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Spherical polars

  1. Nov 6, 2004 #1
    I have two points, one given in spherical polar coordinates and the other in cartesian coordinates. If I want to work out the distance inbetween these points do I need to convert the cartesian into spherical polars?
     
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  3. Nov 6, 2004 #2

    matt grime

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    Unless you happen to be able to visualize these things in your head, then yes, writing the point in sphericals in cartesian coords would seem like a good step, especially as distance is usually calculated in terms of cartesion coordinates.
     
  4. Nov 6, 2004 #3

    StatusX

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    the formula for the distance between two points in spherical coordinates is:

    [tex]d^2=r_1^2+r_2^2-2 r_1 r_2(sin\theta_1 sin\theta_2 cos\phi_1 cos\phi_2 + sin\theta_1 sin\theta_2 sin\phi_1 sin\phi_2 + cos\theta_1 cos\theta_2)[/tex]

    obviously, the cartesian formula is a lot simpler.
     
  5. Nov 7, 2004 #4
    Okay so I have point P with spherical polar coordinates
    [tex](R, \theta , \phi) [/tex]
    and point A with cartesian coordinates
    [tex](0, 0, a)[/tex]

    so A is just the z axis with length a? so in spherical polars:

    A is [tex] (a, 0, 0) [/tex] ? or maybe [tex] (a, \theta, \phi) [/tex] ?

    Im getting confused!
     
  6. Nov 7, 2004 #5

    arildno

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    With [tex]\theta[/tex] being the angle in the horizontal plane, whereas [tex]\phi[/tex] the angle to the z-axis, a given point at distance abs(a) ("a" itself either positive or negative) on the z-axis has the polar representation:
    [tex](abs(a),\theta,\frac{\pi}{2}(1-sign(a))[/tex]
    where sign(a) is 1 when a is positive or -1 when a is negative.
     
  7. Nov 7, 2004 #6
    Thanks, I've now worked out:

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

    I'll see how far I can get now!
     
  8. Nov 7, 2004 #7

    arildno

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    Note:
    In your notation, the angles are interchanged from how I've used them!
     
  9. Nov 7, 2004 #8
    Yeah, sorry forgot to mention that I had interchanged them.

    So now I have two points in cartesian coordinates:

    P [tex] (R\sin\theta\cos\theta , R\sin\theta\sin\phi , R\cos\theta) [/tex]

    A [tex] (0, 0, a) [/tex]

    In order to work out the distance AP do I need to square that awful looking thing?!
     
  10. Nov 7, 2004 #9

    arildno

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    Yep, that's what you need to do.
     
  11. Nov 8, 2004 #10
    Okay so I need:

    [tex] (R\sin\theta\cos\phi)^2 + (R\sin\theta\sin\phi)^2 + (R\cos\theta)^2 [/tex]

    How can I simplify this? In a book I have it just sets a "similar" expression equal to 1 with no intermediate steps:

    [tex] (\sin\theta\cos\phi)^2 + (\sin\theta\sin\phi)^2 + (\cos\theta)^2 = 1[/tex]

    Also this has no 'R' in it.

    edit: sorry had theta where there should have been phi
     
    Last edited: Nov 8, 2004
  12. Nov 9, 2004 #11

    HallsofIvy

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    Did you notice that
    [tex] (R\sin\theta\cos\phi)^2 + (R\sin\theta\sin\phi)^2 + (R\cos\theta)^2 [/tex]
    [tex] =R^2(\sin\theta\cos\phi)^2 + R^2(\sin\theta\sin\phi)^2 + R^2(\cos\theta)^2 [/tex]
    [tex] = R^2((\sin\theta\cos\phi)^2 + (\sin\theta\sin\phi)^2 + (\cos\theta)^2 [/tex])
    ?
     
  13. Nov 9, 2004 #12
    Yeah, I've got the distance AP to be

    [tex](R^2 - 2aR\cos\theta + a^2)^(1/2)[/tex]

    which I think is correct?

    Now writing the triple integral of AP over the sphere R less than or equal to a in terms of spherical polar coords gives:

    [tex]\iiint {\sqrt(R^2 - 2aR\cos\theta + a^2)} R^2 \sin\theta\,dR\,d\theta\,d\phi[/tex]

    with the integration over R between 0 and a
    the integration over [tex] \theta [/tex] between 0 and pi
    the integration over [tex] \phi [/tex] between 0 and 2pi

    correct so far?
     
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