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As indicated in Jackson [1], the components of the electric field can be obtained directly from the multipole moments

\begin{equation}\label{eqn:dipoleFromSphericalMoments:20}

\Phi(\Bx)

= \inv{4 \pi \epsilon_0} \sum \frac{4 \pi}{ (2 l + 1) r^{l + 1} } q_{l m} Y_{l m},

\end{equation}

so for the \( l,m \) contribution to this sum the components of the electric field are

\begin{equation}\label{eqn:dipoleFromSphericalMoments:40}

E_r

=

\inv{\epsilon_0} \sum \frac{l+1}{ (2 l + 1) r^{l + 2} } q_{l m} Y_{l m},

\end{equation}

\begin{equation}\label{eqn:dipoleFromSphericalMoments:60}

E_\theta

= -\inv{\epsilon_0} \sum \frac{1}{ (2 l + 1) r^{l + 2} } q_{l m} \partial_\theta Y_{l m}

\end{equation}

\begin{equation}\label{eqn:dipoleFromSphericalMoments:80}

\begin{aligned}

E_\phi

&= -\inv{\epsilon_0} \sum \frac{1}{ (2 l + 1) r^{l + 2} \sin\theta } q_{l m} \partial_\phi Y_{l m} \\

&= -\inv{\epsilon_0} \sum \frac{j m}{ (2 l + 1) r^{l + 2} \sin\theta } q_{l m} Y_{l m}.

\end{aligned}

\end{equation}

Here I’ve translated from CGS to SI. Let’s calculate the \( l = 1 \) electric field components directly from these expressions and check against the previously calculated results.

\begin{equation}\label{eqn:dipoleFromSphericalMoments:100}

\begin{aligned}

E_r

&=

\inv{\epsilon_0} \frac{2}{ 3 r^{3} }

\lr{

2 \lr{ -\sqrt{\frac{3}{8\pi}} }^2 \textrm{Re} \lr{

(p_x – j p_y) \sin\theta e^{j\phi}

}

+

\lr{ \sqrt{\frac{3}{4\pi}} }^2 p_z \cos\theta

} \\

&=

\frac{2}{4 \pi \epsilon_0 r^3}

\lr{

p_x \sin\theta \cos\phi + p_y \sin\theta \sin\phi + p_z \cos\theta

} \\

&=

\frac{1}{4 \pi \epsilon_0 r^3} 2 \Bp \cdot \rcap.

\end{aligned}

\end{equation}

Note that

\begin{equation}\label{eqn:dipoleFromSphericalMoments:120}

\partial_\theta Y_{11} = -\sqrt{\frac{3}{8\pi}} \cos\theta e^{j \phi},

\end{equation}

and

\begin{equation}\label{eqn:dipoleFromSphericalMoments:140}

\partial_\theta Y_{1,-1} = \sqrt{\frac{3}{8\pi}} \cos\theta e^{-j \phi},

\end{equation}

so

\begin{equation}\label{eqn:dipoleFromSphericalMoments:160}

\begin{aligned}

E_\theta

&=

-\inv{\epsilon_0} \frac{1}{ 3 r^{3} }

\lr{

2 \lr{ -\sqrt{\frac{3}{8\pi}} }^2 \textrm{Re} \lr{

(p_x – j p_y) \cos\theta e^{j\phi}

}

–

\lr{ \sqrt{\frac{3}{4\pi}} }^2 p_z \sin\theta

} \\

&=

-\frac{1}{4 \pi \epsilon_0 r^3}

\lr{

p_x \cos\theta \cos\phi + p_y \cos\theta \sin\phi – p_z \sin\theta

} \\

&=

-\frac{1}{4 \pi \epsilon_0 r^3} \Bp \cdot \thetacap.

\end{aligned}

\end{equation}

For the \(\phicap\) component, the \( m = 0 \) term is killed. This leaves

\begin{equation}\label{eqn:dipoleFromSphericalMoments:180}

\begin{aligned}

E_\phi

&=

-\frac{1}{\epsilon_0} \frac{1}{ 3 r^{3} \sin\theta }

\lr{

j q_{11} Y_{11} – j q_{1,-1} Y_{1,-1}

} \\

&=

-\frac{1}{3 \epsilon_0 r^{3} \sin\theta }

\lr{

j q_{11} Y_{11} – j (-1)^{2m} q_{11}^\conj Y_{11}^\conj

} \\

&=

\frac{2}{\epsilon_0} \frac{1}{ 3 r^{3} \sin\theta }

\textrm{Im} q_{11} Y_{11} \\

&=

\frac{2}{3 \epsilon_0 r^{3} \sin\theta }

\textrm{Im} \lr{

\lr{ -\sqrt{\frac{3}{8\pi}} }^2 (p_x – j p_y) \sin\theta e^{j \phi}

} \\

&=

\frac{1}{ 4 \pi \epsilon_0 r^{3} }

\textrm{Im} \lr{

(p_x – j p_y) e^{j \phi}

} \\

&=

\frac{1}{ 4 \pi \epsilon_0 r^{3} }

\lr{

p_x \sin\phi – p_y \cos\phi

} \\

&=

-\frac{\Bp \cdot \phicap}{ 4 \pi \epsilon_0 r^3}.

\end{aligned}

\end{equation}

That is

\begin{equation}\label{eqn:dipoleFromSphericalMoments:200}

\boxed{

\begin{aligned}

E_r &=

\frac{2}{4 \pi \epsilon_0 r^3}

\Bp \cdot \rcap \\

E_\theta &= –

\frac{1}{4 \pi \epsilon_0 r^3}

\Bp \cdot \thetacap \\

E_\phi &= –

\frac{1}{4 \pi \epsilon_0 r^3}

\Bp \cdot \phicap.

\end{aligned}

}

\end{equation}

These are consistent with equations (4.12) from the text for when \( \Bp \) is aligned with the z-axis.

Observe that we can sum each of the projections of \( \BE \) to construct the total electric field due to this \( l = 1 \) term of the multipole moment sum

\begin{equation}\label{eqn:dipoleFromSphericalMoments:n}

\begin{aligned}

\BE

&=

\frac{1}{4 \pi \epsilon_0 r^3}

\lr{

2 \rcap (\Bp \cdot \rcap)

–

\phicap ( \Bp \cdot \phicap)

–

\thetacap ( \Bp \cdot \thetacap)

} \\

&=

\frac{1}{4 \pi \epsilon_0 r^3}

\lr{

3 \rcap (\Bp \cdot \rcap)

–

\Bp

},

\end{aligned}

\end{equation}

which recovers the expected dipole moment approximation.

# References

[1] JD Jackson. *Classical Electrodynamics*. John Wiley and Sons, 2nd edition, 1975.