dipole moment

Dipole field from multipole moment sum

November 12, 2016 math and physics play , , , , ,

[Click here for a PDF of this post with nicer formatting]

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.

Dipole and Quadropole electrostatic potential moments and coefficents

November 5, 2016 math and physics play , , , ,

<a href=”https://peeterjoot.com/archives/math2016//momentCoeffiecients.pdf”>[Click here for a PDF of this post with nicer formatting]</a>

In class Thursday we calculated the \( q_{1,1} \) coefficient of the electrostatic moment, as covered in [1] chapter 4. Let’s verify the rest, as well as the tensor sum formula for the quadropole moment, and the spherical harmonic sum that yields the dipole moment potential.

The quadropole term of the potential was stated to be

\begin{equation}\label{eqn:momentCoeffiecients:120}
\inv{4 \pi \epsilon_0} \frac{4 \pi}{5 r^3} \sum_{m=-2}^2 \int (r’)^2 \rho(\Bx’) Y_{lm}^\conj(\theta’, \phi’) Y_{lm}(\theta, \phi)
=
\inv{2} \sum_{ij} Q_{ij} \frac{x_i x_j}{r^5},
\end{equation}

where

\begin{equation}\label{eqn:momentCoeffiecients:140}
Q_{i,j} = \int \lr{ 3 x_i’ x_j’ – \delta_{ij} (r’)^2 } \rho(\Bx’) d^3 x’.
\end{equation}

Let’s verify this. First note that

\begin{equation}\label{eqn:momentCoeffiecients:160}
Y_{l,m} = \sqrt{\frac{2 l + 1}{4 \pi} \frac{(l-m)!}{(l+m)!}} P_l^m(\cos\theta) e^{i m \phi},
\end{equation}

and
\begin{equation}\label{eqn:momentCoeffiecients:180}
P_l^{-m}(x) =
(-1)^m \frac{(l-m)!}{(l+m)!} P_l^m(x),
\end{equation}

so
\begin{equation}\label{eqn:momentCoeffiecients:200}
\begin{aligned}
Y_{l,-m}
&= \sqrt{\frac{2 l + 1}{4 \pi} \frac{(l+m)!}{(l-m)!} }
P_l^{-m}(\cos\theta)
e^{-i m \phi} \\
&=
(-1)^m
\sqrt{\frac{2 l + 1}{4 \pi} \frac{(l-m)!}{(l+m)!} }
P_l^m(x)
e^{-i m \phi} \\
&=
(-1)^m Y_{l,m}^\conj.
\end{aligned}
\end{equation}

That means

\begin{equation}\label{eqn:momentCoeffiecients:220}
\begin{aligned}
q_{l,-m}
&=
\int (r’)^l \rho(\Bx’)
Y^\conj_{l,-m}(\theta’, \phi’)
d^3 x’ \\
&=
(-1)^m
\int (r’)^l \rho(\Bx’)
Y_{l,m}(\theta’, \phi’)
d^3 x’ \\
&=
(-1)^m q_{lm}^\conj.
\end{aligned}
\end{equation}

In particular, for \( m \ne 0 \)

\begin{equation}\label{eqn:momentCoeffiecients:320}
(r’)^l Y_{l, m}^\conj (\theta’, \phi’) r^l Y_{l, m}(\theta, \phi)
+ (r’)^l Y_{l, -m}^\conj (\theta’, \phi’) r^l Y_{l, -m}(\theta, \phi)
=
(r’)^l Y_{l, m}^\conj (\theta’, \phi’) r^l Y_{l, m}(\theta, \phi)
+ (r’)^l Y_{l, m} (\theta’, \phi’) r^l Y_{l, m}^\conj(\theta, \phi) ,
\end{equation}

or
\begin{equation}\label{eqn:momentCoeffiecients:340}
(r’)^l Y_{l, m}^\conj (\theta’, \phi’) r^l Y_{l, m}(\theta, \phi)
+ (r’)^l Y_{l, -m}^\conj (\theta’, \phi’) r^l Y_{l, -m}(\theta, \phi)
=
2 \textrm{Re} \lr{ (r’)^l Y_{l, m}^\conj (\theta’, \phi’) r^l Y_{l, m}(\theta, \phi) }.
\end{equation}

To verify the quadropole expansion formula in a compact way it is helpful to compute some intermediate results.

\begin{equation}\label{eqn:momentCoeffiecients:360}
\begin{aligned}
r Y_{1, 1}
&= -r \sqrt{\frac{3}{8 \pi}} \sin\theta e^{i\phi} \\
&= -\sqrt{\frac{3}{8 \pi}} (x + i y),
\end{aligned}
\end{equation}

\begin{equation}\label{eqn:momentCoeffiecients:380}
\begin{aligned}
r Y_{1, 0}
&= r \sqrt{\frac{3}{4 \pi}} \cos\theta \\
&= \sqrt{\frac{3}{4 \pi}} z,
\end{aligned}
\end{equation}

\begin{equation}\label{eqn:momentCoeffiecients:400}
\begin{aligned}
r^2 Y_{2, 2}
&= -r^2 \sqrt{\frac{15}{32 \pi}} \sin^2\theta e^{2 i\phi} \\
&= – \sqrt{\frac{15}{32 \pi}} (x + i y)^2,
\end{aligned}
\end{equation}

\begin{equation}\label{eqn:momentCoeffiecients:420}
\begin{aligned}
r^2 Y_{2, 1}
&= r^2 \sqrt{\frac{15}{8 \pi}} \sin\theta \cos\theta e^{i\phi} \\
&= \sqrt{\frac{15}{8 \pi}} z ( x + i y ),
\end{aligned}
\end{equation}

\begin{equation}\label{eqn:momentCoeffiecients:440}
\begin{aligned}
r^2 Y_{2, 0}
&= r^2 \sqrt{\frac{5}{16 \pi}} \lr{ 3 \cos^2\theta – 1 } \\
&= \sqrt{\frac{5}{16 \pi}} \lr{ 3 z^2 – r^2 }.
\end{aligned}
\end{equation}

Given primed coordinates and integrating the conjugate of each of these with \( \rho(\Bx’) dV’ \), we obtain the \( q_{lm} \) moment coeffients. Those are

\begin{equation}\label{eqn:momentCoeffiecients:460}
q_{11}
= -\sqrt{\frac{3}{8 \pi}} \int d^3 x’ \rho(\Bx’) (x – i y),
\end{equation}

\begin{equation}\label{eqn:momentCoeffiecients:480}
q_{1, 0}
= \sqrt{\frac{3}{4 \pi}} \int d^3 x’ \rho(\Bx’) z’,
\end{equation}

\begin{equation}\label{eqn:momentCoeffiecients:500}
q_{2, 2}
= – \sqrt{\frac{15}{32 \pi}} \int d^3 x’ \rho(\Bx’) (x’ – i y’)^2,
\end{equation}

\begin{equation}\label{eqn:momentCoeffiecients:520}
q_{2, 1}
= \sqrt{\frac{15}{8 \pi}} \int d^3 x’ \rho(\Bx’) z’ ( x’ – i y’ ),
\end{equation}

\begin{equation}\label{eqn:momentCoeffiecients:540}
q_{2, 0}
= \sqrt{\frac{5}{16 \pi}} \int d^3 x’ \rho(\Bx’) \lr{ 3 (z’)^2 – (r’)^2 }.
\end{equation}

For the potential we are interested in

\begin{equation}\label{eqn:momentCoeffiecients:560}
\begin{aligned}
2 \textrm{Re} q_{11} r^2 Y_{11}(\theta, \phi)
&= 2 \frac{3}{8 \pi} \int d^3 x’ \rho(\Bx’) \textrm{Re} \lr{ (x’ – i y’)( x + i y) } \\
&= \frac{3}{4 \pi} \int d^3 x’ \rho(\Bx’) \lr{ x x’ + y y’ },
\end{aligned}
\end{equation}

\begin{equation}\label{eqn:momentCoeffiecients:580}
q_{1, 0} r Y_{1,0}(\theta, \phi)
= \frac{3}{4 \pi} \int d^3 x’ \rho(\Bx’) z’ z,
\end{equation}

\begin{equation}\label{eqn:momentCoeffiecients:600}
\begin{aligned}
2 \textrm{Re} q_{22} r^2 Y_{22}(\theta, \phi)
&= 2 \frac{15}{32 \pi} \int d^3 x’ \rho(\Bx’) \textrm{Re} \lr{
(x’ – i y’)^2
(x + i y)^2
} \\
&= \frac{15}{16 \pi} \int d^3 x’ \rho(\Bx’) \textrm{Re} \lr{
((x’)^2 – 2 i x’ y’ -(y’)^2)
(x^2 + 2 i x y -y^2)
} \\
&= \frac{15}{16 \pi} \int d^3 x’ \rho(\Bx’) \lr{
((x’)^2 -(y’)^2) (x^2 -y^2)
+ 4 x x’ y y’
},
\end{aligned}
\end{equation}

\begin{equation}\label{eqn:momentCoeffiecients:620}
\begin{aligned}
2 \textrm{Re} q_{21} r^2 Y_{21}(\theta, \phi)
&= 2 \frac{15}{8 \pi} \int d^3 x’ \rho(\Bx’) z \textrm{Re} \lr{ ( x’ – i y’ ) (x + i y) } \\
&= \frac{15}{4 \pi} \int d^3 x’ \rho(\Bx’) z \lr{ x x’ + y y’ },
\end{aligned}
\end{equation}

and
\begin{equation}\label{eqn:momentCoeffiecients:640}
q_{2, 0} r^2 Y_{20}(\theta, \phi)
= \frac{5}{16 \pi} \int d^3 x’ \rho(\Bx’) \lr{ 3 (z’)^2 – (r’)^2 } \lr{ 3 z^2 – r^2 }.
\end{equation}

The dipole term of the potential is

\begin{equation}\label{eqn:momentCoeffiecients:660}
\begin{aligned}
\inv{ 4 \pi \epsilon_0 } \frac{4 \pi}{3 r^3}
\lr{
\frac{3}{4 \pi} \int d^3 x’ \rho(\Bx’) \lr{ x x’ + y y’ }
+
\frac{3}{4 \pi} \int d^3 x’ \rho(\Bx’) z’ z
} \\
&=
\inv{ 4 \pi \epsilon_0 r^3}
\Bx \cdot \int d^3 x’ \rho(\Bx’) \Bx’ \\
&=
\frac{\Bx \cdot \Bp}{ 4 \pi \epsilon_0 r^3},
\end{aligned}
\end{equation}

as obtained directly when a strict dipole approximation was used.

Summing all the terms for the quadrople gives

\begin{equation}\label{eqn:momentCoeffiecients:680}
\begin{aligned}
\inv{ 4 \pi \epsilon r^5 } \frac{ 4 \pi }{5}
\Bigl(
&\frac{15}{16 \pi} \int d^3 x’ \rho(\Bx’) \lr{
((x’)^2 -(y’)^2) (x^2 -y^2)
+ 4 x x’ y y’
} \\
&+
\frac{15}{4 \pi} \int d^3 x’ \rho(\Bx’) z z’ \lr{ x x’ + y y’ } \\
&+
\frac{5}{16 \pi} \int d^3 x’ \rho(\Bx’) \lr{ 3 (z’)^2 – (r’)^2 } \lr{ 3 z^2 – r^2 }
\Bigr) \\
=
\inv{ 4 \pi \epsilon r^5 }
\int d^3 x’ \rho(\Bx’)
\inv{4}
\Bigl(
&3
\lr{
((x’)^2 -(y’)^2) (x^2 -y^2)
+ 4 x x’ y y’
} \\
&+
12
z z’ \lr{ x x’ + y y’ } \\
&+
\lr{ 3 (z’)^2 – (r’)^2 } \lr{ 3 z^2 – r^2 }
\Bigr).
\end{aligned}
\end{equation}

The portion in brackets is

\begin{equation}\label{eqn:momentCoeffiecients:700}
\begin{aligned}
3
&\lr{
((x’)^2 -(y’)^2) (x^2 -y^2)
+ 4 x x’ y y’
} \\
+
12
& z z’ \lr{ x x’ + y y’ } \\
+
&\lr{ 2 (z’)^2 – (x’)^2 – (y’)^2} \lr{ 2 z^2 – x^2 -y^2 } \\
=
x^2 &\lr{
3 (x’)^2 – 3(y’)^2
–
\lr{ 2 (z’)^2 – (x’)^2 – (y’)^2}
} \\
+
y^2 &\lr{
-3 (x’)^2 + 3 (y’)^2
–
\lr{ 2 (z’)^2 – (x’)^2 – (y’)^2}
} \\
+
2 z^2 &\lr{
2 (z’)^2 – (x’)^2 – (y’)^2
} \\
+
&12{ x x’ y y’ + x x’ z z’ + y y’ z z’ } \\
=
2 x^2 &\lr{
2 (x’)^2 – (y’)^2 – (z’)^2
} \\
+
2 y^2 &\lr{
2 (y’)^2 – (x’)^2 – (z’)^2
} \\
+
2 z^2 &\lr{
2 (z’)^2 – (x’)^2 – (y’)^2
} \\
+
&12{ x x’ y y’ + x x’ z z’ + y y’ z z’ }.
\end{aligned}
\end{equation}

The quadopole sum can now be written as
\begin{equation}\label{eqn:momentCoeffiecients:720}
\inv{2}
\inv{ 4 \pi \epsilon r^5 }
\int d^3 x’ \rho(\Bx’)
\biglr{
x^2 \lr{ 3 (x’)^2 – (r’)^2 }
+y^2 \lr{ 3 (y’)^2 – (r’)^2 }
+z^2 \lr{ 3 (z’)^2 – (r’)^2 }
+
3 \lr{
x y x’ y’
+y x y’ x’
+x z x’ z’
+z x z’ x’
+y z y’ z’
+z y z’ y’
}
},
\end{equation}

which is precisely \ref{eqn:momentCoeffiecients:120}, the quadropole potential stated in the text and class notes.

<h1>References</h1>

[1] JD Jackson. <em>Classical Electrodynamics</em>. John Wiley and Sons, 2nd edition, 1975.

Updated notes for ece1229 antenna theory

March 16, 2015 ece1229 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ,

I’ve now posted a first update of my notes for the antenna theory course that I am taking this term at UofT.

Unlike most of the other classes I have taken, I am not attempting to take comprehensive notes for this class. The class is taught on slides which go by faster than I can easily take notes for (and some of which match the textbook closely). In class I have annotated my copy of textbook with little details instead. This set of notes contains musings of details that were unclear, or in some cases, details that were provided in class, but are not in the text (and too long to pencil into my book), as well as some notes Geometric Algebra formalism for Maxwell’s equations with magnetic sources (something I’ve encountered for the first time in any real detail in this class).

The notes compilation linked above includes all of the following separate notes, some of which have been posted separately on this blog: