Concept of dipole - You may ask why do we need to study about these dipoles, quadruples or etc. ?
All of the objects around us are neutral in Nature but we know that everything is made of positive and negative charges. That is it...
A system with total charge zero but having separate positive and negative charge is what we need to study. You can think of quadrupole, octopole, etc. Simplest of those system is a dipole.
But the real mathematical definition of dipole is not made from this physical fact. It is a completely mathematical abstraction derived from the so called Multipole expansion.
Note: To get a feel, analyze the following sentence,
A combination of three or more or any volume distribution of charges possess dipole moment from its definition.
All our definitions are more general. It has nothing to do with dipole or quadrupole that we used to imagine in our lower classes. This derivation is applicable for all 1/r potential.
For general discussion, let us consider the potential for an arbitrary volume charge distribution $\rho(r')$ at a point r from the origin and r' is the distance to the source from origin.
Coulomb potential $$ V(\vec{r}) = \frac{1}{4\pi \epsilon_0} \int_{V'} \frac{\rho(\vec{r'})}{|\vec{r}-\vec{r'}|} \,dV' \ldots.. eq.(1)$$ but we know that from the vector addition,$$|\vec{r}-\vec{r'}|=R=(r^2-2\vec{r}\cdotp\vec{r'}+r'^2)^{\frac{1}{2}} = r \left(1-2\frac{\hat{r} \cdotp\vec{r'}}{r}+\frac{r'^2}{r^2}\right)^{\frac{1}{2}}$$
Hence $$ \frac{1}{R} = \frac{1}{r} \left[\frac{1}{(1-2\frac{\hat{r}\cdotp\vec{r'}}{r}+\frac{r'^2}{r^2})^{\frac{1}{2}}}\right].... eq.(2)$$
From special functions, the generating function of Legendre polynomials are given by,
$$\frac{1}{(1-2xz+z^2)^\frac{1}{2}} = \sum_{n=0}^\infty P_n(x) z^n \ldots.. eq.(3) \\~\\ \\~\\ for -1\leq x \leq1 and |z|<1$$
In eq.(2) we know that $\hat{r}\cdotp\vec{r'} = r' cos\theta$ where $-1 \leq cos\theta \leq 1$ and we assumed here that source charges are near to origin and the potential is comparatively calculated very far from the origin and so, $r>>r' or \frac{r'}{r}<< 1$.
Thus comparing eq(2) and (3) we get the general expansion of 1/R in terms of legendre polynomial as, $$\frac{1}{R} = \frac{1}{r} \sum_{n=0}^\infty P_n(cos\theta) (\frac{r'}{r})^n \ldots.. eq.(4)$$
Substituting this at our primary eq(1) we get, $$ V(r) = \frac{1}{4\pi {\epsilon}_0} \int_{V'} \frac{\rho(\vec{r'})}{r} \sum_{n=0}^\infty P_n(cos\theta) (\frac{r'}{r})^n \,dV' \ldots... eq.(5)$$which is the general expansion for multipole expansion.
But what else this expansion tells us..?
I think it could be examined by looking at the first few terms of the expansion of eq.(5).. $$ V(\vec{r})= \frac{1}{4\pi\epsilon_0} \left[ \frac{1}{r} \int_{V'} \rho(\vec{r'}) \,dV' + \frac{1}{r^2} \int_{V'} r' cos\theta' \rho(\vec{r'}) \,dV' + \\~\\ \frac{1}{r^3} \int_{V'} r'^2 \left(\frac{3}{2}cos^2\theta'-\frac{1}{2}\right) \rho(\vec{r'}) \,dV' + \ldots.. \right] \ldots.. eq.(6)$$
The first term is just the familiar Electric monopole term from the definition. Where else the second term is the dipole and the third is the quadrupole term and so on. The higher order terms are useful for better approximations.
[We know it is a dipole and the quadrupole term because it is possible to arrive at the same resultant potential by separately taking the case of a single dipole or quadruple system].
But wait.. It is quite mysterious because in the whole derivation I haven't used any distinct physical input about dipoles or quadrupoles or octopoles, etc. Then how it comes into play automatically? What is the physical significance and from where it arise into this picture?
I think the reason maybe as simple as follows,
1 = 1 + [1 - 1] + [2 - 2] + [4 - 4] + ...
The reason for the non existance of tripole term in the expansion is due to its lack of physical symmetry in its distribution [because dipole moments follows vector addition and it plays the role as the integrand].
It looks there is a much more mathematical connection than anticipated with the physical structure of these multi poles in Nature.
All of the objects around us are neutral in Nature but we know that everything is made of positive and negative charges. That is it...
A system with total charge zero but having separate positive and negative charge is what we need to study. You can think of quadrupole, octopole, etc. Simplest of those system is a dipole.
But the real mathematical definition of dipole is not made from this physical fact. It is a completely mathematical abstraction derived from the so called Multipole expansion.
Note: To get a feel, analyze the following sentence,
A combination of three or more or any volume distribution of charges possess dipole moment from its definition.
All our definitions are more general. It has nothing to do with dipole or quadrupole that we used to imagine in our lower classes. This derivation is applicable for all 1/r potential.
For general discussion, let us consider the potential for an arbitrary volume charge distribution $\rho(r')$ at a point r from the origin and r' is the distance to the source from origin.
Coulomb potential $$ V(\vec{r}) = \frac{1}{4\pi \epsilon_0} \int_{V'} \frac{\rho(\vec{r'})}{|\vec{r}-\vec{r'}|} \,dV' \ldots.. eq.(1)$$ but we know that from the vector addition,$$|\vec{r}-\vec{r'}|=R=(r^2-2\vec{r}\cdotp\vec{r'}+r'^2)^{\frac{1}{2}} = r \left(1-2\frac{\hat{r} \cdotp\vec{r'}}{r}+\frac{r'^2}{r^2}\right)^{\frac{1}{2}}$$
Hence $$ \frac{1}{R} = \frac{1}{r} \left[\frac{1}{(1-2\frac{\hat{r}\cdotp\vec{r'}}{r}+\frac{r'^2}{r^2})^{\frac{1}{2}}}\right].... eq.(2)$$
From special functions, the generating function of Legendre polynomials are given by,
$$\frac{1}{(1-2xz+z^2)^\frac{1}{2}} = \sum_{n=0}^\infty P_n(x) z^n \ldots.. eq.(3) \\~\\ \\~\\ for -1\leq x \leq1 and |z|<1$$
In eq.(2) we know that $\hat{r}\cdotp\vec{r'} = r' cos\theta$ where $-1 \leq cos\theta \leq 1$ and we assumed here that source charges are near to origin and the potential is comparatively calculated very far from the origin and so, $r>>r' or \frac{r'}{r}<< 1$.
Thus comparing eq(2) and (3) we get the general expansion of 1/R in terms of legendre polynomial as, $$\frac{1}{R} = \frac{1}{r} \sum_{n=0}^\infty P_n(cos\theta) (\frac{r'}{r})^n \ldots.. eq.(4)$$
Substituting this at our primary eq(1) we get, $$ V(r) = \frac{1}{4\pi {\epsilon}_0} \int_{V'} \frac{\rho(\vec{r'})}{r} \sum_{n=0}^\infty P_n(cos\theta) (\frac{r'}{r})^n \,dV' \ldots... eq.(5)$$which is the general expansion for multipole expansion.
But what else this expansion tells us..?
I think it could be examined by looking at the first few terms of the expansion of eq.(5).. $$ V(\vec{r})= \frac{1}{4\pi\epsilon_0} \left[ \frac{1}{r} \int_{V'} \rho(\vec{r'}) \,dV' + \frac{1}{r^2} \int_{V'} r' cos\theta' \rho(\vec{r'}) \,dV' + \\~\\ \frac{1}{r^3} \int_{V'} r'^2 \left(\frac{3}{2}cos^2\theta'-\frac{1}{2}\right) \rho(\vec{r'}) \,dV' + \ldots.. \right] \ldots.. eq.(6)$$
The first term is just the familiar Electric monopole term from the definition. Where else the second term is the dipole and the third is the quadrupole term and so on. The higher order terms are useful for better approximations.
[We know it is a dipole and the quadrupole term because it is possible to arrive at the same resultant potential by separately taking the case of a single dipole or quadruple system].
But wait.. It is quite mysterious because in the whole derivation I haven't used any distinct physical input about dipoles or quadrupoles or octopoles, etc. Then how it comes into play automatically? What is the physical significance and from where it arise into this picture?
I think the reason maybe as simple as follows,
1 = 1 + [1 - 1] + [2 - 2] + [4 - 4] + ...
The reason for the non existance of tripole term in the expansion is due to its lack of physical symmetry in its distribution [because dipole moments follows vector addition and it plays the role as the integrand].
It looks there is a much more mathematical connection than anticipated with the physical structure of these multi poles in Nature.
