math and physics play

Center of mass on line of symmetry: questions from a little guy

August 27, 2026 math and physics play No comments , ,

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Motivation

My friend’s son is an exceptional little guy. I think he’s about 12 or so, but he’s already half way through our old 1st year calculus book. When they visited, he was intensely dissatisfied with an “obvious” statement that the center of mass necessarily lies on the axis of symmetry (when there is such a symmetry). I didn’t take the time when he was here to reason through that statement in a systematic way. It turns out that it’s pretty simple to do so.

A uniform mass with symmetry

Suppose that we have equal density thin plate, with surface density \( \rho \), with rotational symmetry about some axis, as illustrated in fig. 1.

fig. 1. An flat mass with rotational symmetry

In the plot example, I’ve plotted a positive function \( g(x) = \Abs{f(x)} \), and rotated it, plotting \( -g(a-x) \) below the axis, with the whole things bounded, in this case by \( x \in [-2, 5] \). We don’t actually need those bounds if the function is integrable everywhere (we just need a finite mass, i.e.: a finite area).

The problem of computing the center of mass coordinates is to find the values
\begin{equation}\label{eqn:centerOfMassRotationalSymmetry:20}
\begin{aligned}
\bar{x} M &= \int \sigma x dA \\
\bar{y} M &= \int \sigma y dA,
\end{aligned}
\end{equation}
where \( M = \int \sigma dA \). We are also only interested in \( \bar{y} \), since we want to show that the center of mass lies on the axis of symmetry. In symbols, with the surface mass density \( \sigma \) constant, that is
\begin{equation}\label{eqn:centerOfMassRotationalSymmetry:40}
\bar{y}
= \inv{A} \int y dx dy.
\end{equation}
Let’s ignore the bounds in the example, and perform the integral for all \( x \). If the area is bounded, then that integral will be zero outside of some region. That is
\begin{equation}\label{eqn:centerOfMassRotationalSymmetry:60}
\begin{aligned}
\bar{y}
&= \inv{A} \int_{x = -\infty}^\infty dx \int_{y = -g(a-x)}^{g(x)} y dy \\
&= \inv{A} \int_{x = -\infty}^\infty dx \inv{2} \evalrange{ y^2 }{-g(a-x)}{g(x)} \\
&= \inv{2 A} \int_{x = -\infty}^\infty dx \lr{ g^2(x) – \lr{ -g(a-x) }^2 } \\
&= \inv{2 A} \int_{x = -\infty}^\infty g^2(x) dx – \inv{2 A} \int_{x = -\infty}^\infty g^2(a-x) dx
\end{aligned}
\end{equation}
A change of variables \( u = a – x \) in the second integral gives
\begin{equation}\label{eqn:centerOfMassRotationalSymmetry:80}
\begin{aligned}
\inv{2 A} \int_{x = -\infty}^\infty g^2(a-x) dx
&=
\inv{2 A} \int_{u = \infty}^{-\infty} g^2(u) (-du) \\
&=
\inv{2 A} \int_{u = -\infty}^{\infty} g^2(u) du,
\end{aligned}
\end{equation}
which is exactly the first integral (other than a change of dummy variables.) That means
\begin{equation}\label{eqn:centerOfMassRotationalSymmetry:100}
\bar{y} = 0.
\end{equation}
This wasn’t the most general function, as we used a well behaved function (not a relation), that isn’t multivalued for any value of \( x \), but this argument would still work (requiring superposition arguments to generalize it.)

Some line integral examples of the Fundamental theorem of geometric calculus

January 20, 2026 math and physics play , , , , , , , ,

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On my discord server, Frank asked about his attempt to demonstrate an example line integral computation of the fundamental theorem of geometric calculus.

Before working through his example, and some others, it is first worth restating the
line integral specialization of the \textit{Fundamental theorem of geometric calculus}:

Theorem 1.1: Fundamental theorem of geometric calculus (line integral version.)

Given multivectors \(F, G \), a single variable parameterization \( \Bx = \Bx(u) \), with line element \( d\Bx = du \Bx_u \), \( \Bx_u = \PDi{u}{\Bx} \), \( \boldpartial = \Bx^u \PDi{u}{} \), and \( \Bx^u \cdot \Bx_u = 1 \), then
the line integral is related to the boundary by
\begin{equation*}
\int F d\Bx \boldpartial G = \evalbar{F G}{\Delta u},
\end{equation*}
(with the \( \boldpartial \) acting bidirectionally on \( F, G \).)

It is very important to point out that the derivative operator here is the vector derivative, and not the gradient. Roughly speaking, the vector derivative is the projection of the gradient onto the tangent space. In this case, the tangent space is just the line in the direction \( \Bx_u \), which may vary along the parameterized path.

Here are some examples of some one variable parameterizations, all in two dimensions

  1. \( \Bx = u \Be_1 + y_0 \Be_2 \).
    We compute
    \begin{equation}\label{eqn:lineintegralExamples:20}
    \begin{aligned}
    \Bx_u &= \PD{\Bx}{u} = \Be_1 \\
    \Bx^u &= \Be_1 \\
    d\Bx &= du \Be_1 \\
    \boldpartial &= \Be_1 \PD{u}{}.
    \end{aligned}
    \end{equation}
    and \( d\Bx \boldpartial = \PDi{u}{} \).
    The fundamental theorem is really just a statement that
    \begin{equation}\label{eqn:lineintegralExamples:40}
    \int \PD{u}{} \lr{ F G } du = \evalbar{ F G }{\Delta u}.
    \end{equation}

  2. \( \Bx = \alpha u \Be_1 + \beta u \Be_2 \), where \( \alpha, \beta \) are constants. i.e.: a line, but not necessarily on the horizontal this time.
    This time, we compute
    \begin{equation}\label{eqn:lineintegralExamples:60}
    \begin{aligned}
    \Bx_u &= \alpha \Be_1 + \beta \Be_2 \\
    \Bx^u &= \inv{\Bx_u} = \frac{\alpha \Be_1 + \beta \Be_2}{\alpha^2 + \beta^2} \\
    d\Bx &= du \lr{ \alpha \Be_1 + \beta \Be_2 } \\
    \boldpartial &= \inv{\alpha \Be_1 + \beta \Be_2} \PD{u}{}.
    \end{aligned}
    \end{equation}
    Again, we have \( d\Bx \boldpartial = \PDi{u}{} \), and the story repeats.

  3. \( \Bx = R \Be_1 e^{i\theta}, i = \Be_1 \Be_2 \). This time we are going along a circular arc.

    Let \( \rcap = \Be_1 e^{i\theta} \), and \(\thetacap = \Be_2 e^{i\theta} \). We can compute
    \begin{equation}\label{eqn:lineintegralExamples:80}
    \begin{aligned}
    \Bx_\theta &= R \Be_2 e^{i\theta} = R \thetacap \\
    \Bx^\theta &= \inv{\Bx_\theta} = \inv{ R \Be_2 e^{i\theta} } = \inv{R} \thetacap \\
    d\Bx &= d\theta \thetacap \\
    \boldpartial &= \frac{\thetacap}{R} \PD{\theta}{}.
    \end{aligned}
    \end{equation}
    This time, probably to no suprise, we have \( d\Bx \boldpartial = \PDi{\theta}{} \), so the fundamental theorem for this parameterization is a statement that
    \begin{equation}\label{eqn:lineintegralExamples:100}
    \int \PD{\theta}{} \lr{ F G } d\theta = \evalbar{ F G }{\Delta \theta}.
    \end{equation}

  4. \( \Bx = r e^{i\theta_0} \), where \( \theta_0 \) is a constant. We’ve already computed this above with a Cartesian representation of a line, but can do it again this time with an explicitly radial parameterization. We compute
    \begin{equation}\label{eqn:lineintegralExamples:120}
    \begin{aligned}
    \Bx_r &= \Be_1 e^{i \theta_0} \\
    \Bx^r &= \inv{\Bx_r} = \Be_1 e^{i \theta_0} \\
    d\Bx &= dr \Be_1 e^{i \theta_0} \\
    \boldpartial &= e^{i \theta_0} \PD{r}{}.
    \end{aligned}
    \end{equation}
    This time, \( d\Bx \boldpartial = \PDi{r}{} \), and the fundamental theorem for this parameterization is a statement that
    \begin{equation}\label{eqn:lineintegralExamples:140}
    \int \PD{r}{} \lr{ F G } dr = \evalbar{ F G }{\Delta r}.
    \end{equation}

Observe that we do not get the same result if we use the gradient instead of the vector derivative. We may only make a gradient substitution for the vector derivative when the dimension of the hypervolume integral equals the dimension of the vector space itself. For a line integral that would mean we are restricting the domain of the underlying vector space to \(\mathbb{R}^1\), which isn’t a very interesting case for geometric algebra.

In Frank’s example, he was working with a generating vector space of \(\mathbb{R}^2\), with the horizontal parameterization \( \Bx = u \Be_1 + y_0 \Be_2 \) that we used in the first example (with \( F = 1, G = x y i \), where \( i = \Be_1 \Be_2 \), the pseudoscalar for the space).

Let’s see what happens if we compute a similar integral, but swapping out the vector derivative with the gradient
\begin{equation}\label{eqn:lineintegralExamples:160}
\begin{aligned}
\int d\Bx \spacegrad x y i
&=
\int du \Be_1 \lr{ \Be_1 \partial_x + \Be_2 \partial_y } ( x y i ) \\
&=
\int du \Be_1 \lr{ \Be_1 y + \Be_2 x } i \\
&=
\int du \lr{ y + i x } i \\
&=
\int du \lr{ y_0 + i u } i \\
&=
\lr{\Delta x} y_0 i – \frac{x_1^2}{2} + \frac{x_0^2}{2}.
\end{aligned}
\end{equation}
As well as the pseudoscalar term that we had when evaluating the fundamental theorem integral, this time we have an extra scalar term, a contribution that goes back to the \( y \) component of the gradient. There is nothing wrong with performing such an integral, but it’s not an instance of the fundamental theorem, and the same tidy answer should not be expected. In Frank’s original example, he also didn’t put the \( \Bx \) adjacent to the differential operator, which is required to get the perfect cancelation of the tangent space vectors that we’ve seen in the evaluations above.

Curl of Curl. Tensor and GA expansion, and GA equivalent identity.

November 12, 2025 math and physics play , , , , , , , , , , ,

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In this blog post, we will expand \(\spacegrad \cross \lr{ \spacegrad \cross \Bf } = -\spacegrad^2 \Bf + \spacegrad \lr{ \spacegrad \cdot \Bf } \) two different ways, using tensor index gymnastics and using geometric algebra.

The tensor way.

To expand the curl using a tensor expansion, let’s first expand the cross product in coordinates
\begin{equation}\label{eqn:curlcurl2:20}
\begin{aligned}
\Ba \cross \Bb
&=
\lr{ \Be_r \cross \Be_s } a_r b_s \\
&=
\Be_t \cdot \lr{ \Be_r \cross \Be_s } \Be_t a_r b_s \\
&=
\epsilon_{rst} a_r b_s \Be_t.
\end{aligned}
\end{equation}
Here \( \epsilon_{rst} \) is the completely antisymmetric (Levi-Civita) tensor, and allows us to compactly express the geometrical nature of the triple product.

We can then expand the curl of the curl by applying this twice
\begin{equation}\label{eqn:curlcurl2:40}
\begin{aligned}
\spacegrad \cross \lr{ \spacegrad \cross \Bf }
&=
\epsilon_{rst} \partial_r \lr{ \spacegrad \cross \Bf }_s \Be_t \\
&=
\epsilon_{rst} \partial_r \lr{ \epsilon_{uvw} \partial_u f_v \Be_w }_s \Be_t \\
&=
\epsilon_{rst} \partial_r \epsilon_{uvs} \partial_u f_v \Be_t.
\end{aligned}
\end{equation}

It turns out that there’s a nice identity to reduce the single index contraction of a pair of Levi-Civita tensors.
\begin{equation}\label{eqn:curlcurl2:60}
\epsilon_{abt} \epsilon_{cdt} = \delta_{ac} \delta_{bd} – \delta_{ad} \delta_{bc}.
\end{equation}
To show this, consider the \( t = 1 \) term of this sum \( \epsilon_{ab1} \epsilon_{cd1} \). This is non-zero only for \( a,b,c,d \in \setlr{2,3} \). If \( a,b = c,d \), this is one, and if \( a,b = d,c \), this is minus one. We may summarize that as
\begin{equation}\label{eqn:curlcurl2:80}
\epsilon_{ab1} \epsilon_{cd1} = \delta_{ac} \delta_{bd} – \delta_{ad} \delta_{bc},
\end{equation}
but this holds for \( t = 2,3 \) too, so \ref{eqn:curlcurl2:60} holds generally.

We may now contract the tensors to find
\begin{equation}\label{eqn:curlcurl2:100}
\begin{aligned}
\spacegrad \cross \lr{ \spacegrad \cross \Bf }
&=
\epsilon_{rst} \epsilon_{uvs} \Be_t \partial_r \partial_u f_v \\
&=
-\epsilon_{rts} \epsilon_{uvs} \Be_t \partial_r \partial_u f_v \\
&=
-\lr{ \delta_{ru} \delta_{tv} – \delta_{rv} \delta_{tu} } \Be_t \partial_r \partial_u f_v \\
&=
– \Be_v \partial_u \partial_u f_v
+ \Be_u \partial_v \partial_u f_v \\
&=
-\spacegrad^2 \Bf + \spacegrad \lr{ \spacegrad \cdot \Bf }.
\end{aligned}
\end{equation}

Using geometric algebra.

Now let’s pull out the GA toolbox. We start with introducing a no-op grade-1 selection, and using the identity \( \Ba \cross \Bb = -I \lr{ \Ba \wedge \Bb } \)
\begin{equation}\label{eqn:curlcurl2:120}
\begin{aligned}
\spacegrad \cross \lr{ \spacegrad \cross \Bf }
&=
\gpgradeone{
\spacegrad \cross \lr{ \spacegrad \cross \Bf }
} \\
&=
\gpgradeone{
-I \lr{ \spacegrad \wedge \lr{ \spacegrad \cross \Bf } }
} \\
\end{aligned}
\end{equation}
We can now expand \( \Ba \wedge \Bb = \Ba \Bb – \Ba \cdot \Bb \)
\begin{equation}\label{eqn:curlcurl2:140}
\spacegrad \cross \lr{ \spacegrad \cross \Bf }
=
\gpgradeone{
-I \spacegrad \lr{ \spacegrad \cross \Bf }
+I \lr{ \spacegrad \cdot \lr{ \spacegrad \cross \Bf } }
}
\end{equation}
but that dot product is a scalar, leaving just a pseudoscalar, which has a zero grade-1 selection. This leaves
\begin{equation}\label{eqn:curlcurl2:160}
\begin{aligned}
\spacegrad \cross \lr{ \spacegrad \cross \Bf }
&=
\gpgradeone{
-I \spacegrad \lr{ -I \lr{ \spacegrad \wedge \Bf } }
} \\
&=
-\gpgradeone{
\spacegrad \lr{ \spacegrad \wedge \Bf }
}.
\end{aligned}
\end{equation}
We use \( \Ba \wedge \Bb = \Ba \Bb – \Ba \cdot \Bb \) once more
\begin{equation}\label{eqn:curlcurl2:180}
\begin{aligned}
\spacegrad \cross \lr{ \spacegrad \cross \Bf }
&=
-\gpgradeone{
\spacegrad \lr{ \spacegrad \Bf }
-\spacegrad \lr{ \spacegrad \cdot \Bf }
}
\\
&=
-\spacegrad^2 \Bf
+\spacegrad \lr{ \spacegrad \cdot \Bf }.
\end{aligned}
\end{equation}

GA identity.

It’s also worth noting that there’s a natural GA formulation of the curl of a curl. From the Laplacian and divergence relationship that we ended up with, we need only factor out the gradient
\begin{equation}\label{eqn:curlcurl2:200}
\begin{aligned}
\spacegrad \cross \lr{ \spacegrad \cross \Bf }
&=
-\spacegrad^2 \Bf +\spacegrad \lr{ \spacegrad \cdot \Bf } \\
&=
-\spacegrad \lr{ \spacegrad \Bf – \spacegrad \cdot \Bf } \\
&=
-\spacegrad \lr{ \spacegrad \wedge \Bf }.
\end{aligned}
\end{equation}
Because \( \spacegrad \wedge \lr{ \spacegrad \wedge \Bf } = 0 \), we may also write this as
\begin{equation}\label{eqn:curlcurl2:220}
\boxed{
\spacegrad \cdot \lr{ \spacegrad \wedge \Bf } = -\spacegrad \cross \lr{ \spacegrad \cross \Bf }.
}
\end{equation}
From the GA LHS, we see by inspection that
\begin{equation}\label{eqn:curlcurl2:240}
\spacegrad \cdot \lr{ \spacegrad \wedge \Bf } = \spacegrad^2 \Bf – \spacegrad \lr{ \spacegrad \cdot \Bf }.
\end{equation}

A fun application of Green’s functions and geometric algebra: Residue calculus

November 2, 2025 math and physics play , , , , , , , , , , , , , , , , , ,

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Motivation.

A fun application of both Green’s functions and geometric algebra is to show how the Cauchy integral equation can be expressed in terms of the Green’s function for the 2D gradient. This is covered, almost as an aside, in [1]. I found that treatment a bit hard to understand, so I am going to work through it here at my own pace.

Complex numbers in geometric algebra.

Anybody who has studied geometric algebra is likely familiar with a variety of ways to construct complex numbers from geometric objects. For example, complex numbers can be constructed for any plane. If \( \Be_1, \Be_2 \) is a pair of orthonormal vectors for some plane in \(\mathbb{R}^N\), then any vector in that plane has the form
\begin{equation}\label{eqn:residueGreens:20}
\Bf = \Be_1 u + \Be_2 v,
\end{equation}
has an associated complex representation, by simply multiplying that vector one of those basis vectors. For example, if we pre-multiply \( \Bf \) by \( \Be_1 \), forming
\begin{equation}\label{eqn:residueGreens:40}
\begin{aligned}
z
&= \Be_1 \Bf \\
&= \Be_1 \lr{ \Be_1 u + \Be_2 v } \\
&= u + \Be_1 \Be_2 v.
\end{aligned}
\end{equation}

We may identify the unit bivector \( \Be_1 \Be_2 \) as an imaginary, designed by \( i \), since it has the expected behavior
\begin{equation}\label{eqn:residueGreens:60}
\begin{aligned}
i^2 &=
\lr{\Be_1 \Be_2}^2 \\
&=
\lr{\Be_1 \Be_2}
\lr{\Be_1 \Be_2} \\
&=
\Be_1 \lr{\Be_2
\Be_1} \Be_2 \\
&=
-\Be_1 \lr{\Be_1
\Be_2} \Be_2 \\
&=
-\lr{\Be_1 \Be_1}
\lr{\Be_2 \Be_2} \\
&=
-1.
\end{aligned}
\end{equation}

Complex numbers are seen to be isomorphic to even grade multivectors in a planar subspace. The imaginary is the grade-two pseudoscalar, and geometrically is an oriented unit area (bivector.)

Cauchy-equations in terms of the gradient.

It is natural to wonder about the geometric algebra equivalents of various complex-number relationships and identities. Of particular interest for this discussion is the geometric algebra equivalent of the Cauchy equations that specify required conditions for a function to be differentiable.

If a complex function \( f(z) = u(z) + i v(z) \) is differentiable, then we must be able to find the limit of
\begin{equation}\label{eqn:residueGreens:80}
\frac{\Delta f(z_0)}{\Delta z} = \frac{f(z_0 + h) – f(z_0)}{h},
\end{equation}
for any complex \( h \rightarrow 0 \), for any possible trajectory of \( z_0 + h \) toward \( z_0 \). In particular, for real \( h = \epsilon \),
\begin{equation}\label{eqn:residueGreens:100}
\lim_{\epsilon \rightarrow 0} \frac{u(x_0 + \epsilon, y_0) + i v(x_0 + \epsilon, y_0) – u(x_0, y_0) – i v(x_0, y_0)}{\epsilon}
=
\PD{x}{u(z_0)} + i \PD{x}{v(z_0)},
\end{equation}
and for imaginary \( h = i \epsilon \)
\begin{equation}\label{eqn:residueGreens:120}
\lim_{\epsilon \rightarrow 0} \frac{u(x_0, y_0 + \epsilon) + i v(x_0, y_0 + \epsilon) – u(x_0, y_0) – i v(x_0, y_0)}{i \epsilon}
=
-i\lr{ \PD{y}{u(z_0)} + i \PD{y}{v(z_0)} }.
\end{equation}
Equating real and imaginary parts, we see that existence of the derivative requires
\begin{equation}\label{eqn:residueGreens:140}
\begin{aligned}
\PD{x}{u} &= \PD{y}{v} \\
\PD{x}{v} &= -\PD{y}{u}.
\end{aligned}
\end{equation}
These are the Cauchy equations. When the derivative exists in a given neighbourhood, we say that the function is analytic in that region. If we use a bivector interpretation of the imaginary, with \( i = \Be_1 \Be_2 \), the Cauchy equations are also satisfied if the gradient of the complex function is zero, since
\begin{equation}\label{eqn:residueGreens:160}
\begin{aligned}
\spacegrad f
&=
\lr{ \Be_1 \partial_x + \Be_2 \partial_y } \lr{ u + \Be_1 \Be_2 v } \\
&=
\Be_1 \lr{ \partial_x u – \partial_y v } + \Be_2 \lr{ \partial_y u + \partial_x v }.
\end{aligned}
\end{equation}
We see that the geometric algebra equivalent of the Cauchy equations is simply
\begin{equation}\label{eqn:residueGreens:200}
\spacegrad f = 0.
\end{equation}
Roughly speaking, we may say that a function is analytic in a region, if the Cauchy equations are satisfied, or the gradient is zero, in a neighbourhood of all points in that region.

A special case of the fundamental theorem of geometric calculus.

Given an even grade multivector \( \psi \in \mathbb{R}^2 \) (i.e.: a complex number), we can show that
\begin{equation}\label{eqn:residueGreens:220}
\int_A \spacegrad \psi d^2\Bx = \oint_{\partial A} d\Bx \psi.
\end{equation}
Let’s get an idea why this works by expanding the area integral for a rectangular parameterization
\begin{equation}\label{eqn:residueGreens:240}
\begin{aligned}
\int_A \spacegrad \psi d^2\Bx
&=
\int_A \lr{ \Be_1 \partial_1 + \Be_2 \partial_2 } \psi I dx dy \\
&=
\int \Be_1 I dy \evalrange{\psi}{x_0}{x_1}
+
\int \Be_2 I dx \evalrange{\psi}{y_0}{y_1} \\
&=
\int \Be_2 dy \evalrange{\psi}{x_0}{x_1}
–
\int \Be_1 dx \evalrange{\psi}{y_0}{y_1} \\
&=
\int d\By \evalrange{\psi}{x_0}{x_1}
–
\int d\Bx \evalrange{\psi}{y_0}{y_1}.
\end{aligned}
\end{equation}
We took advantage of the fact that the \(\mathbb{R}^2\) pseudoscalar commutes with \( \psi \). The end result, is illustrated in fig. 1, shows pictorially that the remaining integral is an oriented line integral.

fig. 1. Oriented multivector line integral.

 

If we want to approximate a more general area, we may do so with additional tiles, as illustrated in fig. 2. We may evaluate the area integral using the line integral over just the exterior boundary using such a tiling, as any overlapping opposing boundary contributions cancel exactly.

fig. 2. A crude circular tiling approximation.

 

The reason that this is interesting is that it allows us to re-express a complex integral as a corresponding multivector area integral. With \( d\Bx = \Be_1 dz \), we have
\begin{equation}\label{eqn:residueGreens:260}
\oint dz\, \psi = \Be_1 \int \spacegrad \psi d^2\Bx.
\end{equation}

The Cauchy kernel as a Green’s function.

We’ve previously derived the Green’s function for the 2D Laplacian, and found
\begin{equation}\label{eqn:residueGreens:280}
\tilde{G}(\Bx, \Bx’) = \inv{2\pi} \ln \Abs{\lr{\Bx – \Bx’}},
\end{equation}
which satisfies
\begin{equation}\label{eqn:residueGreens:300}
\delta^2(\Bx – \Bx’) = \spacegrad^2 \tilde{G}(\Bx, \Bx’) = \spacegrad \lr{ \spacegrad \tilde{G}(\Bx, \Bx’) }.
\end{equation}
This means that \( G(\Bx, \Bx’) = \spacegrad \tilde{G}(\Bx, \Bx’) \) is the Green’s function for the gradient. That Green’s function is
\begin{equation}\label{eqn:residueGreens:320}
\begin{aligned}
G(\Bx, \Ba)
&= \inv{2 \pi} \frac{\spacegrad \Abs{\Bx – \Ba}}{\Abs{\Bx – \Ba}} \\
&= \inv{2 \pi} \frac{\Bx – \Ba}{\Abs{\Bx – \Ba}^2}.
\end{aligned}
\end{equation}
We may cast this Green’s function into complex form with \( z = \Be_1 \Bx, a = \Be_1 \Ba \). In particular
\begin{equation}\label{eqn:residueGreens:340}
\begin{aligned}
\inv{z – a}
&=
\frac{(z – a)^\conj}{\Abs{z – a}^2} \\
&=
\frac{(z – a)^\conj}{\Abs{z – a}^2} \\
&=
\frac{\Bx – \Ba}{\Abs{\Bx – \Ba}^2} \Be_1 \\
&=
2 \pi G(\Bx, \Ba) \Be_1.
\end{aligned}
\end{equation}

Cauchy’s integral.

With
\begin{equation}\label{eqn:residueGreens:360}
\psi = \frac{f(z)}{z – a},
\end{equation}
using \ref{eqn:residueGreens:260}, we can now evaluate
\begin{equation}\label{eqn:residueGreens:265}
\begin{aligned}
\oint dz\, \frac{f(z)}{z – a}
&= \Be_1 \int \spacegrad \frac{f(z)}{z – a} d^2\Bx \\
&= \Be_1 \int \lr{ \frac{\spacegrad f(z)}{z – a} + \lr{ \spacegrad \inv{z – a}} f(z) } I dA \\
&= \Be_1 \int f(z) \spacegrad 2 \pi G(\Bx – \Ba) \Be_1 I dA \\
&= 2 \pi \Be_1 \int \delta^2(\Bx – \Ba) \Be_1 f(\Bx) I dA \\
&= 2 \pi \Be_1^2 f(\Ba) I \\
&= 2 \pi I f(a),
\end{aligned}
\end{equation}
where we’ve made use of the analytic condition \( \spacegrad f = 0 \), and the fact that \( f \) and \( 1/(z-a) \), both even multivectors, commute.

The Cauchy integral equation
\begin{equation}\label{eqn:residueGreens:380}
f(a) = \inv{2 \pi I} \oint dz\, \frac{f(z)}{z – a},
\end{equation}
falls out naturally. This sort of residue calculation always seemed a bit miraculous. By introducing a geometric algebra encoding of complex numbers, we get a new and interesting interpretation. In particular,

  1. the imaginary factor in the geometric algebra formulation of this identity is an oriented unit area coming directly from the area element,
  2. the factor of \( 2 \pi \) comes directly from the Green’s function for the gradient,
  3. the fact that this particular form of integral picks up only the contribution at the point \( z = a \) is no longer mysterious seeming. This is directly due to delta-function filtering.

Also, if we are looking for an understanding of how to generalize the Cauchy equation to more general multivector functions, we now also have a good clue how that would be done.

References

[1] C. Doran and A.N. Lasenby. Geometric algebra for physicists. Cambridge University Press New York, Cambridge, UK, 1st edition, 2003.

Summary of some gradient related Green’s functions

October 28, 2025 math and physics play , , , , , ,

[Click here for a PDF version of this post]

Here is a summary of Green’s functions for a number of gradient related differential operators (many of which are of interest for electrodynamics, and most of them have been derived recently in blog posts.) These Green’s functions all satisfy
\begin{equation}\label{eqn:deltaFunctions:120}
\delta(\Bx – \Bx’) = L G(\Bx, \Bx’).
\end{equation}

Let \( \Br = \Bx – \Bx’ \), \( r = \Norm{\Br} \), \( \mathbf{\hat{r}} = \Br/r \), and \( \tau = t – t’ \), then

  1. Gradient operator, \( L = \spacegrad \), in 1D, 2D and 3D respectively
    \begin{equation}\label{eqn:deltaFunctions:25}
    \begin{aligned}
    G\lr{ \Bx, \Bx’ } &= \frac{\mathbf{\hat{r}}}{2} \\
    G\lr{ \Bx, \Bx’ } &= \frac{1}{2 \pi} \frac{\mathbf{\hat{r}}}{r} \\
    G\lr{ \Bx, \Bx’ } &= \inv{4 \pi} \frac{\mathbf{\hat{r}}}{r^2}.
    \end{aligned}
    \end{equation}

  2. Laplacian operator, \( L = \spacegrad^2 \), in 1D, 2D and 3D respectively
    \begin{equation}\label{eqn:deltaFunctions:20}
    \begin{aligned}
    G\lr{ \Bx, \Bx’ } &= \frac{r}{2} \\
    G\lr{ \Bx, \Bx’ } &= \frac{1}{2 \pi} \ln r \\
    G\lr{ \Bx, \Bx’ } &= -\frac{1}{4 \pi r}.
    \end{aligned}
    \end{equation}

  3. Second order Helmholtz operator, \( L = \spacegrad^2 + k^2 \) for 1D, 2D and 3D respectively
    \begin{equation}\label{eqn:deltaFunctions:60}
    \begin{aligned}
    G\lr{ \Bx, \Bx’ } &= \pm \frac{1}{2 j k} e^{\pm j k r} \\
    G(\Bx, \Bx’) &= \frac{1}{4 j} H_0^{(1)}(\pm k r) \\
    G\lr{ \Bx, \Bx’ } &= -\frac{1}{4 \pi} \frac{e^{\pm j k r }}{r}.
    \end{aligned}
    \end{equation}

  4. First order Helmholtz operator, \( L = \spacegrad + j k \), in 1D, 2D and 3D respectively

    \begin{equation}\label{eqn:deltaFunctions:80}
    \begin{aligned}
    G\lr{ \Bx, \Bx’ } &= \frac{j}{2} \lr{ \mathbf{\hat{r}} \mp 1 } e^{\pm j k r} \\
    G\lr{ \Bx, \Bx’ } &= \frac{k}{4} \lr{ \pm j \mathbf{\hat{r}} H_1^{(1)}(\pm k r) – H_0^{(1)}(\pm k r) } \\
    G\lr{ \Bx, \Bx’ } &= \frac{e^{\pm j k r}}{4 \pi r} \lr{ jk \lr{ 1 \mp \mathbf{\hat{r}} } + \frac{\mathbf{\hat{r}}}{r} }.
    \end{aligned}
    \end{equation}

    This is also the Green’s function for a left acting operator \( G(\Bx, \Bx’) \lr{ – \lspacegrad + j k } = \delta(\Bx – \Bx’) \).

  5. Wave equation, \( \spacegrad^2 – (1/c^2) \partial_{tt} \), in 1D, 2D and 3D respectively
    \begin{equation}\label{eqn:deltaFunctions:140}
    \begin{aligned}
    G(\Br, \tau) &= -\frac{c}{2} \Theta( \pm \tau – r/c ) \\
    G(\Br, \tau) &= -\inv{2 \pi \sqrt{ \tau^2 – r^2/c^2 } } \Theta( \pm \tau – r/c ) \\
    G(\Br, \tau) &= -\inv{4 \pi r} \delta( \pm \tau – r/c ),
    \end{aligned}
    \end{equation}
    The positive sign is for the retarded solution, and the negative for advancing.

  6. Spacetime gradient \( L = \spacegrad + (1/c) \partial_t \), satisfying \( L G(\Bx – \Bx’, t – t’) = \delta(\Bx – \Bx’) \delta(t – t’) \), in 1D, 2D, and 3D respectively
    \begin{equation}\label{eqn:deltaFunctions:100}
    \begin{aligned}
    G(\Br, \tau)
    &= \inv{2} \lr{ \mathbf{\hat{r}} \pm 1 } \delta(\pm \tau – r/c) \\
    G(\Br, \tau)
    &=
    \frac{
    \lr{\tau^2 – r^2/c^2}^{-3/2}
    }{2 \pi c^2}
    \lr{
    c \lr{ \mathbf{\hat{r}} \pm 1 }
    \lr{\tau^2 – r^2/c^2}
    \delta(\pm \tau – r/c)
    -\lr{ \Br + c \tau }
    \Theta(\pm \tau – r/c)
    }
    \\
    G(\Br, \tau)
    &= \inv{4 \pi r} \delta(\pm \tau – r/c)
    \lr{
    \frac{\mathbf{\hat{r}}}{r}
    +
    \lr{ \mathbf{\hat{r}} \pm 1} \inv{c} \PD{t’}{}
    }
    \end{aligned}
    \end{equation}
    The plus sign is for the retarded solution, and negative for advanced.