Multipole Expansion

From Course@DGIST
Revision as of 22:47, 25 August 2019 by Kjlee (talk | contribs)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Jump to navigation Jump to search
Multipole Expansion
관련코스 전기역학
소분류 물리
선행 키워드 Dipole Moment
연관 키워드

Electric dipole moment in freshmen physics

Definition
[math]\mathbf{p} = q\mathbf{d}[/math]

when +q and -q are separated by distance d.

Coulomb potential due to charge distribution

If we think about a potential due to an arbitrary charge distribution [math]\rho(\textbf{r})[/math], it is given as below:

[math]V(\textbf{r}) = \frac{1}{4\pi\epsilon_0}\int\frac{\rho(\textbf{r}')}{R} d\tau'[/math]

where [math]R\equiv|\textbf{r}-\textbf{r}'|[/math].

Here [math]|\mathbf{r}-\mathbf{r}'|^{-1}=(r^2-2\mathbf{r}\cdot\mathbf{r}'+r'^2)^{-1}[/math] can be expanded in powers of [math]r'[/math], resulting in

[math]V(\textbf{r}) = \frac{1}{4\pi\epsilon_0}\int\left( \frac{1}{r} + \frac{\mathbf{r}\cdot\mathbf{r}'}{r^3} + \frac{1}{2}\left[ \frac{3(\mathbf{r}\cdot\mathbf{r}')^2}{r^5}-\frac{r'^2}{r^3} \right] + \cdot\cdot\cdot \right) \rho(\textbf{r}')d\tau'[/math]

After arranging so all the unprimed variables come out of the integral, we have

[math]V(\textbf{r}) = \frac{1}{4\pi\epsilon_0}\left[ \frac{1}{r}\int_V \rho(\mathbf{r}')d\tau' + \frac{\mathbf{r}}{r^3}\cdot\int_V\mathbf{r}'\rho(\mathbf{r}')d\tau' + \sum\limits_{i=1}^3\sum\limits_{j=1}^3 \frac{1}{2} \frac{x_i x_j}{r^5}\int_V (3x_i'x_j'-\delta_{ij}r'^2) \rho(\textbf{r}')d\tau' +\cdot\cdot\cdot\right][/math]

Now we can define dipole moment and quadrupole moment as below:

[math]Q = \int_V\rho(\mathbf{r}')d\tau'[/math]
[math]\mathbf{p} = \int_V\mathbf{r}'\rho(\mathbf{r}')d\tau'[/math]
[math]Q_{ij} = \int_V (3x_i'x_j'-\delta_{ij}r'^2) \rho(\textbf{r}')d\tau'[/math]

so that potential function can be rewritten as

[math]V(\textbf{r}) = \frac{1}{4\pi\epsilon_0} \left[ \frac{Q}{r} + \frac{\mathbf{p}\cdot\mathbf{r}}{r^3} + \sum\limits_{i=1}^3\sum\limits_{j=1}^3 \frac{1}{2} \frac{x_i x_j}{r^5}Q_{ij} + \cdot\cdot\cdot \right][/math]

Expansion using Legendre polynomial as basis

As we have following identity

[math]\frac{1}{R}=\frac{1}{|\mathbf{r}-\mathbf{r}'|}=\frac{1}{r} \sum\limits_{n=0}^\infty \left(\frac{r'}{r}\right)^n P_n(\cos\alpha)[/math]

the potential due to local charge density [math]\rho(\mathbf{r}')[/math] can be expressed as

[math]V(\textbf{r}) = \frac{1}{4\pi\epsilon_0} \sum\limits_{n=0}^\infty \frac{1}{r^{n+1}}\int r'^n P_n(\cos\alpha)\rho(\mathbf{r}')d\tau'[/math]

where [math]\alpha[/math] refers to the angle between [math]\mathbf{r}[/math] and [math]\mathbf{r}'[/math], and [math]P_n(\cos\alpha)[/math] is Legendre polynomial.

General Formulation

The Taylor expansion of an arbitrary function v(r-r') around the origin r = 0 is

[math] v(\mathbf{r}- \mathbf{r'}) = v(\mathbf{r'}) - \sum_{i=x,y,z} r_i v_i(\mathbf{r'}) +\frac{1}{2} \sum_{i=x,y,z}\sum_{j=x,y,z} r_i r_j v_{ij}(\mathbf{r'}) -\cdots+\cdots [/math]

with

[math] v_i(\mathbf{r'}) \equiv\left( \frac{\partial v(\mathbf{r}-\mathbf{r'}) }{\partial r_i}\right)_{\mathbf{r}= \mathbf0}\quad\hbox{and} \quad v_{ij}(\mathbf{r'}) \equiv\left( \frac{\partial^2 v(\mathbf{r}-\mathbf{r'}) }{\partial r_{i}\partial r_{j}}\right)_{\mathbf{r}= \mathbf0} . [/math]

If v(r-r') satisfies the Laplace equation

[math] \left(\nabla^2 v(\mathbf{r}- \mathbf{r'})\right)_{\mathbf{r}=\mathbf0} = \sum_{i=x,y,z} v_{ii}(\mathbf{r'}) = 0 [/math]

then the expansion can be rewritten in terms of the components of a traceless Cartesian second rank tensor:

[math] \sum_{i=x,y,z}\sum_{j=x,y,z} r_i r_j v_{ij}(\mathbf{r'}) = \frac{1}{3} \sum_{i=x,y,z}\sum_{j=x,y,z} (3r_i r_j - \delta_{ij} r^2) v_{ij}(\mathbf{r'}) , [/math]

where δij is the Kronecker delta and r2 ≡ |r|2. Removing the trace is common, because it takes the rotationally invariant r2 out of the second rank tensor.

Case of 1/R potential

Now let's assume the following form of v(r-r'):

[math] v(\mathbf{r}- \mathbf{r'}) \equiv \frac{1}{|\mathbf{r}- \mathbf{r'}|} . [/math]

Then by direct differentiation it follows that

[math] v(\mathbf{R}) = \frac{1}{R},\quad v_i(\mathbf{R})= -\frac{R_i}{R^3},\quad \hbox{and}\quad v_{ij}(\mathbf{R}) = \frac{3R_i R_j- \delta_{ij}R^2}{R^5} . [/math]

where we defined [math]\mathbf{R} \equiv \mathbf{r}- \mathbf{r'}[/math].

Define a monopole, dipole, and (traceless) quadrupole by, respectively,

[math] q_\mathrm{tot} \equiv \sum_{k=1}^N q_k , \quad P_i \equiv\sum_{k=1}^N q_k r_{ki} , \quad \hbox{and}\quad Q_{ij} \equiv \sum_{k=1}^N q_k (3r_{ki} r_{kj} - \delta_{ij} r_k^2) , [/math]

and we obtain finally the first few terms of the multipole expansion of the total potential, which is the sum of the Coulomb potentials of the separate charges:

[math] 4\pi\varepsilon_0 V(\mathbf{r'}) \equiv \sum_{i=1}^N q_i v(\mathbf{r}_i-\mathbf{r'}) [/math]
[math] = \frac{q_\mathrm{tot}}{R} + \frac{1}{R^3}\sum_{\alpha=x,y,z} P_\alpha R_\alpha + \frac{1}{2 R^5}\sum_{\alpha,\beta=x,y,z} Q_{\alpha\beta} R_\alpha R_\beta +\cdots [/math]

This expansion of the potential of a discrete charge distribution is very similar to the one in real solid harmonics given below.