Mathematical Methods of Physics/The multipole expansion

Tensors are useful in all physical situations that involve complicated dependence on directions. Here, we consider one such example, the multipole expansion of the potential of a charge distribution.

Introduction

Consider an arbitrary charge distribution $\rho (\mathbf {r} ')$$\rho (\mathbf {r} ')$. We wish to find the electrostatic potential due to this charge distribution at a given point $\mathbf {r}$${\mathbf {r}}$. We assume that this point is at a large distance from the charge distribution, that is if $\mathbf {r} '$$\mathbf {r} '$ varies over the charge distribution, then $\mathbf {r} >>\mathbf {r} '$$\mathbf {r} >>\mathbf {r} '$

Now, the coulomb potential for a charge distribution is given by $V(\mathbf {r} )={\frac {1}{4\pi \epsilon _{0}}}\int _{V'}{\frac {\rho (\mathbf {r} ')}{|\mathbf {r} -\mathbf {r'} |}}dV'$$V(\mathbf {r} )={\frac {1}{4\pi \epsilon _{0}}}\int _{V'}{\frac {\rho (\mathbf {r} ')}{|\mathbf {r} -\mathbf {r'} |}}dV'$

Here, $|\mathbf {r} -\mathbf {r'} |=|r^{2}-2\mathbf {r} \cdot \mathbf {r} '+r'^{2}|^{\frac {1}{2}}=r\left|1-2{\frac {{\hat {\mathbf {r} }}\cdot \mathbf {r} '}{r}}+\left({\frac {r'}{r}}\right)^{2}\right|^{\frac {1}{2}}$$|\mathbf {r} -\mathbf {r'} |=|r^{2}-2\mathbf {r} \cdot \mathbf {r} '+r'^{2}|^{\frac {1}{2}}=r\left|1-2{\frac {{\hat {\mathbf {r} }}\cdot \mathbf {r} '}{r}}+\left({\frac {r'}{r}}\right)^{2}\right|^{\frac {1}{2}}$, where ${\hat {\mathbf {r} }}\triangleq \mathbf {r} /r$${\hat {\mathbf {r} }}\triangleq \mathbf {r} /r$

Thus, using the fact that $\mathbf {r}$${\mathbf {r}}$ is much larger than $\mathbf {r} '$$\mathbf {r} '$, we can write ${\frac {1}{|\mathbf {r} -\mathbf {r'} |}}={\frac {1}{r}}{\frac {1}{\left|1-2{\frac {{\hat {\mathbf {r} }}\cdot \mathbf {r} '}{r}}+\left({\frac {r'}{r}}\right)^{2}\right|^{\frac {1}{2}}}}$${\frac {1}{|\mathbf {r} -\mathbf {r'} |}}={\frac {1}{r}}{\frac {1}{\left|1-2{\frac {{\hat {\mathbf {r} }}\cdot \mathbf {r} '}{r}}+\left({\frac {r'}{r}}\right)^{2}\right|^{\frac {1}{2}}}}$, and using the binomial expansion,

${\frac {1}{\left|1-2{\frac {{\hat {\mathbf {r} }}\cdot \mathbf {r} '}{r}}+\left({\frac {r'}{r}}\right)^{2}\right|^{\frac {1}{2}}}}=1-{\frac {{\hat {\mathbf {r} }}\cdot \mathbf {r} '}{r}}+{\frac {1}{2r^{2}}}\left(3({\hat {\mathbf {r} }}\cdot \mathbf {r} '\right)^{2}-r'^{2})+O\left({\frac {r'}{r}}\right)^{3}$${\frac {1}{\left|1-2{\frac {{\hat {\mathbf {r} }}\cdot \mathbf {r} '}{r}}+\left({\frac {r'}{r}}\right)^{2}\right|^{\frac {1}{2}}}}=1-{\frac {{\hat {\mathbf {r} }}\cdot \mathbf {r} '}{r}}+{\frac {1}{2r^{2}}}\left(3({\hat {\mathbf {r} }}\cdot \mathbf {r} '\right)^{2}-r'^{2})+O\left({\frac {r'}{r}}\right)^{3}$ (we neglect the third and higher order terms).

The multipole expansion

Thus, the potential can be written as $V(\mathbf {r} )={\frac {1}{4\pi \epsilon _{0}r}}\int _{V'}\rho (\mathbf {r} ')\left(1-{\frac {{\hat {\mathbf {r} }}\cdot \mathbf {r} '}{r}}+{\frac {1}{2r^{2}}}\left(3({\hat {\mathbf {r} }}\cdot \mathbf {r} '\right)^{2}-r'^{2})+O\left({\frac {r'}{r}}\right)^{3}\right)dV'$$V(\mathbf {r} )={\frac {1}{4\pi \epsilon _{0}r}}\int _{V'}\rho (\mathbf {r} ')\left(1-{\frac {{\hat {\mathbf {r} }}\cdot \mathbf {r} '}{r}}+{\frac {1}{2r^{2}}}\left(3({\hat {\mathbf {r} }}\cdot \mathbf {r} '\right)^{2}-r'^{2})+O\left({\frac {r'}{r}}\right)^{3}\right)dV'$

We write this as $V(\mathbf {r} )=V_{\text{mon}}(\mathbf {r} )+V_{\text{dip}}(\mathbf {r} )+V_{\text{quad}}(\mathbf {r} )+\ldots$$V(\mathbf {r} )=V_{\text{mon}}(\mathbf {r} )+V_{\text{dip}}(\mathbf {r} )+V_{\text{quad}}(\mathbf {r} )+\ldots$, where,

$V_{\text{mon}}(\mathbf {r} )={\frac {1}{4\pi \epsilon _{0}r}}\int _{V'}\rho (\mathbf {r} ')dV'$$V_{\text{mon}}(\mathbf {r} )={\frac {1}{4\pi \epsilon _{0}r}}\int _{V'}\rho (\mathbf {r} ')dV'$

$V_{\text{dip}}(\mathbf {r} )=-{\frac {1}{4\pi \epsilon _{0}r^{2}}}\int _{V'}\rho (\mathbf {r} ')\left({\hat {\mathbf {r} }}\cdot \mathbf {r} '\right)dV'$$V_{\text{dip}}(\mathbf {r} )=-{\frac {1}{4\pi \epsilon _{0}r^{2}}}\int _{V'}\rho (\mathbf {r} ')\left({\hat {\mathbf {r} }}\cdot \mathbf {r} '\right)dV'$

$V_{\text{quad}}(\mathbf {r} )={\frac {1}{8\pi \epsilon _{0}r^{3}}}\int _{V'}\rho (\mathbf {r} ')\left(3\left({\hat {\mathbf {r} }}\cdot \mathbf {r} '\right)^{2}-r'^{2}\right)dV'$$V_{\text{quad}}(\mathbf {r} )={\frac {1}{8\pi \epsilon _{0}r^{3}}}\int _{V'}\rho (\mathbf {r} ')\left(3\left({\hat {\mathbf {r} }}\cdot \mathbf {r} '\right)^{2}-r'^{2}\right)dV'$

and so on.

Monopole

Observe that $Q=\int _{V'}\rho (\mathbf {r} ')dV'$$Q=\int _{V'}\rho (\mathbf {r} ')dV'$ is a scalar, (actually the total charge in the distribution) and is called the electric monopole. This term indicates point charge electrical potential.

Dipole

We can write $V_{\text{dip}}(\mathbf {r} )=-{\frac {\hat {\mathbf {r} }}{4\pi \epsilon _{0}r^{2}}}\cdot \int _{V'}\rho (\mathbf {r} ')\mathbf {r} 'dV'$$V_{\text{dip}}(\mathbf {r} )=-{\frac {\hat {\mathbf {r} }}{4\pi \epsilon _{0}r^{2}}}\cdot \int _{V'}\rho (\mathbf {r} ')\mathbf {r} 'dV'$

The vector $\mathbf {p} =\int _{V'}\rho (\mathbf {r} ')\mathbf {r} 'dV'$$\mathbf {p} =\int _{V'}\rho (\mathbf {r} ')\mathbf {r} 'dV'$ is called the electric dipole. And its magnitude is called the dipole moment of the charge distribution. This terms indicates the linear charge distribution geometry of a dipole electrical potential.

Quadrupole

Let ${\hat {\mathbf {r} }}$${\hat {{\mathbf {r}}}}$ and $\mathbf {r} '$$\mathbf {r} '$ be expressed in Cartesian coordinates as $(r_{1},r_{2},r_{3})$$(r_{1},r_{2},r_{3})$ and $(x_{1},x_{2},x_{3})$$(x_{1},x_{2},x_{3})$. Then, $({\hat {\mathbf {r} }}\cdot \mathbf {r} ')^{2}=(r_{i}x_{i})^{2}=r_{i}r_{j}x_{i}x_{j}$$({\hat {\mathbf {r} }}\cdot \mathbf {r} ')^{2}=(r_{i}x_{i})^{2}=r_{i}r_{j}x_{i}x_{j}$

We define a dyad to be the tensor ${\hat {\mathbf {r} }}{\hat {\mathbf {r} }}$${\hat {\mathbf {r} }}{\hat {\mathbf {r} }}$ given by $\left({\hat {\mathbf {r} }}{\hat {\mathbf {r} }}\right)_{ij}=r_{i}r_{j}$$\left({\hat {\mathbf {r} }}{\hat {\mathbf {r} }}\right)_{ij}=r_{i}r_{j}$

Define the Quadrupole tensor as $T=\int _{V'}\rho (\mathbf {r} ')\left(3(\mathbf {r} '\mathbf {r} ')-\mathbf {I} r'^{2}\right)dV'$$T=\int _{V'}\rho (\mathbf {r} ')\left(3(\mathbf {r} '\mathbf {r} ')-\mathbf {I} r'^{2}\right)dV'$

Then, we can write $V_{\text{qua}}$$V_{\text{qua}}$ as the tensor contraction $V_{\text{qua}}(\mathbf {r} )=-{\frac {{\hat {\mathbf {r} }}{\hat {\mathbf {r} }}}{4\pi \epsilon _{0}r^{3}}}::T$$V_{\text{qua}}(\mathbf {r} )=-{\frac {{\hat {\mathbf {r} }}{\hat {\mathbf {r} }}}{4\pi \epsilon _{0}r^{3}}}::T$ this term indicates the three dimensional distribution of a quadruple electrical potential.