Spherical Coordinates


by Hypnotoad
Tags: coordinates, spherical
Hypnotoad
Hypnotoad is offline
#1
Sep3-04, 12:34 AM
P: 35
I'm trying to find the line element in spherical coordinates as well as a velocity element. I know that they are (ds)^2=(dr)^2+r^2(sin(theta))^2(dtheta)^2+r^2(dphi)^2 and sqrt[(dr/dt)^2+r^2(sin theta)^2(dtheta/dt)^2+r^2(dphi/dt)^2].

I know that this should be a quick and easy problem, but I simply can not figure it out. I would really appreciate some help on this one.
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Tide
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#2
Sep3-04, 01:18 AM
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Start from Cartesian coordinates in which [itex]ds^2 = dx^2+dy^2+dz^2[/itex] then calculate the differentials dx, dy and dz using:

[tex]x = r \sin \theta \cos \phi[/tex]
[tex]y = r \sin \theta \sin \phi[/tex]
[tex]z = r \cos \theta[/tex]

Substitute for dx, dy and dz in [itex]ds^2 = dx^2+dy^2+dz^2[/itex] and after a bit of algebra you should get the desired result.
robphy
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#3
Sep3-04, 10:45 AM
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To find the velocity vector, write the position vector as [itex] r(t) \hat r [/itex].
Then [itex]v(t)=\frac{d}{dt} \left[ r(t) \hat r \right] [/itex].
Use the product rule.
You'll have to compute [itex]\frac{d}{dt} \hat r [/itex],
where [tex]\hat r= \sin\theta\cos\phi \hat\imath + \sin\theta\sin\phi \hat\jmath + \cos\theta \hat k[/tex].

To simplify what you get, you might find it useful to know that
[tex]\hat \theta= \cos\theta\cos\phi \hat\imath + \cos\theta\sin\phi \hat\jmath - \sin\theta \hat k[/tex]
and [tex]\hat \phi= -\sin\phi \hat\imath + \cos\phi \hat\jmath [/tex]

You can derive these expressions for the spherical-polar unit vectors if you calculate the vectorial element
[tex]d \vec s = (dx)\hat \imath + (dy)\hat \jmath + (dz)\hat k [/tex]
using Tide's expressions for x, y, and z. [The strategy is to group the terms in [itex] dr[/itex], [itex]d\theta[/itex], and [itex]d\phi[/itex].]


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