General Relativity and the Circumference of a Circle

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?
  • #1
Poirot
94
3

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*}

Homework Equations

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!
 
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  • #2
Does this help? Figure out how ##R## varies with ##\theta_0##.
gr.png
 
  • #3
vela said:
Does this help? Figure out how ##R## varies with ##\theta_0##.
View attachment 111022
Thank you for your reply!
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
Poirot said:
Thank you for your reply!
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.
 
  • #5
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?
 
  • #6
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.
 

Related to General Relativity and the Circumference of a Circle

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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