Skip to main content
MathOverflow

Return to Answer

I'm stupid; deleted 11 characters in body
Source Link
zeb
zeb
  • 8.7k
  • 6
  • 37
  • 50

Edit: sorry, there used to be a completely wrong solution here (I thought that a certain singular curve was a projective line). Now it is fixed.

There is another solution using algebraic geometry. If you consider that set of ordered pairs (point onIdentify the ellipse, with the projective line passingby sending the two points where the line through the point onfoci meets the ellipse to $0$ and at least one focus), you see that this set defines an algebraic curve $C$$\infty$. The natural map from $C$ to the ellipse (that we get by forgetting the line) is two to one at everystarting from a point except for the two points where the line through the foci hitson the ellipse, so bygetting the Riemann-Hurwitz formula $C$ has genus zero, i.e. it is isomorphic tosecond intersection of the projective line.

You have defined a map $f$ from $C$ to itself, and this map can be defined algebraically through it and is clearly invertible$F$ with the ellipse, so as a map fromthen getting the projective line to itself it is a fractional linear transformation. The map $f \circ f$ has two obvious fixed points wheresecond intersection of the line through the foci intersects the ellipse,that point and if we fix an identification of $C$$F'$ with the projective lineellipse is an invertible algebraic map sending the two fixed points$0$ to $0$ and $\infty$, then we see that the map to $f$$\infty$, so it must have the form $f(x) = c/x$$x \mapsto cx$ for some constant $c$ (depending on the eccentricity of the ellipse). Thus as we iterate $f$, $f^n(x)$ tends to $0$ and $\infty$ (in other words, the billiard ball tends to aapproaches the horizontal line), and it does so at an exponential rate.

There is another solution using algebraic geometry. If you consider that set of ordered pairs (point on the ellipse, line passing through the point on the ellipse and at least one focus), you see that this set defines an algebraic curve $C$. The natural map from $C$ to the ellipse (that we get by forgetting the line) is two to one at every point except for the two points where the line through the foci hits the ellipse, so by the Riemann-Hurwitz formula $C$ has genus zero, i.e. it is isomorphic to the projective line.

You have defined a map $f$ from $C$ to itself, and this map can be defined algebraically and is clearly invertible, so as a map from the projective line to itself it is a fractional linear transformation. The map $f \circ f$ has two obvious fixed points where the line through the foci intersects the ellipse, and if we fix an identification of $C$ with the projective line sending the two fixed points to $0$ and $\infty$, then we see that the map $f$ must have the form $f(x) = c/x$ for some constant $c$ (depending on the eccentricity of the ellipse). Thus as we iterate $f$, $f^n(x)$ tends to $0$ and $\infty$ (in other words, the billiard ball tends to a horizontal line), and it does so at an exponential rate.

Edit: sorry, there used to be a completely wrong solution here (I thought that a certain singular curve was a projective line). Now it is fixed.

There is another solution using algebraic geometry. Identify the ellipse with the projective line by sending the two points where the line through the foci meets the ellipse to $0$ and $\infty$. The map we get by starting from a point on the ellipse, getting the second intersection of the line through it and $F$ with the ellipse, then getting the second intersection of the line through that point and $F'$ with the ellipse is an invertible algebraic map sending $0$ to $0$ and $\infty$ to $\infty$, so it must have the form $x \mapsto cx$ for some constant $c$ (depending on the eccentricity). Thus, the billiard ball approaches the horizontal line at an exponential rate.

Source Link
zeb
zeb
  • 8.7k
  • 6
  • 37
  • 50

There is another solution using algebraic geometry. If you consider that set of ordered pairs (point on the ellipse, line passing through the point on the ellipse and at least one focus), you see that this set defines an algebraic curve $C$. The natural map from $C$ to the ellipse (that we get by forgetting the line) is two to one at every point except for the two points where the line through the foci hits the ellipse, so by the Riemann-Hurwitz formula $C$ has genus zero, i.e. it is isomorphic to the projective line.

You have defined a map $f$ from $C$ to itself, and this map can be defined algebraically and is clearly invertible, so as a map from the projective line to itself it is a fractional linear transformation. The map $f \circ f$ has two obvious fixed points where the line through the foci intersects the ellipse, and if we fix an identification of $C$ with the projective line sending the two fixed points to $0$ and $\infty$, then we see that the map $f$ must have the form $f(x) = c/x$ for some constant $c$ (depending on the eccentricity of the ellipse). Thus as we iterate $f$, $f^n(x)$ tends to $0$ and $\infty$ (in other words, the billiard ball tends to a horizontal line), and it does so at an exponential rate.