# General Relativity and the Circumference of a Circle

• Poirot
In summary: 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?
Poirot

## Homework Statement

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*}

## 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!

Does this help? Figure out how ##R## varies with ##\theta_0##.

vela said:
Does this help? Figure out how ##R## varies with ##\theta_0##.
View attachment 111022
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!

Poirot said:
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.
##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.

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!
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.

vela said:
##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.
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?

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.

## 1. What is General Relativity?

General Relativity is a theory developed by Albert Einstein to explain the force of gravity and its effects on space and time. It is considered one of the pillars of modern physics.

## 2. How does General Relativity relate to the circumference of a circle?

In General Relativity, the curvature of space-time is affected by the presence of mass and energy. This curvature can be visualized as a fabric that is stretched or warped by objects with mass. The circumference of a circle is affected by this curvature, causing it to be different than what is predicted by classical Newtonian physics.

## 3. Can General Relativity explain why the circumference of a circle is always the same regardless of its size?

No, General Relativity does not explain why the circumference of a circle remains constant. This is a fundamental property of circles in Euclidean geometry and is not dependent on the theory of General Relativity.

## 4. Does General Relativity affect the shape of a circle?

Yes, General Relativity can affect the shape of a circle due to the curvature of space-time. In extreme cases, such as near a black hole, the curvature can cause the circle to appear distorted or even appear as an ellipse.

## 5. Is General Relativity necessary to understand the circumference of a circle?

No, General Relativity is not necessary to understand the concept of the circumference of a circle. This can be explained using classical Euclidean geometry and does not require the complexities of General Relativity.

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