Dipole term in multipole expansion

  • Thread starter assed
  • Start date
  • #1
27
1
Hi.
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.
 

Answers and Replies

  • #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).
 

Related Threads on Dipole term in multipole expansion

  • Last Post
Replies
2
Views
2K
Replies
2
Views
1K
  • Last Post
Replies
1
Views
4K
Replies
2
Views
5K
Replies
2
Views
767
  • Last Post
Replies
6
Views
5K
Replies
0
Views
3K
Replies
1
Views
3K
Replies
0
Views
2K
  • Last Post
Replies
6
Views
4K
Top