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Spherical Coordinates Question

  1. Nov 25, 2012 #1
    In spherical coordinates we have three axes namely
    r, θ, ∅

    the ranges of these axes are

    0≤r≤∞
    0≤θ≤∏
    0≤∅≤2∏

    what will happen in a physical situation if we allow θ to change from zero to 2∏
     
  2. jcsd
  3. Nov 25, 2012 #2

    lewando

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    The physical situation would be unchanged. The mathematical model of the physical situation would be affected in that you no longer have a unique spherical coordinate for each physical point.
     
  4. Nov 26, 2012 #3

    HallsofIvy

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    If I understand this correctly, as [itex]\theta[/itex] changes from 0 to [itex]2\pi[/itex] you (or whatever point you are tracking) makes one complete circle around the z-axis. When you have completed the [itex]2\pi[/itex] you will be exactly where you were to begin, as welatiger says.

    (By the way, those are coordinates, not "axes". If by "axis" you mean as line on which one coordinate changes, the others being 0, then the the "[itex]\phi[/itex]" and "[itex]\theta[/itex]" axes are both the z-axis, since r= 0, while the "r" axis is the x-axis.)
     
  5. Nov 27, 2012 #4
    so guys i cannot understand what is the consequences of being θ running from 0 to 2∏.
    so if the mathematical model of the physical theory doesn't have a unique coordinate what will happen ?.

    I need a more illustrative example

    Thank you
     
  6. Nov 27, 2012 #5

    HallsofIvy

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    As [itex]\theta[/itex] goes from each point moves through a circle around the z-axis.
    I don't know what more you want.

    Unless you are using the "physics" convention rather than the "mathmatics" convention. Mathematics has [itex]\theta[/itex] measuring the angle the line from the origin to the given point's projection in the xy-plane makes with the positive x-axis while [itex]\phi[/itex] measures the angle the line from the origin to the point itself makes with the z-axis. The "physics" convetion swaps x and y. If that is what you mean then as [itex]\theta[/itex] goes from 0 to [itex]\pi[/itex], a point with the same r and [itex]\phi[/itex] is swung down through a semi-circle, with center at the origin and radius z, from (0, 0, z) to (0, 0, -z). If you then continue to increase [itex]\theta[/itex] from [itex]\pi[/itex] to [itex]2\pi[/itex] the point swings through the other haf of the same circle, from (0, 0, -z) back up to (0, 0, z). Yes, if we allowed [itex]\theta[/itex] to go from [itex]0[/itex] to [itex]2\pi[/itex] we could have the same point with two different sets of coordinates. [itex](r, \phi, \theta+ \pi)[/itex] would be the same point as [itex](r, \phi+ \pi, \theta)[/itex]. Which is why we restrict [itex]\theta[/itex] to be from 0 to [itex]\pi[/itex] rather than [itex]2\pi[/itex].
     
  7. Nov 27, 2012 #6

    LCKurtz

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    Here's an example of what can go wrong. I am going to assume ##\theta## is the same as in cylindrical coordinates and ##\phi## measures the angle from the ##z## axis. Suppose you wish to calculate the volume of a unit sphere:$$
    V =\int_0^{2\pi}\int_0^\pi\int_0^1 1\cdot \rho^2\sin\phi\, d\rho d\phi d\theta$$Here we have let ##\phi## go from ##0## to ##\pi## and ##\theta## go from ##0## to ##2\pi## to cover the whole sphere once. But we could also cover the whole sphere by letting ##\phi## go from ##0## to ##2\pi## and ##\theta## go from ##0## to ##\pi##, giving$$
    V =\int_0^{\pi}\int_0^{2\pi}\int_0^1 1\cdot \rho^2\sin\phi\, d\rho d\phi d\theta$$The first integral gives the correct volume while the second gives a volume of ##0##. Do you see where the error is? While the error can be fixed, it is best to avoid the problem in the first place and restrict ##\phi## to ##[0,\pi]##.
     
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