Editing Multipole expansion (section)

===Expansion in Cartesian coordinates===
Assume {{math|1=''v''('''r''') = ''v''(−'''r''')}} for convenience. The [[Taylor series|Taylor expansion]] of {{math|1=''v''('''r''' − '''R''')}} around the origin {{math|1='''r''' = '''0'''}} can be written as
<math display="block">\begin{align}
v(\mathbf{r}- \mathbf{R}) &= v(\mathbf{-R}) + \sum_{\alpha=x,y,z} r_\alpha v_\alpha(\mathbf{-R}) +\frac{1}{2} \sum_{\alpha=x,y,z}\sum_{\beta=x,y,z} r_\alpha r_\beta v_{\alpha\beta}(\mathbf{-R})
+ \cdots + \cdots\\
&=v(\mathbf{R}) - \sum_{\alpha=x,y,z} r_\alpha v_\alpha(\mathbf{R}) +\frac{1}{2} \sum_{\alpha=x,y,z}\sum_{\beta=x,y,z} r_\alpha r_\beta v_{\alpha\beta}(\mathbf{R})
- \cdots + \cdots
\end{align}</math>
with Taylor coefficients
<math display="block">v_\alpha(\mathbf{R}) \equiv\left( \frac{\partial v(\mathbf{r}-\mathbf{R}) }{\partial r_\alpha}\right)_{\mathbf{r} = \mathbf 0} \quad\text{and} \quad
v_{\alpha\beta}(\mathbf{R}) \equiv\left( \frac{\partial^2 v(\mathbf{r}-\mathbf{R}) }{\partial r_{\alpha}\partial r_{\beta}}\right)_{\mathbf{r}= \mathbf0} .</math>
If {{math|''v''('''r''' − '''R''')}} satisfies the [[Laplace equation]], then by the above expansion we have
<math display="block">\left(\nabla^2 v(\mathbf{r}- \mathbf{R})\right)_{\mathbf{r}=\mathbf0} = \sum_{\alpha=x,y,z} v_{\alpha\alpha}(\mathbf{R}) = 0,</math>
and the expansion can be rewritten in terms of the components of a traceless Cartesian second rank [[tensor]]:
<math display="block">\sum_{\alpha=x,y,z}\sum_{\beta=x,y,z} r_\alpha r_\beta v_{\alpha\beta}(\mathbf{R})
= \frac{1}{3} \sum_{\alpha=x,y,z}\sum_{\beta=x,y,z} \left(3r_\alpha r_\beta - \delta_{\alpha\beta} r^2\right) v_{\alpha\beta}(\mathbf{R}) ,</math>
where {{math|''δ''<sub>''αβ''</sub>}} is the [[Kronecker delta]] and {{math|''r''<sup>2</sup> ≡ {{abs|'''r'''}}<sup>2</sup>}}. Removing the trace is common, because it takes the rotationally invariant {{math|''r''<sup>2</sup>}} out of the second rank tensor.

====Example====

Consider now the following form of {{math|''v''('''r''' − '''R''')}}:
<math display="block">v(\mathbf{r}- \mathbf{R}) \equiv \frac{1}{|\mathbf{r}- \mathbf{R}|} .</math>
Then by direct [[differentiation (mathematics)|differentiation]] it follows that
<math display="block">v(\mathbf{R}) = \frac{1}{R},\quad v_\alpha(\mathbf{R})= -\frac{R_\alpha}{R^3},\quad \hbox{and}\quad v_{\alpha\beta}(\mathbf{R}) = \frac{3R_\alpha R_\beta- \delta_{\alpha\beta}R^2}{R^5} .</math>
Define a monopole, dipole, and (traceless) quadrupole by, respectively,
<math display="block">q_\mathrm{tot} \equiv \sum_{i=1}^N q_i , \quad P_\alpha \equiv\sum_{i=1}^N q_i r_{i\alpha} , \quad \text{and}\quad Q_{\alpha\beta} \equiv \sum_{i=1}^N q_i (3r_{i\alpha} r_{i\beta} - \delta_{\alpha\beta} r_i^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:<ref name="Jackson75">{{cite book| last1=Jackson|first1=John David| title=Classical electrodynamics|date=1975| publisher=Wiley|location=New York| isbn=047143132X|edition=2d|url-access=registration| url=https://archive.org/details/classicalelectro00jack_0}}</ref>{{rp|pages=137–138}}
<math display="block">\begin{align}
4\pi\varepsilon_0 V(\mathbf{R}) &\equiv \sum_{i=1}^N q_i v(\mathbf{r}_i-\mathbf{R}) \\
&= \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
\end{align}</math>

This expansion of the potential of a discrete charge distribution is very similar to the one in real solid harmonics given below. The main difference is that the present one is in terms of linearly dependent quantities, for
<math display="block">\sum_{\alpha} v_{\alpha\alpha} = 0 \quad \hbox{and} \quad \sum_{\alpha} Q_{\alpha\alpha} = 0 .</math>

'''Note:'''
If the charge distribution consists of two charges of opposite sign which are an infinitesimal distance {{mvar|d}} apart, so that {{math|''d''/''R'' ≫ (''d''/''R'')<sup>2</sup>}}, it is easily shown that the dominant term in the expansion is
<math display="block">V(\mathbf{R}) = \frac{1}{4\pi \varepsilon_0 R^3} (\mathbf{P}\cdot\mathbf{R}) ,</math>
the electric [[Dipole#Field from an electric dipole|dipolar potential field]].