spherical coordinates

V0.3.5 of Geometric Algebra for Electrical Engineers (and temp hardcover price drop)

December 16, 2023 Geometric Algebra for Electrical Engineers , , , , ,

Yes, I just published an update last week, but here’s another one.

Temporary price drop on hardcover.

It’s been 4 years since I printed a copy of the book for myself to mark up and edit.  In particular, having added some vector calculus identities and their geometric algebra equivalents to chapter II, it messes up the flow a bit, and I’d like a paper copy to review to help figure out how to sequence it all better.  I may just start with div and curl (in their GA forms) before moving on to curvilinear coordinates, the vector derivative and all the integration theorems, …

While I intend to mark up my copy, I’m going to treat myself to a hardcover version this time, to see what it looks like.

However, to make things a bit cheaper on myself, I’ve reduced the price on the amazon.com marketplace for the hardcover version of the book to the absolute cheapest amazon will let me make it: $16.12 USD.  At that price, amazon will make some profit above printing costs, but I will not.  I believe this will also result in a price drop on the amazon.ca marketplace (unlike the paperback pricing, there is no explicit option to set an amazon.ca price for the hardcover version, so I think it’s just the USD price converted to CAD.)

So if you would like a hardcover print copy for yourself at bargain prices, now is your chance.  The paperback, in contrast, is $15.50 USD, so for only $0.62 USD more, you (and I) can get a hardcover version!  I’ll wait about a week before ordering my copy to make sure that it’s the newest version when I order, and will leave the hardcover price at $16.12 USD until I get my copy in the mail.  After that, the price will go back up, and I’ll make a couple bucks for any hardcover sales again after that.  Note that the PDF version is still available for free, as always.

If you ask “What about author proofs”.  Well, Kindle direct publishing (formerly Createspace) does have a mechanism for ordering author proofs, and if I lived in the USA, I’d use that.  However, for us poor second class Canadians, it costs just about as much to get an author proof with shipping, as just buying a copy.

What changed in this version

V0.3.5 (Dec 15, 2023)

  • Rewrite spherical polar section with the geometry first, not the coordinate representation, nor the CAS stuff.
  • Move the coordinates -> GA derivation to a problem.
  • New figure: sphericalPolarFig2.
  • update spherical polar figure1 with orientation of j.
  • Add to helpful formulas: Vector calculus identities.
  • Note about ambiguity of our curl notation.
  • Add some references to the d’Alembertian (wave equation) operator.
  • New section (chapter 2): Vector calculus identities.
  • Two (wedge) curl examples (vector field) to make things less abstract.
  • bivector field curl examples (problem.)
  • curl with polar form representation of gradient and field (problem.)
  • curl of 3D vector field example.

 

Angular momentum bivector in cylindrical and spherical bases.

September 15, 2022 math and physics play , , , , , , ,

[Click here for a PDF version of this post]

Motivation

In a discord thread on the bivector group (a geometric algebra group chat), MoneyKills posts about trouble he has calculating the correct expression for the angular momentum bivector or it’s dual.

This blog post is a more long winded answer than my bivector response and includes this calculation using both cylindrical and spherical coordinates.

Cylindrical coordinates.

The position vector for any point on a plane can be expressed as
\begin{equation}\label{eqn:amomentum:20}
\Br = r \rcap,
\end{equation}
where \( \rcap = \rcap(\phi) \) encodes all the angular dependence of the position vector, and \( r \) is the length along that direction to our point, as illustrated in fig. 1.

fig. 1. Cylindrical coordinates position vector.

The radial unit vector has a compact GA representation
\begin{equation}\label{eqn:amomentum:40}
\rcap = \Be_1 e^{i\phi},
\end{equation}
where \( i = \Be_1 \Be_2 \).

The velocity (or momentum) will have both \( \rcap \) and \( \phicap \) dependence. By chain rule, that velocity is
\begin{equation}\label{eqn:amomentum:60}
\Bv = \dot{r} \rcap + r \dot{\rcap},
\end{equation}
where
\begin{equation}\label{eqn:amomentum:80}
\begin{aligned}
\dot{\rcap}
&= \Be_1 i e^{i\phi} \dot{\phi} \\
&= \Be_2 e^{i\phi} \dot{\phi} \\
&= \phicap \dot{\phi}.
\end{aligned}
\end{equation}
It is left to the reader to show that the vector designated \( \phicap \), is a unit vector and perpendicular to \( \rcap \) (Hint: compute the grade-0 selection of the product of the two to show that they are perpendicular.)

We can now compute the momentum, which is
\begin{equation}\label{eqn:amomentum:100}
\Bp = m \Bv = m \lr{ \dot{r} \rcap + r \dot{\phi} \phicap },
\end{equation}
and the angular momentum bivector
\begin{equation}\label{eqn:amomentum:120}
\begin{aligned}
L
&= \Br \wedge \Bp \\
&= m \lr{ r \rcap } \wedge \lr{ \dot{r} \rcap + r \dot{\phi} \phicap } \\
&= m r^2 \dot{\phi} \rcap \phicap.
\end{aligned}
\end{equation}

This has the \( m r^2 \dot{\phi} \) magnitude that the OP was seeking.

Spherical coordinates.

In spherical coordinates, our position vector is
\begin{equation}\label{eqn:amomentum:140}
\Br = r \lr{ \Be_1 \sin\theta \cos\phi + \Be_2 \sin\theta \sin\phi + \Be_3 \cos\theta },
\end{equation}
as sketched in fig. 2.

fig. 2. Spherical coordinates.

We can factor this into a more compact representation
\begin{equation}\label{eqn:amomentum:160}
\begin{aligned}
\Br
&= r \lr{ \sin\theta \Be_1 (\cos\phi + \Be_{12} \sin\phi ) + \Be_3 \cos\theta } \\
&= r \lr{ \sin\theta \Be_1 e^{\Be_{12} \phi } + \Be_3 \cos\theta } \\
&= r \Be_3 \lr{ \cos\theta + \sin\theta \Be_3 \Be_1 e^{\Be_{12} \phi } }.
\end{aligned}
\end{equation}

It is useful to name two of the bivector terms above, first, we write \( i \) for the azimuthal plane bivector sketched in fig. 3.

Spherical coordinates, azimuthal plane.

\begin{equation}\label{eqn:amomentum:180}
i = \Be_{12},
\end{equation}
and introduce a bivector \( j \) that encodes the \( \Be_3, \rcap \) plane as sketched in fig. 4.

Spherical coordinates, “j-plane”.

\begin{equation}\label{eqn:amomentum:200}
j = \Be_{31} e^{i \phi}.
\end{equation}

Having done so, we now have a compact representation for our position vector
\begin{equation}\label{eqn:amomentum:220}
\begin{aligned}
\Br
&= r \Be_3 \lr{ \cos\theta + j \sin\theta } \\
&= r \Be_3 e^{j \theta}.
\end{aligned}
\end{equation}

This provides us with a nice compact representation of the radial unit vector
\begin{equation}\label{eqn:amomentum:240}
\rcap = \Be_3 e^{j \theta}.
\end{equation}

Just as was the case in cylindrical coordinates, our azimuthal plane unit vector is
\begin{equation}\label{eqn:amomentum:280}
\phicap = \Be_2 e^{i\phi}.
\end{equation}

Now we want to compute the velocity vector. As was the case in cylindrical coordinates, we have
\begin{equation}\label{eqn:amomentum:300}
\Bv = \dot{r} \rcap + r \dot{\rcap},
\end{equation}
but now we need the spherical representation for the \( \rcap \) derivative, which is
\begin{equation}\label{eqn:amomentum:320}
\begin{aligned}
\dot{\rcap}
&=
\PD{\theta}{\rcap} \dot{\theta} + \PD{\phi}{\rcap} \dot{\phi} \\
&=
\Be_3 e^{j\theta} j \dot{\theta} + \Be_3 \sin\theta \PD{\phi}{j} \dot{\phi} \\
&=
\rcap j \dot{\theta} + \Be_3 \sin\theta j i \dot{\phi}.
\end{aligned}
\end{equation}
We can reduce the second multivector term without too much work
\begin{equation}\label{eqn:amomentum:340}
\begin{aligned}
\Be_3 j i
&=
\Be_3 \Be_{31} e^{i\phi} i \\
&=
\Be_3 \Be_{31} i e^{i\phi} \\
&=
\Be_{33112} e^{i\phi} \\
&=
\Be_{2} e^{i\phi} \\
&= \phicap,
\end{aligned}
\end{equation}
so we have
\begin{equation}\label{eqn:amomentum:360}
\dot{\rcap}
=
\rcap j \dot{\theta} + \sin\theta \phicap \dot{\phi}.
\end{equation}

The velocity is
\begin{equation}\label{eqn:amomentum:380}
\Bv = \dot{r} \rcap + r \lr{ \rcap j \dot{\theta} + \sin\theta \phicap \dot{\phi} }.
\end{equation}

Now we can finally compute the angular momentum bivector, which is
\begin{equation}\label{eqn:amomentum:400}
\begin{aligned}
L &=
\Br \wedge \Bp \\
&=
m r \rcap \wedge \lr{ \dot{r} \rcap + r \lr{ \rcap j \dot{\theta} + \sin\theta \phicap \dot{\phi} } } \\
&=
m r^2 \rcap \wedge \lr{ \rcap j \dot{\theta} + \sin\theta \phicap \dot{\phi} } \\
&=
m r^2 \gpgradetwo{ \rcap \lr{ \rcap j \dot{\theta} + \sin\theta \phicap \dot{\phi} } },
\end{aligned}
\end{equation}
which is just
\begin{equation}\label{eqn:amomentum:420}
L =
m r^2 \lr{ j \dot{\theta} + \sin\theta \rcap \phicap \dot{\phi} }.
\end{equation}

I was slightly surprised by this result, as I naively expected the cylindrical coordinate result. We have a \( m r^2 \rcap \phicap \dot{\phi} \) term, as was the case in cylindrical coordinates, but scaled down with a \( \sin\theta \) factor. However, this result does make sense. Consider for example, some fixed circular motion with \( \theta = \mathrm{constant} \), as sketched in fig. 5.

fig. 5. Circular motion for constant theta

The radius of this circle is actually \( r \sin\theta \), so the total angular momentum for that motion is scaled down to \( m r^2 \sin\theta \dot{\phi} \), smaller than the maximum circular angular momentum of \( m r^2 \dot{\phi} \) which occurs in the \( \theta = \pi/2 \) azimuthal plane. Similarly, if we have circular motion in the “j-plane”, sketched in fig. 6.

fig. 6. Circular motion for constant phi.

where \( \phi = \mathrm{constant} \), then our angular momentum is \( L = m r^2 j \dot{\theta} \).

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.

Vector wave equation in spherical coordinates

November 10, 2016 math and physics play , , ,

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

For a vector \( \BA \) in spherical coordinates, let’s compute the Laplacian

\begin{equation}\label{eqn:vectorWaveEquationSpherical:20}
\spacegrad^2 \BA,
\end{equation}

to see the form of the wave equation. The spherical vector representation has a curvilinear basis
\begin{equation}\label{eqn:vectorWaveEquationSpherical:40}
\BA = \rcap A_r + \thetacap A_\theta + \phicap A_\phi,
\end{equation}

and the spherical Laplacian has been found to have the representation

\begin{equation}\label{eqn:vectorWaveEquationSpherical:60}
\spacegrad^2 \psi
=
\inv{r^2} \PD{r}{} \lr{ r^2 \PD{r}{ \psi} }
+ \frac{1}{r^2 \sin\theta} \PD{\theta}{} \lr{ \sin\theta \PD{\theta}{ \psi } }
+ \frac{1}{r^2 \sin^2\theta} \PDSq{\phi}{ \psi}.
\end{equation}

Evaluating the Laplacian will require the following curvilinear basis derivatives

\begin{equation}\label{eqn:vectorWaveEquationSpherical:80}
\begin{aligned}
\partial_\theta \rcap &= \thetacap \\
\partial_\theta \thetacap &= -\rcap \\
\partial_\theta \phicap &= 0 \\
\partial_\phi \rcap &= S_\theta \phicap \\
\partial_\phi \thetacap &= C_\theta \phicap \\
\partial_\phi \phicap &= -\rcap S_\theta – \thetacap C_\theta.
\end{aligned}
\end{equation}

We’ll need to evaluate a number of derivatives. Starting with the \( \rcap \) components

\begin{equation}\label{eqn:vectorWaveEquationSpherical:120}
\partial_r \lr{ r^2 \partial_r \lr{ \rcap \psi} }
=
\rcap \partial_r \lr{ r^2 \partial_r \psi }
\end{equation}

\begin{equation}\label{eqn:vectorWaveEquationSpherical:140}
\begin{aligned}
\partial_\theta \lr{ S_\theta \partial_\theta \lr{ \rcap \psi } }
&=
\partial_\theta \lr{ S_\theta (\thetacap \psi + \rcap \partial_\theta \psi ) } \\
&=
C_\theta (\thetacap \psi + \rcap \partial_\theta \psi )
+ S_\theta \partial_\theta (\thetacap \psi + \rcap \partial_\theta \psi ) \\
&=
C_\theta (\thetacap \psi + \rcap \partial_\theta \psi )
+ S_\theta \partial_\theta ((\partial_\theta \thetacap) \psi + (\partial_\theta \rcap) \partial_\theta \psi )
+ S_\theta \partial_\theta (\thetacap \partial_\theta \psi + \rcap \partial_{\theta \theta} \psi ) \\
&=
C_\theta (\thetacap \psi + \rcap \partial_\theta \psi )
+ S_\theta ((-\rcap) \psi + (\thetacap) \partial_\theta \psi )
+ S_\theta (\thetacap \partial_\theta \psi + \rcap \partial_{\theta \theta} \psi ) \\
&=
\rcap \lr{
C_\theta \partial_\theta \psi
– S_\theta \psi
+ S_\theta \partial_{\theta \theta} \psi
}
+\thetacap \lr{
C_\theta \psi
+ 2 S_\theta \partial_\theta \psi
}
\end{aligned}
\end{equation}

\begin{equation}\label{eqn:vectorWaveEquationSpherical:160}
\begin{aligned}
\partial_{\phi \phi} \lr{ \rcap \psi}
&=
\partial_\phi \lr{ (\partial_\phi \rcap) \psi + \rcap \partial_\phi \psi } \\
&=
\partial_\phi \lr{ (S_\theta \phicap) \psi + \rcap \partial_\phi \psi } \\
&=
S_\theta \partial_\phi (\phicap \psi)
+ \partial_\phi \lr{ \rcap \partial_\phi \psi } \\
&=
S_\theta (\partial_\phi \phicap) \psi
+ S_\theta \phicap \partial_\phi \psi
+ (\partial_\phi \rcap) \partial_\phi \psi
+ \rcap \partial_{\phi\phi} \psi \\
&=
S_\theta (-S_\theta \rcap – C_\theta \thetacap) \psi
+ S_\theta \phicap \partial_\phi \psi
+ (S_\theta \phicap) \partial_\phi \psi
+ \rcap \partial_{\phi\phi} \psi \\
&=
\rcap \lr{
– S_\theta^2 \psi
+ \partial_{\phi\phi} \psi
}
+
\thetacap \lr{
– S_\theta C_\theta \psi
}
+
\phicap \lr{
2 S_\theta \phicap \partial_\phi \psi
}
\end{aligned}
\end{equation}

This gives

\begin{equation}\label{eqn:vectorWaveEquationSpherical:180}
\begin{aligned}
\spacegrad^2 (\rcap A_r)
&=
\rcap \lr{
\inv{r^2}
\partial_r \lr{ r^2 \partial_r A_r }
+
\inv{r^2 S_\theta}
\lr{
C_\theta \partial_\theta A_r
– S_\theta A_r
+ S_\theta \partial_{\theta \theta} A_r
}
+ \inv{r^2 S_\theta^2}
\lr{
– S_\theta^2 A_r
+ \partial_{\phi\phi} A_r
}
} \\
&\quad +
\thetacap
\lr{
\inv{r^2 S_\theta}
\lr{
C_\theta A_r
+ 2 S_\theta \partial_\theta A_r
}
–
\inv{r^2 S_\theta}
S_\theta C_\theta A_r
} \\
&\quad +
\phicap
\lr{
\inv{r^2 S_\theta^2}
2 S_\theta \partial_\phi A_r
} \\
&=
\rcap \lr{
\spacegrad^2 A_r
-\frac{2}{r^2 } A_r
}
+
\frac{\thetacap}{r^2}
\lr{
\frac{C_\theta}{S_\theta} A_r
+ 2 \partial_\theta A_r
– C_\theta A_r
}
+
\phicap
\frac{2}{r^2 S_\theta} \partial_\phi A_r.
\end{aligned}
\end{equation}

Next, let’s compute the derivatives of the \( \thetacap \) projection.

\begin{equation}\label{eqn:vectorWaveEquationSpherical:220}
\partial_r \lr{ r^2 \partial_r \lr{ \thetacap \psi} }
=
\thetacap \partial_r \lr{ r^2 \partial_r \psi }
\end{equation}

\begin{equation}\label{eqn:vectorWaveEquationSpherical:240}
\begin{aligned}
\partial_\theta \lr{ S_\theta \partial_\theta \lr{ \thetacap \psi } }
&=
\partial_\theta \lr{ S_\theta
\lr{
(\partial_\theta \thetacap ) \psi
+\thetacap \partial_\theta \psi
}
} \\
&=
\partial_\theta
\lr{ S_\theta
\lr{
(-\rcap ) \psi
+\thetacap \partial_\theta \psi
}
} \\
&=
C_\theta \lr{
-\rcap \psi
+\thetacap \partial_\theta \psi
}
+
S_\theta
\lr{
-(\partial_\theta \rcap) \psi
-\rcap \partial_\theta \psi
+(\partial_\theta \thetacap) \partial_\theta \psi
+\thetacap \partial_{\theta \theta} \psi
} \\
&=
C_\theta \lr{
-\rcap \psi
+\thetacap \partial_\theta \psi
}
+
S_\theta
\lr{
-(\thetacap) \psi
-\rcap \partial_\theta \psi
+(-\rcap) \partial_\theta \psi
+\thetacap \partial_{\theta \theta} \psi
} \\
&=
\rcap \lr{
-C_\theta \psi
-2 S_\theta \partial_\theta \psi
}
+
\thetacap \lr{
+C_\theta \partial_\theta \psi
-S_\theta \psi
+S_\theta \partial_{\theta \theta} \psi
} \\
&=
\rcap \lr{
-C_\theta \psi
-2 S_\theta \partial_\theta \psi
}
+
\thetacap \lr{
+\partial_\theta (S_\theta \partial_\theta \psi)
-S_\theta \psi
}
\end{aligned}
\end{equation}

\begin{equation}\label{eqn:vectorWaveEquationSpherical:260}
\begin{aligned}
\partial_{\phi \phi} \lr{ \thetacap \psi}
&=
\partial_{\phi} \lr{
(\partial_\phi \thetacap) \psi
+\thetacap \partial_\phi \psi
} \\
&=
\partial_{\phi} \lr{
(C_\theta \phicap) \psi
+\thetacap \partial_\phi \psi
} \\
&=
C_\theta \partial_{\phi} (\phicap \psi)
+
\partial_{\phi} ( \thetacap \partial_\phi \psi ) \\
&=
C_\theta (\partial_\phi \phicap) \psi
+C_\theta \phicap \partial_{\phi} \psi
+ (\partial_\phi \thetacap) \partial_\phi \psi
+\thetacap \partial_{\phi\phi} \psi \\
&=
C_\theta (-\rcap S_\theta – \thetacap C_\theta) \psi
+C_\theta \phicap \partial_{\phi} \psi
+ (C_\theta \phicap) \partial_\phi \psi
+\thetacap \partial_{\phi\phi} \psi \\
&=
-\rcap C_\theta S_\theta \psi
+\thetacap \lr{
-C_\theta C_\theta \psi
+\partial_{\phi\phi} \psi
}
+2 \phicap C_\theta \partial_\phi \psi,
\end{aligned}
\end{equation}

which gives
\begin{equation}\label{eqn:vectorWaveEquationSpherical:360}
\begin{aligned}
\spacegrad^2 (\thetacap A_\theta)
&=
\rcap
\lr{
\inv{r^2 S_\theta}
\lr{
-C_\theta A_\theta
-2 S_\theta \partial_\theta A_\theta
}
–
\inv{r^2 S_\theta^2}
C_\theta S_\theta A_\theta
} \\
&\quad +
\thetacap \lr{
\inv{r^2} \partial_r \lr{ r^2 \partial_r A_\theta }
+
\inv{r^2 S_\theta}
\lr{
+\partial_\theta (S_\theta \partial_\theta A_\theta)
-S_\theta A_\theta
}
+\inv{r^2 S_\theta^2}
\lr{
-C_\theta C_\theta A_\theta
+\partial_{\phi\phi} A_\theta
}
} \\
&\quad +
\phicap \lr{
\inv{r^2 S_\theta^2}
2 C_\theta \partial_\phi A_\theta
} \\
&=
-2 \rcap
\inv{r^2 S_\theta}
\partial_\theta (S_\theta A_\theta)
+
\thetacap \lr{
\spacegrad^2 A_\theta
-\inv{r^2}
A_\theta
-\inv{r^2 S_\theta^2} C_\theta^2 A_\theta
}
+
2 \phicap \lr{
\inv{r^2 S_\theta^2}
C_\theta \partial_\phi A_\theta
}.
\end{aligned}
\end{equation}

Finally, we can compute the derivatives of the \( \phicap \) projection.

\begin{equation}\label{eqn:vectorWaveEquationSpherical:300}
\partial_r \lr{ r^2 \partial_r \lr{ \phicap \psi} }
=
\phicap \partial_r \lr{ r^2 \partial_r \psi }
\end{equation}

\begin{equation}\label{eqn:vectorWaveEquationSpherical:320}
\partial_\theta \lr{ S_\theta \partial_\theta \lr{ \phicap \psi } }
=
\phicap \partial_\theta \lr{ S_\theta \partial_\theta \psi }
\end{equation}

\begin{equation}\label{eqn:vectorWaveEquationSpherical:340}
\begin{aligned}
\partial_{\phi \phi} \lr{ \phicap \psi}
&=
\partial_{\phi} \lr{
(\partial_\phi \phicap) \psi
+\phicap \partial_\phi \psi
} \\
&=
\partial_{\phi} \lr{
(-\rcap S_\theta – \thetacap C_\theta) \psi
+\phicap \partial_\phi \psi
} \\
&=
-((\partial_\phi \rcap) S_\theta + (\partial_\phi \thetacap) C_\theta) \psi
-(\rcap S_\theta + \thetacap C_\theta) \partial_\phi \psi
+(\partial_\phi \phicap \partial_\phi \psi
+\phicap \partial_{\phi \phi} \psi \\
&=
-((S_\theta \phicap) S_\theta + (C_\theta \phicap) C_\theta) \psi
-(\rcap S_\theta + \thetacap C_\theta) \partial_\phi \psi
+(-\rcap S_\theta – \thetacap C_\theta) \partial_\phi \psi
+\phicap \partial_{\phi \phi} \psi \\
&=
– 2 \rcap S_\theta \partial_\phi \psi
– 2 \thetacap C_\theta \partial_\phi \psi
+ \phicap \lr{
\partial_{\phi \phi} \psi
-\psi
},
\end{aligned}
\end{equation}

which gives
\begin{equation}\label{eqn:vectorWaveEquationSpherical:380}
\begin{aligned}
\spacegrad^2 \lr{ \phicap A_\phi }
&=
-2 \rcap \inv{r^2 S_\theta} \partial_\phi A_\phi
-2 \thetacap \inv{r^2 S_\theta^2} C_\theta \partial_\phi A_\phi \\
&\quad +
\phicap \lr{
\inv{r^2}
\partial_r \lr{ r^2 \partial_r A_\phi }
+
\inv{r^2 S_\theta}
\partial_\theta \lr{ S_\theta \partial_\theta A_\phi }
+
\inv{r^2 S_\theta^2}
\lr{
\partial_{\phi \phi} A_\phi -A_\phi
}
} \\
&=
-2 \rcap \inv{r^2 S_\theta} \partial_\phi A_\phi
-2 \thetacap \inv{r^2 S_\theta^2} C_\theta \partial_\phi A_\phi
+
\phicap \lr{
\spacegrad^2 A_\phi – \inv{r^2} A_\phi
}.
\end{aligned}
\end{equation}

The vector Laplacian resolves into three augmented scalar wave equations, all highly coupled

\begin{equation}\label{eqn:vectorWaveEquationSpherical:420}
\boxed{
\begin{aligned}
\rcap \cdot \lr{ \spacegrad^2 \BA }
&=
\spacegrad^2 A_r
-\frac{2}{r^2 } A_r
– \frac{2}{r^2 S_\theta} \partial_\theta (S_\theta A_\theta)
– \frac{2}{r^2 S_\theta} \partial_\phi A_\phi \\
\thetacap \cdot \lr{ \spacegrad^2 \BA }
&=
\frac{1}{r^2} \frac{C_\theta}{S_\theta} A_r
+ \frac{2}{r^2} \partial_\theta A_r
– \frac{1}{r^2} C_\theta A_r
+ \spacegrad^2 A_\theta
– \inv{r^2} A_\theta
– \inv{r^2 S_\theta^2} C_\theta^2 A_\theta
-2 \inv{r^2 S_\theta^2} C_\theta \partial_\phi A_\phi \\
\phicap \cdot \lr{ \spacegrad^2 \BA }
&=
\frac{2}{r^2 S_\theta} \partial_\phi A_r
+ \frac{2}{r^2 S_\theta^2} C_\theta \partial_\phi A_\theta
+ \spacegrad^2 A_\phi – \inv{r^2} A_\phi.
\end{aligned}
}
\end{equation}

I’d guess one way to decouple these equations would be to impose a constraint that allows all the non-wave equation terms in one of the component equations to be killed, and then substitute that constraint into the remaining equations. Let’s try one such constraint

\begin{equation}\label{eqn:vectorWaveEquationSpherical:480}
A_r
=
– \inv{S_\theta} \partial_\theta (S_\theta A_\theta)
– \inv{S_\theta} \partial_\phi A_\phi.
\end{equation}

This gives

\begin{equation}\label{eqn:vectorWaveEquationSpherical:520}
\begin{aligned}
\rcap \cdot \lr{ \spacegrad^2 \BA }
&=
\spacegrad^2 A_r \\
\thetacap \cdot \lr{ \spacegrad^2 \BA }
&=
\lr{
\frac{1}{r^2} \frac{C_\theta}{S_\theta}
+ \frac{2}{r^2} \partial_\theta
– \frac{1}{r^2} C_\theta
}
\lr{
– \inv{S_\theta} \partial_\theta (S_\theta A_\theta)
– \inv{S_\theta} \partial_\phi A_\phi
} \\
&\quad+ \spacegrad^2 A_\theta
– \inv{r^2} A_\theta
– \inv{r^2 S_\theta^2} C_\theta^2 A_\theta
-\frac{2}{r^2 S_\theta^2} C_\theta \partial_\phi A_\phi \\
\phicap \cdot \lr{ \spacegrad^2 \BA }
&=
– \frac{2}{r^2 S_\theta} \partial_\phi
\lr{
\inv{S_\theta} \partial_\theta (S_\theta A_\theta)
+ \inv{S_\theta} \partial_\phi A_\phi
}
+ \frac{2}{r^2 S_\theta^2} C_\theta \partial_\phi A_\theta
+ \spacegrad^2 A_\phi – \inv{r^2} A_\phi \\
&=
-\frac{2}{r^2 S_\theta} \partial_\theta A_\theta
-\frac{2}{r^2 S_\theta^2} \partial_{\phi\phi} A_\theta
+ \spacegrad^2 A_\phi – \inv{r^2} A_\phi
\end{aligned}
\end{equation}

It looks like some additional cancellations may be had in the \( \thetacap \) projection of this constrained vector Laplacian. I’m not inclined to try to take this reduction any further without a thorough check of all the algebra (using Mathematica to do so would make sense).

I also guessing that such a solution might be how the \( \textrm{TE}^r \) and \( \textrm{TM}^r \) modes were defined, but that doesn’t appear to be the case according to [1]. There the wave equation is formulated in terms of the vector potentials (picking one to be zero and the other to be radial only). The solution obtained from such a potential wave equation then directly defines the \( \textrm{TE}^r \) and \( \textrm{TM}^r \) modes. It would be interesting to see how the modes derived in that analysis transform with application of the vector Laplacian derived above.

References

[1] Constantine A Balanis. Advanced engineering electromagnetics. Wiley New York, 1989.

Spherical gradient, divergence, curl and Laplacian

November 9, 2016 math and physics play , , , , , , , , , ,

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

Unit vectors

Two of the spherical unit vectors we can immediately write by inspection.

\begin{equation}\label{eqn:sphericalLaplacian:20}
\begin{aligned}
\rcap &= \Be_1 \sin\theta \cos\phi + \Be_2 \sin\theta \sin\phi + \Be_3 \cos\theta \\
\phicap &= -\Be_1 \sin\theta + \Be_2 \cos\phi
\end{aligned}
\end{equation}

We can compute \( \thetacap \) by utilizing the right hand triplet property

\begin{equation}\label{eqn:sphericalLaplacian:40}
\begin{aligned}
\thetacap
&=
\phicap \cross \rcap \\
&=
\begin{vmatrix}
\Be_1 & \Be_2 & \Be_3 \\
-S_\phi & C_\phi & 0 \\
S_\theta C_\phi & S_\theta S_\phi & C_\theta \\
\end{vmatrix} \\
&=
\Be_1 \lr{ C_\theta C_\phi }
+\Be_2 \lr{ C_\theta S_\phi }
+\Be_3 \lr{ -S_\theta \lr{ S_\phi^2 + C_\phi^2 } } \\
&=
\Be_1 \cos\theta \cos\phi
+\Be_2 \cos\theta \sin\phi
-\Be_3 \sin\theta.
\end{aligned}
\end{equation}

Here I’ve used \( C_\theta = \cos\theta, S_\phi = \sin\phi, \cdots \) as a convenient shorthand. Observe that with \( i = \Be_1 \Be_2 \), these unit vectors admit a small factorization that makes further manipulation easier

\begin{equation}\label{eqn:sphericalLaplacian:80}
\boxed{
\begin{aligned}
\rcap &= \Be_1 e^{i\phi} \sin\theta + \Be_3 \cos\theta \\
\thetacap &= \cos\theta \Be_1 e^{i\phi} – \sin\theta \Be_3 \\
\phicap &= \Be_2 e^{i\phi}
\end{aligned}
}
\end{equation}

It should also be the case that \( \rcap \thetacap \phicap = I \), where \( I = \Be_1 \Be_2 \Be_3 = \Be_{123}\) is the \R{3} pseudoscalar, which is straightforward to check

\begin{equation}\label{eqn:sphericalLaplacian:60}
\begin{aligned}
\rcap \thetacap \phicap
&=
\lr{ \Be_1 e^{i\phi} \sin\theta + \Be_3 \cos\theta }
\lr{ \cos\theta \Be_1 e^{i\phi} – \sin\theta \Be_3 }
\Be_2 e^{i\phi} \\
&=
\lr{ \sin\theta \cos\theta – \cos\theta \sin\theta + \Be_{31} e^{i\phi} \lr{ \cos^2\theta + \sin^2\theta } }
\Be_2 e^{i\phi} \\
&=
\Be_{31} \Be_2 e^{-i\phi} e^{i\phi} \\
&=
\Be_{123}.
\end{aligned}
\end{equation}

This property could also have been used to compute \(\thetacap\).

Gradient

To compute the gradient, note that the coordinate vectors for the spherical parameterization are
\begin{equation}\label{eqn:sphericalLaplacian:120}
\begin{aligned}
\Bx_r
&= \PD{r}{\Br} \\
&= \PD{r}{\lr{r \rcap}} \\
&= \rcap + r \PD{r}{\rcap} \\
&= \rcap,
\end{aligned}
\end{equation}
\begin{equation}\label{eqn:sphericalLaplacian:140}
\begin{aligned}
\Bx_\theta
&= \PD{\theta}{\lr{r \rcap} } \\
&= r \PD{\theta}{} \lr{ S_\theta \Be_1 e^{i\phi} + C_\theta \Be_3 } \\
&= r \PD{\theta}{} \lr{ C_\theta \Be_1 e^{i\phi} – S_\theta \Be_3 } \\
&= r \thetacap,
\end{aligned}
\end{equation}
\begin{equation}\label{eqn:sphericalLaplacian:160}
\begin{aligned}
\Bx_\phi
&= \PD{\phi}{\lr{r \rcap} } \\
&= r \PD{\phi}{} \lr{ S_\theta \Be_1 e^{i\phi} + C_\theta \Be_3 } \\
&= r S_\theta \Be_2 e^{i\phi} \\
&= r \sin\theta \phicap.
\end{aligned}
\end{equation}

Since these are all normal, the dual vectors defined by \( \Bx^j \cdot \Bx_k = \delta^j_k \), can be obtained by inspection
\begin{equation}\label{eqn:sphericalLaplacian:180}
\begin{aligned}
\Bx^r &= \rcap \\
\Bx^\theta &= \inv{r} \thetacap \\
\Bx^\phi &= \inv{r \sin\theta} \phicap.
\end{aligned}
\end{equation}

The gradient follows immediately
\begin{equation}\label{eqn:sphericalLaplacian:200}
\spacegrad =
\Bx^r \PD{r}{} +
\Bx^\theta \PD{\theta}{} +
\Bx^\phi \PD{\phicap}{},
\end{equation}

or
\begin{equation}\label{eqn:sphericalLaplacian:240}
\boxed{
\spacegrad
=
\rcap \PD{r}{} +
\frac{\thetacap}{r} \PD{\theta}{} +
\frac{\phicap}{r\sin\theta} \PD{\phicap}{}.
}
\end{equation}

More information on this general dual-vector technique of computing the gradient in curvilinear coordinate systems can be found in
[2].

Partials

To compute the divergence, curl and Laplacian, we’ll need the partials of each of the unit vectors \( \PDi{\theta}{\rcap}, \PDi{\phi}{\rcap}, \PDi{\theta}{\thetacap}, \PDi{\phi}{\thetacap}, \PDi{\phi}{\phicap} \).

The \( \thetacap \) partials are

\begin{equation}\label{eqn:sphericalLaplacian:260}
\begin{aligned}
\PD{\theta}{\thetacap}
&=
\PD{\theta}{} \lr{
C_\theta \Be_1 e^{i\phi} – S_\theta \Be_3
} \\
&=
-S_\theta \Be_1 e^{i\phi} – C_\theta \Be_3 \\
&=
-\rcap,
\end{aligned}
\end{equation}
\begin{equation}\label{eqn:sphericalLaplacian:280}
\begin{aligned}
\PD{\phi}{\thetacap}
&=
\PD{\phi}{} \lr{
C_\theta \Be_1 e^{i\phi} – S_\theta \Be_3
} \\
&=
C_\theta \Be_2 e^{i\phi} \\
&=
C_\theta \phicap.
\end{aligned}
\end{equation}

The \( \phicap \) partials are

\begin{equation}\label{eqn:sphericalLaplacian:300}
\begin{aligned}
\PD{\theta}{\phicap}
&=
\PD{\theta}{} \Be_2 e^{i\phi} \\
&=
0.
\end{aligned}
\end{equation}
\begin{equation}\label{eqn:sphericalLaplacian:320}
\begin{aligned}
\PD{\phi}{\phicap}
&=
\PD{\phi}{} \Be_2 e^{i \phi} \\
&=
-\Be_1 e^{i \phi} \\
&=
-\rcap \gpgradezero{ \rcap \Be_1 e^{i \phi} }
– \thetacap \gpgradezero{ \thetacap \Be_1 e^{i \phi} }
– \phicap \gpgradezero{ \phicap \Be_1 e^{i \phi} } \\
&=
-\rcap \gpgradezero{ \lr{
\Be_1 e^{i\phi} S_\theta + \Be_3 C_\theta
} \Be_1 e^{i \phi} }
– \thetacap \gpgradezero{ \lr{
C_\theta \Be_1 e^{i\phi} – S_\theta \Be_3
} \Be_1 e^{i \phi} } \\
&=
-\rcap \gpgradezero{ e^{-i\phi} S_\theta e^{i \phi} }
– \thetacap \gpgradezero{ C_\theta e^{-i\phi} e^{i \phi} } \\
&=
-\rcap S_\theta
– \thetacap C_\theta.
\end{aligned}
\end{equation}

The \( \rcap \) partials are were computed as a side effect of evaluating \( \Bx_\theta \), and \( \Bx_\phi \), and are

\begin{equation}\label{eqn:sphericalLaplacian:340}
\PD{\theta}{\rcap}
=
\thetacap,
\end{equation}
\begin{equation}\label{eqn:sphericalLaplacian:360}
\PD{\phi}{\rcap}
=
S_\theta \phicap.
\end{equation}

In summary
\begin{equation}\label{eqn:sphericalLaplacian:380}
\boxed{
\begin{aligned}
\partial_{\theta}{\rcap} &= \thetacap \\
\partial_{\phi}{\rcap} &= S_\theta \phicap \\
\partial_{\theta}{\thetacap} &= -\rcap \\
\partial_{\phi}{\thetacap} &= C_\theta \phicap \\
\partial_{\theta}{\phicap} &= 0 \\
\partial_{\phi}{\phicap} &= -\rcap S_\theta – \thetacap C_\theta.
\end{aligned}
}
\end{equation}

Divergence and curl.

The divergence and curl can be computed from the vector product of the spherical coordinate gradient and the spherical representation of a vector. That is

\begin{equation}\label{eqn:sphericalLaplacian:400}
\spacegrad \BA
= \spacegrad \cdot \BA + \spacegrad \wedge \BA
= \spacegrad \cdot \BA + I \spacegrad \cross \BA.
\end{equation}

That gradient vector product is

\begin{equation}\label{eqn:sphericalLaplacian:420}
\begin{aligned}
\spacegrad \BA
&=
\lr{
\rcap \partial_{r}
+ \frac{\thetacap}{r} \partial_{\theta}
+ \frac{\phicap}{rS_\theta} \partial_{\phi}
}
\lr{ \rcap A_r + \thetacap A_\theta + \phicap A_\phi} \\
&=
\rcap \partial_{r}
\lr{ \rcap A_r + \thetacap A_\theta + \phicap A_\phi} \\
&+ \frac{\thetacap}{r} \partial_{\theta}
\lr{ \rcap A_r + \thetacap A_\theta + \phicap A_\phi} \\
&+ \frac{\phicap}{rS_\theta} \partial_{\phicap}
\lr{ \rcap A_r + \thetacap A_\theta + \phicap A_\phi} \\
&=
\lr{ \partial_r A_r + \rcap \thetacap \partial_r A_\theta + \rcap \phicap \partial_r A_\phi} \\
&+ \frac{1}{r}
\lr{
\thetacap (\partial_\theta \rcap) A_r + \thetacap (\partial_\theta \thetacap) A_\theta + \thetacap (\partial_\theta \phicap) A_\phi
+\thetacap \rcap \partial_\theta A_r + \partial_\theta A_\theta + \thetacap \phicap \partial_\theta A_\phi
} \\
&+ \frac{1}{rS_\theta}
\lr{
\phicap (\partial_\phi \rcap) A_r + \phicap (\partial_\phi \thetacap) A_\theta + \phicap (\partial_\phi \phicap) A_\phi
+\phicap \rcap \partial_\phi A_r + \phicap \thetacap \partial_\phi A_\theta + \partial_\phi A_\phi
} \\
&=
\lr{ \partial_r A_r + \rcap \thetacap \partial_r A_\theta + \rcap \phicap \partial_r A_\phi} \\
&+ \frac{1}{r}
\lr{
\thetacap (\thetacap) A_r + \thetacap (-\rcap) A_\theta + \thetacap (0) A_\phi
+\thetacap \rcap \partial_\theta A_r + \partial_\theta A_\theta + \thetacap \phicap \partial_\theta A_\phi
} \\
&+ \frac{1}{r S_\theta}
\lr{
\phicap (S_\theta \phicap) A_r + \phicap (C_\theta \phicap) A_\theta – \phicap (\rcap S_\theta + \thetacap C_\theta) A_\phi
+\phicap \rcap \partial_\phi A_r + \phicap \thetacap \partial_\phi A_\theta + \partial_\phi A_\phi
}.
\end{aligned}
\end{equation}

The scalar component of this is the divergence
\begin{equation}\label{eqn:sphericalLaplacian:440}
\begin{aligned}
\spacegrad \cdot \BA
&=
\partial_r A_r
+ \frac{A_r}{r}
+ \inv{r} \partial_\theta A_\theta
+ \frac{1}{r S_\theta}
\lr{ S_\theta A_r + C_\theta A_\theta + \partial_\phi A_\phi
} \\
&=
\partial_r A_r
+ 2 \frac{A_r}{r}
+ \inv{r} \partial_\theta A_\theta
+ \frac{1}{r S_\theta}
C_\theta A_\theta
+ \frac{1}{r S_\theta} \partial_\phi A_\phi \\
&=
\partial_r A_r
+ 2 \frac{A_r}{r}
+ \inv{r} \partial_\theta A_\theta
+ \frac{1}{r S_\theta}
C_\theta A_\theta
+ \frac{1}{r S_\theta} \partial_\phi A_\phi,
\end{aligned}
\end{equation}

which can be factored as
\begin{equation}\label{eqn:sphericalLaplacian:460}
\boxed{
\spacegrad \cdot \BA
=
\inv{r^2} \partial_r (r^2 A_r)
+ \inv{r S_\theta} \partial_\theta (S_\theta A_\theta)
+ \frac{1}{r S_\theta} \partial_\phi A_\phi.
}
\end{equation}

The bivector grade of \( \spacegrad \BA \) is the bivector curl
\begin{equation}\label{eqn:sphericalLaplacian:480}
\begin{aligned}
\spacegrad \wedge \BA
&=
\lr{
\rcap \thetacap \partial_r A_\theta + \rcap \phicap \partial_r A_\phi
} \\
&\quad + \frac{1}{r}
\lr{
\thetacap (-\rcap) A_\theta
+\thetacap \rcap \partial_\theta A_r + \thetacap \phicap \partial_\theta A_\phi
} \\
&\quad +
\frac{1}{r S_\theta}
\lr{
-\phicap (\rcap S_\theta + \thetacap C_\theta) A_\phi
+\phicap \rcap \partial_\phi A_r + \phicap \thetacap \partial_\phi A_\theta
} \\
&=
\lr{
\rcap \thetacap \partial_r A_\theta – \phicap \rcap \partial_r A_\phi
} \\
&\quad + \frac{1}{r}
\lr{
\rcap \thetacap A_\theta
-\rcap \thetacap \partial_\theta A_r + \thetacap \phicap \partial_\theta A_\phi
} \\
&\quad +
\frac{1}{r S_\theta}
\lr{
-\phicap \rcap S_\theta A_\phi + \thetacap \phicap C_\theta A_\phi
+\phicap \rcap \partial_\phi A_r – \thetacap \phicap \partial_\phi A_\theta
} \\
&=
\thetacap \phicap \lr{
\inv{r S_\theta} C_\theta A_\phi
+\frac{1}{r} \partial_\theta A_\phi
-\frac{1}{r S_\theta} \partial_\phi A_\theta
} \\
&\quad +\phicap \rcap \lr{
-\partial_r A_\phi
+
\frac{1}{r S_\theta}
\lr{
-S_\theta A_\phi
+ \partial_\phi A_r
}
} \\
&\quad +\rcap \thetacap \lr{
\partial_r A_\theta
+ \frac{1}{r} A_\theta
– \inv{r} \partial_\theta A_r
} \\
&=
I
\rcap \lr{
\inv{r S_\theta} \partial_\theta (S_\theta A_\phi)
-\frac{1}{r S_\theta} \partial_\phi A_\theta
}
+ I \thetacap \lr{
\frac{1}{r S_\theta} \partial_\phi A_r
-\inv{r} \partial_r (r A_\phi)
}
+ I \phicap \lr{
\inv{r} \partial_r (r A_\theta)
– \inv{r} \partial_\theta A_r
}
\end{aligned}
\end{equation}

This gives
\begin{equation}\label{eqn:sphericalLaplacian:500}
\boxed{
\spacegrad \cross \BA
=
\rcap \lr{
\inv{r S_\theta} \partial_\theta (S_\theta A_\phi)
-\frac{1}{r S_\theta} \partial_\phi A_\theta
}
+ \thetacap \lr{
\frac{1}{r S_\theta} \partial_\phi A_r
-\inv{r} \partial_r (r A_\phi)
}
+ \phicap \lr{
\inv{r} \partial_r (r A_\theta)
– \inv{r} \partial_\theta A_r
}.
}
\end{equation}

This and the divergence result above both check against the back cover of [1].

Laplacian

Using the divergence and curl it’s possible to compute the Laplacian from those, but we saw in cylindrical coordinates that it was much harder to do it that way than to do it directly.

\begin{equation}\label{eqn:sphericalLaplacian:540}
\begin{aligned}
\spacegrad^2 \psi
&=
\lr{
\rcap \partial_{r} +
\frac{\thetacap}{r} \partial_{\theta} +
\frac{\phicap}{r S_\theta} \partial_{\phi}
}
\lr{
\rcap \partial_{r} \psi
+ \frac{\thetacap}{r} \partial_{\theta} \psi
+ \frac{\phicap}{r S_\theta} \partial_{\phi} \psi
} \\
&=
\partial_{rr} \psi
+ \rcap \thetacap \partial_r \lr{ \inv{r} \partial_\theta \psi}
+ \rcap \phicap \inv{S_\theta} \partial_r \lr{ \inv{r} \partial_\phi \psi } \\
&
\quad + \frac{\thetacap}{r} \partial_{\theta} \lr{ \rcap \partial_{r} \psi }
+ \frac{\thetacap}{r^2} \partial_{\theta} \lr{ \thetacap \partial_{\theta} \psi }
+ \frac{\thetacap}{r^2} \partial_{\theta} \lr{ \frac{\phicap}{S_\theta} \partial_{\phi} \psi } \\
&
\quad + \frac{\phicap}{r S_\theta} \partial_{\phi} \lr{ \rcap \partial_{r} \psi }
+ \frac{\phicap}{r^2 S_\theta} \partial_{\phi} \lr{ \thetacap \partial_{\theta} \psi }
+ \frac{\phicap}{r^2 S_\theta^2} \partial_{\phi} \lr{ \phicap \partial_{\phi} \psi } \\
&=
\partial_{rr} \psi
+ \rcap \thetacap \partial_r \lr{ \inv{r} \partial_\theta \psi}
+ \rcap \phicap \inv{S_\theta} \partial_r \lr{ \inv{r} \partial_\phi \psi } \\
&
\quad + \frac{\thetacap\rcap}{r} \partial_{\theta} \lr{ \partial_{r} \psi }
+ \frac{1}{r^2} \partial_{\theta \theta} \psi
+ \frac{\thetacap \phicap}{r^2} \partial_{\theta} \lr{ \frac{1}{S_\theta} \partial_{\phi} \psi } \\
&
\quad + \frac{\phicap \rcap}{r S_\theta} \partial_{\phi r} \psi
+ \frac{\phicap\thetacap}{r^2 S_\theta} \partial_{\phi\theta} \psi
+ \frac{1}{r^2 S_\theta^2} \partial_{\phi \phi} \psi \\
&
\quad + \frac{\thetacap}{r} (\partial_\theta \rcap) \partial_{r} \psi
+ \frac{\thetacap}{r^2} (\partial_\theta \thetacap) \partial_{\theta} \psi
+ \frac{\thetacap}{r^2} (\partial_\theta \phicap) \frac{\phicap}{S_\theta} \partial_{\phi} \psi \\
&
\quad + \frac{\phicap}{r S_\theta} (\partial_\phi \rcap) \partial_{r} \psi
+ \frac{\phicap}{r^2 S_\theta} (\partial_\phi \thetacap) \partial_{\theta} \psi
+ \frac{\phicap}{r^2 S_\theta^2} (\partial_\phi \phicap) \partial_{\phi} \psi \\
&=
\partial_{rr} \psi
+ \rcap \thetacap \partial_r \lr{ \inv{r} \partial_\theta \psi}
+ \rcap \phicap \inv{S_\theta} \partial_r \lr{ \inv{r} \partial_\phi \psi } \\
&
\quad + \frac{\thetacap\rcap}{r} \partial_{\theta} \lr{ \partial_{r} \psi }
+ \frac{1}{r^2} \partial_{\theta \theta} \psi
+ \frac{\thetacap \phicap}{r^2} \partial_{\theta} \lr{ \frac{1}{S_\theta} \partial_{\phi} \psi } \\
&
\quad + \frac{\phicap \rcap}{r S_\theta} \partial_{\phi r} \psi
+ \frac{\phicap\thetacap}{r^2 S_\theta} \partial_{\phi\theta} \psi
+ \frac{1}{r^2 S_\theta^2} \partial_{\phi \phi} \psi \\
&
\quad + \frac{\thetacap}{r} (\thetacap) \partial_{r} \psi
+ \frac{\thetacap}{r^2} (-\rcap) \partial_{\theta} \psi
+ \frac{\thetacap}{r^2} (0) \frac{\phicap}{S_\theta} \partial_{\phi} \psi \\
&
\quad + \frac{\phicap}{r S_\theta} (S_\theta \phicap) \partial_{r} \psi
+ \frac{\phicap}{r^2 S_\theta} (C_\theta \phicap) \partial_{\theta} \psi
+ \frac{\phicap}{r^2 S_\theta^2} (-\rcap S_\theta – \thetacap C_\theta) \partial_{\phi} \psi
\end{aligned}
\end{equation}

All the bivector factors are expected to cancel out, but this should be checked. Those with an \( \rcap \thetacap \) factor are

\begin{equation}\label{eqn:sphericalLaplacian:560}
\partial_r \lr{ \inv{r} \partial_\theta \psi}
– \frac{1}{r} \partial_{\theta r} \psi
+ \frac{1}{r^2} \partial_{\theta} \psi
=
-\inv{r^2} \partial_\theta \psi
+\inv{r} \partial_{r \theta} \psi
– \frac{1}{r} \partial_{\theta r} \psi
+ \frac{1}{r^2} \partial_{\theta} \psi
= 0,
\end{equation}

and those with a \( \thetacap \phicap \) factor are
\begin{equation}\label{eqn:sphericalLaplacian:580}
\frac{1}{r^2} \partial_{\theta} \lr{ \frac{1}{S_\theta} \partial_{\phi} \psi }
– \frac{1}{r^2 S_\theta} \partial_{\phi\theta} \psi
+ \frac{1}{r^2 S_\theta^2} C_\theta \partial_{\phi} \psi
=
– \frac{1}{r^2} \frac{C_\theta}{S_\theta^2} \partial_{\phi} \psi
+ \frac{1}{r^2 S_\theta} \partial_{\theta \phi} \psi
– \frac{1}{r^2 S_\theta} \partial_{\phi\theta} \psi
+ \frac{1}{r^2 S_\theta^2} C_\theta \partial_{\phi} \psi
= 0,
\end{equation}

and those with a \( \phicap \rcap \) factor are
\begin{equation}\label{eqn:sphericalLaplacian:600}
– \inv{S_\theta} \partial_r \lr{ \inv{r} \partial_\phi \psi }
+ \frac{1}{r S_\theta} \partial_{\phi r} \psi
– \frac{1}{r^2 S_\theta^2} S_\theta \partial_{\phi} \psi
=
\inv{S_\theta} \frac{1}{r^2} \partial_\phi \psi
– \inv{r S_\theta} \partial_{r \phi} \psi
+ \frac{1}{r S_\theta} \partial_{\phi r} \psi
– \frac{1}{r^2 S_\theta} \partial_{\phi} \psi
= 0.
\end{equation}

This leaves
\begin{equation}\label{eqn:sphericalLaplacian:620}
\spacegrad^2 \psi
=
\partial_{rr} \psi
+ \frac{2}{r} \partial_{r} \psi
+ \frac{1}{r^2} \partial_{\theta \theta} \psi
+ \frac{1}{r^2 S_\theta} C_\theta \partial_{\theta} \psi
+ \frac{1}{r^2 S_\theta^2} \partial_{\phi \phi} \psi.
\end{equation}

This factors nicely as

\begin{equation}\label{eqn:sphericalLaplacian:640}
\boxed{
\spacegrad^2 \psi
=
\inv{r^2} \PD{r}{} \lr{ r^2 \PD{r}{ \psi} }
+ \frac{1}{r^2 \sin\theta} \PD{\theta}{} \lr{ \sin\theta \PD{\theta}{ \psi } }
+ \frac{1}{r^2 \sin\theta^2} \PDSq{\phi}{ \psi}
,
}
\end{equation}

which checks against the back cover of Jackson. Here it has been demonstrated explicitly that this operator expression is valid for multivector fields \( \psi \) as well as scalar fields \( \psi \).

References

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

[2] A. Macdonald. Vector and Geometric Calculus. CreateSpace Independent Publishing Platform, 2012.