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Spherical gradient, divergence, curl and Laplacian

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

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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.

Gradient, divergence, curl and Laplacian in cylindrical coordinates

November 6, 2016 math and physics play , , , , , , , , , , , ,

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

In class it was suggested that the identity

\begin{equation}\label{eqn:laplacianCylindrical:20}
\spacegrad^2 \BA =
\spacegrad \lr{ \spacegrad \cdot \BA }
-\spacegrad \cross \lr{ \spacegrad \cross \BA },
\end{equation}

can be used to compute the Laplacian in non-rectangular coordinates. Is that the easiest way to do this?

How about just sequential applications of the gradient on the vector? Let’s start with the vector product of the gradient and the vector. First recall that the cylindrical representation of the gradient is

\begin{equation}\label{eqn:laplacianCylindrical:80}
\spacegrad = \rhocap \partial_\rho + \frac{\phicap}{\rho} \partial_\phi + \zcap \partial_z,
\end{equation}

where
\begin{equation}\label{eqn:laplacianCylindrical:100}
\begin{aligned}
\rhocap &= \Be_1 e^{\Be_1 \Be_2 \phi} \\
\phicap &= \Be_2 e^{\Be_1 \Be_2 \phi} \\
\end{aligned}
\end{equation}

Taking \( \phi \) derivatives of \ref{eqn:laplacianCylindrical:100}, we have

\begin{equation}\label{eqn:laplacianCylindrical:120}
\begin{aligned}
\partial_\phi \rhocap &= \Be_1 \Be_1 \Be_2 e^{\Be_1 \Be_2 \phi} = \Be_2 e^{\Be_1 \Be_2 \phi} = \phicap \\
\partial_\phi \phicap &= \Be_2 \Be_1 \Be_2 e^{\Be_1 \Be_2 \phi} = -\Be_1 e^{\Be_1 \Be_2 \phi} = -\rhocap.
\end{aligned}
\end{equation}

The gradient of a vector \( \BA = \rhocap A_\rho + \phicap A_\phi + \zcap A_z \) is

\begin{equation}\label{eqn:laplacianCylindrical:60}
\begin{aligned}
\spacegrad \BA
&=
\lr{ \rhocap \partial_\rho + \frac{\phicap}{\rho} \partial_\phi + \zcap \partial_z }
\lr{ \rhocap A_\rho + \phicap A_\phi + \zcap A_z } \\
&=
\quad \rhocap \partial_\rho \lr{ \rhocap A_\rho + \phicap A_\phi + \zcap A_z } \\
&\quad + \frac{\phicap}{\rho} \partial_\phi \lr{ \rhocap A_\rho + \phicap A_\phi + \zcap A_z } \\
&\quad + \zcap \partial_z \lr{ \rhocap A_\rho + \phicap A_\phi + \zcap A_z } \\
&=
\quad \rhocap \lr{ \rhocap \partial_\rho A_\rho + \phicap \partial_\rho A_\phi + \zcap \partial_\rho A_z } \\
&\quad + \frac{\phicap}{\rho} \lr{ \partial_\phi(\rhocap A_\rho) + \partial_\phi(\phicap A_\phi) + \zcap \partial_\phi A_z } \\
&\quad + \zcap \lr{ \rhocap \partial_z A_\rho + \phicap \partial_z A_\phi + \zcap \partial_z A_z } \\
&=
\quad \partial_\rho A_\rho + \rhocap \phicap \partial_\rho A_\phi + \rhocap \zcap \partial_\rho A_z \\
&\quad +\frac{1}{\rho} \lr{ A_\rho + \phicap \rhocap \partial_\phi A_\rho – \phicap \rhocap A_\phi + \partial_\phi A_\phi + \phicap \zcap \partial_\phi A_z } \\
&\quad + \zcap \rhocap \partial_z A_\rho + \zcap \phicap \partial_z A_\phi + \partial_z A_z \\
&=
\quad \partial_\rho A_\rho + \frac{1}{\rho} \lr{ A_\rho + \partial_\phi A_\phi } + \partial_z A_z \\
&\quad +
\zcap \rhocap \lr{
\partial_z A_\rho
-\partial_\rho A_z
} \\
&\quad +
\phicap \zcap \lr{
\inv{\rho} \partial_\phi A_z
– \partial_z A_\phi
} \\
&\quad +
\rhocap \phicap \lr{
\partial_\rho A_\phi
– \inv{\rho} \lr{ \partial_\phi A_\rho – A_\phi }
},
\end{aligned}
\end{equation}

As expected, we see that the gradient splits nicely into a dot and curl

\begin{equation}\label{eqn:laplacianCylindrical:160}
\begin{aligned}
\spacegrad \BA
&= \spacegrad \cdot \BA + \spacegrad \wedge \BA \\
&= \spacegrad \cdot \BA + \rhocap \phicap \zcap (\spacegrad \cross \BA ),
\end{aligned}
\end{equation}

where the cylindrical representation of the divergence is seen to be

\begin{equation}\label{eqn:laplacianCylindrical:140}
\spacegrad \cdot \BA
=
\inv{\rho} \partial_\rho (\rho A_\rho) + \frac{1}{\rho} \partial_\phi A_\phi + \partial_z A_z,
\end{equation}

and the cylindrical representation of the curl is

\begin{equation}\label{eqn:laplacianCylindrical:180}
\spacegrad \cross \BA
=
\rhocap
\lr{
\inv{\rho} \partial_\phi A_z
– \partial_z A_\phi
}
+
\phicap
\lr{
\partial_z A_\rho
-\partial_\rho A_z
}
+
\inv{\rho} \zcap \lr{
\partial_\rho ( \rho A_\phi )
– \partial_\phi A_\rho
}.
\end{equation}

Should we want to, it is now possible to evaluate the Laplacian of \( \BA \) using
\ref{eqn:laplacianCylindrical:20}
, which will have the following components

\begin{equation}\label{eqn:laplacianCylindrical:220}
\begin{aligned}
\rhocap \cdot \lr{ \spacegrad^2 \BA }
&=
\partial_\rho
\lr{
\inv{\rho} \partial_\rho (\rho A_\rho) + \frac{1}{\rho} \partial_\phi A_\phi + \partial_z A_z
}
–
\lr{
\inv{\rho} \partial_\phi \lr{
\inv{\rho} \lr{
\partial_\rho ( \rho A_\phi ) – \partial_\phi A_\rho
}
}
– \partial_z \lr{
\partial_z A_\rho -\partial_\rho A_z
}
} \\
&=
\partial_\rho \lr{ \inv{\rho} \partial_\rho (\rho A_\rho)}
+ \partial_\rho \lr{ \frac{1}{\rho} \partial_\phi A_\phi}
+ \partial_{\rho z} A_z
– \inv{\rho^2}\partial_{\phi \rho} ( \rho A_\phi )
+ \inv{\rho^2}\partial_{\phi\phi} A_\rho
+ \partial_{zz} A_\rho
– \partial_{z\rho} A_z \\
&=
\partial_\rho \lr{ \inv{\rho} \partial_\rho (\rho A_\rho)}
+ \inv{\rho^2}\partial_{\phi\phi} A_\rho
+ \partial_{zz} A_\rho
– \frac{1}{\rho^2} \partial_\phi A_\phi
+ \frac{1}{\rho} \partial_{\rho\phi} A_\phi
– \inv{\rho^2}\partial_{\phi} A_\phi
– \inv{\rho}\partial_{\phi\rho} A_\phi \\
&=
\partial_\rho \lr{ \inv{\rho} \partial_\rho (\rho A_\rho)}
+ \inv{\rho^2}\partial_{\phi\phi} A_\rho
+ \partial_{zz} A_\rho
– \frac{2}{\rho^2} \partial_\phi A_\phi \\
&=
\inv{\rho} \partial_\rho \lr{ \rho \partial_\rho A_\rho}
+ \inv{\rho^2}\partial_{\phi\phi} A_\rho
+ \partial_{zz} A_\rho
– \frac{A_\rho}{\rho^2}
– \frac{2}{\rho^2} \partial_\phi A_\phi,
\end{aligned}
\end{equation}

\begin{equation}\label{eqn:laplacianCylindrical:240}
\begin{aligned}
\phicap \cdot \lr{ \spacegrad^2 \BA }
&=
\inv{\rho} \partial_\phi
\lr{
\inv{\rho} \partial_\rho (\rho A_\rho) + \frac{1}{\rho} \partial_\phi A_\phi + \partial_z A_z
}
–
\lr{
\lr{
\partial_z \lr{
\inv{\rho} \partial_\phi A_z – \partial_z A_\phi
}
-\partial_\rho \lr{
\inv{\rho} \lr{ \partial_\rho ( \rho A_\phi ) – \partial_\phi A_\rho}
}
}
} \\
&=
\inv{\rho^2} \partial_{\phi\rho} (\rho A_\rho)
+ \frac{1}{\rho^2} \partial_{\phi\phi} A_\phi
+ \inv{\rho}\partial_{\phi z} A_z
– \inv{\rho} \partial_{z\phi} A_z
+ \partial_{z z} A_\phi
+\partial_\rho \lr{ \inv{\rho} \partial_\rho ( \rho A_\phi ) }
– \partial_\rho \lr{ \inv{\rho} \partial_\phi A_\rho} \\
&=
\partial_\rho \lr{ \inv{\rho} \partial_\rho ( \rho A_\phi ) }
+ \frac{1}{\rho^2} \partial_{\phi\phi} A_\phi
+ \partial_{z z} A_\phi
+ \inv{\rho^2} \partial_{\phi\rho} (\rho A_\rho)
+ \inv{\rho}\partial_{\phi z} A_z
– \inv{\rho} \partial_{z\phi} A_z
– \partial_\rho \lr{ \inv{\rho} \partial_\phi A_\rho} \\
&=
\partial_\rho \lr{ \inv{\rho} \partial_\rho ( \rho A_\phi ) }
+ \frac{1}{\rho^2} \partial_{\phi\phi} A_\phi
+ \partial_{z z} A_\phi
+ \inv{\rho^2} \partial_{\phi} A_\rho
+ \inv{\rho} \partial_{\phi\rho} A_\rho
+ \inv{\rho^2} \partial_\phi A_\rho
– \inv{\rho} \partial_{\rho\phi} A_\rho \\
&=
\partial_\rho \lr{ \inv{\rho} \partial_\rho ( \rho A_\phi ) }
+ \frac{1}{\rho^2} \partial_{\phi\phi} A_\phi
+ \partial_{z z} A_\phi
+ \frac{2}{\rho^2} \partial_{\phi} A_\rho \\
&=
\inv{\rho} \partial_\rho \lr{ \rho \partial_\rho A_\phi }
+ \frac{1}{\rho^2} \partial_{\phi\phi} A_\phi
+ \partial_{z z} A_\phi
+ \frac{2}{\rho^2} \partial_{\phi} A_\rho
– \frac{A_\phi}{\rho^2},
\end{aligned}
\end{equation}

\begin{equation}\label{eqn:laplacianCylindrical:260}
\begin{aligned}
\zcap \cdot \lr{ \spacegrad^2 \BA }
&=
\partial_z
\lr{
\inv{\rho} \partial_\rho (\rho A_\rho) + \frac{1}{\rho} \partial_\phi A_\phi + \partial_z A_z
}
–
\inv{\rho} \lr{
\partial_\rho \lr{ \rho \lr{
\partial_z A_\rho -\partial_\rho A_z
}
}
– \partial_\phi \lr{
\inv{\rho} \partial_\phi A_z – \partial_z A_\phi
}
} \\
&=
\inv{\rho} \partial_{z\rho} (\rho A_\rho)
+ \frac{1}{\rho} \partial_{z\phi} A_\phi
+ \partial_{zz} A_z
– \inv{\rho}\partial_\rho \lr{ \rho \partial_z A_\rho }
+ \inv{\rho}\partial_\rho \lr{ \rho \partial_\rho A_z }
+ \inv{\rho^2} \partial_{\phi\phi} A_z
– \inv{\rho} \partial_{\phi z} A_\phi \\
&=
\inv{\rho}\partial_\rho \lr{ \rho \partial_\rho A_z }
+ \inv{\rho^2} \partial_{\phi\phi} A_z
+ \partial_{zz} A_z
+ \inv{\rho} \partial_{z} A_\rho
+\partial_{z\rho} A_\rho
+ \frac{1}{\rho} \partial_{z\phi} A_\phi
– \inv{\rho}\partial_z A_\rho
– \partial_{\rho z} A_\rho
– \inv{\rho} \partial_{\phi z} A_\phi \\
&=
\inv{\rho}\partial_\rho \lr{ \rho \partial_\rho A_z }
+ \inv{\rho^2} \partial_{\phi\phi} A_z
+ \partial_{zz} A_z
\end{aligned}
\end{equation}

Evaluating these was a fairly tedious and mechanical job, and would have been better suited to a computer algebra system than by hand as done here.

Explicit cylindrical Laplacian

Let’s try this a different way. The most obvious potential strategy is to just apply the Laplacian to the vector itself, but we need to include the unit vectors in such an operation

\begin{equation}\label{eqn:laplacianCylindrical:280}
\spacegrad^2 \BA =
\spacegrad^2 \lr{ \rhocap A_\rho + \phicap A_\phi + \zcap A_z }.
\end{equation}

First we need to know the explicit form of the cylindrical Laplacian. From the painful expansion, we can guess that it is

\begin{equation}\label{eqn:laplacianCylindrical:300}
\spacegrad^2 \psi
=
\inv{\rho}\partial_\rho \lr{ \rho \partial_\rho \psi }
+ \inv{\rho^2} \partial_{\phi\phi} \psi
+ \partial_{zz} \psi.
\end{equation}

Let’s check that explicitly. Here I use the vector product where \( \rhocap^2 = \phicap^2 = \zcap^2 = 1 \), and these vectors anticommute when different

\begin{equation}\label{eqn:laplacianCylindrical:320}
\begin{aligned}
\spacegrad^2 \psi
&=
\lr{ \rhocap \partial_\rho + \frac{\phicap}{\rho} \partial_\phi + \zcap \partial_z }
\lr{ \rhocap \partial_\rho \psi + \frac{\phicap}{\rho} \partial_\phi \psi + \zcap \partial_z \psi } \\
&=
\rhocap \partial_\rho
\lr{ \rhocap \partial_\rho \psi + \frac{\phicap}{\rho} \partial_\phi \psi + \zcap \partial_z \psi }
+ \frac{\phicap}{\rho} \partial_\phi
\lr{ \rhocap \partial_\rho \psi + \frac{\phicap}{\rho} \partial_\phi \psi + \zcap \partial_z \psi }
+ \zcap \partial_z
\lr{ \rhocap \partial_\rho \psi + \frac{\phicap}{\rho} \partial_\phi \psi + \zcap \partial_z \psi } \\
&=
\partial_{\rho\rho} \psi
+ \rhocap \phicap \partial_\rho \lr{ \frac{1}{\rho} \partial_\phi \psi}
+ \rhocap \zcap \partial_{\rho z} \psi
+ \frac{\phicap}{\rho} \partial_\phi \lr{ \rhocap \partial_\rho \psi }
+ \frac{\phicap}{\rho} \partial_\phi \lr{ \frac{\phicap}{\rho} \partial_\phi \psi }
+ \frac{\phicap \zcap }{\rho} \partial_{\phi z} \psi
+ \zcap \rhocap \partial_{z\rho} \psi
+ \frac{\zcap \phicap}{\rho} \partial_{z\phi} \psi
+ \partial_{zz} \psi \\
&=
\partial_{\rho\rho} \psi
+ \inv{\rho} \partial_\rho \psi
+ \frac{1}{\rho^2} \partial_{\phi \phi} \psi
+ \partial_{zz} \psi
+ \rhocap \phicap
\lr{
-\frac{1}{\rho^2} \partial_\phi \psi
+\frac{1}{\rho} \partial_{\rho \phi} \psi
-\inv{\rho} \partial_{\phi \rho} \psi
+ \frac{1}{\rho^2} \partial_\phi \psi
}
+ \zcap \rhocap \lr{
-\partial_{\rho z} \psi
+ \partial_{z\rho} \psi
}
+ \phicap \zcap \lr{
\inv{\rho} \partial_{\phi z} \psi
– \inv{\rho} \partial_{z\phi} \psi
} \\
&=
\partial_{\rho\rho} \psi
+ \inv{\rho} \partial_\rho \psi
+ \frac{1}{\rho^2} \partial_{\phi \phi} \psi
+ \partial_{zz} \psi,
\end{aligned}
\end{equation}

so the Laplacian operator is

\begin{equation}\label{eqn:laplacianCylindrical:340}
\boxed{
\spacegrad^2
=
\inv{\rho} \PD{\rho}{} \lr{ \rho \PD{\rho}{} }
+ \frac{1}{\rho^2} \PDSq{\phi}{}
+ \PDSq{z}{}.
}
\end{equation}

All the bivector grades of the Laplacian operator are seen to explicitly cancel, regardless of the grade of \( \psi \), just as if we had expanded the scalar Laplacian as a dot product
\( \spacegrad^2 \psi = \spacegrad \cdot \lr{ \spacegrad \psi} \).
Unlike such a scalar expansion, this derivation is seen to be valid for any grade \( \psi \). We know now that we can trust this result when \( \psi \) is a scalar, a vector, a bivector, a trivector, or even a multivector.

Vector Laplacian

Now that we trust that the typical scalar form of the Laplacian applies equally well to multivectors as it does to scalars, that cylindrical coordinate operator can now be applied to a
vector. Consider the projections onto each of the directions in turn

\begin{equation}\label{eqn:laplacianCylindrical:360}
\spacegrad^2 \lr{ \rhocap A_\rho }
=
\rhocap \inv{\rho} \partial_\rho \lr{ \rho \partial_\rho A_\rho }
+ \frac{1}{\rho^2} \partial_{\phi\phi} \lr{\rhocap A_\rho}
+ \rhocap \partial_{zz} A_\rho
\end{equation}

\begin{equation}\label{eqn:laplacianCylindrical:380}
\begin{aligned}
\partial_{\phi\phi} \lr{\rhocap A_\rho}
&=
\partial_\phi \lr{ \phicap A_\rho + \rhocap \partial_\phi A_\rho } \\
&=
-\rhocap A_\rho
+\phicap \partial_\phi A_\rho
+ \phicap \partial_\phi A_\rho
+ \rhocap \partial_{\phi\phi} A_\rho \\
&=
\rhocap \lr{ \partial_{\phi\phi} A_\rho -A_\rho }
+ 2 \phicap \partial_\phi A_\rho
\end{aligned}
\end{equation}

so this component of the vector Laplacian is

\begin{equation}\label{eqn:laplacianCylindrical:400}
\begin{aligned}
\spacegrad^2 \lr{ \rhocap A_\rho }
&=
\rhocap
\lr{
\inv{\rho} \partial_\rho \lr{ \rho \partial_\rho A_\rho }
+ \inv{\rho^2} \partial_{\phi\phi} A_\rho
– \inv{\rho^2} A_\rho
+ \partial_{zz} A_\rho
}
+
\phicap
\lr{
2 \inv{\rho^2} \partial_\phi A_\rho
} \\
&=
\rhocap \lr{
\spacegrad^2 A_\rho
– \inv{\rho^2} A_\rho
}
+
\phicap
\frac{2}{\rho^2} \partial_\phi A_\rho
.
\end{aligned}
\end{equation}

The Laplacian for the projection of the vector onto the \( \phicap \) direction is

\begin{equation}\label{eqn:laplacianCylindrical:420}
\spacegrad^2 \lr{ \phicap A_\phi }
=
\phicap \inv{\rho} \partial_\rho \lr{ \rho \partial_\rho A_\phi }
+ \frac{1}{\rho^2} \partial_{\phi\phi} \lr{\phicap A_\phi}
+ \phicap \partial_{zz} A_\phi,
\end{equation}

Again, since the unit vectors are \( \phi \) dependent, the \( \phi \) derivatives have to be treated carefully

\begin{equation}\label{eqn:laplacianCylindrical:440}
\begin{aligned}
\partial_{\phi\phi} \lr{\phicap A_\phi}
&=
\partial_{\phi} \lr{-\rhocap A_\phi + \phicap \partial_\phi A_\phi} \\
&=
-\phicap A_\phi
-\rhocap \partial_\phi A_\phi
– \rhocap \partial_\phi A_\phi
+ \phicap \partial_{\phi \phi} A_\phi \\
&=
– 2 \rhocap \partial_\phi A_\phi
+
\phicap
\lr{
\partial_{\phi \phi} A_\phi
– A_\phi
},
\end{aligned}
\end{equation}

so the Laplacian of this projection is
\begin{equation}\label{eqn:laplacianCylindrical:460}
\begin{aligned}
\spacegrad^2 \lr{ \phicap A_\phi }
&=
\phicap
\lr{
\inv{\rho} \partial_\rho \lr{ \rho \partial_\rho A_\phi }
+ \phicap \partial_{zz} A_\phi,
\inv{\rho^2} \partial_{\phi \phi} A_\phi
– \frac{A_\phi }{\rho^2}
}
– \rhocap \frac{2}{\rho^2} \partial_\phi A_\phi \\
&=
\phicap \lr{
\spacegrad^2 A_\phi
– \frac{A_\phi}{\rho^2}
}
– \rhocap \frac{2}{\rho^2} \partial_\phi A_\phi.
\end{aligned}
\end{equation}

Since \( \zcap \) is fixed we have

\begin{equation}\label{eqn:laplacianCylindrical:480}
\spacegrad^2 \zcap A_z
=
\zcap \spacegrad^2 A_z.
\end{equation}

Putting all the pieces together we have
\begin{equation}\label{eqn:laplacianCylindrical:500}
\boxed{
\spacegrad^2 \BA
=
\rhocap \lr{
\spacegrad^2 A_\rho
– \inv{\rho^2} A_\rho
– \frac{2}{\rho^2} \partial_\phi A_\phi
}
+\phicap \lr{
\spacegrad^2 A_\phi
– \frac{A_\phi}{\rho^2}
+ \frac{2}{\rho^2} \partial_\phi A_\rho
}
+
\zcap \spacegrad^2 A_z.
}
\end{equation}

This matches the results of \ref{eqn:laplacianCylindrical:220}, …, from the painful expansion of
\( \spacegrad \lr{ \spacegrad \cdot \BA } – \spacegrad \cross \lr{ \spacegrad \cross \BA } \).