Pre-Quantum Electrodynamics

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

Remember our expansion for the inverse distance (which we made good use of in the multipole expansion of the electrostatic field), equation 1or_Leg \[ \frac{1}{|{\bf r} - {\bf r}'|} = \frac{1}{r} \sum_{l=0}^{\infty} \left(\frac{r'}{r}\right)^l P_l (\hat{\bf r} \cdot \hat{\bf r}'). \] The expansion for the vector potential for a current loop carrying current \(I\) over path \({\cal P}\) can thus be written \[ {\bf A} ({\bf r}) = \frac{\mu_0 I}{4\pi} \sum_{l=0}^{\infty} \frac{1}{|{\bf r}|^{l+1}} \oint_{\cal P} d{\bf l}_s |{\bf r}_s|^l P_l (\hat{\bf r} \cdot \hat{\bf r}_s) \rho({\bf r}_s), \hspace{1cm} |{\bf r}| > |{\bf r}_s| ~\forall~ {\bf r}_s \in {\cal P}. \] Explicitly, the first few terms are (letting \(\hat{\bf r} \cdot \hat{\bf r}_s = \cos \theta\))

  • Gr (5.80)
\begin{align} {\bf A}({\bf r}) = \frac{\mu_0 I}{4\pi} \left[ \frac{1}{r} \oint_{\cal P} d{\bf l}_s + \frac{1}{r^2} \oint d{\bf l}_s r_s \cos \theta + \frac{1}{r^3} \oint d{\bf l}_s (r_s)^2 \left( \frac{3}{2} \cos^2 \theta - \frac{1}{2} \right) + ... \right] \tag{A_Leg}\label{A_Leg} \end{align}

Again, these are known as the magnetic monopole, dipole, quadrupole terms.

Note: the magnetic monopole term always vanishes. This is simply because the total vector displacement on a closed loop is zero, \(\oint d{\bf l}_s = 0\), or in other words: there are no magnetic monopoles (also from Maxwell's equation divB0 \({\boldsymbol \nabla} \cdot {\bf B} = 0\)).

The dominant term is thus the dipole,

  • Gr (5.85)

\[ {\bf A}_{di} ({\bf r}) = \frac{\mu_0 I}{4\pi} \frac{1}{r^2} \oint d{\bf l}_s (\hat{\bf r} \cdot {\bf r}_s) \tag{A_di1}\label{A_di1} \] Using the identity (see Gr Problem 1.62),

  • Gr (1.108)

\[ \oint_{\cal P} d{\bf l} ({\bf c} \cdot {\bf r}) = {\bf a} \times {\bf c}, \hspace{1cm} {\bf a} = \int_{\cal S} d{\bf a} \tag{intcr}\label{intcr} \] with \({\bf c} = \hat {\bf r}\),

Parenthesis: derivation of intcr

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

Back to our problem:

\[ \oint d{\bf l}_s (\hat{\bf r} \cdot {\bf r}_s) = -\hat{\bf r} \times \int_{\cal S} d{\bf a}_s \] and defining the

magnetic dipole moment

  • Gr (5.86)

\[ {\bf m} \equiv I \int_{\cal S} d{\bf a} = I {\bf a} \tag{magdim}\label{magdim} \]

we obtain the convenient expression for the

dipole term of the vector potential

  • Gr (5.85)

\[ {\bf A}_{di} ({\bf r}) = \frac{\mu_0}{4\pi} \frac{{\bf m} \times \hat{\bf r}}{r^2} \tag{A_di}\label{A_di} \]

Example: bookend shape

Task: find the magnetic dipole moment of a bookend shape current-carrying loop (c.f. Gr Fig. 5.52). All sides have length \(w\) and carry current \(I\).

Solution: combine two loops, use magdim \[ {\bf m} = I w^2 \hat{\bf y} + I w^2 \hat{\bf z} \]

Note: the magnetic dipole moment is independent of the choice of origin.

Note: does there exist a 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: the dipole approximation often good enough when far away from source on a scale of the source's current loops.




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Author: Jean-Sébastien Caux

Created: 2022-03-24 Thu 08:42