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Spherical coordinates - phi vs theta

  1. Mar 30, 2015 #1
    1. The problem statement, all variables and given/known data
    My textbook states that when ##\theta = c ## where c is the constant angle with respect to the x-axis, the graph is a "half-plane". However, when ##\phi = c ## it is a half-cone. The only difference I see is that ##\phi## is the angle with respect to the z-axis, rather than the x-axis. Why is it that one ends up being a cone and the other a plane?

    2. Relevant equations
    n/a

    3. The attempt at a solution
    That is the only part I don't understand. It is clear how ##\rho = c ## is a sphere, but why are ##\phi## and ##\theta## so different?
     
  2. jcsd
  3. Mar 30, 2015 #2

    LCKurtz

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    Look at the geometry. If you pick a point where ##\theta = c## and vary ##r>0## and ##\phi## do you see why you get a half plane? Now pick a point where ##\phi = c##. What happens to that point as you vary ##\theta##? When you very ##r##? It might help to be looking at http://en.wikipedia.org/wiki/Spherical_coordinate_system when you answer. Use the second picture there as it has the usual notation used in math.
     
  4. Apr 2, 2015 #3

    HallsofIvy

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    You are mistaken. For one thing, [itex]\theta[/itex] can range from 0 to [itex]2\pi[/itex] while [itex]\phi[/itex] only ranges from 0 to [itex]\pi[/itex]. Do you see why that is true?
    ("Mathematics notation" and "physics notation" reverse [itex]\theta[/itex] and [itex]\phi[/itex]. I am assuming that you are using "mathematics notation".)

     
  5. Apr 2, 2015 #4

    BvU

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    How about some physical experience ?

    Stand up straight and lift one arm straight up. Let the direction your feet are pointing in represent ##\theta## and the angle your arm makes with the vertical when you lower it represent ##\phi## ("math notation" - I didn't know it existed). Fortunately they have r in common :smile: imagine a very long arm....

    Moving your arm up and down (##[0,\pi]##) with feet fixed defines a half plane.
    And rotating on your feet (##[0,2\pi]##) with arm angle fixed gives you the cone. (I take it the mistake Ivy refers to is that it's a cone surface, not half a cone).

    And now I will revert to "physics notation" in order not to confuse myself :wink: -- after all this is PF and not MF !
     
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