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General Relativity and the Circumference of a Circle

  1. Jan 2, 2017 #1
    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. jcsd
  3. Jan 2, 2017 #2

    vela

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    Does this help? Figure out how ##R## varies with ##\theta_0##.
    gr.png
     
  4. Jan 3, 2017 #3
    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!
     
  5. Jan 3, 2017 #4

    vela

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    ##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.
     
  6. Jan 4, 2017 #5
    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?
     
  7. Jan 4, 2017 #6

    vela

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    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.
     
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