Electrostatics and the Gauss–Lucas Theorem

Say you know the roots of a polynomial P and you want to know the roots of its derivative. You can do it using physics! Namely, electrostatics in 2d space, viewed as the complex plane.

To keep things simple, let us assume P does not have repeated roots. Then the procedure works as follows.

Put equal point charges at each root of P, then see where the resulting electric field vanishes. Those are the roots of P’.

I’ll explain why this is true a bit later. But first, we use this trick to see something cool.

There’s no way the electric field can vanish outside the convex hull of your set of point charges. After all, if all the charges are positive, the electric field must point out of that region. So, the roots of P’ must lie in the convex hull of the roots of P!

This cool fact is called the Gauss–Lucas theorem. It always seemed mysterious to me. Now, thanks to this ‘physics proof’, it seems completely obvious!

Of course, it relies on my first claim: that if we put equal point
charges at the roots of P, the electric field they generate will vanish at the roots of P’. Why is this true?

By multiplying by a constant if necessary, we can assume

Thus

This function is the electric potential created by equal point charges at the points a_i in the complex plane. The corresponding electric field is minus the gradient of the potential, so it vanishes at the critical points of this function. Equivalently, it vanishes at the critical points of the exponential of this function, namely |P|. Apart from one possible exception, these points are the same as the critical points of P, namely the roots of P’. So, we’re almost done!

The exception occurs when P has a critical point where P vanishes. |P| is not smooth where P vanishes, so in this case we cannot say the critical point of P is a critical point of |P|.

However, when P has a critical point where P vanishes, then this point is a repeated root of P, and I already said I’m assuming P has no repeated roots. So, we’re done—given this assumption.

Everything gets a bit more complicated when our polynomial has repeated roots. Greg Egan explored this, and also the case where its derivative has repeated roots.

And this is what it looks like if the original polynomial has a repeated root: the blue disk in the upper left is both a root of the derivative and the repeated root of the original polynomial. pic.twitter.com/eeNg3f0PFV

— Greg Egan (@gregeganSF) May 24, 2021

However, the Gauss–Lucas theorem still applies to polynomials with repeated roots, and this proof explains why:

• Wikipedia, Gauss–Lucas theorem.

Alternatively, it should be possible to handle the case of a polynomial with repeated roots by thinking of it as a limit of polynomials without repeated roots.

By the way, in my physics proof of the Gauss–Lucas theorem I said the electric field generated by a bunch of positive point charges cannot vanish outside the convex hull of these point charges because the field ‘points out’ of this region. Let me clarify that.

It’s true even if the positive point charges aren’t all equal; they just need to have the same sign. The rough idea is that each charge creates an electric field that points radially outward, so these electric fields can’t cancel at a point that’s not ‘between’ several charges—in other words, at a point that’s not in the convex hull of the charges.

But let’s turn this idea into a rigorous argument.

Suppose z is some point outside the convex hull of the points a_i. Then, by the hyperplane separation theorem, we can draw a line with z on one side and all the points a_i on the other side. Let v be a vector normal to this line and pointing toward the z side. Then

for all i. Since the electric field created by the ith point charge is a positive multiple of z – a_i at the point z, the total electric field at z has a positive dot product with v. So, it can’t be zero!

Credits

The picture of a convex hull is due to Robert Laurini.