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Taking legendre polynomials outside the integral in a multipole expansion

  1. Oct 11, 2009 #1
    1. The problem statement, all variables and given/known data

    A chare +Q is distributed uniformly along the z axis from z=-a to z=+a. Find the multipole expansion.

    2. Relevant equations

    eimg182.gif

    Here rho has been changed to lambda, which is just Q/2a and d^3r to dz.

    3. The attempt at a solution

    I have solved the problem correctly (confirmed since the answer is given in the book.) I could not figure out how to solve the problem. Just to see what happened, I took the legendre polynomial factor outside the integral, and I stumbled onto the correct answer. Was my attempt incorrect, and my correct solution a lucky coincidence, or was my method correct? If my method was correct, why are you allowed to take the legendre polynomial outside the integral? It seems to me that theta must certainly depend on z (I was convinced of this based on fig. 3.28 in Griffiths). It can also be seen from:

    eimg178.gif

    where r corresponds to z in my problem. Certainly in the figure above if you changed r theta would also change.
     

    Attached Files:

  2. jcsd
  3. Oct 11, 2009 #2

    D H

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    You got lucky. That equation from ScienceWorld isn't even right. There should be a [itex]P_n(\cos\,\theta)[/itex] outside the integral and a [itex]P_n(\cos\,\theta\,')[/itex] inside the integral, where [itex]\theta\,'[/itex] is one of the terms integrated over by [itex]d^3\mathbf r[/itex].
     
  4. Oct 11, 2009 #3

    gabbagabbahey

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    No, there's nothing wrong with his equation; it results from a straight forward expansion of [itex]|\textbf{R}-\textbf{r}|^{-1}=(R^2+r^2-2rR\cos\theta)^{-1/2}[/itex] in the case [itex]R>r[/itex]

    It might help you to visualize what's going on here if you represent [itex]\rho[/itex] in terms of delta functions and explicitly reduce the integral to a single one over [itex]z[/itex]....

    Careful, since [itex]\theta[/itex] is the angle between [itex]\textbf{r}[/itex] and [itex]\textbf{R}[/itex], it will certainly change as [itex]\textbf{r}[/itex] changes in direction, but changing the length of [itex]\textbf{r}[/itex] will have no effect on the angle between it and [itex]\textbf{R}[/itex].

    When you integrate over [itex]x[/itex] and [itex]y[/itex], the direction of [itex]\textbf{r}[/itex] will change, but the charge density is zero except for [itex]x=y=0[/itex], and hence contributes nothing to the integral. The only non-zero contribution comes when [itex]\textbf{r}=z\mathbf{\hat{z}}[/itex], and this vector points in the same direction throughout the integration over [itex]z[/itex], so [itex]\theta[/itex] will be be a constant in your integral.
     
  5. Oct 11, 2009 #4

    D H

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    In general I disagree. In this case, it happens to work. That equation is from a wolfram article, which should have known better.

    Another way to look at this is that the spherical harmonics form an orthogonal set of functions over the square integrable functions. One can represent a vector v in N-space in terms of any set of orthogonal vectors ui

    [tex]\aligned
    \mathbf v &= \sum_n a_n \mathbf u_n \\
    \mathbf a_n &= \frac {\mathbf v \cdot \mathbf u_n}{||\mathbf u_n||}
    \endaligned[/tex]

    This concept carries over to function space spanned by a set of orthogonal polynomials, with the inner product defined in terms of an integral. The above finite sum becomes an infinite sum.

    The expansion of [itex]1/||\mathbf r - \mathbf r'||[/itex] in terms of the spherical harmonics leads to the multipole expansion of the electrostatic potential. A couple of articles:
    http://www.cartage.org.lb/en/themes/Sciences/Physics/Electromagnetism/Electrostatics/ElectrostaticPotential/Multipole/Multipole.htm [Broken]
    http://www.uni-regensburg.de/Fakult...Chemie/Homeier/preprint/TC-QM-97-5/node2.html

    In the special case of axial symmetry, the sectorial and tesseral harmonics terms vanish, leaving only the zonal harmonics. These can be rewritten in terms of the Legendre polynomials:

    [tex]\aligned
    \Phi(r,\theta) &= \sum_{n=0}^{\infty} \frac{a_n P_n(\cos\,\theta)}{r^{n+1}} \\
    a_n &= \int d\mathbf r^{\prime}\, \rho(\mathbf r^{\prime}) r^{\prime n} P_n(\cos\,\theta')
    \endaligned[/tex]

    In the very special case of a charge restricted to the z axis, those [itex]P_n(\cos\,\theta')[/itex] terms become 1 or -1; they nicely just drop out.

    This is what I meant by he got lucky.
     
    Last edited by a moderator: May 4, 2017
  6. Oct 11, 2009 #5

    gabbagabbahey

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    I maintain my position that there is still nothing wrong with his equation

    [tex]V(R)=\sum_{n=0}^{\infty} \frac{1}{R^{n+1}}\int r^n P_n(\cos\theta)\rho(\textbf{r})d^3\textbf{r}[/tex]

    even in the general case, so long as [itex]R>r[/itex].

    In the above equation, [itex]\theta[/itex] is defined as the angle between the vectors [itex]\textbf{R}[/itex] (the position of the field point relative to the origin) and [itex]\textbf{r}[/itex] ((the position of each source point [itex]dq=\rho(\textbf{r})d^3\textbf{r}[/itex] relative to the origin), not as the polar angle of [itex]\textbf{R}[/itex] or [itex]\textbf{r}[/itex] in general. This definition should be clear from the OP's diagram.
     
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