normal

Shortest distance between lines

December 14, 2024 math and physics play 2 comments , , , , , , ,

[Click here for a PDF version of this post]

Motivation.

Having just worked one of Karl’s problems for the distance between two parallel planes, I recalled that when I was in high school, I had trouble with the intuition for the problem of shortest distance between two lines.

In this post, I’ll revisit that ancient trouble and finally come to terms with it.

Distance from origin to a line.

Before doing the two lines problem, let’s look at a similar simpler problem, the shortest distance from the origin to a line. The parametric representation of a line going through point \( \Bp \) with (unit) direction vector \( \mathbf{\hat{d}} \) is
\begin{equation}\label{eqn:shortestDistanceBetweenLines:40}
L: \Bx(r) = \Bp + r \mathbf{\hat{d}}.
\end{equation}
If we want to solve this the dumb but mechanical way, we have only to minimize a length-squared function for the distance to a point on the line
\begin{equation}\label{eqn:shortestDistanceBetweenLines:60}
\begin{aligned}
D
&= \Norm{\Bx}^2 \\
&= \Bp^2 + r^2 + 2 r \Bp \cdot \mathbf{\hat{d}}.
\end{aligned}
\end{equation}
The minimum will occur where the first derivative is zero
\begin{equation}\label{eqn:shortestDistanceBetweenLines:80}
\begin{aligned}
0
&= \PD{r}{D} \\
&= 2 r + 2 \Bp \cdot \mathbf{\hat{d}},
\end{aligned}
\end{equation}
so at the minimum we have
\begin{equation}\label{eqn:shortestDistanceBetweenLines:100}
r = -\Bp \cdot \mathbf{\hat{d}}.
\end{equation}
This means the vector to the nearest point on the line is
\begin{equation}\label{eqn:shortestDistanceBetweenLines:120}
\begin{aligned}
\Bx
&= \Bp – \lr{ \Bp \cdot \mathbf{\hat{d}} }\mathbf{\hat{d}} \\
&= \Bp \mathbf{\hat{d}}^2 – \lr{ \Bp \cdot \mathbf{\hat{d}}}\mathbf{\hat{d}} \\
&= \lr{ \Bp \wedge \mathbf{\hat{d}} } \mathbf{\hat{d}},
\end{aligned}
\end{equation}
or for \(\mathbb{R}^3\)
\begin{equation}\label{eqn:shortestDistanceBetweenLines:140}
\Bx
= -\lr{ \Bp \cross \mathbf{\hat{d}} } \cross \mathbf{\hat{d}}.
\end{equation}

fig. 1. Directed distance from origin to line.

The geometry of the problem is illustrated in fig. 1.

Observe that all the calculation above was superfluous, as observation of the geometry shows that we just wanted the rejection of \( \mathbf{\hat{d}} \) (green) from \( \Bp \) (blue), and could have stated the result directly.

Distance between two lines.

Suppose that we have two lines, specified with a point and direction vector for each line, as illustrated in fig. 2
\begin{equation}\label{eqn:shortestDistanceBetweenLines:20}
\begin{aligned}
L_1&: \Bx(r) = \Bp_1 + r \mathbf{\hat{d}}_1 \\
L_2&: \Bx(s) = \Bp_2 + s \mathbf{\hat{d}}_2 \\
\end{aligned}
\end{equation}

fig. 2. Two lines, not intersecting.

 

The geometry of the problem becomes more clear if we augment this figure, adding a line with the direction \( \mathbf{\hat{d}}_2 \) to \( L_1 \), and adding a line with the direction \( \mathbf{\hat{d}}_1 \) to \( L_2 \), as illustrated in fig. 3, and then visually extending each of those lines to planes that pass through the respective points. We can now rotate our viewpoint so that we look at those planes on edge, which shows that the problem to solve is really just the distance between two parallel planes.

fig 3a. Augmenting the lines with alternate direction vectors to form planes.

fig 3b. Rotation to view planes edge on.

The equations of these planes are just
\begin{equation}\label{eqn:shortestDistanceBetweenLines:160}
\begin{aligned}
P_1&: \Bx(r, s) = \Bp_1 + r \mathbf{\hat{d}}_1 + s \mathbf{\hat{d}}_2 \\
P_2&: \Bx(t, u) = \Bp_2 + t \mathbf{\hat{d}}_2 + u \mathbf{\hat{d}}_1,
\end{aligned}
\end{equation}
or with
\begin{equation}\label{eqn:shortestDistanceBetweenLines:180}
\mathbf{\hat{n}} = \frac{\mathbf{\hat{d}}_1 \cross \mathbf{\hat{d}}_2}{\Norm{\mathbf{\hat{d}}_1 \cross \mathbf{\hat{d}}_2}},
\end{equation}
the normal form for these planes are
\begin{equation}\label{eqn:shortestDistanceBetweenLines:200}
\begin{aligned}
P_1&: \Bx \cdot \mathbf{\hat{n}} = \Bp_1 \cdot \mathbf{\hat{n}} \\
P_2&: \Bx \cdot \mathbf{\hat{n}} = \Bp_2 \cdot \mathbf{\hat{n}}.
\end{aligned}
\end{equation}
Now we can state the distance between the planes trivially, using the results from our previous posts, finding
\begin{equation}\label{eqn:shortestDistanceBetweenLines:220}
d = \Abs{ \Bp_2 \cdot \mathbf{\hat{n}} – \Bp_1 \cdot \mathbf{\hat{n}} },
\end{equation}
or, encoding the triple product in it’s determinant form, the shortest distance between the lines (up to a sign), is
\begin{equation}\label{eqn:shortestDistanceBetweenLines:240}
\boxed{
d = \frac{
\begin{vmatrix}
\Bp_2 – \Bp_1 & \mathbf{\hat{d}}_1 & \mathbf{\hat{d}}_2
\end{vmatrix}
}{
\Norm{ \mathbf{\hat{d}}_1 \cross \mathbf{\hat{d}}_2 }
}.
}
\end{equation}

As a minimization problem.

Trying this as a minimization problem is actually kind of fun, albeit messy. Doing that calculation will give us the directed shortest distance between the lines.
We can form any number of vectors \( \Bm \) connecting the two lines
\begin{equation}\label{eqn:shortestDistanceBetweenLines:260}
\Bp_1 + s \mathbf{\hat{d}}_1 + \Bm = \Bp_2 + t \mathbf{\hat{d}}_2,
\end{equation}
or, with \( \Delta = \Bp_2 – \Bp_1 \), that directed distance is
\begin{equation}\label{eqn:shortestDistanceBetweenLines:280}
\Bm = \Delta + t \mathbf{\hat{d}}_2 – s \mathbf{\hat{d}}_1.
\end{equation}
We seek to minimize the squared length of this vector
\begin{equation}\label{eqn:shortestDistanceBetweenLines:300}
D = \Norm{\Bm}^2 = \Delta^2 + \lr{ t \mathbf{\hat{d}}_2 – s \mathbf{\hat{d}}_1 }^2 + 2 \Delta \cdot \lr{ t \mathbf{\hat{d}}_2 – s \mathbf{\hat{d}}_1 }.
\end{equation}
The minimization constraints are
\begin{equation}\label{eqn:shortestDistanceBetweenLines:320}
\begin{aligned}
0 &= \PD{s}{D} = 2 \lr{ t \mathbf{\hat{d}}_2 – s \mathbf{\hat{d}}_1 } \cdot \lr{ -\mathbf{\hat{d}}_1 } + 2 \Delta \cdot \lr{ – \mathbf{\hat{d}}_1 } \\
0 &= \PD{t}{D} = 2 \lr{ t \mathbf{\hat{d}}_2 – s \mathbf{\hat{d}}_1 } \cdot \lr{ \mathbf{\hat{d}}_2 } + 2 \Delta \cdot \lr{ \mathbf{\hat{d}}_2 }.
\end{aligned}
\end{equation}
With \( \alpha = \mathbf{\hat{d}}_1 \cdot \mathbf{\hat{d}}_2 \), our solution for \( s, t \) is
\begin{equation}\label{eqn:shortestDistanceBetweenLines:340}
\begin{bmatrix}
s \\
t
\end{bmatrix}
=
{
\begin{bmatrix}
1 & -\alpha \\
-\alpha & 1
\end{bmatrix}
}^{-1}
\begin{bmatrix}
\Delta \cdot \mathbf{\hat{d}}_1 \\
-\Delta \cdot \mathbf{\hat{d}}_2 \\
\end{bmatrix},
\end{equation}
so
\begin{equation}\label{eqn:shortestDistanceBetweenLines:360}
\begin{aligned}
t \mathbf{\hat{d}}_2 – s \mathbf{\hat{d}}_1
&=
\begin{bmatrix}
-\mathbf{\hat{d}}_1 & \mathbf{\hat{d}}_2
\end{bmatrix}
\begin{bmatrix}
s \\
t
\end{bmatrix} \\
&=
\inv{1 – \alpha^2}
\begin{bmatrix}
-\mathbf{\hat{d}}_1 & \mathbf{\hat{d}}_2
\end{bmatrix}
\begin{bmatrix}
1 & \alpha \\
\alpha & 1
\end{bmatrix}
\begin{bmatrix}
\Delta \cdot \mathbf{\hat{d}}_1 \\
-\Delta \cdot \mathbf{\hat{d}}_2 \\
\end{bmatrix} \\
&=
\inv{1 – \alpha^2}
\begin{bmatrix}
-\mathbf{\hat{d}}_1 & \mathbf{\hat{d}}_2
\end{bmatrix}
\begin{bmatrix}
\Delta \cdot \mathbf{\hat{d}}_1 – \alpha \Delta \cdot \mathbf{\hat{d}}_2 \\
\alpha \Delta \cdot \mathbf{\hat{d}}_1 -\Delta \cdot \mathbf{\hat{d}}_2 \\
\end{bmatrix} \\
&=
\inv{1 – \alpha^2}
\lr{
-\lr{ \Delta \cdot \mathbf{\hat{d}}_1} \mathbf{\hat{d}}_1 + \alpha \lr{ \Delta \cdot \mathbf{\hat{d}}_2} \mathbf{\hat{d}}_1
+
\alpha \lr{ \Delta \cdot \mathbf{\hat{d}}_1} \mathbf{\hat{d}}_2 – \lr{ \Delta \cdot \mathbf{\hat{d}}_2} \mathbf{\hat{d}}_2
} \\
&=
\inv{1 – \alpha^2}
\lr{
\lr{ \Delta \cdot \mathbf{\hat{d}}_1} \lr{
-\mathbf{\hat{d}}_1 + \alpha \mathbf{\hat{d}}_2
}
+
\lr{ \Delta \cdot \mathbf{\hat{d}}_2} \lr{
-\mathbf{\hat{d}}_2 + \alpha \mathbf{\hat{d}}_1
}
}.
\end{aligned}
\end{equation}
This might look a bit hopeless to simplify, but note that
\begin{equation}\label{eqn:shortestDistanceBetweenLines:380}
\begin{aligned}
\lr{ \mathbf{\hat{d}}_1 \wedge \mathbf{\hat{d}}_2}^2
&=
\lr{ \mathbf{\hat{d}}_1 \wedge \mathbf{\hat{d}}_2} \cdot \lr{ \mathbf{\hat{d}}_1 \wedge \mathbf{\hat{d}}_2} \\
&=
\mathbf{\hat{d}}_1 \cdot \lr{ \mathbf{\hat{d}}_2 \cdot \lr{ \mathbf{\hat{d}}_1 \wedge \mathbf{\hat{d}}_2} } \\
&=
\mathbf{\hat{d}}_1 \cdot \lr{ \alpha \mathbf{\hat{d}}_2 – \mathbf{\hat{d}}_1 } \\
&=
\alpha^2 – 1,
\end{aligned}
\end{equation}
and
\begin{equation}\label{eqn:shortestDistanceBetweenLines:400}
\begin{aligned}
\mathbf{\hat{d}}_2 \cdot \lr{ \mathbf{\hat{d}}_1 \wedge \mathbf{\hat{d}}_2} &= \alpha \mathbf{\hat{d}}_2 – \mathbf{\hat{d}}_1 \\
-\mathbf{\hat{d}}_1 \cdot \lr{ \mathbf{\hat{d}}_1 \wedge \mathbf{\hat{d}}_2} &= -\mathbf{\hat{d}}_2 + \alpha \mathbf{\hat{d}}_1.
\end{aligned}
\end{equation}
Let’s write \( B = \mathbf{\hat{d}}_1 \wedge \mathbf{\hat{d}}_2 \), and plug in our bivector expressions
\begin{equation}\label{eqn:shortestDistanceBetweenLines:420}
\begin{aligned}
t \mathbf{\hat{d}}_2 – s \mathbf{\hat{d}}_1
&=
-\inv{B^2}
\lr{
\lr{ \Delta \cdot \mathbf{\hat{d}}_1} \mathbf{\hat{d}}_2 \cdot B
-\lr{ \Delta \cdot \mathbf{\hat{d}}_2} \mathbf{\hat{d}}_1 \cdot B
} \\
&=
-\inv{B^2}
\lr{
\lr{ \Delta \cdot \mathbf{\hat{d}}_1} \mathbf{\hat{d}}_2
-\lr{ \Delta \cdot \mathbf{\hat{d}}_2} \mathbf{\hat{d}}_1
}
\cdot B \\
&=
-\inv{B^2}
\lr{
\Delta \cdot \lr{ \mathbf{\hat{d}}_1 \wedge \mathbf{\hat{d}}_2}
}
\cdot B \\
&=
-\inv{B^2} \lr{ \Delta \cdot B } \cdot B \\
&=
-\inv{B^2} \gpgradeone{ \lr{ \Delta \cdot B } B } \\
&=
-\inv{B^2} \gpgradeone{ \lr{ \Delta B – \Delta \wedge B } B } \\
&=
-\Delta + \inv{B^2} \lr{ \Delta \wedge B } \cdot B.
\end{aligned}
\end{equation}
The minimum directed distance between the lines is now reduced to just
\begin{equation}\label{eqn:shortestDistanceBetweenLines:440}
\begin{aligned}
\Bm
&= \Delta + t \mathbf{\hat{d}}_2 – s \mathbf{\hat{d}}_1 \\
&= \lr{ \Delta \wedge B } \inv{B}.
\end{aligned}
\end{equation}
Again, had we used the geometry effectively, illustrated in fig. 4, we could have skipped directly to this result. This is the rejection of the plane \( B \) from \( \Delta \), that is, rejection of both \( \mathbf{\hat{d}}_1, \mathbf{\hat{d}}_2 \) from the difference \( \Bp_2 – \Bp_1 \), leaving just the perpendicular shortest connector between the lines.

fig. 4. Projection onto the normal to the parallel planes.

We can also conceptualize this as computing the trivector volume (parallelepiped with edges \( \Bp_2 – \Bp_1, \mathbf{\hat{d}}_1, \mathbf{\hat{d}}_2 \)), and then dividing out the bivector (parallelogram with edges \(\mathbf{\hat{d}}_1, \mathbf{\hat{d}}_2 \)), to find the vector (height) of the parallelepiped.

Observe that rejecting from the plane, is equivalent to projecting onto the normal, so for \(\mathbb{R}^3\) we may translate this to conventional vector algebra, as a projection
\begin{equation}\label{eqn:shortestDistanceBetweenLines:500}
\Bm = \lr{\Delta \cdot \mathbf{\hat{n}}} \mathbf{\hat{n}},
\end{equation}
where \( \mathbf{\hat{n}} = \lr{ \mathbf{\hat{d}}_1 \cross \mathbf{\hat{d}}_2 }/\Norm{\mathbf{\hat{d}}_1 \cross \mathbf{\hat{d}}_2 } \).

If we want the magnitude of this vector, it’s just
\begin{equation}\label{eqn:shortestDistanceBetweenLines:480}
\begin{aligned}
\Norm{\Bm}^2
&=
\inv{B^4} \lr{ \lr{ \Delta \wedge B } B } \lr{ B \lr{ B \wedge \Delta } } \\
&=
\inv{B^2} \lr{ \Delta \wedge B }^2,
\end{aligned}
\end{equation}
or
\begin{equation}\label{eqn:shortestDistanceBetweenLines:460}
\boxed{
\Norm{\Bm} = \frac{ \Norm{ \lr{\Bp_2 – \Bp_1} \wedge \mathbf{\hat{d}}_1 \wedge \mathbf{\hat{d}}_2 } }{ \Norm{\mathbf{\hat{d}}_1 \wedge \mathbf{\hat{d}}_2} },
}
\end{equation}
which, for \(\mathbb{R}^3\), is equivalent to the triple product result we found above.

Shortest distance between two parallel planes.

December 13, 2024 math and physics play No comments , , ,

[Click here for a PDF version of this post]

The problem.

Helping Karl with his linear algebra exam prep, he asked me about this problem

Problem:

Find the shortest distance between the two parallel planes, \( P_1 \), and \( P_2 \), with respective equations:
\begin{equation*}
\begin{aligned}
x – y + 2 z &= -3 \\
3 x – 3 y + 6 z &= 1.
\end{aligned}
\end{equation*}

A numerical way to tackle the problem.

A fairly straightforward way to tackle this problem is illustrated in the sketch of fig. 1. If we can find a point in the first plane, we can follow the normal to the plane to the next, and compute the length of that connecting vector.

fig. 1. Distance between two planes.

fig. 1. Distance between two planes.

For this problem, let
\begin{equation}\label{eqn:distanceBetweenPlanes:20}
\Bn = (1,-1,2),
\end{equation}
and rescale the two plane equations to use the same normal. That is
\begin{equation}\label{eqn:distanceBetweenPlanes:40}
\begin{aligned}
\Bx_1 \cdot \Bn &= -3 \\
\Bx_2 \cdot \Bn &= \inv{3},
\end{aligned}
\end{equation}
where \( \Bx_1 \) are vectors in the first plane, and \( \Bx_2 \) are vectors in the second plane. Finding a vector in one of the planes isn’t hard. Suppose, for example, that \( \Bx_0 = (\alpha, \beta, \gamma) \) is a vector in the first plane, then
\begin{equation}\label{eqn:distanceBetweenPlanes:60}
\alpha – \beta + 2 \gamma = -3.
\end{equation}
One solution is \( \alpha = -3, \beta = 0, \gamma = 0 \), or \( \Bx_0 = (-3, 0, 0) \). We can follow the normal from that point to the closest point in the second plane by forming
\begin{equation}\label{eqn:distanceBetweenPlanes:80}
\By_0 = \Bx_0 + k \Bn,
\end{equation}
where \( k \) is to be determined. If \( \By_0 \) is a point in the second plane, we must have
\begin{equation}\label{eqn:distanceBetweenPlanes:100}
\begin{aligned}
\inv{3}
&=
\By_0 \cdot \Bn \\
&=
\lr{ \Bx_0 + k \Bn } \cdot \Bn \\
&=
(-3, 0, 0 ) \cdot (1,-1,2) + k (1,-1,2) \cdot (1,-1,2) \\
&=
-3 + 6 k,
\end{aligned}
\end{equation}
or
\begin{equation}\label{eqn:distanceBetweenPlanes:120}
k = \frac{10}{18} = \frac{5}{9}.
\end{equation}
This means the point in plane two closest to \( \Bx_0 = (-3,0,0) \) is
\begin{equation}\label{eqn:distanceBetweenPlanes:140}
\begin{aligned}
\By_0
&= (-3, 0, 0 ) + \frac{5}{9} (1,-1,2) \\
&= \inv{9} (-27 + 5, -5, 10) \\
&= \inv{9} (-22, -5, 10),
\end{aligned}
\end{equation}
and the vector distance between the planes is
\begin{equation}\label{eqn:distanceBetweenPlanes:160}
\begin{aligned}
\By_0 – \Bx_0
&= \inv{9} (-22, -5, 10) – (-3, 0, 0 ) \\
&= \inv{9} (-22 + 27, -5, 10) \\
&= \inv{9} (5, -5, 10).
\end{aligned}
\end{equation}
This vector’s length is \( \sqrt{150}/9 = (5/9) \sqrt{6} \), which is the shortest distance between the planes.

A symbolic approach.

Generally, we get more clarity if we avoid plugging in numbers until the very end, so let’s try a generalization of this problem.

Problem:

Find the shortest distance between the two parallel planes, \( P_1 \), and \( P_2 \), with respective equations:
\begin{equation*}
\begin{aligned}
\Bx_1 \cdot \Bn_1 &= d_1 \\
\Bx_2 \cdot \Bn_2 &= d_2.
\end{aligned}
\end{equation*}

We can use the same approach, but first, let’s rescale the two normals. Let
\begin{equation}\label{eqn:distanceBetweenPlanes:180}
\Bn_2 = t \Bn_1,
\end{equation}
or
\begin{equation}\label{eqn:distanceBetweenPlanes:200}
\Bn_1 \cdot \Bn_2 = t \Bn_1^2,
\end{equation}
so
\begin{equation}\label{eqn:distanceBetweenPlanes:220}
\Bn_2 = \frac{\Bn_1 \cdot \Bn_2}{\Bn_1^2} \Bn_1,
\end{equation}
which means that our plane equations are
\begin{equation}\label{eqn:distanceBetweenPlanes:240}
\begin{aligned}
\Bx_1 \cdot \Bn_1 &= d_1 \\
\Bx_2 \cdot \Bn_1 &= \frac{\Bn_1^2}{\Bn_1 \cdot \Bn_2} d_2,
\end{aligned}
\end{equation}
We can further streamline our plane equation representation, setting \( \ncap = \Bn_1/\Norm{\Bn_1} \), which gives us
\begin{equation}\label{eqn:distanceBetweenPlanes:260}
\begin{aligned}
\Bx_1 \cdot \ncap &= \frac{d_1}{\Norm{\Bn_1}} \\
\Bx_2 \cdot \ncap &= \frac{d_2}{\ncap \cdot \Bn_2}.
\end{aligned}
\end{equation}

This time, let’s assume that we can find a point \( \Bx_0 \) in the first plane, but not actually try to find it. We can still follow the normal to the second plane from that point
\begin{equation}\label{eqn:distanceBetweenPlanes:280}
\By_0 = \Bx_0 + k \ncap,
\end{equation}
but since we only care about the vector distance between the planes, we seek
\begin{equation}\label{eqn:distanceBetweenPlanes:300}
\By_0 -\Bx_0 = k \ncap.
\end{equation}
Now, the constant \( k \), once we find it, is exactly the distance between the planes that we seek. Plugging \( \By_0 \) into the \( P_2 \) equation, we find
\begin{equation}\label{eqn:distanceBetweenPlanes:320}
\begin{aligned}
\frac{d_2}{\ncap \cdot \Bn_2}
&=
\lr{ \Bx_0 + k \ncap } \cdot \ncap \\
&=
\Bx_0 \cdot \ncap + k \\
&=
\frac{d_1}{\Norm{\Bn_1}} + k,
\end{aligned}
\end{equation}
or
\begin{equation}\label{eqn:distanceBetweenPlanes:340}
\boxed{
\Abs{k} = \Norm{\By_0 – \Bx_0} = \Abs{ \frac{d_2}{\ncap \cdot \Bn_2} – \frac{d_1}{\ncap \cdot \Bn_1} }.
}
\end{equation}
If \( \Bn_2 = \Bn_1 = \Bn \), then we have
\begin{equation}\label{eqn:distanceBetweenPlanes:360}
\begin{aligned}
\Norm{\By_0 – \Bx_0} &=
\Abs{
\frac{d_2}{\Bn_1^2/\Norm{\Bn_1}} – \frac{d_1}{\Norm{\Bn_1}}
} \\
&=
\frac{\Abs{d_2 – d_1}}{\Norm{\Bn}},
\end{aligned}
\end{equation}
and if \( \Bn \) is a unit normal, this further reduces to just \( \Abs{d_2 – d_1} \).

Let’s try this for the specific problem originally given. We have \( \Bn_1 = \Bn_2 \), so the distance between the planes is
\begin{equation}\label{eqn:distanceBetweenPlanes:380}
\begin{aligned}
\Norm{\By_0 – \Bx_0}
&= \frac{\Abs{1/3 + 3}}{\sqrt{6}} \\
&= \frac{10}{3 \times 6} \sqrt{6} \\
&= \frac{5}{9} \sqrt{6},
\end{aligned}
\end{equation}
as previously calculated.

Unpacking the fundamental theorem of multivector calculus in two dimensions

January 18, 2021 math and physics play , , , , , , , , , , , , , , , , , , ,

Notes.

Due to limitations in the MathJax-Latex package, all the oriented integrals in this blog post should be interpreted as having a clockwise orientation. [See the PDF version of this post for more sophisticated formatting.]

Guts.

Given a two dimensional generating vector space, there are two instances of the fundamental theorem for multivector integration
\begin{equation}\label{eqn:unpackingFundamentalTheorem:20}
\int_S F d\Bx \lrpartial G = \evalbar{F G}{\Delta S},
\end{equation}
and
\begin{equation}\label{eqn:unpackingFundamentalTheorem:40}
\int_S F d^2\Bx \lrpartial G = \oint_{\partial S} F d\Bx G.
\end{equation}
The first case is trivial. Given a parameterizated curve \( x = x(u) \), it just states
\begin{equation}\label{eqn:unpackingFundamentalTheorem:60}
\int_{u(0)}^{u(1)} du \PD{u}{}\lr{FG} = F(u(1))G(u(1)) – F(u(0))G(u(0)),
\end{equation}
for all multivectors \( F, G\), regardless of the signature of the underlying space.

The surface integral is more interesting. Let’s first look at the area element for this surface integral, which is
\begin{equation}\label{eqn:unpackingFundamentalTheorem:80}
d^2 \Bx = d\Bx_u \wedge d \Bx_v.
\end{equation}
Geometrically, this has the area of the parallelogram spanned by \( d\Bx_u \) and \( d\Bx_v \), but weighted by the pseudoscalar of the space. This is explored algebraically in the following problem and illustrated in fig. 1.

fig. 1. 2D vector space and area element.

Problem: Expansion of 2D area bivector.

Let \( \setlr{e_1, e_2} \) be an orthonormal basis for a two dimensional space, with reciprocal frame \( \setlr{e^1, e^2} \). Expand the area bivector \( d^2 \Bx \) in coordinates relating the bivector to the Jacobian and the pseudoscalar.

Answer

With parameterization \( x = x(u,v) = x^\alpha e_\alpha = x_\alpha e^\alpha \), we have
\begin{equation}\label{eqn:unpackingFundamentalTheorem:120}
\Bx_u \wedge \Bx_v
=
\lr{ \PD{u}{x^\alpha} e_\alpha } \wedge
\lr{ \PD{v}{x^\beta} e_\beta }
=
\PD{u}{x^\alpha}
\PD{v}{x^\beta}
e_\alpha
e_\beta
=
\PD{(u,v)}{(x^1,x^2)} e_1 e_2,
\end{equation}
or
\begin{equation}\label{eqn:unpackingFundamentalTheorem:160}
\Bx_u \wedge \Bx_v
=
\lr{ \PD{u}{x_\alpha} e^\alpha } \wedge
\lr{ \PD{v}{x_\beta} e^\beta }
=
\PD{u}{x_\alpha}
\PD{v}{x_\beta}
e^\alpha
e^\beta
=
\PD{(u,v)}{(x_1,x_2)} e^1 e^2.
\end{equation}
The upper and lower index pseudoscalars are related by
\begin{equation}\label{eqn:unpackingFundamentalTheorem:180}
e^1 e^2 e_1 e_2 =
-e^1 e^2 e_2 e_1 =
-1,
\end{equation}
so with \( I = e_1 e_2 \),
\begin{equation}\label{eqn:unpackingFundamentalTheorem:200}
e^1 e^2 = -I^{-1},
\end{equation}
leaving us with
\begin{equation}\label{eqn:unpackingFundamentalTheorem:140}
d^2 \Bx
= \PD{(u,v)}{(x^1,x^2)} du dv\, I
= -\PD{(u,v)}{(x_1,x_2)} du dv\, I^{-1}.
\end{equation}
We see that the area bivector is proportional to either the upper or lower index Jacobian and to the pseudoscalar for the space.

We may write the fundamental theorem for a 2D space as
\begin{equation}\label{eqn:unpackingFundamentalTheorem:680}
\int_S du dv \, \PD{(u,v)}{(x^1,x^2)} F I \lrgrad G = \oint_{\partial S} F d\Bx G,
\end{equation}
where we have dispensed with the vector derivative and use the gradient instead, since they are identical in a two parameter two dimensional space. Of course, unless we are using \( x^1, x^2 \) as our parameterization, we still want the curvilinear representation of the gradient \( \grad = \Bx^u \PDi{u}{} + \Bx^v \PDi{v}{} \).

Problem: Standard basis expansion of fundamental surface relation.

For a parameterization \( x = x^1 e_1 + x^2 e_2 \), where \( \setlr{ e_1, e_2 } \) is a standard (orthogonal) basis, expand the fundamental theorem for surface integrals for the single sided \( F = 1 \) case. Consider functions \( G \) of each grade (scalar, vector, bivector.)

Answer

From \ref{eqn:unpackingFundamentalTheorem:140} we see that the fundamental theorem takes the form
\begin{equation}\label{eqn:unpackingFundamentalTheorem:220}
\int_S dx^1 dx^2\, F I \lrgrad G = \oint_{\partial S} F d\Bx G.
\end{equation}
In a Euclidean space, the operator \( I \lrgrad \), is a \( \pi/2 \) rotation of the gradient, but has a rotated like structure in all metrics:
\begin{equation}\label{eqn:unpackingFundamentalTheorem:240}
I \grad
=
e_1 e_2 \lr{ e^1 \partial_1 + e^2 \partial_2 }
=
-e_2 \partial_1 + e_1 \partial_2.
\end{equation}

  • \( F = 1 \) and \( G \in \bigwedge^0 \) or \( G \in \bigwedge^2 \). For \( F = 1 \) and scalar or bivector \( G \) we have
    \begin{equation}\label{eqn:unpackingFundamentalTheorem:260}
    \int_S dx^1 dx^2\, \lr{ -e_2 \partial_1 + e_1 \partial_2 } G = \oint_{\partial S} d\Bx G,
    \end{equation}
    where, for \( x^1 \in [x^1(0),x^1(1)] \) and \( x^2 \in [x^2(0),x^2(1)] \), the RHS written explicitly is
    \begin{equation}\label{eqn:unpackingFundamentalTheorem:280}
    \oint_{\partial S} d\Bx G
    =
    \int dx^1 e_1
    \lr{ G(x^1, x^2(1)) – G(x^1, x^2(0)) }
    – dx^2 e_2
    \lr{ G(x^1(1),x^2) – G(x^1(0), x^2) }.
    \end{equation}
    This is sketched in fig. 2. Since a 2D bivector \( G \) can be written as \( G = I g \), where \( g \) is a scalar, we may write the pseudoscalar case as
    \begin{equation}\label{eqn:unpackingFundamentalTheorem:300}
    \int_S dx^1 dx^2\, \lr{ -e_2 \partial_1 + e_1 \partial_2 } g = \oint_{\partial S} d\Bx g,
    \end{equation}
    after right multiplying both sides with \( I^{-1} \). Algebraically the scalar and pseudoscalar cases can be thought of as identical scalar relationships.
  • \( F = 1, G \in \bigwedge^1 \). For \( F = 1 \) and vector \( G \) the 2D fundamental theorem for surfaces can be split into scalar
    \begin{equation}\label{eqn:unpackingFundamentalTheorem:320}
    \int_S dx^1 dx^2\, \lr{ -e_2 \partial_1 + e_1 \partial_2 } \cdot G = \oint_{\partial S} d\Bx \cdot G,
    \end{equation}
    and bivector relations
    \begin{equation}\label{eqn:unpackingFundamentalTheorem:340}
    \int_S dx^1 dx^2\, \lr{ -e_2 \partial_1 + e_1 \partial_2 } \wedge G = \oint_{\partial S} d\Bx \wedge G.
    \end{equation}
    To expand \ref{eqn:unpackingFundamentalTheorem:320}, let
    \begin{equation}\label{eqn:unpackingFundamentalTheorem:360}
    G = g_1 e^1 + g_2 e^2,
    \end{equation}
    for which
    \begin{equation}\label{eqn:unpackingFundamentalTheorem:380}
    \lr{ -e_2 \partial_1 + e_1 \partial_2 } \cdot G
    =
    \lr{ -e_2 \partial_1 + e_1 \partial_2 } \cdot
    \lr{ g_1 e^1 + g_2 e^2 }
    =
    \partial_2 g_1 – \partial_1 g_2,
    \end{equation}
    and
    \begin{equation}\label{eqn:unpackingFundamentalTheorem:400}
    d\Bx \cdot G
    =
    \lr{ dx^1 e_1 – dx^2 e_2 } \cdot \lr{ g_1 e^1 + g_2 e^2 }
    =
    dx^1 g_1 – dx^2 g_2,
    \end{equation}
    so \ref{eqn:unpackingFundamentalTheorem:320} expands to
    \begin{equation}\label{eqn:unpackingFundamentalTheorem:500}
    \int_S dx^1 dx^2\, \lr{ \partial_2 g_1 – \partial_1 g_2 }
    =
    \int
    \evalbar{dx^1 g_1}{\Delta x^2} – \evalbar{ dx^2 g_2 }{\Delta x^1}.
    \end{equation}
    This coordinate expansion illustrates how the pseudoscalar nature of the area element results in a duality transformation, as we end up with a curl like operation on the LHS, despite the dot product nature of the decomposition that we used. That can also be seen directly for vector \( G \), since
    \begin{equation}\label{eqn:unpackingFundamentalTheorem:560}
    dA (I \grad) \cdot G
    =
    dA \gpgradezero{ I \grad G }
    =
    dA I \lr{ \grad \wedge G },
    \end{equation}
    since the scalar selection of \( I \lr{ \grad \cdot G } \) is zero.In the grade-2 relation \ref{eqn:unpackingFundamentalTheorem:340}, we expect a pseudoscalar cancellation on both sides, leaving a scalar (divergence-like) relationship. This time, we use upper index coordinates for the vector \( G \), letting
    \begin{equation}\label{eqn:unpackingFundamentalTheorem:440}
    G = g^1 e_1 + g^2 e_2,
    \end{equation}
    so
    \begin{equation}\label{eqn:unpackingFundamentalTheorem:460}
    \lr{ -e_2 \partial_1 + e_1 \partial_2 } \wedge G
    =
    \lr{ -e_2 \partial_1 + e_1 \partial_2 } \wedge G
    \lr{ g^1 e_1 + g^2 e_2 }
    =
    e_1 e_2 \lr{ \partial_1 g^1 + \partial_2 g^2 },
    \end{equation}
    and
    \begin{equation}\label{eqn:unpackingFundamentalTheorem:480}
    d\Bx \wedge G
    =
    \lr{ dx^1 e_1 – dx^2 e_2 } \wedge
    \lr{ g^1 e_1 + g^2 e_2 }
    =
    e_1 e_2 \lr{ dx^1 g^2 + dx^2 g^1 }.
    \end{equation}
    So \ref{eqn:unpackingFundamentalTheorem:340}, after multiplication of both sides by \( I^{-1} \), is
    \begin{equation}\label{eqn:unpackingFundamentalTheorem:520}
    \int_S dx^1 dx^2\,
    \lr{ \partial_1 g^1 + \partial_2 g^2 }
    =
    \int
    \evalbar{dx^1 g^2}{\Delta x^2} + \evalbar{dx^2 g^1 }{\Delta x^1}.
    \end{equation}

As before, we’ve implicitly performed a duality transformation, and end up with a divergence operation. That can be seen directly without coordinate expansion, by rewriting the wedge as a grade two selection, and expanding the gradient action on the vector \( G \), as follows
\begin{equation}\label{eqn:unpackingFundamentalTheorem:580}
dA (I \grad) \wedge G
=
dA \gpgradetwo{ I \grad G }
=
dA I \lr{ \grad \cdot G },
\end{equation}
since \( I \lr{ \grad \wedge G } \) has only a scalar component.

 

fig. 2. Line integral around rectangular boundary.

Theorem 1.1: Green’s theorem [1].

Let \( S \) be a Jordan region with a piecewise-smooth boundary \( C \). If \( P, Q \) are continuously differentiable on an open set that contains \( S \), then
\begin{equation*}
\int dx dy \lr{ \PD{y}{P} – \PD{x}{Q} } = \oint P dx + Q dy.
\end{equation*}

Problem: Relationship to Green’s theorem.

If the space is Euclidean, show that \ref{eqn:unpackingFundamentalTheorem:500} and \ref{eqn:unpackingFundamentalTheorem:520} are both instances of Green’s theorem with suitable choices of \( P \) and \( Q \).

Answer

I will omit the subtleties related to general regions and consider just the case of an infinitesimal square region.

Start proof:

Let’s start with \ref{eqn:unpackingFundamentalTheorem:500}, with \( g_1 = P \) and \( g_2 = Q \), and \( x^1 = x, x^2 = y \), the RHS is
\begin{equation}\label{eqn:unpackingFundamentalTheorem:600}
\int dx dy \lr{ \PD{y}{P} – \PD{x}{Q} }.
\end{equation}
On the RHS we have
\begin{equation}\label{eqn:unpackingFundamentalTheorem:620}
\int \evalbar{dx P}{\Delta y} – \evalbar{ dy Q }{\Delta x}
=
\int dx \lr{ P(x, y_1) – P(x, y_0) } – \int dy \lr{ Q(x_1, y) – Q(x_0, y) }.
\end{equation}
This pair of integrals is plotted in fig. 3, from which we see that \ref{eqn:unpackingFundamentalTheorem:620} can be expressed as the line integral, leaving us with
\begin{equation}\label{eqn:unpackingFundamentalTheorem:640}
\int dx dy \lr{ \PD{y}{P} – \PD{x}{Q} }
=
\oint dx P + dy Q,
\end{equation}
which is Green’s theorem over the infinitesimal square integration region.

For the equivalence of \ref{eqn:unpackingFundamentalTheorem:520} to Green’s theorem, let \( g^2 = P \), and \( g^1 = -Q \). Plugging into the LHS, we find the Green’s theorem integrand. On the RHS, the integrand expands to
\begin{equation}\label{eqn:unpackingFundamentalTheorem:660}
\evalbar{dx g^2}{\Delta y} + \evalbar{dy g^1 }{\Delta x}
=
dx \lr{ P(x,y_1) – P(x, y_0)}
+
dy \lr{ -Q(x_1, y) + Q(x_0, y)},
\end{equation}
which is exactly what we found in \ref{eqn:unpackingFundamentalTheorem:620}.

End proof.

 

fig. 3. Path for Green’s theorem.

We may also relate multivector gradient integrals in 2D to the normal integral around the boundary of the bounding curve. That relationship is as follows.

Theorem 1.2: 2D gradient integrals.

\begin{equation*}
\begin{aligned}
\int J du dv \rgrad G &= \oint I^{-1} d\Bx G = \int J \lr{ \Bx^v du + \Bx^u dv } G \\
\int J du dv F \lgrad &= \oint F I^{-1} d\Bx = \int J F \lr{ \Bx^v du + \Bx^u dv },
\end{aligned}
\end{equation*}
where \( J = \partial(x^1, x^2)/\partial(u,v) \) is the Jacobian of the parameterization \( x = x(u,v) \). In terms of the coordinates \( x^1, x^2 \), this reduces to
\begin{equation*}
\begin{aligned}
\int dx^1 dx^2 \rgrad G &= \oint I^{-1} d\Bx G = \int \lr{ e^2 dx^1 + e^1 dx^2 } G \\
\int dx^1 dx^2 F \lgrad &= \oint G I^{-1} d\Bx = \int F \lr{ e^2 dx^1 + e^1 dx^2 }.
\end{aligned}
\end{equation*}
The vector \( I^{-1} d\Bx \) is orthogonal to the tangent vector along the boundary, and for Euclidean spaces it can be identified as the outwards normal.

Start proof:

Respectively setting \( F = 1 \), and \( G = 1\) in \ref{eqn:unpackingFundamentalTheorem:680}, we have
\begin{equation}\label{eqn:unpackingFundamentalTheorem:940}
\int I^{-1} d^2 \Bx \rgrad G = \oint I^{-1} d\Bx G,
\end{equation}
and
\begin{equation}\label{eqn:unpackingFundamentalTheorem:960}
\int F d^2 \Bx \lgrad I^{-1} = \oint F d\Bx I^{-1}.
\end{equation}
Starting with \ref{eqn:unpackingFundamentalTheorem:940} we find
\begin{equation}\label{eqn:unpackingFundamentalTheorem:700}
\int I^{-1} J du dv I \rgrad G = \oint d\Bx G,
\end{equation}
to find \( \int dx^1 dx^2 \rgrad G = \oint I^{-1} d\Bx G \), as desireed. In terms of a parameterization \( x = x(u,v) \), the pseudoscalar for the space is
\begin{equation}\label{eqn:unpackingFundamentalTheorem:720}
I = \frac{\Bx_u \wedge \Bx_v}{J},
\end{equation}
so
\begin{equation}\label{eqn:unpackingFundamentalTheorem:740}
I^{-1} = \frac{J}{\Bx_u \wedge \Bx_v}.
\end{equation}
Also note that \( \lr{\Bx_u \wedge \Bx_v}^{-1} = \Bx^v \wedge \Bx^u \), so
\begin{equation}\label{eqn:unpackingFundamentalTheorem:760}
I^{-1} = J \lr{ \Bx^v \wedge \Bx^u },
\end{equation}
and
\begin{equation}\label{eqn:unpackingFundamentalTheorem:780}
I^{-1} d\Bx
= I^{-1} \cdot d\Bx
= J \lr{ \Bx^v \wedge \Bx^u } \cdot \lr{ \Bx_u du – \Bx_v dv }
= J \lr{ \Bx^v du + \Bx^u dv },
\end{equation}
so the right acting gradient integral is
\begin{equation}\label{eqn:unpackingFundamentalTheorem:800}
\int J du dv \grad G =
\int
\evalbar{J \Bx^v G}{\Delta v} du + \evalbar{J \Bx^u G dv}{\Delta u},
\end{equation}
which we write in abbreviated form as \( \int J \lr{ \Bx^v du + \Bx^u dv} G \).

For the \( G = 1 \) case, from \ref{eqn:unpackingFundamentalTheorem:960} we find
\begin{equation}\label{eqn:unpackingFundamentalTheorem:820}
\int J du dv F I \lgrad I^{-1} = \oint F d\Bx I^{-1}.
\end{equation}
However, in a 2D space, regardless of metric, we have \( I a = – a I \) for any vector \( a \) (i.e. \( \grad \) or \( d\Bx\)), so we may commute the outer pseudoscalars in
\begin{equation}\label{eqn:unpackingFundamentalTheorem:840}
\int J du dv F I \lgrad I^{-1} = \oint F d\Bx I^{-1},
\end{equation}
so
\begin{equation}\label{eqn:unpackingFundamentalTheorem:850}
-\int J du dv F I I^{-1} \lgrad = -\oint F I^{-1} d\Bx.
\end{equation}
After cancelling the negative sign on both sides, we have the claimed result.

To see that \( I a \), for any vector \( a \) is normal to \( a \), we can compute the dot product
\begin{equation}\label{eqn:unpackingFundamentalTheorem:860}
\lr{ I a } \cdot a
=
\gpgradezero{ I a a }
=
a^2 \gpgradezero{ I }
= 0,
\end{equation}
since the scalar selection of a bivector is zero. Since \( I^{-1} = \pm I \), the same argument shows that \( I^{-1} d\Bx \) must be orthogonal to \( d\Bx \).

End proof.

Let’s look at the geometry of the normal \( I^{-1} \Bx \) in a couple 2D vector spaces. We use an integration volume of a unit square to simplify the boundary term expressions.

  • Euclidean: With a parameterization \( x(u,v) = u\Be_1 + v \Be_2 \), and Euclidean basis vectors \( (\Be_1)^2 = (\Be_2)^2 = 1 \), the fundamental theorem integrated over the rectangle \( [x_0,x_1] \times [y_0,y_1] \) is
    \begin{equation}\label{eqn:unpackingFundamentalTheorem:880}
    \int dx dy \grad G =
    \int
    \Be_2 \lr{ G(x,y_1) – G(x,y_0) } dx +
    \Be_1 \lr{ G(x_1,y) – G(x_0,y) } dy,
    \end{equation}
    Each of the terms in the integrand above are illustrated in fig. 4, and we see that this is a path integral weighted by the outwards normal.

    fig. 4. Outwards oriented normal for Euclidean space.

  • Spacetime: Let \( x(u,v) = u \gamma_0 + v \gamma_1 \), where \( (\gamma_0)^2 = -(\gamma_1)^2 = 1 \). With \( u = t, v = x \), the gradient integral over a \([t_0,t_1] \times [x_0,x_1]\) of spacetime is
    \begin{equation}\label{eqn:unpackingFundamentalTheorem:900}
    \begin{aligned}
    \int dt dx \grad G
    &=
    \int
    \gamma^1 dt \lr{ G(t, x_1) – G(t, x_0) }
    +
    \gamma^0 dx \lr{ G(t_1, x) – G(t_1, x) } \\
    &=
    \int
    \gamma_1 dt \lr{ -G(t, x_1) + G(t, x_0) }
    +
    \gamma_0 dx \lr{ G(t_1, x) – G(t_1, x) }
    .
    \end{aligned}
    \end{equation}
    With \( t \) plotted along the horizontal axis, and \( x \) along the vertical, each of the terms of this integrand is illustrated graphically in fig. 5. For this mixed signature space, there is no longer any good geometrical characterization of the normal.

    fig. 5. Orientation of the boundary normal for a spacetime basis.

  • Spacelike:
    Let \( x(u,v) = u \gamma_1 + v \gamma_2 \), where \( (\gamma_1)^2 = (\gamma_2)^2 = -1 \). With \( u = x, v = y \), the gradient integral over a \([x_0,x_1] \times [y_0,y_1]\) of this space is
    \begin{equation}\label{eqn:unpackingFundamentalTheorem:920}
    \begin{aligned}
    \int dx dy \grad G
    &=
    \int
    \gamma^2 dx \lr{ G(x, y_1) – G(x, y_0) }
    +
    \gamma^1 dy \lr{ G(x_1, y) – G(x_1, y) } \\
    &=
    \int
    \gamma_2 dx \lr{ -G(x, y_1) + G(x, y_0) }
    +
    \gamma_1 dy \lr{ -G(x_1, y) + G(x_1, y) }
    .
    \end{aligned}
    \end{equation}
    Referring to fig. 6. where the elements of the integrand are illustrated, we see that the normal \( I^{-1} d\Bx \) for the boundary of this region can be characterized as inwards.

    fig. 6. Inwards oriented normal for a Dirac spacelike basis.

References

[1] S.L. Salas and E. Hille. Calculus: one and several variables. Wiley New York, 1990.

Fundamental theorem of geometric calculus for line integrals (relativistic.)

December 16, 2020 math and physics play , , , , , , , , , , , , , , , , , , , , , , , , , , ,

[This post is best viewed in PDF form, due to latex elements that I could not format with wordpress mathjax.]

Background for this particular post can be found in

  1. Curvilinear coordinates and gradient in spacetime, and reciprocal frames, and
  2. Lorentz transformations in Space Time Algebra (STA)
  3. A couple more reciprocal frame examples.

Motivation.

I’ve been slowly working my way towards a statement of the fundamental theorem of integral calculus, where the functions being integrated are elements of the Dirac algebra (space time multivectors in the geometric algebra parlance.)

This is interesting because we want to be able to do line, surface, 3-volume and 4-volume space time integrals. We have many \(\mathbb{R}^3\) integral theorems
\begin{equation}\label{eqn:fundamentalTheoremOfGC:40a}
\int_A^B d\Bl \cdot \spacegrad f = f(B) – f(A),
\end{equation}
\begin{equation}\label{eqn:fundamentalTheoremOfGC:60a}
\int_S dA\, \ncap \cross \spacegrad f = \int_{\partial S} d\Bx\, f,
\end{equation}
\begin{equation}\label{eqn:fundamentalTheoremOfGC:80a}
\int_S dA\, \ncap \cdot \lr{ \spacegrad \cross \Bf} = \int_{\partial S} d\Bx \cdot \Bf,
\end{equation}
\begin{equation}\label{eqn:fundamentalTheoremOfGC:100a}
\int_S dx dy \lr{ \PD{y}{P} – \PD{x}{Q} }
=
\int_{\partial S} P dx + Q dy,
\end{equation}
\begin{equation}\label{eqn:fundamentalTheoremOfGC:120a}
\int_V dV\, \spacegrad f = \int_{\partial V} dA\, \ncap f,
\end{equation}
\begin{equation}\label{eqn:fundamentalTheoremOfGC:140a}
\int_V dV\, \spacegrad \cross \Bf = \int_{\partial V} dA\, \ncap \cross \Bf,
\end{equation}
\begin{equation}\label{eqn:fundamentalTheoremOfGC:160a}
\int_V dV\, \spacegrad \cdot \Bf = \int_{\partial V} dA\, \ncap \cdot \Bf,
\end{equation}
and want to know how to generalize these to four dimensions and also make sure that we are handling the relativistic mixed signature correctly. If our starting point was the mess of equations above, we’d be in trouble, since it is not obvious how these generalize. All the theorems with unit normals have to be handled completely differently in four dimensions since we don’t have a unique normal to any given spacetime plane.
What comes to our rescue is the Fundamental Theorem of Geometric Calculus (FTGC), which has the form
\begin{equation}\label{eqn:fundamentalTheoremOfGC:40}
\int F d^n \Bx\, \lrpartial G = \int F d^{n-1} \Bx\, G,
\end{equation}
where \(F,G\) are multivectors functions (i.e. sums of products of vectors.) We’ve seen ([2], [1]) that all the identities above are special cases of the fundamental theorem.

Do we need any special care to state the FTGC correctly for our relativistic case? It turns out that the answer is no! Tangent and reciprocal frame vectors do all the heavy lifting, and we can use the fundamental theorem as is, even in our mixed signature space. The only real change that we need to make is use spacetime gradient and vector derivative operators instead of their spatial equivalents. We will see how this works below. Note that instead of starting with \ref{eqn:fundamentalTheoremOfGC:40} directly, I will attempt to build up to that point in a progressive fashion that is hopefully does not require the reader to make too many unjustified mental leaps.

Multivector line integrals.

We want to define multivector line integrals to start with. Recall that in \(\mathbb{R}^3\) we would say that for scalar functions \( f\), the integral
\begin{equation}\label{eqn:fundamentalTheoremOfGC:180b}
\int d\Bx\, f = \int f d\Bx,
\end{equation}
is a line integral. Also, for vector functions \( \Bf \) we call
\begin{equation}\label{eqn:fundamentalTheoremOfGC:200}
\int d\Bx \cdot \Bf = \inv{2} \int d\Bx\, \Bf + \Bf d\Bx.
\end{equation}
a line integral. In order to generalize line integrals to multivector functions, we will allow our multivector functions to be placed on either or both sides of the differential.

Definition 1.1: Line integral.

Given a single variable parameterization \( x = x(u) \), we write \( d^1\Bx = \Bx_u du \), and call
\begin{equation}\label{eqn:fundamentalTheoremOfGC:220a}
\int F d^1\Bx\, G,
\end{equation}
a line integral, where \( F,G \) are arbitrary multivector functions.

We must be careful not to reorder any of the factors in the integrand, since the differential may not commute with either \( F \) or \( G \). Here is a simple example where the integrand has a product of a vector and differential.

Problem: Circular parameterization.

Given a circular parameterization \( x(\theta) = \gamma_1 e^{-i\theta} \), where \( i = \gamma_1 \gamma_2 \), the unit bivector for the \(x,y\) plane. Compute the line integral
\begin{equation}\label{eqn:fundamentalTheoremOfGC:100}
\int_0^{\pi/4} F(\theta)\, d^1 \Bx\, G(\theta),
\end{equation}
where \( F(\theta) = \Bx^\theta + \gamma_3 + \gamma_1 \gamma_0 \) is a multivector valued function, and \( G(\theta) = \gamma_0 \) is vector valued.

Answer

The tangent vector for the curve is
\begin{equation}\label{eqn:fundamentalTheoremOfGC:60}
\Bx_\theta
= -\gamma_1 \gamma_1 \gamma_2 e^{-i\theta}
= \gamma_2 e^{-i\theta},
\end{equation}
with reciprocal vector \( \Bx^\theta = e^{i \theta} \gamma^2 \). The differential element is \( d^1 \Bx = \gamma_2 e^{-i\theta} d\theta \), so the integrand is
\begin{equation}\label{eqn:fundamentalTheoremOfGC:80}
\begin{aligned}
\int_0^{\pi/4} \lr{ \Bx^\theta + \gamma_3 + \gamma_1 \gamma_0 } d^1 \Bx\, \gamma_0
&=
\int_0^{\pi/4} \lr{ e^{i\theta} \gamma^2 + \gamma_3 + \gamma_1 \gamma_0 } \gamma_2 e^{-i\theta} d\theta\, \gamma_0 \\
&=
\frac{\pi}{4} \gamma_0 + \lr{ \gamma_{32} + \gamma_{102} } \inv{-i} \lr{ e^{-i\pi/4} – 1 } \gamma_0 \\
&=
\frac{\pi}{4} \gamma_0 + \inv{\sqrt{2}} \lr{ \gamma_{32} + \gamma_{102} } \gamma_{120} \lr{ 1 – \gamma_{12} } \\
&=
\frac{\pi}{4} \gamma_0 + \inv{\sqrt{2}} \lr{ \gamma_{310} + 1 } \lr{ 1 – \gamma_{12} }.
\end{aligned}
\end{equation}
Observe how care is required not to reorder any terms. This particular end result is a multivector with scalar, vector, bivector, and trivector grades, but no pseudoscalar component. The grades in the end result depend on both the function in the integrand and on the path. For example, had we integrated all the way around the circle, the end result would have been the vector \( 2 \pi \gamma_0 \) (i.e. a \( \gamma_0 \) weighted unit circle circumference), as all the other grades would have been killed by the complex exponential integrated over a full period.

Problem: Line integral for boosted time direction vector.

Let \( x = e^{\vcap \alpha/2} \gamma_0 e^{-\vcap \alpha/2} \) represent the spacetime curve of all the boosts of \( \gamma_0 \) along a specific velocity direction vector, where \( \vcap = (v \wedge \gamma_0)/\Norm{v \wedge \gamma_0} \) is a unit spatial bivector for any constant vector \( v \). Compute the line integral
\begin{equation}\label{eqn:fundamentalTheoremOfGC:240}
\int x\, d^1 \Bx.
\end{equation}

Answer

Observe that \( \vcap \) and \( \gamma_0 \) anticommute, so we may write our boost as a one sided exponential
\begin{equation}\label{eqn:fundamentalTheoremOfGC:260}
x(\alpha) = \gamma_0 e^{-\vcap \alpha} = e^{\vcap \alpha} \gamma_0 = \lr{ \cosh\alpha + \vcap \sinh\alpha } \gamma_0.
\end{equation}
The tangent vector is just
\begin{equation}\label{eqn:fundamentalTheoremOfGC:280}
\Bx_\alpha = \PD{\alpha}{x} = e^{\vcap\alpha} \vcap \gamma_0.
\end{equation}
Let’s get a bit of intuition about the nature of this vector. It’s square is
\begin{equation}\label{eqn:fundamentalTheoremOfGC:300}
\begin{aligned}
\Bx_\alpha^2
&=
e^{\vcap\alpha} \vcap \gamma_0
e^{\vcap\alpha} \vcap \gamma_0 \\
&=
-e^{\vcap\alpha} \vcap e^{-\vcap\alpha} \vcap (\gamma_0)^2 \\
&=
-1,
\end{aligned}
\end{equation}
so we see that the tangent vector is a spacelike unit vector. As the vector representing points on the curve is necessarily timelike (due to Lorentz invariance), these two must be orthogonal at all points. Let’s confirm this algebraically
\begin{equation}\label{eqn:fundamentalTheoremOfGC:320}
\begin{aligned}
x \cdot \Bx_\alpha
&=
\gpgradezero{ e^{\vcap \alpha} \gamma_0 e^{\vcap \alpha} \vcap \gamma_0 } \\
&=
\gpgradezero{ e^{-\vcap \alpha} e^{\vcap \alpha} \vcap (\gamma_0)^2 } \\
&=
\gpgradezero{ \vcap } \\
&= 0.
\end{aligned}
\end{equation}
Here we used \( e^{\vcap \alpha} \gamma_0 = \gamma_0 e^{-\vcap \alpha} \), and \( \gpgradezero{A B} = \gpgradezero{B A} \). Geometrically, we have the curious fact that the direction vectors to points on the curve are perpendicular (with respect to our relativistic dot product) to the tangent vectors on the curve, as illustrated in fig. 1.

fig. 1. Tangent perpendicularity in mixed metric.

Perfect differentials.

Having seen a couple examples of multivector line integrals, let’s now move on to figure out the structure of a line integral that has a “perfect” differential integrand. We can take a hint from the \(\mathbb{R}^3\) vector result that we already know, namely
\begin{equation}\label{eqn:fundamentalTheoremOfGC:120}
\int_A^B d\Bl \cdot \spacegrad f = f(B) – f(A).
\end{equation}
It seems reasonable to guess that the relativistic generalization of this is
\begin{equation}\label{eqn:fundamentalTheoremOfGC:140}
\int_A^B dx \cdot \grad f = f(B) – f(A).
\end{equation}
Let’s check that, by expanding in coordinates
\begin{equation}\label{eqn:fundamentalTheoremOfGC:160}
\begin{aligned}
\int_A^B dx \cdot \grad f
&=
\int_A^B d\tau \frac{dx^\mu}{d\tau} \partial_\mu f \\
&=
\int_A^B d\tau \frac{dx^\mu}{d\tau} \PD{x^\mu}{f} \\
&=
\int_A^B d\tau \frac{df}{d\tau} \\
&=
f(B) – f(A).
\end{aligned}
\end{equation}
If we drop the dot product, will we have such a nice result? Let’s see:
\begin{equation}\label{eqn:fundamentalTheoremOfGC:180}
\begin{aligned}
\int_A^B dx \grad f
&=
\int_A^B d\tau \frac{dx^\mu}{d\tau} \gamma_\mu \gamma^\nu \partial_\nu f \\
&=
\int_A^B d\tau \frac{dx^\mu}{d\tau} \PD{x^\mu}{f}
+
\int_A^B
d\tau
\sum_{\mu \ne \nu} \gamma_\mu \gamma^\nu
\frac{dx^\mu}{d\tau} \PD{x^\nu}{f}.
\end{aligned}
\end{equation}
This scalar component of this integrand is a perfect differential, but the bivector part of the integrand is a complete mess, that we have no hope of generally integrating. It happens that if we consider one of the simplest parameterization examples, we can get a strong hint of how to generalize the differential operator to one that ends up providing a perfect differential. In particular, let’s integrate over a linear constant path, such as \( x(\tau) = \tau \gamma_0 \). For this path, we have
\begin{equation}\label{eqn:fundamentalTheoremOfGC:200a}
\begin{aligned}
\int_A^B dx \grad f
&=
\int_A^B \gamma_0 d\tau \lr{
\gamma^0 \partial_0 +
\gamma^1 \partial_1 +
\gamma^2 \partial_2 +
\gamma^3 \partial_3 } f \\
&=
\int_A^B d\tau \lr{
\PD{\tau}{f} +
\gamma_0 \gamma^1 \PD{x^1}{f} +
\gamma_0 \gamma^2 \PD{x^2}{f} +
\gamma_0 \gamma^3 \PD{x^3}{f}
}.
\end{aligned}
\end{equation}
Just because the path does not have any \( x^1, x^2, x^3 \) component dependencies does not mean that these last three partials are neccessarily zero. For example \( f = f(x(\tau)) = \lr{ x^0 }^2 \gamma_0 + x^1 \gamma_1 \) will have a non-zero contribution from the \( \partial_1 \) operator. In that particular case, we can easily integrate \( f \), but we have to know the specifics of the function to do the integral. However, if we had a differential operator that did not include any component off the integration path, we would ahve a perfect differential. That is, if we were to replace the gradient with the projection of the gradient onto the tangent space, we would have a perfect differential. We see that the function of the dot product in \ref{eqn:fundamentalTheoremOfGC:140} has the same effect, as it rejects any component of the gradient that does not lie on the tangent space.

Definition 1.2: Vector derivative.

Given a spacetime manifold parameterized by \( x = x(u^0, \cdots u^{N-1}) \), with tangent vectors \( \Bx_\mu = \PDi{u^\mu}{x} \), and reciprocal vectors \( \Bx^\mu \in \textrm{Span}\setlr{\Bx_\nu} \), such that \( \Bx^\mu \cdot \Bx_\nu = {\delta^\mu}_\nu \), the vector derivative is defined as
\begin{equation}\label{eqn:fundamentalTheoremOfGC:240a}
\partial = \sum_{\mu = 0}^{N-1} \Bx^\mu \PD{u^\mu}{}.
\end{equation}
Observe that if this is a full parameterization of the space (\(N = 4\)), then the vector derivative is identical to the gradient. The vector derivative is the projection of the gradient onto the tangent space at the point of evaluation.Furthermore, we designate \( \lrpartial \) as the vector derivative allowed to act bidirectionally, as follows
\begin{equation}\label{eqn:fundamentalTheoremOfGC:260a}
R \lrpartial S
=
R \Bx^\mu \PD{u^\mu}{S}
+
\PD{u^\mu}{R} \Bx^\mu S,
\end{equation}
where \( R, S \) are multivectors, and summation convention is implied. In this bidirectional action,
the vector factors of the vector derivative must stay in place (as they do not neccessarily commute with \( R,S\)), but the derivative operators apply in a chain rule like fashion to both functions.

Noting that \( \Bx_u \cdot \grad = \Bx_u \cdot \partial \), we may rewrite the scalar line integral identity \ref{eqn:fundamentalTheoremOfGC:140} as
\begin{equation}\label{eqn:fundamentalTheoremOfGC:220}
\int_A^B dx \cdot \partial f = f(B) – f(A).
\end{equation}
However, as our example hinted at, the fundamental theorem for line integrals has a multivector generalization that does not rely on a dot product to do the tangent space filtering, and is more powerful. That generalization has the following form.

Theorem 1.1: Fundamental theorem for line integrals.

Given multivector functions \( F, G \), and a single parameter curve \( x(u) \) with line element \( d^1 \Bx = \Bx_u du \), then
\begin{equation}\label{eqn:fundamentalTheoremOfGC:280a}
\int_A^B F d^1\Bx \lrpartial G = F(B) G(B) – F(A) G(A).
\end{equation}

Start proof:

Writing out the integrand explicitly, we find
\begin{equation}\label{eqn:fundamentalTheoremOfGC:340}
\int_A^B F d^1\Bx \lrpartial G
=
\int_A^B \lr{
\PD{\alpha}{F} d\alpha\, \Bx_\alpha \Bx^\alpha G
+
F d\alpha\, \Bx_\alpha \Bx^\alpha \PD{\alpha}{G }
}
\end{equation}
However for a single parameter curve, we have \( \Bx^\alpha = 1/\Bx_\alpha \), so we are left with
\begin{equation}\label{eqn:fundamentalTheoremOfGC:360}
\begin{aligned}
\int_A^B F d^1\Bx \lrpartial G
&=
\int_A^B d\alpha\, \PD{\alpha}{(F G)} \\
&=
\evalbar{F G}{B}
–
\evalbar{F G}{A}.
\end{aligned}
\end{equation}

End proof.

More to come.

In the next installment we will explore surface integrals in spacetime, and the generalization of the fundamental theorem to multivector space time integrals.

References

[1] Peeter Joot. Geometric Algebra for Electrical Engineers. Kindle Direct Publishing, 2019.

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

Helmholtz theorem

October 1, 2016 math and physics play , , , , , , , , , , , , , ,

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

This is a problem from ece1228. I attempted solutions in a number of ways. One using Geometric Algebra, one devoid of that algebra, and then this method, which combined aspects of both. Of the three methods I tried to obtain this result, this is the most compact and elegant. It does however, require a fair bit of Geometric Algebra knowledge, including the Fundamental Theorem of Geometric Calculus, as detailed in [1], [3] and [2].

Question: Helmholtz theorem

Prove the first Helmholtz’s theorem, i.e. if vector \(\BM\) is defined by its divergence

\begin{equation}\label{eqn:helmholtzDerviationMultivector:20}
\spacegrad \cdot \BM = s
\end{equation}

and its curl
\begin{equation}\label{eqn:helmholtzDerviationMultivector:40}
\spacegrad \cross \BM = \BC
\end{equation}

within a region and its normal component \( \BM_{\textrm{n}} \) over the boundary, then \( \BM \) is
uniquely specified.

Answer

The gradient of the vector \( \BM \) can be written as a single even grade multivector

\begin{equation}\label{eqn:helmholtzDerviationMultivector:60}
\spacegrad \BM
= \spacegrad \cdot \BM + I \spacegrad \cross \BM
= s + I \BC.
\end{equation}

We will use this to attempt to discover the relation between the vector \( \BM \) and its divergence and curl. We can express \( \BM \) at the point of interest as a convolution with the delta function at all other points in space

\begin{equation}\label{eqn:helmholtzDerviationMultivector:80}
\BM(\Bx) = \int_V dV’ \delta(\Bx – \Bx’) \BM(\Bx’).
\end{equation}

The Laplacian representation of the delta function in \R{3} is

\begin{equation}\label{eqn:helmholtzDerviationMultivector:100}
\delta(\Bx – \Bx’) = -\inv{4\pi} \spacegrad^2 \inv{\Abs{\Bx – \Bx’}},
\end{equation}

so \( \BM \) can be represented as the following convolution

\begin{equation}\label{eqn:helmholtzDerviationMultivector:120}
\BM(\Bx) = -\inv{4\pi} \int_V dV’ \spacegrad^2 \inv{\Abs{\Bx – \Bx’}} \BM(\Bx’).
\end{equation}

Using this relation and proceeding with a few applications of the chain rule, plus the fact that \( \spacegrad 1/\Abs{\Bx – \Bx’} = -\spacegrad’ 1/\Abs{\Bx – \Bx’} \), we find

\begin{equation}\label{eqn:helmholtzDerviationMultivector:720}
\begin{aligned}
-4 \pi \BM(\Bx)
&= \int_V dV’ \spacegrad^2 \inv{\Abs{\Bx – \Bx’}} \BM(\Bx’) \\
&= \gpgradeone{\int_V dV’ \spacegrad^2 \inv{\Abs{\Bx – \Bx’}} \BM(\Bx’)} \\
&= -\gpgradeone{\int_V dV’ \spacegrad \lr{ \spacegrad’ \inv{\Abs{\Bx – \Bx’}}} \BM(\Bx’)} \\
&= -\gpgradeone{\spacegrad \int_V dV’ \lr{
\spacegrad’ \frac{\BM(\Bx’)}{\Abs{\Bx – \Bx’}}
-\frac{\spacegrad’ \BM(\Bx’)}{\Abs{\Bx – \Bx’}}
} } \\
&=
-\gpgradeone{\spacegrad \int_{\partial V} dA’
\ncap \frac{\BM(\Bx’)}{\Abs{\Bx – \Bx’}}
}
+\gpgradeone{\spacegrad \int_V dV’
\frac{s(\Bx’) + I\BC(\Bx’)}{\Abs{\Bx – \Bx’}}
} \\
&=
-\gpgradeone{\spacegrad \int_{\partial V} dA’
\ncap \frac{\BM(\Bx’)}{\Abs{\Bx – \Bx’}}
}
+\spacegrad \int_V dV’
\frac{s(\Bx’)}{\Abs{\Bx – \Bx’}}
+\spacegrad \cdot \int_V dV’
\frac{I\BC(\Bx’)}{\Abs{\Bx – \Bx’}}.
\end{aligned}
\end{equation}

By inserting a no-op grade selection operation in the second step, the trivector terms that would show up in subsequent steps are automatically filtered out. This leaves us with a boundary term dependent on the surface and the normal and tangential components of \( \BM \). Added to that is a pair of volume integrals that provide the unique dependence of \( \BM \) on its divergence and curl. When the surface is taken to infinity, which requires \( \Abs{\BM}/\Abs{\Bx – \Bx’} \rightarrow 0 \), then the dependence of \( \BM \) on its divergence and curl is unique.

In order to express final result in traditional vector algebra form, a couple transformations are required. The first is that

\begin{equation}\label{eqn:helmholtzDerviationMultivector:800}
\gpgradeone{ \Ba I \Bb } = I^2 \Ba \cross \Bb = -\Ba \cross \Bb.
\end{equation}

For the grade selection in the boundary integral, note that

\begin{equation}\label{eqn:helmholtzDerviationMultivector:740}
\begin{aligned}
\gpgradeone{ \spacegrad \ncap \BX }
&=
\gpgradeone{ \spacegrad (\ncap \cdot \BX) }
+
\gpgradeone{ \spacegrad (\ncap \wedge \BX) } \\
&=
\spacegrad (\ncap \cdot \BX)
+
\gpgradeone{ \spacegrad I (\ncap \cross \BX) } \\
&=
\spacegrad (\ncap \cdot \BX)
–
\spacegrad \cross (\ncap \cross \BX).
\end{aligned}
\end{equation}

These give

\begin{equation}\label{eqn:helmholtzDerviationMultivector:721}
\boxed{
\begin{aligned}
\BM(\Bx)
&=
\spacegrad \inv{4\pi} \int_{\partial V} dA’ \ncap \cdot \frac{\BM(\Bx’)}{\Abs{\Bx – \Bx’}}
–
\spacegrad \cross \inv{4\pi} \int_{\partial V} dA’ \ncap \cross \frac{\BM(\Bx’)}{\Abs{\Bx – \Bx’}} \\
&-\spacegrad \inv{4\pi} \int_V dV’
\frac{s(\Bx’)}{\Abs{\Bx – \Bx’}}
+\spacegrad \cross \inv{4\pi} \int_V dV’
\frac{\BC(\Bx’)}{\Abs{\Bx – \Bx’}}.
\end{aligned}
}
\end{equation}

References

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

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

[3] Garret Sobczyk and Omar Le’on S’anchez. Fundamental theorem of calculus. Advances in Applied Clifford Algebras, 21:221–231, 2011. URL https://arxiv.org/abs/0809.4526.