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Laplacian of f equals zero and spherical harmonics equation

  1. Dec 3, 2009 #1
    Lets consider the equation:
    [tex]\nabla^2 f=0[/tex]
    I know that in spherical coordinates this equation may be decomposed into two equations,
    first which depends only on r, and the second one which has the form of spherical harmonics equation except that the [tex]l(l+1)[/tex] is an arbitrary constant, lets say C (and of course the same constant is present in the first equation).
    I do not understand why we consider (in literature for example) only the case of
    [tex]C=l(l+1)[/tex]

    What if we have an equation of form:
    [tex]\left(\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\sin\theta\frac{\partial}{\partial\theta}+\frac{1}{\sin^2\theta}\frac{\partial^2}{\partial\phi^2}+C\right)u=0[/tex]

    is any way to decompose [tex]u[/tex] into spherical harmonics or to transform this equation into standard spherical harmonic equation?
     
  2. jcsd
  3. Dec 3, 2009 #2

    HallsofIvy

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    The reason we only consider [itex]l(l+1)[/itex] is that, when we separate variables, we get two ordinary differential equations, one in [itex]\theta[/itex], the other in [itex]\phi[/itex]. Basically, we get a function in [itex]\theta[/itex] only equal to a function of [itex]\phi[/itex] only. Those must each be equal to a constant, C. And, since [itex]\theta[/itex] goes all the way from 0 to [itex]2\pi[/itex] the function in [itex]\theta[/itex] must be periodic with period [itex]2\pi[/itex]. That only happens when [itex]C= l(l+1)[/itex].
     
  4. Dec 3, 2009 #3
    The differential equations are:

    [tex]\frac{1}{\Phi}\frac{d^2 \Phi}{d \phi^2}=-m^2
    [/tex]

    and

    [tex]
    (1-x^2)\frac{d^2\Theta}{dx^2}-2x\frac{d\Theta}{dx}+[l(l+1)-\frac{m^2}{1-x^2}]\Theta=0
    [/tex]

    where [tex]x=cos(\theta) [/tex].

    The solution of the first equation requires that "m" is an integer for the solution [tex] e^{im\phi}[/tex] to be periodic.

    However, I don't see why the second equation requires l to be an integer. If l is an integer then it's nice as you can find solutions that are polynomial (the equation is the associated Legendre equation). But does l have to be integral?
     
  5. Dec 3, 2009 #4
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