Length of Circle on a Sphere

Question

I want to calculate the length of the circle $\theta=\pi/4$, where $\theta$ is the latitude, on the unit sphere.

I know that the length of a curve $\gamma (t), t \in [0,T]$ on a manifold is given by $\int_{0}^{T} \sqrt{g_{ij} \frac{d\gamma ^i}{dt}\frac{d\gamma ^j}{dt}} dt$ (using Einstein's summation convention).

I also know that the metric on spherical coordinates is given by $g=\begin{pmatrix} 1 & 0 & 0\\ 0 & r^2 & 0 \\ 0 & 0 & r^2 \sin{\theta} \end{pmatrix}$

Using this information the only non zero term in the root is $r^2 \sin^2{\theta} \cdot 1 \cdot 1 = \frac{1}{2}$ and the whole integral gives $\sqrt{2} \pi$ as the result for the length.

Is this correct? I have the feeling that the result should be slightly bigger. Where did I get it wrong?

Answer 1

You can prove that your result is correct simply noting that the radius of the circle at latitude $\pi/4$ is $\cos (\pi/4)=\sqrt{2}/2$. So the leght is $\sqrt{2}\pi$.

Answer 2

Using high school level method gives the same result. What is the length of circumference of circle?

$$L=2\pi R$$

and how much is the radius of $\theta=\frac{\pi}4$?

$$R=1 \times cos(\frac{\pi}4)=\frac{\sqrt{2}}2$$

So your answer is correct:

$$L=\sqrt{2} \pi$$

Answer 3

Draw a diagram. You will find that the circle of constant latitude $\pi/4$ has radius $1/\sqrt{2}$, and so length $\sqrt{2}\; \pi$, as expected.