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Spherical coordinates doubt

  1. Aug 9, 2014 #1
    1. The problem statement, all variables and given/known data

    In spherical coordinates (ρ,θ,ø); I understood the ranges of ρ, and θ. But ø, still eludes my understanding. Why is ø only from 0 to π, why not 0 to 2π??
     
  2. jcsd
  3. Aug 9, 2014 #2

    Ray Vickson

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    Look at a diagram to see why.
     
  4. Aug 9, 2014 #3

    AlephZero

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    Values of ø between from 0 to π cover the whole surface of the sphere. On maps of the earth latitude is measured from -90 to + 90 degrees not 0 and 180, and longitude from -180 to +180 not 0 to 360, but the basic idea is the same.

    652px-Latitude_and_Longitude_of_the_Earth.svg.png
     
  5. Aug 9, 2014 #4
    ## \phi = 0 ## is directly overhead, ## \phi = \pi ## is directly beneath your feet, where would ## \phi = 2\pi ## be?
     
  6. Aug 9, 2014 #5
    If seeing the diagram would have had helped, then I would not have asked the question in the first place.
     
  7. Aug 9, 2014 #6
    Okay, I have uploaded two attachments.

    When I view from side, ø = π, covers only half the circle (see the picture). When I try to think of it as a clock, ø = π, covers 12 to 6. Now shouldn't ø = 1.5π cover 12 to 9, and ø = 2π cover the whole circle, and reach the same point as π = 0.

    Also, in ø = π, the 3d section appears to me as a hemisphere. Shouldn't it be a total sphere??

    Or maybe, I am confusing spherical coordinates with polar or cartesian coordinates??
     

    Attached Files:

  8. Aug 9, 2014 #7
    So what you have shown is a coloured half-disk. For every point on that disk, Θ = 0. If you vary Θ from 0 to 2π the half-disk will sweep out a complete sphere.
     
  9. Aug 9, 2014 #8
    Please explain further, I can't seem to understand.

    EDIT: Okay, I think, I got a little idea of why ø = π would work. Basically, The up and down thing was exactly right. I think, I confused myself, when I added a sense of left and right. It is a total different axis, and the coordinates seem to do exactly the same thing, by putting θ = π to 2π.
     
  10. Aug 9, 2014 #9
    Okay, this is the link,

    http://mathinsight.org/spherical_coordinates
     
  11. Aug 9, 2014 #10

    HallsofIvy

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    Since [itex]\theta[/itex] goes from [itex]0[/itex] to [itex]2\pi[/itex], if we allowed [itex]\phi[/itex] to go also from [itex]0[itex] to [itex]2\pi[/itex] some points would have two descriptions. For example, [itex]\theta= 3\pi/2[/itex], [itex]\phi= \pi/4[/itex] and [itex]\theta= \pi/2[/itex], [itex]\phi= 7\pi/4[/itex], [itex]\rho[/itex] and fixed value, say 1, designate the same point.

    You can see that by converting to Cartesian coordinates: [itex]x= \rho cos(\theta) sin(\phi)[/itex], [itex]y= \rho sin(\theta) sin(\phi)[/itex], [itex]z= \rho cos(\phi)[/itex].

    [itex]\rho= 1[/itex], [itex]\theta= 3\pi/2[/itex], [itex]\phi= \pi/4[/itex] gives [itex]x= 1(0)(\sqrt{2}/2)= 0[/itex], [itex]y= 1(-1)(\sqrt{2}/2)= -\sqrt{2}/2[/itex] and [itex]z= 1(\sqrt{2}/2)= \sqrt{2}/2[/itex].

    [itex]\rho= 1[/itex], [itex]\theta= \pi/2[/itex], [itex]\phi= 7\pi/4[/itex] gives [itex]x= 1(0)(-\sqrt{2}/2)= 0[/itex], [itex]y= 1(1)(-\sqrt{2}/2)= -\sqrt{2}/2[/itex], and [itex]z= 1(\sqrt{2}/2)= \sqrt{2}/2[/itex].
     
  12. Aug 9, 2014 #11

    Ray Vickson

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    If I was a mind-reader I would have known that. I had no way to know what you have, or have not looked at already.
     
  13. Aug 9, 2014 #12
    Phi is the angle between the axis of the sphere and a line drawn through the center of the sphere to a given latitude, measured from the North Pole. It is equal to 90 degrees (i.e., ∏/2) minus the latitude. So, ø =0 represents a line drawn from the center of the sphere through the North pole, ø = ∏/2 represents a line drawn through the center of the sphere to any point on the equator, and ø =∏ represents a line drawn through the center of the sphere to the South pole.

    Chet
     
  14. Aug 10, 2014 #13
    Thanks, got it.
     
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