The zeros a handful of charges can make

The field before the lines were drawn on it listed what is lost when a vector field is drawn as lines instead of arrows, and one of the losses was the null points: the places where the field is zero, where a line drawing has nothing to show and an arrow drawing has an arrow of no length. Field lines are a choice called the null point between two like charges “a good test of whether the picture is being read as physics or as decoration.” The two essays after them followed the field outwards, into multipoles and shapes, and left the zeros where they were.

This essay goes back to them. They turn out to be the hardest thing about an electrostatic field to count. For two charges there is one. For three there can be one, two or four, depending on the arrangement. In two dimensions a theorem from the algebra of polynomials gives the count exactly, and it is always one fewer than the number of charges. In three dimensions nobody knows the largest number n charges can make. Maxwell raised the question in his Treatise in 1873, conjectured an answer, and a century and a half later it has not been proved even for three charges.

Four points of cancellation #

Between two equal positive charges the field points away from each, and at the midpoint the two contributions are equal and opposite. That midpoint is a zero, and it is the only one. Three equal charges at the corners of an equilateral triangle make more.

The figure traces the field lines in the plane of the three charges and marks every point where the field vanishes, found by searching from hundreds of starting points and merging what the searches converge on. There are four. One is at the centre, where the three contributions cancel by symmetry. The other three lie between the centre and the middle of each side, 0.164 of a side out from the centre: at each, the two charges at the ends of that side push inward across it, the third charge pushes outward, and the balance falls just inside the triangle.

The field lines show how the zeros organise the picture. Each charge sends out lines in every direction, and the lines from different charges never cross. The zeros are where the territories meet. Every line passing close to a zero is deflected sharply by it, and a few special lines run into it and out of it, the separatrices, which divide the plane into the regions each charge’s lines claim. That structure is a statement about the whole field, not about any point in it, and it is what the zeros encode.

Four is also a notable number. It is $(n-1)^2$ for $n = 3$, which is the largest number of zeros Maxwell conjectured three charges could ever make. The equilateral triangle, the most symmetric arrangement of three, reaches the bound exactly.

Why no zero can hold a charge #

Every one of these zeros is the same kind of point.

The figure draws the field around the zero between two equal charges, as arrows. Along the line joining the charges it points back towards the zero: displace a positive test charge that way and it is pushed back. Across the line it points away: displace it sideways and it is pushed further out. The field near the zero grows linearly with distance, and along the three natural directions its rates of growth are $-4$, $2$ and $2$, in units of the charge divided by the cube of half the separation.

Those three numbers add to zero, and they have to. Near a zero the field is linear, $\mathbf{E} \approx J\mathbf{r}$, with a matrix $J$ whose trace is the divergence of the field, which is zero where there is no charge, and which is symmetric because the field has no curl. A symmetric matrix with zero trace has real eigenvalues adding to zero, so at least one is positive and at least one negative, unless all three vanish. Every zero of an electrostatic field in empty space is therefore a saddle: some directions push a test charge in and others push it out.

That is the theorem that nothing can be held still by a static field, Earnshaw’s result, read off a single point. The earlier essay derived it from the mean-value property of the potential. Here it appears as a fact about the zeros themselves: a zero is exactly the place a charge would have to be held, and every zero is unstable. The same fact appears in the potential as the place random walkers stop, where a harmonic function has no interior maximum or minimum, only saddles.

The three rates also say something about the zero’s shape. The field lines approach along the axis, from the two charges, and leave in the perpendicular plane. In the triangle’s plane the four zeros are each such a crossing, and it is this local saddle structure, repeated, that the separatrices in the first figure are built from.

Two dimensions, where algebra decides #

In two dimensions the count can be derived, and the derivation is one of the prettiest connections between electrostatics and algebra.

The figure uses parallel line charges seen end on, which is the two-dimensional version of electrostatics: the field of a line charge falls as one over the distance, not one over its square. Place $n$ equal line charges at positions $z_1, \dots, z_n$ in the complex plane. Then the field at $z$, written as a complex number, is the complex conjugate of

$$\sum_{i=1}^{n} \frac{1}{z - z_i} = \frac{P'(z)}{P(z)}, \qquad P(z) = \prod_{i=1}^{n}(z - z_i).$$

The field vanishes exactly where $P'$ does. $P'$ is a polynomial of degree $n-1$, so it has exactly $n - 1$ roots, counted with multiplicity. Five line charges make four zeros, always. The figure computes the roots of $P'$ for five charges placed at random and checks that the field vanishes at each.

There is more. The Gauss–Lucas theorem, from the nineteenth century, says that the roots of a polynomial’s derivative lie inside the convex hull of the polynomial’s roots — the smallest convex region containing them. The figure draws the hull, and all four zeros lie inside it. In electrostatic terms the statement is obvious once said: outside the hull, every charge is on one side, so every contribution to the field has a component pointing away from the hull, and they cannot cancel. The algebraic theorem and the physical argument are the same fact.

The count in two dimensions has a topological backing too. The field’s direction, followed round a large circle enclosing all the charges, turns once, like the field of a single charge. Round a small circle enclosing one charge it also turns once. Each saddle-shaped zero, followed round, turns once the opposite way. The total turning must match, so $n$ charges and one far-field turn require $n - 1$ saddles. The index argument counts zeros with a sign and cannot tell the difference between a single degenerate zero and several simple ones: three equal charges on a triangle make one zero in two dimensions, at the centre, where $P'(z) = 3z^2$ has a double root.

A count that topology does fix #

Three dimensions do not leave the count completely free. The zeros of the field are the places where the potential is flat — its critical points — and one number for every point, the potential, is a smooth function whose shape is constrained by topology in a way the field alone does not make obvious.

For $n$ positive charges the potential rises to infinity at each charge and falls to zero far away. Think of the charges as $n$ peaks and of the distant region as a single basin. Between them the landscape has passes, and in three dimensions a pass comes in two kinds. At one kind, a zero with one inward direction and two outward, a test charge is pushed back along one line and away across a plane: the midpoint between two charges is of this kind. At the other, with two inward directions and one outward, it is pushed back within a plane and away along the line through it. Morse theory, which relates the critical points of a function to the shape of the space it lives on, requires that the passes of the first kind outnumber those of the second by exactly $n - 1$, whatever the arrangement.

The triangle obeys it. Its centre is a zero of the second kind: displaced within the plane, a charge is pushed back to the centre, and displaced perpendicular to the plane it is pushed away. The three other zeros are of the first kind, pushing back along the line towards the centre and away across it. Three minus one is two, which is $n - 1$ for three charges. The pair has one zero of the first kind and none of the second, and one is $n - 1$ for two. Every count in the next two figures obeys the same rule, and a count can grow only by adding one zero of each kind together.

That is exactly what the topology cannot forbid, and why it cannot bound the total. Pairs of opposite-kind zeros can be created or annihilated as charges move, at arrangements where two zeros merge into a degenerate one and vanish, the three-dimensional counterpart of the crossing that never happens run in reverse. Maxwell’s conjecture is a claim that there is a limit to how many such pairs $n$ charges can sustain. The topology fixes the difference and leaves the sum to geometry.

Gravity offers a contrast that makes the saddles clearer. The combined pull of two bodies in orbit, seen in the frame that rotates with them, also has five zeros — the Lagrange points — and two of them, the Trojan points, are stable. That is not a violation of the saddle rule. In the rotating frame the Coriolis force, which does no work and is not the gradient of anything, turns a slide off the summit of the potential into a circle around it, as the hilltop that holds the Trojans showed. Every zero of a static field is a saddle; a zero can hold something only when a force that is not part of the field is there to help.

Three dimensions, where nobody knows #

The same three charges in three dimensions, point charges with an inverse-square field, make four zeros rather than one. The topological argument no longer fixes the number, because in three dimensions zeros come in two kinds of saddle — with one inward direction and two outward, or two inward and one outward — whose indices have opposite signs and cancel in the count. Any number of extra pairs can appear without violating the topology.

The figure counts zeros for five symmetric arrangements of equal point charges. Two charges make one. Three in a line make two. Three on a triangle make four, the conjectured maximum. A square of four makes five: one at the centre and four near the middles of the sides. A regular pentagon of five makes six. The counts exceed the two-dimensional value in every case beyond two charges, and they fall short of Maxwell’s $(n-1)^2$ in all but one.

Maxwell’s conjecture has an argument behind it. Zeros of the field satisfy three equations in three unknowns, and after clearing the square roots each equation can be written as a polynomial whose degree grows with the number of charges; Bézout’s theorem bounds the number of solutions of such a system by the product of the degrees. That bound is enormous. Maxwell’s $(n-1)^2$ is a guess at the true answer. In 2007 Gabrielov, Novikov and Shapiro proved that the number of isolated zeros is always finite and gave an explicit bound. It is astronomically larger than $(n-1)^2$, and closing the gap has proved hard.

A search cannot settle the question, and the figure shows why. It places three, four and five equal charges at random in a cube forty times each, and counts the zeros each arrangement makes. Most random arrangements of $n$ charges make $n - 1$ zeros, the two-dimensional number, and a few make more. The largest counts found are 4 for three charges, 3 for four and 6 for five, far below the conjectured 9 and 16 for four and five. That is typical: random arrangements are rarely the extreme ones, and the arrangements that make many zeros are special and symmetric. A search can show that a count occurs. It can never show that a larger count does not, which is the whole difficulty of the conjecture.

Why zeros matter #

The question looks like a mathematical curiosity, and it is more than that, because zeros of fields organise physical phenomena wherever fields do.

In a plasma, the zeros of a magnetic field are where field lines can break and reconnect. The knot the field cannot untie described a squashed neutral point carrying a current sheet: that is a zero of the magnetic field, and the topology of the field’s zeros decides where solar flares release their energy. In a molecule, the zeros of the gradient of the electron density — bond critical points — are used to define where one atom ends and the next begins, and their count is constrained by a topological rule of exactly the kind the two-dimensional argument used. In ion traps, which the essay on static fields showed must be dynamic, the zero of the oscillating field is where the ion sits. And in gravity, the zeros of the combined pull of several masses are Lagrange points, some of which — unlike any electrostatic zero — are stable because rotation adds forces that are not a gradient.

The multipole picture of the shape that takes five numbers gives a partial reason the counts vary. Far away, $n$ charges look like one charge, and closer in their dipole and quadrupole moments distort the field. Zeros appear where the distortion is strong enough to reverse the far-field direction locally, and symmetric arrangements, which cancel their low moments, let the higher ones compete on equal terms in several places at once. The triangle has no dipole moment about its centre, and its four zeros come from a competition in which the quadrupole matters as much as the monopole.

What the counting leaves out #

The figures count isolated, nondegenerate zeros of fields of equal positive charges, and each of those words excludes a case.

Degenerate zeros. Arrangements of high symmetry can make zeros at which the field vanishes faster than linearly. Four equal charges at the corners of a regular tetrahedron make one at the centre, where the linear part cancels entirely. Such a zero is counted once but can split into several ordinary zeros when the arrangement is perturbed, which is one reason the maximum count is so hard to pin down.

Charges of both signs. With positive and negative charges the field can also vanish on curves rather than at points, and the counting changes character. Maxwell’s conjecture is usually stated for charges of arbitrary signs and positions, with the zeros required to be isolated.

Numerical search is not proof. The counts in the last two figures come from searching, and a search can miss a zero whose basin of attraction is small. The counts for symmetric arrangements are checked against the symmetry; the random ones are lower bounds on the true numbers.

Still open: Maxwell’s conjecture #

The conjecture remains open in general and has not been proved even for three charges, where the conjectured maximum of four is attained by the triangle but no argument rules out five. Partial results bound the number of zeros for special cases — charges in a plane, or collinear charges, where the count is easier — and the general finiteness result of 2007 gives a bound that grows faster than exponentially. Whether $(n-1)^2$ is the true maximum, whether it is achieved for every $n$, and which arrangements achieve it are unknown. The problem is simple to state, involves nothing but Coulomb’s law, and sits at the boundary of what current methods for counting solutions of real polynomial systems can do.

The habit worth carrying away is to count the places where a field says nothing. A field’s zeros are where its structure is decided, every electrostatic zero is a saddle, and in two dimensions their number is fixed by algebra while in three it is set by the arrangement in a way nobody has bounded. A drawing of field lines hides them; a list of charges determines them; and between those two descriptions lies a question Maxwell asked and left for others, which is still waiting.