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Integrating the metric in 3-D Spherical coordinates |
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| Jun26-12, 03:52 AM | #1 |
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Integrating the metric in 3-D Spherical coordinates
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! |
| Jun26-12, 05:13 AM | #2 |
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Hi GRstudent!
![]() (try using the X2 button just above the Reply box )for example, if the curve is constant ψ, then ds = dθ ("longitude") if the curve is constant θ, then ds = sinθdψ ("latitude") |
| Jun26-12, 06:00 AM | #3 |
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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!
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| Jun26-12, 06:37 AM | #4 |
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Integrating the metric in 3-D Spherical coordinates
I would like to get an integral solution of spherical coordinates metric.
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| Jun26-12, 07:21 AM | #5 |
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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] |
| Jun26-12, 07:33 AM | #6 |
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Thanks!
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