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Integrating the metric in 3-D Spherical coordinates

  1. Jun 26, 2012 #1
    Guys,

    I read that integrating the ds gives the arc length along the curved manifold. So in this case, I have a unit sphere and its metric is ds^2=dθ^2+sin(θ)^2*dψ^2. So how to integrate it? What is the solution for S?

    Note, it also is known as ds^2=dΩ^2

    Thanks!
     
  2. jcsd
  3. Jun 26, 2012 #2

    tiny-tim

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    Hi GRstudent! :smile:

    (try using the X2 button just above the Reply box :wink:)
    integrate ds "dot" the expression for the curve

    for example, if the curve is constant ψ, then ds = dθ ("longitude")

    if the curve is constant θ, then ds = sinθdψ ("latitude")
     
  4. Jun 26, 2012 #3
    and if the curve is neither, then you need to parametrize, for example by giving θ = θ(t), ψ=ψ(t), write ds2 = (something that depends on t) dt2 and then integrate!
     
  5. Jun 26, 2012 #4
    I would like to get an integral solution of spherical coordinates metric.
     
  6. Jun 26, 2012 #5
    Did you have a particular path in mind, whose arc length you want to calculate? Let's take an example: Suppose the path is defined by [itex] \phi = 2 \theta [/itex], [itex] \theta \in [0, \pi] [/itex] and the arc length is
    [tex] ds^2 = d\theta^2 + \sin^2 \theta d\phi^2 [/tex]
    Plugging the curve parametrization into this formula, and taking the square root, you find
    [tex]ds = \sqrt{d\theta^2 + \sin^2 \theta (2 d\theta)^2} = \sqrt{1+4\sin^2\theta} d\theta [/tex]
    and to find the arc length, you just integrate
    [tex] s = \int_\gamma ds = \int_0^\pi \sqrt{1+4\sin^2\theta} d\theta [/tex]
     
  7. Jun 26, 2012 #6
    Thanks!
     
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