# General Relativity and the Circumference of a Circle

1. Jan 2, 2017

### Poirot

1. The problem statement, all variables and given/known data
Calculate the circumference of the circle θ = θ0 (a constant) in the spatial geometry

\begin{eqnarray*}
dS^2 = a^2(d\theta^2 + sin^2\theta cos^2\theta d\phi^2)
\end{eqnarray*}
Hence, (by finding R(z)) sketch the cross section of the surface embedded in three dimensions via
\begin{eqnarray*}
(x, y, z) = (Rsin\theta cos\phi, Rsin\theta sin\phi, Rcos\theta)
\end{eqnarray*}
2. Relevant equations

3. The attempt at a solution
Calculating the circum:
Parameterising the path set φ = τ, where τ1 = 0 and τ2 =2π, and θ = θ0 (const.)
so
\begin{eqnarray*}
circumference = \int_{0}^{2\pi} \frac{dS}{d\tau}d\tau = \int_{0}^{2\pi} a((\frac{d\theta}{d\tau})^2 + sin^2\theta cos^2\theta (\frac{d\phi}{d\tau})^2)^{1/2} d\tau
\end{eqnarray*}
using the parametrisation we get:
\begin{eqnarray*}
circumference: \int_{0}^{2\pi} a sin\theta_0 cos\theta_0 d\tau = 2\pi a sin\theta_0 cos\theta_0 = \pi a sin2\theta_0
\end{eqnarray*}

I'm not at all sure what the next part is getting at. I've played around with rearranging z=Rcosθ and plugging this in for theta in metric but I don't think that's right.

Any help would be great appreciated, thank you!

2. Jan 2, 2017

### vela

Staff Emeritus
Does this help? Figure out how $R$ varies with $\theta_0$.

3. Jan 3, 2017

### Poirot

I found the radius R to be R=aθ0 by parameterising the theta part this time and integrating up to θ0. And I think by using the z=Rcosθ definition we get that at θ=0 z=R, and at θ=θ0 z=Rcosθ0.

I may be wrong but I think the radius perpendicular = a/2 sin 2θ0 from dividing the circumference by 2π so there's a relation that
\begin{eqnarray*}
z^2 + r(perp)^2 = R^2
\end{eqnarray*}
I may be barking up the wrong tree here, thanks again!

4. Jan 3, 2017

### vela

Staff Emeritus
$R$ is the distance from the origin to a point on the surface whereas $a\theta_0$ is the arc length of between $\theta=0$ and $\theta=\theta_0$ on a curve of constant $\varphi$ on the surface. It doesn't make sense to set those two quantities equal.

That's right. If you use $z=R\cos\theta_0$, everything will be in terms of $R$ and $\theta_0$, so you can get an expression for $R$ in terms of $\theta_0$. Ideally, you'll recognize what shape it describes. If not, try plotting it or reviewing curves in polar coordinates.

5. Jan 4, 2017

### Poirot

If I plug all the things in I get R = a cosθ0, but i'm not sure how this helps. I am probably wrong but I think it may be like an infinity shape on the z axis, similar to the spherical harmonics in atoms?

6. Jan 4, 2017

### vela

Staff Emeritus
Yes, that's right. Did you make a polar plot of $(R,\theta_0)$? You could also multiply both sides by $R$ and then convert to Cartesian coordinates to verify your hunch.