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From spherical to cylindrical

  1. Apr 26, 2012 #1
    1. The problem statement, all variables and given/known data
    Hi

    I have an expression on the form
    [tex]
    df(v, \theta, \phi) = v e^{-v^2/C}\cos(\theta)v^2\sin(\theta)\,dv\,d\theta\,d\phi
    [/tex]
    and I am trying to write it in cylindrical coordinates. Note that θ runs from 0..π, v is a velocity and C a real constant. So I wish to write it in terms of a radial and axial velocity, vr and vz.

    First I thought of integrating out θ and [itex]\phi[/itex]. [itex]\phi[/itex] is easily done, and it yields 2π. However, the integral over θ yields 0. Am I doing this wrong?

    The result should yield something proportional to

    [tex]
    v_re^{-v_r^2/C^2}v_ze^{-v_z^2/C^2}\,dv_r\,dv_z
    [/tex]

    Best,
    Niles.
     
  2. jcsd
  3. Apr 26, 2012 #2

    marcusl

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    In spherical coordinates the integral over theta runs 0 to pi.
     
  4. Apr 26, 2012 #3
    Hi

    Thanks for replying. The integral of cos(theta)sin(theta) over 0..pi gives 0.
     
  5. Apr 26, 2012 #4

    marcusl

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    Gold Member

    Oh, right. I looked at the sin(theta) and didn't see the other term.
     
  6. May 4, 2012 #5
    OK, I am taking a different approach now. Now I use [itex]v_r = v\cos(\theta)[/itex] and [itex]v_z = v\sin(\theta)[/itex]. This gives me
    [tex]
    v_rv_zv\,dv\,d\theta
    [/tex]
    Then I use (from http://en.wikipedia.org/wiki/Spherical_coordinate_system) that [itex]\arctan(v_r/v_z)[/itex] and [itex]v=\sqrt(v_r^2+v_z^2)[/itex] to find dv/dvr and dθ/dvz. However, don't get the desired results.

    Is my approach wrong?

    Best,
    Niles.
     
  7. May 4, 2012 #6

    sharks

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    Gold Member

    Basically, what you're trying to do is express a complete sphere with radius = v, in terms of cylindrical coordinates, with density [itex]v e^{-v^2/C}\cos(\theta)[/itex]. So, forget about the density for now, and focus on the conversion to cylindrical coordinates. Afterwards, plug in the density.
     
  8. May 4, 2012 #7
    Thanks for taking the time to help me. For the conversion I get
    [tex]
    v e^{-v^2/C}\cos(\theta)v^2\sin(\theta) = v_rv_zv e^{-\sqrt{v_r^2+v_z^2}/C}
    [/tex]
    I have used the relations I linked to above. Can I get a hint to how I should take care of the density?

    Best wishes,
    Niles.
     
  9. May 4, 2012 #8
    Ahh, I think I get it now. So what you mean is that
    [tex]
    v\,dvd\theta = dv_r\,dv_z
    [/tex]

    Then it adds up!
     
  10. May 4, 2012 #9

    HallsofIvy

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    In cylindrical coordinates r measures the distance from the origin to the foot of the projection from the point to the xy-plane. The line from the origin to the point in 3 dimensions, the line from the origin to the point at the foot of the projection and the projection form a straight line with hypotenuse of length [itex]\rho[/itex], angle [itex]\pi/2- \phi[/itex], "opposite side" of length z, and "near side" of length r. That is, the point that has spherical coordinates [itex]\rho[/itex], [itex]\theta[/itex], and [itex]\phi[/itex] has cylindrical coordinates [itex]r= \rho sin(\phi)[/itex], [itex]\theta= \theta[/itex] and [itex]z= \rho cos(\phi)[/itex].

    Here, I have used the mathematics convention that [itex]\theta[/itex] is the "longitude" and [itex]\phi[/itex] is the "co-latitude". The physics convention reverses that.
     
  11. May 4, 2012 #10

    sharks

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    Gold Member

    My understanding of the problem is in the attached graph below.
    In this problem, [itex]\phi[/itex] is substituted by [itex]\theta[/itex], [itex]\theta[/itex] is replaced by [itex]\phi[/itex], and [itex]\rho[/itex] is replaced by v.

    So, a description of the sphere would be:
    For [itex]\phi[/itex] and [itex]\theta[/itex] fixed, [itex]v[/itex] varies from 0 to v.
    For [itex]\phi[/itex] fixed, [itex]\theta[/itex] varies from 0 to ∏.
    [itex]\phi[/itex] varies from 0 to 2∏.

    The corresponding description in terms of cylindrical coordinates would be:
    For [itex]\phi[/itex] and [itex]r[/itex] fixed, [itex]z[/itex] varies from [itex]-\sqrt{v^2-v^2\sin^2\theta}[/itex] to [itex]\sqrt{v^2-v^2\sin^2\theta}[/itex], where [itex]\theta[/itex] is [itex]\arctan ({\frac {r}{z}})[/itex].
    For [itex]\phi[/itex] fixed, [itex]r[/itex] varies from 0 to v.
    [itex]\phi[/itex] varies from 0 to 2∏.

    For the integrand:
    [tex]v e^{-v^2/C}\cos(\theta)[/tex]Replace [itex]\theta[/itex] by [itex]\arctan ({\frac {r}{z}})[/itex] and [itex]v[/itex] by [itex]\sqrt{r^2+z^2}[/itex].
     

    Attached Files:

    Last edited: May 4, 2012
  12. May 4, 2012 #11
    Thanks for the help, both of you.

    Best,
    Niles.
     
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