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

\displaystyle{   P(z) = \prod_{i = 1}^n  (z - a_i) }

Thus

\displaystyle{  \ln |P(z)| = \sum_{i = 1}^n \ln|z - a_i| }

This function is the electric potential created by equal point charges at the points ai 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.

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 ai. Then, by the hyperplane separation theorem, we can draw a line with z on one side and all the points ai on the other side. Let v be a vector normal to this line and pointing toward the z side. Then

v \cdot (z - a_i) > 0

for all i. Since the electric field created by the ith point charge is a positive multiple of z – ai 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.

10 Responses to Electrostatics and the Gauss–Lucas Theorem

  1. Toby Bartels says:

    It doesn't seem much more complicated when there are repeated roots. You've shown that the roots of |P|′ lie inside the convex hull of the roots of P, and you want that the roots of P′ lie inside this convex hull. Well, every root of P′ is either a root of |P|′ or a root of P, and of course the roots of P lie within the convex hull of the roots of P. So we're done.

  2. John Baez says:

    Okay, that’s nice.

  3. SteveB says:

    Surely it is the logarithmic derivative of P that is electric potential of point charges, rather than the log of P?

    • John Baez says:

      Remember that in 2d space a point charge at ai creates a potential proportional to

      \ln|z - a_i|

      so a collection of equal point charges at points a1, … , an creates a potential that’s the sum of such logarithms, which is the logarithm of the absolute value of

      P(z) = \prod_{i = 1}^n (z - a_i)

  4. SteveB says:

    I think a point charge at a_{i} creates a potential proportional to

    \displaystyle{ \frac{1}{z-a_{i}} }

    Then

    \displaystyle{ \frac{d}{dz} (\log(P(z)) = \frac{P'(z)}{P(z)} = \sum_i \frac{1}{z-a_{i}} }

    • John Baez says:

      Remember, I’m doing electrostatics in 2d space in this post.

      In 3-dimensional space the potential of a point charge is proportional to 1/r, but in 2-dimensional space the potential is proportional to the logarithm of distance.

      The reason for this is that in 3-dimensional space the electric field of a point charge is proportional to 1/r2, while in 2-dimensional space the electric field is proportional to 1/r.

      And the reason for this is that in 3-dimensonal space the area of a sphere is proportional to r2, while in 2-dimensional space the circumference of a circle is proportional to r.

      • The way I was taught this bit of electrostatics is not as 2d but as a cylindrical geometry in 3d. The contributions of field strength further along the cylinder axis generate the 1/r dependence radially outward from the axis.

      • John Baez says:

        Yes, I mentioned that perspective in my previous article:

        Parallel line masses and Marden’s theorem.

        where I used Newtonian gravity instead of electrostatics. I needed both “line masses” whose potential goes like log(r) and also point masses whose potential goes like -1/r.

  5. Toby Bartels says:

    In n dimensions, if the flux of the electric field out of a simple closed hypersurface is proportional to the total charge inside, and if the field of a point charge is symmetric, then the field of a point charge must be proportional to \lvert z - a _ i \rvert ^ { 1 - n } (which you can check by calculating the flux). Then by integrating this, the potential is proportional to \lvert z - a _ i \rvert ^ { 2 - n } … except if 2 − n = 0, when it's \ln \lvert z - a _ i \rvert instead.

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