# Pre-Quantum Electrodynamics
Author: Jean-Sébastien Caux

##### Multipole Expansion of the Vector Potential
ems.ms.vp.me

Remember our expansion for the electrostatic field (\ref{Gr(3.94)}) \[ \frac{1}{|{\bf r} - {\bf r}'|} = \frac{1}{[r^2 + (r')^2 - 2r r' \cos \theta']^{1/2}} = \frac{1}{r} \sum_{l=0}^{\infty} \left(\frac{r'}{r}\right)^l P_l (\cos \theta'). \label{Gr(5.77)} \] The expansion for the vector potential for a current loop carrying current \(I\) over path \({\cal P}\) can thus be written \[ {\bf A} ({\bf r}) = \frac{\mu_0 I}{4\pi} \sum_{l=0}^{\infty} \frac{1}{r^{l+1}} \oint_{\cal P} d{\bf l}' (r')^l P_l (\cos \theta') \label{Gr(5.78)} \] or (in my notations) \[ {\bf A} ({\bf r}) = \frac{\mu_0 I}{4\pi} \sum_{l=0}^{\infty} \frac{1}{|{\bf r}|^{l+1}} \oint_{\cal P} d{\bf l}_s |{\bf r}_s|^l P_l (\hat{\bf r} \cdot \hat{\bf r}_s) \rho({\bf r}_s), \hspace{1cm} |{\bf r}| > |{\bf r}_s| ~\forall~ {\bf r}_s \in {\cal P}. \] Explicitly, the first few terms are \[ {\bf A}({\bf r}) = \frac{\mu_0 I}{4\pi} \left[ \frac{1}{r} \oint_{\cal P} d{\bf l}'

- \frac{1}{r^2} \oint d{\bf l}' r' cos θ'
- \frac{1}{r^3} \oint d{\bf l}' (r')^2 \left( \frac{3}{2} cos^2 θ' - \frac{1}{2} \right) + … \right]

\label{Gr(5.79)} \] Again, these are known as the {\bf monopole}, {\bf dipole}, {\bf quadrupole} terms.

{\bf Note:} {\bf the magnetic monopole term always vanishes.} This is simply because the total vector displacement on a closed loop is zero, \(\oint d{\bf l}' = 0\), or in other words: there are no magnetic monopoles (also from Maxwell's equation \({\boldsymbol \nabla} \cdot {\bf B} = 0\)).

The dominant term is thus the dipole, \[ {\bf A}_{di} ({\bf r}) = \frac{\mu_0 I}{4\pi} \frac{1}{r^2} \oint d{\bf l}' (\hat{\bf r} \cdot {\bf r}') \label{Gr(5.81)} \] Using equation \ref{Gr(1.108)} from Problem 1.61, \[ \oint_{\cal P} ({\bf c} \cdot {\bf r}) d{\bf l} = {\bf a} \times {\bf c}, \hspace{1cm} {\bf a} = \int_{\cal S} d{\bf a} \label{Gr(1.108)} \] with \({\bf c} = \hat {\bf r}\),

\paragraph{Parenthesis: Problem 1.61 (e)} Start from Stokes' theorem, \[ \int_{\cal S} {\boldsymbol \nabla} \times {\bf V} \cdot d{\bf a} = \oint_{\cal P} {\bf V} \cdot d{\bf l} \] Let \({\bf V} = {\bf c} T\), where \({\bf c}\) is constant. On the left-hand side: \[ LHS = \int_{\cal S} T ({\boldsymbol \nabla} \times c) \cdot d{\bf a} - \int_{\cal S} {\bf c} \times ({\boldsymbol \nabla} T) \cdot d{\bf a} \] The first term is zero since \({\bf c}\) is constant. For the second term: use vector identity nr 1, \(({\bf c} \times ({\boldsymbol \nabla} T)) \cdot d{\bf a} = {\bf c} \cdot ({\boldsymbol \nabla}T \times d{\bf a})\). The second term thus becomes \[

- ∫_{\cal S} {\bf c} × ({\boldsymbol ∇} T) ⋅ d{\bf a} = -{\bf c} ⋅ ∫_{\cal S} {\boldsymbol ∇}T × d{\bf a}

\] Treating the right-hand side of the original equation now, \[ RHS = {\bf c} \cdot \oint_{\cal P} T d{\bf l} \] so we get (since this is valid for any \({\bf c}\)) \[ \int_{\cal S} {\boldsymbol \nabla} T \times d{\bf a} = -\oint_{\cal P} T d{\bf l}. \] Now put \(T = {\bf c} \cdot {\bf r}\) in this: conclusion is \ref{Gr(1.108)}.

Back to our problem: \[ \oint d{\bf l}' (\hat{\bf r} \cdot {\bf r}') = -\hat{\bf r} \times \int_{\cal S} d{\bf a}' \label{Gr(5.82)} \] and defining the

magnetic dipole moment \[ {\bf m} \equiv I \int_{\cal S} d{\bf a} = I {\bf a} \label{Gr(5.84)} \]

we obtain the convenient expression for the

{\bf dipole term of the vector potential} \[ {\bf A}_{di} ({\bf r}) = \frac{\mu_0}{4\pi} \frac{{\bf m} \times \hat{\bf r}}{r^2} \label{Gr(5.83)} \]

\paragraph{Example 5.13:} find magnetic dipole moment of bookend shape of Gr. Fig. 5.52. All sides have length \(w\) and carry current \(I\). \paragraph{Solution:} combine two loops, use \ref{Gr(5.83)} \[ {\bf m} = I w^2 \hat{\bf y} + I w^2 \hat{\bf z} \]

{\bf Note:} {\it the magnetic dipole moment is independent of the choice of origin.}

{\bf Note:} does there exist a {\it pure magnetic dipole} ? Well, yes, but it's an infinitely small loop carrying an infinitely large current, so that the dipole term is finite.

In practice: dipole approximation often good enough when far away from source on a scale of the source's current loops.

Except where otherwise noted, all content is licensed under a Creative Commons Attribution 4.0 International License.

Author: Jean-Sébastien Caux

Created: 2022-02-09 Wed 22:40