit, Gaussian-style	absent in SI; $4\pi \rho$ in Gaussian
Coulomb's law	$E\propto q/r^2$, no $4\pi$	$q/4\pi {\epsilon }_0{r}^2$ in SI; $q/r^2$ in Gaussian
logical status	derived, near the end, from (112)	usually the first of the four to be taught

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Next: The constitutive relations · Previous: No magnetic monopoles

In the paper: Prop. XIV · Prop. XV · See also: The displacement current