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Dipole term in multipole expansion

  1. May 21, 2013 #1
    I'm having some difficult in understanding something about the dipole term in a multipole expansion. Griffiths writes the expansion as a sum of terms in Legendre polynomials, so the dipole term in the potential is writen

    [itex]\frac{1}{4 \pi \epsilon r^{2}}\int r^{'}cos\theta^{'}\rho dv^{'}[/itex]

    Then, by defining

    [itex]\vec{p}=\int \vec{r}^{'}\rho dv^{'}[/itex]

    he writes

    [itex]V=\frac{1}{4 \pi \epsilon}\frac{\vec{p}\cdot\hat{r}}{r^{2}}[/itex]

    I understood how thats done. My problem is: using the dipole moment vector and doing the scalar product it will usually appear the cossine of an angle in the potential, but it will never appear using the first definition , that is, calculating directly the integral. Maybe I understood something wrong but I cant figure out what. Hope someone helps me. Thanks.
  2. jcsd
  3. May 21, 2013 #2
    I'm not sure I understood the question. The cosine is already there in the original definition: it's why the dot product is used below.

    ##\vec{p}\cdot\hat{r} = \int \vec{r}^{'}\rho dv^{'} \cdot \hat{r} = \int \vec{r}^{'}\cdot \hat{r} \rho dv^{'}##

    Griffiths has defined ##\theta'## as the angle between ##\vec{r}## and ##\vec{r}'##, so ##\vec{r}^{'}\cdot \hat{r} = |\vec{r}^{'}|\hat{r}|\cos(\theta') = r'\cos(\theta')##, giving back the original expression. Are you misinterpreting ##\theta'## as the polar angle? Look again at the diagram (Fig. 3.28 in my edition).
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