math and physics play

Poynting theorem

November 7, 2016 math and physics play , , , ,

Poynting relationship
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Problem:

Given
\begin{equation}\label{eqn:poynting:20}
\spacegrad \cross \BE
= -\BM_i – \PD{t}{\BB},
\end{equation}

and
\begin{equation}\label{eqn:poynting:40}
\spacegrad \cross \BH
= \BJ_i + \BJ_c + \PD{t}{\BD},
\end{equation}

expand the divergence of \( \BE \cross \BH \) to find the form of the Poynting theorem.

Solution:

First we need the chain rule for of this sort of divergence. Using primes to indicate the scope of the gradient operation

\begin{equation}\label{eqn:poynting:60}
\begin{aligned}
\spacegrad \cdot \lr{ \BE \cross \BH }
&=
\spacegrad’ \cdot \lr{ \BE’ \cross \BH }
–
\spacegrad’ \cdot \lr{ \BH’ \cross \BE } \\
&=
\BH \cdot \lr{ \spacegrad’ \cross \BE’ }
–
\BH \cdot \lr{ \spacegrad’ \cross \BH’ } \\
&=
\BH \cdot \lr{ \spacegrad \cross \BE }
–
\BE \cdot \lr{ \spacegrad \cross \BH }.
\end{aligned}
\end{equation}

In the second step, cyclic permutation of the triple product was used.
This checks against the inside front cover of Jackson [1]. Now we can plug in the Maxwell equation cross products.

\begin{equation}\label{eqn:poynting:80}
\begin{aligned}
\spacegrad \cdot \lr{ \BE \cross \BH }
&=
\BH \cdot \lr{ -\BM_i – \PD{t}{\BB} }
–
\BE \cdot \lr{ \BJ_i + \BJ_c + \PD{t}{\BD} } \\
&=
-\BH \cdot \BM_i
-\mu \BH \cdot \PD{t}{\BH}
–
\BE \cdot \BJ_i
–
\BE \cdot \BJ_c
–
\epsilon \BE \cdot \PD{t}{\BE},
\end{aligned}
\end{equation}

or

\begin{equation}\label{eqn:poynting:120}
\boxed{
0
=
\spacegrad \cdot \lr{ \BE \cross \BH }
+ \frac{\epsilon}{2} \PD{t}{} \Abs{ \BE }^2
+ \frac{\mu}{2} \PD{t}{} \Abs{ \BH }^2
+ \BH \cdot \BM_i
+ \BE \cdot \BJ_i
+ \sigma \Abs{\BE}^2.
}
\end{equation}

References

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

Gradient, divergence, curl and Laplacian in cylindrical coordinates

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

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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
\( \spacegrad \lr{ \spacegrad \cdot \BA } – \spacegrad \cross \lr{ \spacegrad \cross \BA } \).

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.

Line charge field and potential.

October 26, 2016 math and physics play , , , , , , , , , ,

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When computing the most general solution of the electrostatic potential in a plane, Jackson [1] mentions that \( -2 \lambda_0 \ln \rho \) is the well known potential for an infinite line charge (up to the unit specific factor). Checking that statement, since I didn’t recall what that potential was offhand, I encountered some inconsistencies and non-convergent integrals, and thought it was worthwhile to explore those a bit more carefully. This will be done here.

Using Gauss’s law.

For an infinite length line charge, we can find the radial field contribution using Gauss’s law, imagining a cylinder of length \( \Delta l \) of radius \( \rho \) surrounding this charge with the midpoint at the origin. Ignoring any non-radial field contribution, we have

\begin{equation}\label{eqn:lineCharge:20}
\int_{-\Delta l/2}^{\Delta l/2} \ncap \cdot \BE (2 \pi \rho) dl = \frac{\lambda_0}{\epsilon_0} \Delta l,
\end{equation}

or

\begin{equation}\label{eqn:lineCharge:40}
\BE = \frac{\lambda_0}{2 \pi \epsilon_0} \frac{\rhocap}{\rho}.
\end{equation}

Since

\begin{equation}\label{eqn:lineCharge:60}
\frac{\rhocap}{\rho} = \spacegrad \ln \rho,
\end{equation}

this means that the potential is

\begin{equation}\label{eqn:lineCharge:80}
\phi = -\frac{2 \lambda_0}{4 \pi \epsilon_0} \ln \rho.
\end{equation}

Finite line charge potential.

Let’s try both these calculations for a finite charge distribution. Gauss’s law looses its usefulness, but we can evaluate the integrals directly. For the electric field

\begin{equation}\label{eqn:lineCharge:100}
\BE
= \frac{\lambda_0}{4 \pi \epsilon_0} \int \frac{(\Bx – \Bx’)}{\Abs{\Bx – \Bx’}^3} dl’.
\end{equation}

Using cylindrical coordinates with the field point \( \Bx = \rho \rhocap \) for convience, the charge point \( \Bx’ = z’ \zcap \), and a the charge distributed over \( [a,b] \) this is

\begin{equation}\label{eqn:lineCharge:120}
\BE
= \frac{\lambda_0}{4 \pi \epsilon_0} \int_a^b \frac{(\rho \rhocap – z’ \zcap)}{\lr{\rho^2 + (z’)^2}^{3/2}} dz’.
\end{equation}

When the charge is uniformly distributed around the origin \( [a,b] = b[-1,1] \) the \( \zcap \) component of this field is killed because the integrand is odd. This justifies ignoring such contributions in the Gaussing cylinder analysis above. The general solution to this integral is found to be

\begin{equation}\label{eqn:lineCharge:140}
\BE
=
\frac{\lambda_0}{4 \pi \epsilon_0}
\evalrange{
\lr{
\frac{z’ \rhocap }{\rho \sqrt{ \rho^2 + (z’)^2 } }
+\frac{\zcap}{ \sqrt{ \rho^2 + (z’)^2 } }
}
}{a}{b},
\end{equation}

or
\begin{equation}\label{eqn:lineCharge:240}
\boxed{
\BE
=
\frac{\lambda_0}{4 \pi \epsilon_0}
\lr{
\frac{\rhocap }{\rho}
\lr{
\frac{b}{\sqrt{ \rho^2 + b^2 } }
-\frac{a}{\sqrt{ \rho^2 + a^2 } }
}
+ \zcap
\lr{
\frac{1}{ \sqrt{ \rho^2 + b^2 } }
-\frac{1}{ \sqrt{ \rho^2 + a^2 } }
}
}.
}
\end{equation}

When \( b = -a = \Delta l/2 \), this reduces to

\begin{equation}\label{eqn:lineCharge:160}
\BE
=
\frac{\lambda_0}{4 \pi \epsilon_0}
\frac{\rhocap }{\rho}
\frac{\Delta l}{\sqrt{ \rho^2 + (\Delta l/2)^2 } },
\end{equation}

which further reduces to \ref{eqn:lineCharge:40} when \( \Delta l \gg \rho \).

Finite line charge potential. Wrong but illuminating.

Again, putting the field point at \( z’ = 0 \), we have

\begin{equation}\label{eqn:lineCharge:180}
\phi(\rho)
= \frac{\lambda_0}{4 \pi \epsilon_0} \int_a^b \frac{dz’}{\lr{\rho^2 + (z’)^2}^{1/2}},
\end{equation}

which integrates to
\begin{equation}\label{eqn:lineCharge:260}
\phi(\rho)
= \frac{\lambda_0}{4 \pi \epsilon_0 }
\ln \frac{ b + \sqrt{ \rho^2 + b^2 }}{ a + \sqrt{\rho^2 + a^2}}.
\end{equation}

With \( b = -a = \Delta l/2 \), this approaches

\begin{equation}\label{eqn:lineCharge:200}
\phi
\approx
\frac{\lambda_0}{4 \pi \epsilon_0 }
\ln \frac{ (\Delta l/2) }{ \rho^2/2\Abs{\Delta l/2}}
=
\frac{-2 \lambda_0}{4 \pi \epsilon_0 } \ln \rho
+
\frac{\lambda_0}{4 \pi \epsilon_0 }
\ln \lr{ (\Delta l)^2/2 }.
\end{equation}

Before \( \Delta l \) is allowed to tend to infinity, this is identical (up to a difference in the reference potential) to \ref{eqn:lineCharge:80} found using Gauss’s law. It is, strictly speaking, singular when \( \Delta l \rightarrow \infty \), so it does not seem right to infinity as a reference point for the potential.

There’s another weird thing about this result. Since this has no \( z \) dependence, it is not obvious how we would recover the non-radial portion of the electric field from this potential using \( \BE = -\spacegrad \phi \)? Let’s calculate the elecric field from \ref{eqn:lineCharge:180} explicitly

\begin{equation}\label{eqn:lineCharge:220}
\begin{aligned}
\BE
&=
-\frac{\lambda_0}{4 \pi \epsilon_0}
\spacegrad
\ln \frac{ b + \sqrt{ \rho^2 + b^2 }}{ a + \sqrt{\rho^2 + a^2}} \\
&=
-\frac{\lambda_0 \rhocap}{4 \pi \epsilon_0 }
\PD{\rho}{}
\ln \frac{ b + \sqrt{ \rho^2 + b^2 }}{ a + \sqrt{\rho^2 + a^2}} \\
&=
-\frac{\lambda_0 \rhocap}{4 \pi \epsilon_0}
\lr{
\inv{ b + \sqrt{ \rho^2 + b^2 }} \frac{ \rho }{\sqrt{ \rho^2 + b^2 }}
-\inv{ a + \sqrt{ \rho^2 + a^2 }} \frac{ \rho }{\sqrt{ \rho^2 + a^2 }}
} \\
&=
-\frac{\lambda_0 \rhocap}{4 \pi \epsilon_0 \rho}
\lr{
\frac{ -b + \sqrt{ \rho^2 + b^2 }}{\sqrt{ \rho^2 + b^2 }}
-\frac{ -a + \sqrt{ \rho^2 + a^2 }}{\sqrt{ \rho^2 + a^2 }}
} \\
&=
\frac{\lambda_0 \rhocap}{4 \pi \epsilon_0 \rho}
\lr{
\frac{ b }{\sqrt{ \rho^2 + b^2 }}
-\frac{ a }{\sqrt{ \rho^2 + a^2 }}
}.
\end{aligned}
\end{equation}

This recovers the radial component of the field from \ref{eqn:lineCharge:240}, but where did the \( \zcap \) component go? The required potential appears to be

\begin{equation}\label{eqn:lineCharge:340}
\phi(\rho, z)
=
\frac{\lambda_0}{4 \pi \epsilon_0 }
\ln \frac{ b + \sqrt{ \rho^2 + b^2 }}{ a + \sqrt{\rho^2 + a^2}}
–
\frac{z \lambda_0}{4 \pi \epsilon_0 }
\lr{ \frac{1}{\sqrt{\rho^2 + b^2}}
-\frac{1}{\sqrt{\rho^2 + a^2}}
}.
\end{equation}

When computing the electric field \( \BE(\rho, \theta, z) \), it was convienent to pick the coordinate system so that \( z = 0 \). Doing this with the potential gives the wrong answers. The reason for this appears to be that this kills the potential term that is linear in \( z \) before taking its gradient, and we need that term to have the \( \zcap \) field component that is expected for a charge distribution that is non-symmetric about the origin on the z-axis!

Finite line charge potential. Take II.

Let the point at which the potential is evaluated be

\begin{equation}\label{eqn:lineCharge:360}
\Bx = \rho \rhocap + z \zcap,
\end{equation}

and the charge point be
\begin{equation}\label{eqn:lineCharge:380}
\Bx’ = z’ \zcap.
\end{equation}

This gives

\begin{equation}\label{eqn:lineCharge:400}
\begin{aligned}
\phi(\rho, z)
&= \frac{\lambda_0}{4\pi \epsilon_0} \int_a^b \frac{dz’}{\Abs{\rho^2 + (z – z’)^2 }} \\
&= \frac{\lambda_0}{4\pi \epsilon_0} \int_{a-z}^{b-z} \frac{du}{ \Abs{\rho^2 + u^2} } \\
&= \frac{\lambda_0}{4\pi \epsilon_0}
\evalrange{\ln \lr{ u + \sqrt{ \rho^2 + u^2 }}}{b-z}{a-z} \\
&=
\frac{\lambda_0}{4\pi \epsilon_0}
\ln \frac
{ b-z + \sqrt{ \rho^2 + (b-z)^2 }}
{ a-z + \sqrt{ \rho^2 + (a-z)^2 }}.
\end{aligned}
\end{equation}

The limit of this potential \( a = -\Delta/2 \rightarrow -\infty, b = \Delta/2 \rightarrow \infty \) doesn’t exist in any strict sense. If we are cavilier about the limits, as in \ref{eqn:lineCharge:200}, this can be evaluated as

\begin{equation}\label{eqn:lineCharge:n}
\phi \approx
\frac{\lambda_0}{4\pi \epsilon_0} \lr{ -2 \ln \rho + \textrm{constant} }.
\end{equation}

however, the constant (\( \ln \Delta^2/2 \)) is infinite, so there isn’t really a good justification for using that constant as the potential reference point directly.

It seems that the “right” way to calculate the potential for the infinite distribution, is to

  • Calculate the field from the potential.
  • Take the PV limit of that field with the charge distribution extending to infinity.
  • Compute the corresponding potential from this limiting value of the field.

Doing that doesn’t blow up. That field calculation, for the finite case, should include a \( \zcap \) component. To verify, let’s take the respective derivatives

\begin{equation}\label{eqn:lineCharge:420}
\begin{aligned}
-\PD{z}{} \phi
&=
-\frac{\lambda_0}{4\pi \epsilon_0}
\lr{
\frac{ -1 + \frac{z – b}{\sqrt{ \rho^2 + (b-z)^2 }} }{
b-z + \sqrt{ \rho^2 + (b-z)^2 }
}
–
\frac{ -1 + \frac{z – a}{\sqrt{ \rho^2 + (a-z)^2 }} }{
a-z + \sqrt{ \rho^2 + (a-z)^2 }
}
} \\
&=
\frac{\lambda_0}{4\pi \epsilon_0}
\lr{
\frac{ 1 + \frac{b – z}{\sqrt{ \rho^2 + (b-z)^2 }} }{
b-z + \sqrt{ \rho^2 + (b-z)^2 }
}
–
\frac{ 1 + \frac{a – z}{\sqrt{ \rho^2 + (a-z)^2 }} }{
a-z + \sqrt{ \rho^2 + (a-z)^2 }
}
} \\
&=
\frac{\lambda_0}{4\pi \epsilon_0}
\lr{
\inv{\sqrt{ \rho^2 + (b-z)^2 }}
-\inv{\sqrt{ \rho^2 + (a-z)^2 }}
},
\end{aligned}
\end{equation}

and

\begin{equation}\label{eqn:lineCharge:440}
\begin{aligned}
-\PD{\rho}{} \phi
&=
-\frac{\lambda_0}{4\pi \epsilon_0}
\lr{
\frac{ \frac{\rho}{\sqrt{ \rho^2 + (b-z)^2 }} }{
b-z + \sqrt{ \rho^2 + (b-z)^2 }
}
–
\frac{ \frac{\rho}{\sqrt{ \rho^2 + (a-z)^2 }} }{
a-z + \sqrt{ \rho^2 + (a-z)^2 }
}
} \\
&=
-\frac{\lambda_0}{4\pi \epsilon_0}
\lr{
\frac{\rho \lr{
-(b-z) + \sqrt{ \rho^2 + (b-z)^2 }
}}{ \rho^2 \sqrt{ \rho^2 + (b-z)^2 } }
–
\frac{\rho \lr{
-(a-z) + \sqrt{ \rho^2 + (a-z)^2 }
}}{ \rho^2 \sqrt{ \rho^2 + (a-z)^2 } }
} \\
&=
\frac{\lambda_0}{4\pi \epsilon_0 \rho}
\lr{
\frac{b-z}{\sqrt{ \rho^2 + (b-z)^2 }}
-\frac{a-z}{\sqrt{ \rho^2 + (a-z)^2 }}
}
.
\end{aligned}
\end{equation}

Putting the pieces together, the electric field is
\begin{equation}\label{eqn:lineCharge:460}
\BE =
\frac{\lambda_0}{4\pi \epsilon_0}
\lr{
\frac{\rhocap}{\rho} \lr{
\frac{b-z}{\sqrt{ \rho^2 + (b-z)^2 }}
-\frac{a-z}{\sqrt{ \rho^2 + (a-z)^2 }}
}
+
\zcap \lr{
\inv{\sqrt{ \rho^2 + (b-z)^2 }}
-\inv{\sqrt{ \rho^2 + (a-z)^2 }}
}
}.
\end{equation}

For has a PV limit of \ref{eqn:lineCharge:40} at \( z = 0 \), and also for the finite case, has the \( \zcap \) field component that was obtained when the field was obtained by direct integration.

Conclusions

  • We have to evaluate the potential at all points in space, not just on the axis that we evaluate the field on (should we choose to do so).
  • In this case, we found that it was not directly meaningful to take the limit of a potential distribution. We can, however, compute the field from a potential for a finite charge distribution,
    take the limit of that field, and then calculate the corresponding potential for the infinite distribution.

Is there a more robust mechanism that can be used to directly calculate the potential for an infinite charge distribution, instead of calculating the potential from the field of such an infinite distribution?

I think that were things go wrong is that the integral of \ref{eqn:lineCharge:180} does not apply to charge distributions that are not finite on the infinite range \( z \in [-\infty, \infty] \). That solution was obtained by utilizing an all-space Green’s function, and the boundary term in that Green’s analysis was assumed to tend to zero. That isn’t the case when the charge distribution is \( \lambda_0 \delta( z ) \).

References

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

Magnetostatic force and torque

October 18, 2016 math and physics play , , , , , , ,

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In Jackson [1], the following equations for the vector potential, magnetostatic force and torque are derived

\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:20}
\Bm = \inv{2} \int \Bx’ \cross \BJ(\Bx’) d^3 x’
\end{equation}
\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:40}
\BF = \spacegrad( \Bm \cdot \BB ),
\end{equation}
\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:60}
\BN = \Bm \cross \BB,
\end{equation}

where \( \BB \) is an applied external magnetic field and \( \Bm \) is the magnetic dipole for the current in question. These results (and a similar one derived earlier for the vector potential \( \BA \)) all follow from
an analysis of localized current densities \( \BJ \), evaluated far enough away from the current sources.

For the force and torque, the starting point for the force is one that had me puzzled a bit. Namely

\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:80}
\BF = \int \BJ(\Bx) \cross \BB(\Bx) d^3 x
\end{equation}

This is clearly the continuum generalization of the point particle Lorentz force equation, which for \( \BE = 0 \) is:

\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:100}
\BF = q \Bv \cross \BB
\end{equation}

For the point particle, this is the force on the particle when it is in the external field \( BB \). i.e. this is the force at the position of the particle. My question is what does it mean to sum all the forces on the charge distribution over all space.
How can a force be applied over all, as opposed to a force applied at a single point, or against a surface?

In the special case of a localized current density, this makes some sense. Considering the other half of the force equation \( \BF = \ddt{}\int \rho_m \Bv dV \), where \( \rho_m \) here is mass density of the charged particles making up the continuous current distribution. The other half of this \( \BF = m\Ba \) equation is also an average phenomena, so we have an average of sorts on both the field contribution to the force equation and the mass contribution to the force equation. There is probably a centre-of-mass and centre-of-current density interpretation that would make a bit more sense of this continuum force description.

It’s kind of funny how you can work through all the detailed mathematical steps in a book like Jackson, but then go right back to the beginning and say “Hey, what does that even mean”?

Force

Moving on from the pondering of the meaning of the equation being manipulated, let’s do the easy part, the derivation of the results that Jackson comes up with.

Writing out \ref{eqn:magnetostaticsJacksonNotesForceAndTorque:80} in coordinates

\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:320}
\BF = \epsilon_{ijk} \Be_i \int J_j B_k d^3 x.
\end{equation}

To first order, a slowly varying (external) magnetic field can be expanded around a point of interest

\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:120}
\BB(\Bx) = \BB(\Bx_0) + \lr{ \Bx – \Bx_0 } \cdot \spacegrad \BB,
\end{equation}

where the directional derivative is evaluated at the point \( \Bx_0 \) after the gradient operation. Setting the origin at this point \( \Bx_0 \) gives

\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:340}
\begin{aligned}
\BF
&= \epsilon_{ijk} \Be_i
\lr{
\int J_j(\Bx’) B_k(0) d^3 x’
+
\int J_j(\Bx’) (\Bx’ \cdot \spacegrad) B_k(0) d^3 x’
} \\
&=
\epsilon_{ijk} \Be_i
\Bk_0 \int J_j(\Bx’) d^3 x’
+
\epsilon_{ijk} \Be_i
\int J_j(\Bx’) (\Bx’ \cdot \spacegrad) B_k(0) d^3 x’.
\end{aligned}
\end{equation}

We found

earlier
that the first integral can be written as a divergence

\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:140}
\int J_j(\Bx’) d^3 x’
=
\int \spacegrad’ \cdot \lr{ \BJ(\Bx’) x_j’ } dV’,
\end{equation}

which is zero when the integration surface is outside of the current localization region. We also found

that

\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:160}
\int (\Bx \cdot \Bx’) \BJ
= -\inv{2} \Bx \cross \int \Bx’ \cross \BJ = \Bm \cross \Bx.
\end{equation}

so
\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:180}
\begin{aligned}
\int (\spacegrad B_k(0) \cdot \Bx’) J_j
&= -\inv{2} \lr{ \spacegrad B_k(0) \cross \int \Bx’ \cross \BJ}_j \\
&= \lr{ \Bm \cross (\spacegrad B_k(0)) }_j.
\end{aligned}
\end{equation}

This gives

\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:200}
\begin{aligned}
\BF
&= \epsilon_{ijk} \Be_i \lr{ \Bm \cross (\spacegrad B_k(0)) }_j \\
&= \epsilon_{ijk} \Be_i \lr{ \Bm \cross \spacegrad }_j B_k(0) \\
&= (\Bm \cross \spacegrad) \cross \BB(0) \\
&= -\BB(0) \cross (\Bm \cross \lspacegrad) \\
&= (\BB(0) \cdot \Bm) \lspacegrad – (\BB \cdot \lspacegrad) \Bm \\
&= \spacegrad (\BB(0) \cdot \Bm) – \Bm (\spacegrad \cdot \BB(0)).
\end{aligned}
\end{equation}

The second term is killed by the magnetic Gauss’s law, leaving to first order

\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:220}
\BF = \spacegrad \lr{\Bm \cdot \BB}.
\end{equation}

Torque

For the torque we have a similar quandary at the starting point. About what point is a continuum torque integral of the following form

\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:240}
\BN = \int \Bx’ \cross (\BJ(\Bx’) \cross \BB(\Bx’)) d^3 x’?
\end{equation}

Ignoring that detail again, assuming the answer has something to do with the centre of mass and parallel axis theorem, we can proceed with a constant approximation of the magnetic field

\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:260}
\begin{aligned}
\BN
&= \int \Bx’ \cross (\BJ(\Bx’) \cross \BB(0)) d^3 x’ \\
&=
-\int (\Bx’ \cdot \BJ(\Bx’)) \BB(0) d^3 x’
+\int (\Bx’ \cdot \BB(0)) \BJ(\Bx’) d^3 x’ \\
&=
-\BB(0) \int (\Bx’ \cdot \BJ(\Bx’)) d^3 x’
+\int (\Bx’ \cdot \BB(0)) \BJ(\Bx’) d^3 x’.
\end{aligned}
\end{equation}

Jackson’s trick for killing the first integral is to transform it into a divergence by evaluating

\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:280}
\begin{aligned}
\spacegrad \cdot \lr{ \BJ \Abs{\Bx}^2 }
&=
(\spacegrad \cdot \BJ) \Abs{\Bx}^2
+
\BJ \cdot \spacegrad \Abs{\Bx}^2 \\
&=
\BJ \cdot \Be_i \partial_i x_m x_m \\
&=
2 \BJ \cdot \Be_i \delta_{im} x_m \\
&=
2 \BJ \cdot \Bx,
\end{aligned}
\end{equation}

so

\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:300}
\begin{aligned}
\BN
&=
-\inv{2} \BB(0) \int \spacegrad’ \cdot \lr{ \BJ(\Bx’) \Abs{\Bx’}^2 } d^3 x’
+\int (\Bx’ \cdot \BB(0)) \BJ(\Bx’) d^3 x’ \\
&=
-\inv{2} \BB(0) \oint \Bn \cdot \lr{ \BJ(\Bx’) \Abs{\Bx’}^2 } d^3 x’
+\int (\Bx’ \cdot \BB(0)) \BJ(\Bx’) d^3 x’.
\end{aligned}
\end{equation}

Again, the localized current density assumption kills the surface integral. The second integral can be evaluated with \ref{eqn:magnetostaticsJacksonNotesForceAndTorque:160}, so to first order we have

\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:360}
\BN
=
\Bm \cross \BB.
\end{equation}

References

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