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Spherical coordinates

  1. Jun 21, 2015 #1
    My question is really about converting between spherical coordinates and cartesian coordinates.
    Suppose that ##\phi## and ##\theta## are defined as follows:
    ##\phi## is the angle between the position vector of a point and the ##z##-axis. ##\theta## is the angle between the projection of that vector onto the ##xy##-plane and the ##x##-axis.
    Why exactly do we write ##x = r \sin{\phi} \cos{\theta}##?
    I know that we first project the position vector onto the ##xy##-plane before projecting that projection onto the ##x##-axis. But if that were the case, shouldn't we write ##x = |r \sin{\phi}| \cos{\theta}##?
    I'm confused because the quantity ##r \sin{\phi}## is, on occasion, negative. I thought we're only allowed to project "lengths" or "positive scalars". I am fully aware of the fact that the projection can itself be negative, but the whole notion that the length (not the final projection) which is to be projected can be negative isn't so intuitive.
    Take for example a circle of radius ##r## centred at the origin. I'll define ##t## as the angle made with the ##x##-axis. ##x = r \cos{\theta}##. ##x## is allowed to be positive or negative, but ##r## is always positive.
     
  2. jcsd
  3. Jun 21, 2015 #2
    ##\phi## is only defined on the domain ##0<\phi<\pi/2## so ##\sin(\phi) \ge 0##
     
  4. Jun 21, 2015 #3
    Isn't the choice of the interval somewhat arbitrary?
    When dealing with polar coordinates for instance, we work on either ##(-\pi, \pi]## or ##[0, 2\pi)##.
     
  5. Jun 21, 2015 #4
    Well, that's the standard convention. You can choose another convention if you want, but you'll have to deal with ##\sin(\phi) < 0## somehow, so the standard convention makes a lot of sense.
     
  6. Jun 21, 2015 #5
    Thanks!
    But am I correct in assuming that projecting a "negative length" is meaningless?
    Also, I believe the domain you are talking about is ##0 \leq \phi \leq \pi##, not ##0 \leq \phi \leq \frac{\pi}{2}##.
     
  7. Jun 21, 2015 #6
    Oops!

    In some situations, you can interpret a negative length as a length in the opposite direction.
     
  8. Jun 21, 2015 #7
    Well, yeah. But not in this particular example, right?
     
  9. Jun 25, 2015 #8

    HallsofIvy

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    Given the spherical coordinates as you say, the z coordinate is easy. If we draw a line from the given point perpendicular to the z-axis, [itex]\phi[/itex] is the angle in a right triangle having the straight line from the origin to the point, of length [itex]\rho[/itex], as hypotenuse and the height above the xy-plane as "near side": so [itex]z= \rho cos(\phi)[/itex].

    Now consider the projection of that point in the xy-plane, (x, y, 0). The straight line from the origin to (x, y, 0) is the "opposite side" to [itex]\phi[/itex] of a right triangle having [itex]\rho[/itex] as hypotenuse- its length is [itex]\rho sin(\phi)[/itex]. And that line is the hypotenuse of a right triangle in the xy-plane with angle [itex]\theta[/itex] and "near side" of length x, "opposite side" of length y. [itex]x/r= cos(\theta)[/itex] so [itex]x= r cos(\theta)= \rho sin(\phi) cos(\theta)[/itex] and [itex]y/r= sin(\theta)[/itex] so [itex]y= r sin(\theta)= \rho sin(\phi)sin(\theta)[/itex].
     
  10. Jun 25, 2015 #9
    And the length ##\rho \sin{\phi}## is always positive, right?
     
  11. Jun 25, 2015 #10

    HallsofIvy

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    [itex]\rho[/itex] itself is a distance so always positive. [itex]\phi[/itex] is between 0 and [itex]\pi[/itex] so [itex]sin(\phi)[/itex] is positive. Yes, their product, the product of positive numbers, is positive.
     
  12. Jun 25, 2015 #11

    Mark44

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    Per wikipedia, "always" is too strong.
    -- https://en.wikipedia.org/wiki/Spherical_coordinate_system

    Note that for some reason they are using r instead of ##\rho##, which is more commonly used in spherical coordinates.
     
  13. Jun 25, 2015 #12
    So unlike polar coordinates, ##\phi## is always restricted to be between 0 and ##\pi##.
     
  14. Jun 25, 2015 #13

    Mark44

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    Not according to the same article whose link I posted earlier.
     
  15. Jun 25, 2015 #14
    Which brings me to my initial question; how can we project a "negative length" onto a plane?
     
  16. Jun 25, 2015 #15

    Mark44

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    You can project a positive length onto a plane, right? The projection of a "negative length" would be just as long but would point in the opposite direction.
     
  17. Jun 26, 2015 #16
    So, is it wrong to write ##x = |\rho \sin{\phi}| \cos{\theta}## (where ##\phi## is the angle between the position vector of a point and the ##z##-axis)?
     
  18. Jun 26, 2015 #17

    HallsofIvy

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    Yes, it is. What is correct is that [itex]x= \rho sin(\phi) cos(\theta)[/itex]. There should be no absolute value since x, which is a coordinate, not a distance, can be negative.
     
  19. Jun 26, 2015 #18
    It could still be negative if ##x = |\rho \sin{\phi}| \cos{θ}##; there are no absolute value bars around ##\cos{θ}##.
     
  20. Jun 26, 2015 #19

    HallsofIvy

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    Right. I should have said that the absolute value bars are not necessary here. [itex]\rho[/itex], a distance, is always positive and since [itex]\phi[/itex] lies between 0 and [itex]\pi[/itex], [itex]sin(\phi)[/itex] is always positive [itex]\rho sin(\phi)[/itex] cannot be negative..
     
  21. Jun 26, 2015 #20

    Mark44

    Staff: Mentor

    For what it's worth, and to repeat what I said in post #11, wikipedia disagrees that ##\rho## is always positive -- https://en.wikipedia.org/wiki/Spherical_coordinate_system
     
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