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I'm studying on the Griffiths' book Introduction to Electrodynamics and a doubt came to me reading about the multipole expansion. In chapter 2.3.4 this formula is shown $$ V(\vec{r}) = \frac{1}{4\pi\epsilon_0}\int\frac{\rho(\vec{r}')}{r}d\tau' $$ This is also given as the "solution" to Poisson equation.

But in chapter 3.4, when Griffiths talks about the multipole expansion via approximation of the term $1/r$, it is said that the exact formula is $$ V(\vec{r})= \frac{1}{4\pi\epsilon_0}\sum_{n=0}^\infty\frac{1}{r^{n+1}}\int(r')^nP_n(\cos\theta')\rho(\vec{r}')d\tau' $$ So my question on these two formulas: is the former more correct then the latter? What is going on? I personally prefere to "use" the latter, I feel this as more complete than the other, but i can't figured out what is missing in the first formula. I mean: why in chapter 2.3.4 there is no need to talk about the $1/r$ term? Probably something about $\int\rho d\tau$?

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  • $\begingroup$ I don't understand. The second is derived from the first, so they are the same thing. Note the the $r$ in your first equation is not the same $r$ in the second one. Griffiths uses the script r in the first equation, which is $r-r'$ $\endgroup$ – Aaron Stevens Jan 15 at 15:33
  • $\begingroup$ @AaronStevens, I thought that the first equation is the first term in the summation of the second $\endgroup$ – Matteo Brini Jan 15 at 15:36
  • $\begingroup$ @AaronStevens, so it is all about the r-writing style notation that makes the difference in the meaning and so the equivalence between the two equations? $\endgroup$ – Matteo Brini Jan 15 at 15:37
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    $\begingroup$ In your first equation that should be $r-r'$ in the denominator of the integrand, not $r$ $\endgroup$ – Aaron Stevens Jan 15 at 15:39
  • $\begingroup$ @AaronStevens, ok that's clear, my fault. Thanks a lot. Should i close or delete the question? $\endgroup$ – Matteo Brini Jan 15 at 15:40
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Thanks to the comments below; my doubt was due to a misunderstanding of notation that Griffiths uses in the book. The right way to look at the first formula is $$ V(\vec{r}) = \frac{1}{4\pi\epsilon_0}\int\frac{\rho(\vec{r}')}{|r-r'|}d\tau' $$ From here it is possible to derive the multipole expansion.

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