Today, I want to write about a pretty simple problem: finding the line where two planes intersect. At the end of the day, it boils down to a simple formula that's easy to find elsewhere online. However, I find the derivation interesting, and it serves as a nice introduction to some powerful techniques that can be used on harder problems like intersecting conic sections.

But first, in case you just want the formula: the planes \(\langle n_1, x\rangle + d_1 = 0\) and \(\langle n_2, x \rangle + d_2 = 0\) intersect along the line \(r_o + t r_d\), where \[\begin{aligned} r_d &:= n_1 \times n_2,\\ r_o &:= \frac {r_d \times(d_2n_1 - d_1n_2)} {\|r_d\|^2}. \end{aligned}\]

Prelude: intersecting two lines in the plane

When solving these sorts of problems, it is helpful to work in homogeneous coordinates. So we represent a point \((x, y) \in \mathbb{R}^2\) by the vector \((x, y, 1) \in \mathbb{R}^3\) (or any scalar multiple of this vector). And similarly, we represent the line \(ax + by + c = 0\) using the vector \((a, b, c) \in \mathbb{R}^3\) (or any scalar multiple of this vector). It's no coincidence that both points and lines share the same representation here, but that's a discussion for another day.

Suppose now that we want to find the intersection between two lines \(\ell_1\) and \(\ell_2\) (which are represented as vectors in \(\mathbb{R}^3\). A point \(p\) lies on lines \(\ell_1\) if \(\langle p, \ell_1\rangle = 0\), and similarly for \(\ell_2\), so their intersection must be a vector \(p \in \mathbb{R}^3\) which is simultaneously orthogonal to both \(\ell_1\) and \(\ell_2\). We can construct such a vector easily by taking the cross product \(\ell_1 \times \ell_2\).

Visually, each of our lines in \(\mathbb{R}^2\) becomes a plane passing through the origin when represented in homogeneous coordinates in \(\mathbb{R}^3\) and the point in which the lines intersect in \(\mathbb{R}^2\) becomes the line passing through the origin in \(\mathbb{R}^3\). This line can easily be found by taking the cross product of the planes' normal vectors, and yields homogeneous coordinates for the intersection point in \(\mathbb{R}^2\). Computing the intersection of these planes is particularly simple since both planes pass through the origin; computing the intersection between general planes will take us a bit more work later on.

Before we move on, though, I'll mention that this exact same construction also works to compute the line that connects two points. If we have points \(p_1, p_2\), then the line passing through these two points must be given by a vector \(\ell \in \mathbb{R}^3\) which is simultaneously orthogonal to \(p_2\) and \(p_2\), i.e. \(\ell = p_1 \times p_2\).

Lines in 3D

Now let's move to 3D. Since we've moved up a dimensions, points and planes are represented by vectors in \(\mathbb{R}^4\). But what about lines? One natural representation of lines is given by Plücker coordinates. If we consider the line traced out by \(r(t) = r_o + t r_d \in \mathbb{R^3}\), its Plücker coordinates are given by the vector \[(r_o \times r_d, r_d) \in \mathbb{R}^6.\] This formula looks a little strange at first, but it has many nice properties: for instance, if we had picked some other point on the line as the base point, say \(r_o + \lambda r_d\), then we would still get the same Plücker coordinates since \((r_o + \lambda r_d) \times r_d = r_o \times r_d\). Similarly, if we scale the line direction \(r_d\) by a constant \(\lambda\), then our Plücker coordinates also get scaled by \(\lambda\), but still represent the same line. And moreover, we can recover an equation from the line by its Plücker coordinates. We can recover the direction from the last 3 coordinates, and we can find a point on the line by applying the identity that \(r_d \times (r_o \times r_d) = \|r_d\|^2 r_o - \langle r_o, r_d\rangle r_d\), yielding the point on our line which is closest to the origin.

What if we're given two points \(p_1, p_2\) and we want the line \(\ell\) connecting them? To start with let's represent out points using ordinary coordinate vectors in \(x_1, x_2 \in \mathbb{R}^3\). We can think of our line as starting at \(x_1\), and proceeding in direction \(x_2 - x_1\), so we obtain Plücker coordinates \[\ell = ((x_2 - x_1) \times x_1, x_2-x_1) = (x_2 \times x_1, x_2 - x_1).\] But if you stare at this formula for a minute, and are familiar with wedge products, you may notice that this formula can be simplified to \(\ell = (x_2, 1) \wedge (x_1, 1).\) Indeed, if we write each \(p_i\) using homogeneous coordinates in \(\mathbb{R}^4\), the equation becomes \[\ell = p_2 \wedge p_1.\]

Planes in 3D

What about the plane \(P\) passing through three points \(p_1, p_2, p_3\)? We can construct its homogeneous coordinates as \(P = *(p_1 \wedge p_2 \wedge p_3)\), noting that \[\langle P, p_i \rangle = p_1 \wedge p_2 \wedge p_3 \wedge p_i = 0,\] for each of our points \(p_i\). Similarly, the plane passing through a point \(p\) and a line \(\ell\) is given by \(P = *(\ell \wedge p)\).

This formula will help us determine when a line \(\ell\) is contained in a plane \(P\): the line \(\ell\) lies in \(P\) if there is some point \(p\) such that \(P = *(\ell \wedge p)\). Using another wedge product identity, we can write this equation as \(P = -\iota_{p^\flat} *\ell\), which in turn is true if and only if \(P \wedge *\ell = 0\).

Intersecting planes in 3D

Now we can finally find the formula to intersect two planes in 3D. If we are given two planes \(P_1\) and \(P_2\), the line \(\ell\) contained in their intersection must satisfy the equations \[\begin{aligned}P_1 \wedge *\ell &= 0,\\P_2 \wedge *\ell &= 0.\end{aligned}\] It turns out that this is precisely the system of equations that we solved above to find the intersection of lines in 2D, just written in exterior algebra! And, it has the same solution (written in exterior algebra): \(\ell = *(P_1 \wedge P_2)\).

All that remains is to unpack all of the exterior algebra a formula in terms of more familiar vector algebra operations. Let's denote our planes as \(P_i = (n_i, d_i)\). Then, as we saw above, their wedge product is given by \[P_1 \wedge P_2 = (n_1 \times n_2, d_2 n_1 - d_1 n_2).\] Then the Hodge star simply swaps the components of this vectors, yielding \[\ell = *(P_1 \wedge P_2) = (d_2 n_1 - d_1 n_2, n_1 \times n_2).\] Finally, we can use the formula from earlier to convert these Plücker coordinates into a point on our line and a direction. Putting everything together, the planes \(\langle n_1, x\rangle + d_1 = 0\) and \(\langle n_2, x \rangle + d_2 = 0\) intersect along the line \(r_o + t r_d\), where \[\begin{aligned} r_d &:= n_1 \times n_2,\\ r_o &:= \frac {r_d \times(d_2n_1 - d_1n_2)} {\|r_d\|^2}. \end{aligned}\]

A more symmetric form

I'll quickly note at the end that everything I covered here is arguably cleaner if you use exterior algebra from the beginning. Rather than representing both points and hyperplanes using vectors in \(\mathbb{R}^{n+1}\), we can represent points as vectors in \(p \in \mathbb{R}^{n+1}\) and hyperplanes as vectors in the dual space \(P \in \Lambda^n \mathbb{R}^{n+1}\). And, rather than looking at the inner product between \(p\) and \(P\) (which is no longer well-defined), we can consider the primal-dual pairing \(\langle P, p\rangle = *(p \wedge P).\) One nice aspect of version of the theory is that then our formula for the line \(\ell\) between two points \(p_1, p_2\) is always given by \(\ell = p_1 \wedge p_2\), regardless of whether we are considering \(\mathbb{R}^2\), \(\mathbb{R}^2\), or \(\mathbb{R}^n\).

Today, I want to take Maxwell's equations and write them in the language of differential forms. The resulting equations are clearly covariant (i.e. they look the same after you apply a Lorentz transformation), and look a lot simpler than Maxwell's equations in vector notation. This is one of my favorite examples of how differential forms can make life easier.

Maxwell's Equations

In CGS units, Maxwell's equations are given by

1. $\nabla \cdot E = 4 \pi \rho$
2. $\nabla \cdot B = 0$
3. $\nabla \times E = -\frac 1 c \frac{\partial B}{\partial t}$
4. $\nabla \times B = \frac{4\pi}c J + \frac 1 c \frac{\partial E}{\partial t}$

$E$ is the electric field, $B$ is the magnetic field, $J$ is the electric current density, and $\rho$ is the electric charge density. $E, B$ and $J$ are vector fields and $\rho$ is a scalar field.

If we want to write Maxwell's equations with differential forms, we need to decide what type of forms will represent $E, B, J$ and $\rho$. Following Stern et al we will decide this based on how these fields are used in various equations.

First, we consider Faraday's law \[ \oint_C E \cdot d\ell = -\frac d {dt} \int_S B \cdot dA \]

$E$ is integrated over a curve, so $E$ naturally corresponds to a 1-form (Since one-forms are objects that you can integrate along curves). We will write the 1-form associated to $E$ as $\eta = E^\flat$. Meanwhile $B$ is integrated over a surface, so $B$ naturally corresponds to a 2-form. We will write this 2-form as $\beta = \star B^\flat$ as the 2-form associated to $B$. $\beta$ can be thought of as a function that measures the flux of $B$ through oriented parallelograms. $J$, the current density, is integrated over surfaces to find the current passing through the surface, so $J$ is naturally a 2-form. We will write this forms as $\mathscr J = \star J^\flat$. Finally, $\rho$ is integrated over volumes to find the enclosed charge, so $\rho$ is naturally a 3-form, and we will call the 3-form $\rho$.

We recall the following rules for translating from vector calculus to differential forms: \[\begin{aligned} (\nabla \cdot v)^\flat &= \star d \star v^\flat\\ (\nabla \times v)^\flat &= \star d v^\flat \end{aligned}\] We can use these rules to write Maxwell's equations in terms of $\eta, \beta, \mathscr J,$ and $\rho$.

We will start with the first equation. On the left hand side, $\nabla \cdot E$ becomes $\star d \star \eta$, which is a 0-form. We want to set it equal to the 3-form $4\pi \rho$. So we use the Hodge star to turn $\star d \star \eta$ into a 3 form and find that $\star \star d \star \eta = 4\pi \rho$. In 3D, $\star\star = 1$, so our equation is just

1'.  $d\star \eta = 4\pi \rho$

Now, we move on to the second equation. $\nabla \cdot B$ becomes $\star d \star (\star \beta)$. Since $\star\star = 1$, this just becomes $\star d \beta$. So our equation is $\star d \beta = 0$. Applying $\star$ to both sides gives $d\beta = 0$.

2'.  $d\beta = 0$

For the third equation, our substitutions gives us $\star d \eta = -\frac 1 c \frac{\partial\star \beta} {\partial t}$. Applying $\star$ to both sides yields $d\eta = -\frac 1 c \star \frac{\partial\star \beta} {\partial t}$. But we can pull the $\star$ inside the derivative to get

3'.  $d \eta = -\frac 1 c \frac{\partial \beta}{\partial t}$

Finally, we translate the last equation. Our substitution rules give us

4'.  $\star d \star \beta = \frac {4\pi} c \star \mathscr J + \frac 1 c \frac{\partial \eta}{\partial t}$


Putting all of the equations together, we have

1'.	$d \star \eta = 4 \pi \rho$
2'.	$d \beta = 0$
3'.	$d \eta = - \frac 1 c \frac{\partial \beta}{\partial t}$
4'.	$\star d \star \beta = \frac{4\pi}{c} \star \mathscr J + \frac 1 c \frac {\partial \eta}{\partial t}$

Now, we have written the equations using differential forms. But the equations still don't look very relativistic yet - we still have a big distinction between space and time derivatives. For our next step, we will stop thinking about forms on space that change over time, and instead think about forms on $3+1$-dimensional spacetime (i.e. spacetime with 3 spatial dimensions and 1 time dimension). In spacetime, we have both the spatial exterior derivative, the spatial Hodge star, the spacetime exterior derivative, and the spacetime Hodge star. We will denote the spatial operators $d_s, \star_s$ respectively, and we will denote the spacetime operators $d$ and $\star$.

The Exterior Derivative and Hodge Star in Spacetime

Before we can write Maxwell's equations in spacetime, we have to learn about how our spatial operators are related to our spacetime operators. We will use the convention that coordinates in spacetime are written like \[ (x^0, x^1, x^2, x^3) = (ct, x, y, z) \]

The Exterior Derivative

Now, we will look at the relationship between $d$ and $d_s$. Let $\omega = \sum_I \omega_I dx^I$ be a spatial differential form (i.e. no component of $\omega$ involves a $dx^0$). We can compute $d\omega$ as follows \[\begin{aligned} d\omega &= \sum_I d \omega_i \wedge dx^I\\ &= (dx^0 \wedge \partial_0 + d_s)\omega \end{aligned}\]

So we see that $d = dx^0 \wedge \partial_0 + d_s$. The spacetime exterior derivative is just the spatial exterior derivative with an extra term related to the time derivative.

The Hodge Star of Spatial Forms

Now, we will relate $\star$ and $\star_s$. Let $\omega$ be a spatial $k$-form. The spacetime Hodge star is defined by the property that \[ \omega \wedge \star \omega = \left\langle \omega, \omega\right\rangle \mu \] Here $\mu = dx^0 \wedge dx^1 \wedge dx^2 \wedge dx^3$ is the spacetime volume form. Let $\mu_s = dx^1 \wedge dx^2 \wedge dx^3$ be the spatial volume form. Clearly $\mu = dx^0 \wedge \mu_s$. Furthermore, we know that $\omega \wedge \star_s \omega = \left\langle \omega, \omega\right\rangle \mu_s$. Therefore, \[\begin{aligned} \omega \wedge \star \omega &= \langle \omega, \omega \rangle \mu\\ &= dx^0 \wedge \langle \omega, \omega \rangle \mu_s\\ &= dx^0 \wedge \omega \wedge \star_s\omega\\ &= \omega \wedge (-1)^k (dx^0 \wedge \star_s \omega) \end{aligned}\] So $\star \omega = (-1)^{k} \; dx^0 \wedge \star_s \omega$ when $\omega$ is purely a spatial $k$-form.

The Hodge Star of Forms with a Time Component

Now, suppose that $\omega = dx^0 \wedge \omega_s$ where $\omega_s$ is a spatial form. Then $\left\langle \omega, \omega\right\rangle = -\left\langle \omega_s,\omega_s\right\rangle$, so we need $\omega \wedge \star \omega = -\left\langle\omega_s,\omega_s\right\rangle\mu$. We know that $\omega_s \wedge \star_s \omega_s = \left\langle\omega_s,\omega_s\right\rangle \mu_s$. Thus, \[\omega \wedge \star_s \omega_s = dx^0 \wedge \omega_s \wedge \star_s \omega_s = dx^0 \wedge \mu_s = \mu \] So, $\star \omega = -\star_s \omega$ when $\omega$ is the wedge product of $dx^0$ and a spatial form.

Finally, we note for completeness that $\star dx^0 = -\mu_s$.

Covariant formulation of Maxwell's equations

Finally, we've developed all of the tools we need to write Maxwell's equations in spacetime. We will begin with the homogeneous equations (equations two and three). Maxwell's second equation tells us that $d\beta = 0$ and Maxwell's third equation tells us that $d_s \eta + \frac 1 c \frac{\partial \beta}{\partial t} = 0$. Because, we write coordinates in spacetime as \[ (x^0, x^1, x^2, x^3) = (ct, x, y, z) \] it turns out that $\frac 1 c \frac{\partial \beta}{\partial t} = \frac{\partial \beta}{\partial x^0} =: \partial_0 \beta$. So we can write Maxwell's third equation as $d_s \eta + \partial_0 \beta = 0$

The second equation is an equation of 3-forms, and the third equation is an equation of 2-forms. We can make them both equations of 3-forms by wedging the third equation with $dx^0$. Then we have $d_s \beta = 0$ and $dx^0 \wedge(d_s \eta + \partial_0\beta) = 0$. We can add these together to get $d_s \beta + dx^0 \wedge \partial_0 \beta + dx^0 \wedge d_s\eta = 0$. We note that because we are adding together forms of different types, their sum is 0 if and only if the individual terms in the sum are 0. So this equation expresses both Maxwell's second law and his third law. Inspecting the sum, we see that the first two terms are our expression for $d\beta$! Furthermore, the last term is $d(\eta \wedge dx^0)$. \[\begin{aligned} d(\eta\wedge dx^0) &= d\eta \wedge dx^0\\ &= (d_s \eta + dx^0 \wedge \partial_0 c\eta) \wedge dx^0 \\ &= d_s \eta \wedge dx^0\\ &= dx^0 \wedge d_s \eta \end{aligned}\] Therefore, we can write Maxwell's second and third equations as $d\beta + d(\eta \wedge dx^0) = 0$, or $d(\beta + \eta \wedge dx^0) = 0$. To simplify even more, we call $F := \beta + \eta \wedge dx^0$ the Faraday tensor, and simply write $dF = 0$.

Now, we consider $d \star F$. We'll start by computing $\star F$. \[ \star F = \star(\beta + \eta \wedge dx^0) = \star \beta + \star(\eta \wedge dx^0) \] Since $\beta$ is a spatial 2-form, $\star \beta = dx^0 \wedge \star_s \beta$. Since $\eta$ is a spatial 1-form, \[ \star(\eta \wedge dx^0) = -\star(dx^0 \wedge \eta) = -(- \star_s \eta) = \star_s \eta\] Putting this together shows us that $\star F = dx^0 \wedge \star_s \beta + \star_s \eta$. Now, we can take the exterior derivative. \[\begin{aligned} d \star F &= d(dx^0 \wedge \star_s \beta + \star_s \eta)\\ &= d_s \star_s \eta + dx^0 \wedge (\partial_0 \star_s \eta - d_s \star_s d_s \beta) \end{aligned}\]

Maxwell's first equation tells us that $d_s \star_s \eta = 4\pi \rho$. Maxwell's fourth equation tells us that $\partial_0\eta - \star_s d_s \star_s \beta = -\frac{4\pi} c \star_s \mathscr J$. Therefore, \[\begin{aligned} d\star F &= 4\pi \star \rho\;dx^0 -\frac {4\pi} c dx^0 \wedge \mathscr J\\ \end{aligned}\] We define $c\rho - dx^0 \wedge \mathscr J = \mathfrak J$ to be the four-current. Now, our equation reads $d \star F = \frac {4\pi} c \mathfrak J$. This lets us finally express Maxwell's equations (in cgs units) as \[\begin{aligned} dF = 0 \quad\text{and}\quad d \star F = \frac{4\pi} c \mathfrak J \end{aligned}\]

Final Thoughts

I think this form of Maxwell's equations is very pretty. Because $F$ and $\mathfrak J$ are coordinate-independent objects, you can tell these equations must also be Lorentz covariant. And the equations don't distinguish between space and time coordinates.

Noether's theorem tells us that conserved quantities come from symmetries of physical systems. For example, momentum is conserved because the laws of physics are translation invariant.
This deep insight is helpful for understanding when quantities should be conserved. A mass falling off of a building is allowed to gain momentum because the system is not translation invariant - as you move vertically, the gravitational potential changes. However, a train moving along its tracks should conserve momentum because no relevant physical quantity changes as you move around the surface of the earth.

Proving Noether's Theorem

In the system of Hamiltonian Mechanics, the proof of Noether's theorem is surprisingly simple and elegant. First, we need to set up some machinery. Recall Hamilton's equations of motion \[\begin{aligned} \dot p_i &= \frac{\partial H} {\partial q_i}\\ -\dot q_i &= \frac{\partial H} {\partial p_i} \end{aligned}\] Hamilton's equations define a vector field $X_H = (\dot q, \dot p)$ on phase space that describes how a particle evolves over time. The trajectory of a particle starting at position $q$ with momentum $p$ is the integral curve of $X_H$ passing through point $(q,p)$.
We can express Hamilton's equations more simply using a symplectic form. A symplectic form is a closed, nondegenerate differential 2-form. Using $\Omega$, Hamilton's equations become \[ dH = \iota_{X_H} \Omega \] Where $d$ is the exterior derivative and $\iota_{X_H} \Omega$ is the interior product, a one-form defined by $(\iota_{X_H} \Omega)(X_1) = \Omega(X_H, X_1)$.
Now, let $X_G$ be an infinitesimal symmetry transformation. Then $\mathcal{L}_{X_G}H = 0$. That is to say, if we move space a small amount in the $X_G$ direction, the Hamiltonian stays the same. This is exactly what we mean by a symmetry. Furthermore, let $X_G$ be the symplectic gradient of some potential function $U_G$. i.e. $dU_G = \iota_{X_G} \Omega$. Then \[\begin{aligned} 0 &= \mathcal{L}_{X_G} H\\ &= \iota_{X_G} dH + d \iota_{X_G} H &&\text{Cartan's magic formula}\\ &= \iota_{X_G} dH + 0 &&H\;\text{doesn't take arguments, so}\; \iota_{X_G}H = 0\\ &= \iota_{X_G} \iota_{X_H} \Omega &&\text{definition of}\;X_H \\ &= \Omega(X_H, X_G) && \text{definition of the}\; \iota \; \text{operation}\\ &= -\Omega(X_G, X_H) && \Omega\;\text{is antisymmetric}\\ &= -\iota_{X_H} \iota_{X_G} \Omega && \text{definition of the}\;\iota\;\text{operation}\\ &= -\iota_{X_H} dU_G && \text{definition of}\; X_G\\ &= -\iota_{X_H} dU_G + d\iota_{X_H} U_G &&U_G\;\text{doesn't take arguments, so}\; \iota_{X_H}U_G = 0\\ &= -\mathcal{L}_{X_H} U_G && \text{Cartan's magic formula}\\ \end{aligned}\] Therefore, the quantity $U_G$ does not change when we flow along the vector field $X_H$. But flow along $X_H$ is time evolution! So $U_G$ is a conserved quantity over time!

Examples

Translation in One Dimension

Suppose we have a one-dimensional physical system. Furthermore, suppose our Hamiltonian is invariant under translation. The vector field that infinitesimally moves things in the $x$ direction is the vector field that points in the $x$ direction. We need to express this vector field as a symplectic gradient. So we want a function $U(x, p)$ that satisfies \[\begin{aligned} \frac{\partial U(x, p)} {\partial x} &= 0\\ \frac{\partial U(x, p)} {\partial p} &= -1 \end{aligned}\] Clearly, $U(x, p) = -p$. Satisfies this condition. So the quantity $-p$ (and therefore $p$ as well), is conserved in this physical system! Just like that, we have shown the conservation of momentum!

Rotation in Two Dimensions

Suppose we have a two-dimensional physical system that is invariant under rotation. A rotation by an angle $\theta$ is given by the matrix \[\begin{pmatrix} \cos \theta & -\sin \theta\\\sin \theta & \cos \theta\end{pmatrix}\] For very small $\theta$, this matrix is approximately \[\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\] Rotation affects both position and momentum the same way. So an infinitesimal rotation is given by the transformation $\dot x = -y, \dot y = x, \dot p_x = -p_y, \dot p_y = p_x$. Now, to express this vector field as a symplectic gradient, we need a function $U(x, y, p_x, p_y)$ satisfying \[\begin{aligned} \frac {\partial U} {\partial x} &= \dot p_x = -p_y & \frac {\partial U} {\partial y} &= \dot p_y = p_x\\ \frac {\partial U} {\partial p_x} &= -\dot x = y & \frac {\partial U} {\partial p_y} &= -\dot y = -x \end{aligned}\] To satisfy these conditions, we pick $U(x, y, p_x, p_y) = yp_x - xp_y$, which is the angular momentum!