Parallel Line Masses and Marden’s Theorem

Here’s an idea I got from Albert Chern on Twitter. He did all the hard work, and I think he also drew the picture I’m going to use. I’ll just express the idea in a different way.

Here’s a strange fact about Newtonian gravity.

Consider three parallel ‘line masses’ that have a constant mass per length—the same constant for each one. Choose a plane orthogonal to these lines. There will typically be two points on this plane, say a and b, where a mass can sit in equilibrium, with the gravitational pull from all three lines masses cancelling out. This will be an unstable equilibrium.

Put a mass at point a. Remove the three line masses—but keep in mind the triangle they formed where they pierced your plane!

You can now orbit a test particle in an elliptical orbit around the mass at a in such a way that:

• one focus of this ellipse is a,
• the other focus is b, and
• the ellipse fits inside the triangle, just touching the midpoint of each side of the triangle.

Even better, this ellipse has the largest possible area of any ellipse contained in the triangle!

Here is Chern’s picture:



 

The triangle’s corners are the three points where the line masses pierce your chosen plane. These line masses create a gravitational potential, and the contour lines are level curves of this potential.

You can see that the points a and b are at saddle points of the potential. Thus, a mass placed at either a and b will be in an unstable equilibrium.

You can see the ellipse with a and b as its foci, snugly fitting into the triangle.

You can sort of see that the ellipse touches the midpoints of the triangle’s edges.

What you can’t see is that this ellipse has the largest possible area for any ellipse fitting into the triangle!

Now let me explain the math. While the gravitational potential of a point mass in 3d space is proportional to 1/r, the gravitational potential of a line mass in 3d space is proportional to \log r, which is also the gravitational potential of a point mass in 2d space.

So, if we have three equal line masses, which are parallel and pierce an orthogonal plane at points p_1, p_2 and p_3, then their gravitational potential, as a function on this plane, will be proportional to

\phi(z) = \log|z - p_1| + \log|z - p_2| + \log|z - p_3|

Here I’m using z as our name for an arbitrary point on this plane, because the next trick is to think of this plane as the complex plane!

Where are the critical points (in fact saddle points) of this potential? They are just points where the gradient of \phi vanishes. To find these points, we can just take the exponential of \phi and see where the gradient of that vanishes. This is a nice idea because

e^{\phi(z)} = |(z-p_1)(z-p_2)(z-p_3)|

The gradient of this function will vanish whenever

P'(z) = 0

where

P(z) = (z-p_1)(z-p_2)(z-p_3)

Since P is a cubic polynomial, P' is a quadratic, hence proportional to

(z - a)(z - b)

for some a and b. Now we use

Marden’s theorem. Suppose the zeros p_1, p_2, p_3 of a cubic polynomial P are non-collinear. Then there is a unique ellipse inscribed in the triangle with vertices p_1, p_2, p_3 and tangent to the sides at their midpoints. The foci of this ellipse are the zeroes of the derivative of P.

For a short proof of this theorem go here:

Carlson’s proof of Marden’s theorem.

This ellipse is called the Steiner inellipse of the triangle:

• Wikipedia, Steiner inellipse.

The proof that it has the largest area of any ellipse inscribed in the triangle goes like this. Using a linear transformation of the plane you can map any triangle to an equilateral triangle. It’s obvious that there’s a circle inscribed in any equilateral triangle, touching each of the triangle’s midpoints. It’s at least very plausible that that this circle is the ellipse of largest area contained in the triangle. If we can prove this we’re done.

Why? Because linear transformations map circles to ellipses, and map midpoints of line segments to midpoints of line segments, and simply rescale areas by a constant fact. So applying the inverse linear transformation to the circle inscribed in the equilateral triangle, we get an ellipse inscribed in our original triangle, which will touch this triangle’s midpoints, and have the maximum possible area of any ellipse contained in this triangle!

7 Responses to Parallel Line Masses and Marden’s Theorem

  1. Take three non-overlapping circles of different sizes and place them in arbitrary positions on a plane. For each pair, draw two lines, each of which is tangent to both circles. They will intersect somewhere. The points of intersection of the three pairs of lines are collinear.

    Mathematics is full of things like this, which, which I find fascinating.

    • Toby Bartels says:

      Nice! For anyone who doesn't click on all the pictures (or perhaps has trouble following the Spanish), there is one fact here that's not in John's post: the root of the second derivative is the centre of the ellipse. (It should be well known that the root of the derivative of a quadratic polynomial is the midpoint between the roots of the quadratic, so the root of the second derivative of a cubic is the midpoint between the roots of the first derivative. But it's still nice to see how this fits into the overall picture.)

  2. For this work, would “point masses” work just as well as “line masses”? Is there any reason why we need “line masses” to complicate a scenario restricted to a plane?

    • John Baez says:

      In 3 dimensional space, a line mass gives rise to a potential proportional to log(r) ,which is what we need for this part of the proof to work (examine it). In 3 dimensional space a point mass gives rise to a potential proportional to -1/r, and that wouldn’t work here.

      In 2 dimensional space, a point mass gives rise to a potential proportional to log(r). However, we don’t get elliptical orbits from the gravitational field of a point mass in 2 dimensions—only in 3 dimensions.

      So, I needed to use a carefully chosen combination of point masses and line masses in 3 dimensions, to get everything to work. The fact that it works at all is sort of exciting.

  3. amarashiki says:

    Nice piece. However, I like the most the stuff concerning the Lagrange points and quintic equations. Usually, we use approximations to get Lagrangian points, but I wonder if there is some paper or book talking about lagrangian points from the viewpoint of the 3-body problem, algebra and quintinc equations, and symmetries.

You can use Markdown or HTML in your comments. You can also use LaTeX, like this: $latex E = m c^2 $. The word 'latex' comes right after the first dollar sign, with a space after it.

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

This site uses Akismet to reduce spam. Learn how your comment data is processed.