Is there any ellipse with nonzero rational minor and major axis lengths $a$ and $b$ such that the circumference of the ellipse is rational too? (or the weaker variant: ... such that the circumference is algebraic?)
Note that the surface area $S=\pi ab$ is of course always transcendental in such a situation.

  • 3
    $\begingroup$ This is probably related to the transcendence of periods of elliptic curves defined over the rationals, which you can find, e.g., in Baker's book Transcendental number theory. I am not sure it is exactly the same thing though, you'd have to look at some old book on elliptic integrals. $\endgroup$ – Felipe Voloch Jun 12 '13 at 17:05

No, there is no such ellipse. This is exactly theorem $6.5$ of Alan Baker's book TRANSCENDENTAL NUMBER THEORY, as pointed out by Felipe Voloch.

Let $\omega$ be a primitive period of a $\wp$-function with algebraic invariants $g_2, g_3$ and let $\eta=2\zeta(\omega /2)$ be the associated quasi-period of the Weierstrass $\zeta$-function satisfying $\zeta'(z) = -\wp (z)$. Then we have

Theorem $6.5$. Any linear combination of $\omega$ and $\eta$ with algebraic coefficients, not both zero, is transcendental.

Because $\omega$ and $\eta$ can be represented as elliptic integrals of the first and second kinds respectively, Theorem 6.5 implies that the circumference of any ellipse with algebraic axes-lengths is transcendental.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.