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.

Question: The Lorentz force law says that the magnetic force on a charged particle depends on its speed. This seems pretty weird. Why does it happen?

Answer: Explicitly, The Lorentz force law says that \[\vec{F} = q(\vec{v} \times \vec{B} + \vec{E})\] where $q$ is the charge of the particle, $\vec v$ is its velocity, $\vec B$ is the magnetic field and $E$ is the electric field. The formula tells us that magnetic fields only affect moving particles, which is very strange. As it turns out, this happens as a result of Special Relativity! If you want an answer to the question, read Section 2.2. If you want to read other tangentially related ramblings about cool physics, feel free to read this whole thing.

1 Special Relativity

1.1 The Big Idea

The basic idea behind special relativity is that light always travels at the same speed no matter what frame of reference you are in. Which is weird. At some point in physics class, you might have learned about frames of reference and how you can just add velocities together. Imagine you are in a moving car and you throw a ball. Say you throw the ball at velocity $v_{bc}$ relative to a car, and the car is moving with velocity $v_{cg}$ relative to the ground. Classically, you would think that $v_{bg}$, the velocity of the ball relative to the ground is just $v_{bc} + v_{cg}$. It turns out this isn't quite true. If you're in a car and you throw a photon at the speed of light, the speed of the photon relative to the car is $c$, and the speed of the photon relative to the ground is also $c$! In order for this to be true, you need some weird math to transition between frames. The important thing for us is that when things move faster, they get shorter. This is called length contraction.

Specifically, things get shorter by a factor of $\gamma$, the Lorentz factor [same guy as the Lorentz force law]. \[\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\] where $v$ is the velocity of the frame you're switching into relative to the frame you're currently in. It is also convenient to define $\beta = \frac{v}{c}$, so you get \[\gamma = \frac{1}{\sqrt{1 - \beta^2}} \] The other weird thing about special relativity is that as you move faster, time gets slower. Time slows down by a factor of $\gamma$ as well. This is called time dilation.

1.2 Relativistic Velocity Addition

As I said above, velocity addition is slightly more complicated than it looks. Taking relativity into account, the formula for adding two parallel vectors is as follows:

\[u + v = \frac{ u + v}{1 + \frac{uv}{c^2}}\]

Notice that using this formula, $c + c = \frac{2c}{2} = c$. Nothing can go faster than the speed of light.

1.3 4-vectors

One other weird thing about special relativity is that vectors have 4 components (and so are called 4-vectors). The displacement 4 vector, for example, has $x, y$ and $z$ components, but also a time component. It is written $\tilde{x} = (ct, x, y, z)$. The factor of $c$ in with time is to make sure the whole vector is in units of distance. Also, $\|\tilde{x}\|^2 = -(c^2t^2) + x^2 + y^2 + z^2$. The formula for magnitude is not all that important, but is pretty cool.

The useful thing about 4-vectors is that since they have space and time components, we can apply length contraction and time dilation to them in one operation. This operation is super cool, and is called a Lorentz Transformation (specifically, it is called a boost. There are other Lorentz Transformations as well). It's like a rotation in 4 dimensional space. For movement in the $x$ direction, the boost can be written as

\[ \begin{pmatrix} ct'\\x'\\y'\\z' \end{pmatrix} = \begin{pmatrix} \gamma & -\beta\gamma & 0 & 0 \\ -\beta\gamma & \gamma & 0 & 0\\0 & 0 &1&0\\0&0&0&1 \end{pmatrix} \begin{pmatrix} ct\\x\\y\\z \end{pmatrix} \]

It turns out matrices are really useful. Anyway, that's all the math you need to be able to figure out some stuff about magnetism.

2 Magnetic Force

2.1 Random Calculus and Relation to Electric Force

The electric and magnetic forces and fields are very closely related. If you remember learning about electric potential in physics class at some point, you might know that \[V = \frac{kQ}{r}\] where $V$ is the potential due to a point charge $Q$ at a distance $r$ away. This looks really similar to the formula for the electric field since the electric field is the derivative of potential. In the same way, there is a magnetic potential $\vec{A}$ where the magnetic field is sort of the derivative of $\vec{A}$. If you're interested, \[\vec{B} = \nabla \times \vec{A} \] $\vec{B}$ is the 'curl' of $\vec{A}$. $\times$ is the cross product, and $\nabla$ is a weird derivative thing that's not really a vector called 'del'. Anyway, this is important because you can combine $V$ and $\vec{A}$ to get a 4-vector. This is called the electromagnetic four-potential. And since it is a 4-vector, it gives you a feeling that there's a connection between relativity and electromagnetism.

2.2 Why Currents Push Things

So this is actually sort of the answer to our big question question. In this section, I'll show why a moving charge feels a force from a current-carrying wire. The math gets kind of crazy, so I'll explain the idea here and do the math in the next section.

Imagine a charge $q$ with velocity $\vec{v}$ running next to a wire with current $\vec{I}$.

Note that $\vec{I}$, the conventional current, corresponds to negatively charged particles moving in the other direction. Now, switch to the particle's frame of reference. Suddenly, the velocity of the positive particles $\vec{u}_+ \approx \vec{v}$ and the velocity of the negative particles, $\vec{u}_- \approx \vec{v} - \vec{v_0}$. The positive particles are moving faster. So the distance between them decreases more. So the wire gains a positive charge in this frame of reference and repels the point charge $q$. If you do the right hand rule on a point charge moving opposite the direction of current, you see that it gets repelled! So the effects of the magnetic force can be thought of as the effects of the electric force in a different reference frame! In the frame in the picture, the wire is electrically neutral, so it does not exert an electric force. It exerts a magnetic force instead. But this sort of frame shifting only works when the particle is moving. So the force only affects moving particles! And because the charge on the wire depends on length contraction, which depends on $\vec{v}$, it affects faster particles more.

2.3 So What About Uniform $\vec{B}$ Fields?

So how do you get a uniform $\vec{B}$ field? The easiest place is inside a solenoid. Imagine you have a particle moving around inside a giant loop of wire. If the particle moves, then it is repelled by the wire and pushed towards the center. This leads to the particle going in circles, which is the behavior we learn about.

Picture the particle being pushed by a wire generating the magnetic field. If it moves, it feels a force perpendicular to the direction of motion since the length contraction only makes a charge buildup to its side.

3 Math

3.1 Calculus

If you don't know any calculus, you might want to read the following sections. There are a few things that will be useful to do the actual math about wires.

3.2 Derivatives

A derivative is an extension of the idea of slope to lines that are curved. Slope is $\frac{\Delta y}{\Delta x}$. To figure out slope, you see how much a line rises over some $x$ distance. A derivative is $\frac{dy}{dx}$. This just means you do the same thing, but over a really short distance. Think of measuring velocity. You measure how far something travels over a short period of time and divide the tiny distance by the tiny time. In fact, $v = \frac{dx}{dt}$. Also, $a = \frac{dv}{dt} = \frac{d}{dt}\frac{dx}{dt}$. There are a bunch of formulas for derivatives of common functions. For example, $\frac{d x^n}{dx} = nx^{n-1}$. So $\frac{dx^2}{dx} = 2x$. An important thing to note is that you can pull constants out of a derivative. So $\frac{d (ky)}{dx} = k\frac{dy}{dx}$

3.2 Integrals

An integral is the opposite of a derivative. It can be interpreted as an area under a curve. For example, $\int_{0}^2 x\;dx$ gives the area under the curve $y = x$ from the point $x = 0$ to the point $x = 2$. This area is a triangle of base 2 and height 2, so the area is 2.

When I said that the integral is the opposite of a derivative, what I meant was that $\int f(x)\;dx$ can be seen as asking ''What function F(x) has derivative $f(x)$?''. It just happens to work out by mathemagic that if we find such a function $F(x)$, then $\int_a^bf(x)\;dx = F(b) - F(a)$. Going back to our example of $\int_0^2 x\;dx$, we note that $\frac{d\;\frac{1}{2}x^2}{dx} = \frac{1}{2}\cdot2\cdot x = x$. Therefore, we say that our function is $F(x) = x$ and the solution to $\int_0^2 f(x)\;dx = \int_0^2 x \;dx $ is $ F(2) - F(0) = 2 - 0 = 2$.

Another useful way of looking at an integral is as a sort of multiplication. You multiply the function $f$ by the $dx$ thing. Remember that the integral can be seen as taking in a line and finding the area beneath it? It takes in the height and gives the area by ''multiplying'' by width.

In the case where $f(x)$ is constant, we really can use multiplication. $\int_a^b f(x) \;dx$ for constant $f(x)$ is $f(x)(b-a)$.

One last note about integration: it is also the opposite of a derivative in another sense. If you take the derivative of an integral $\frac{d}{dx}\int_a^b f(x)\;dx$, you just get $f(x)$ back as long as the $dx$'s match.

3.3 Gauss' Law

Gauss's Law is a statement about electric fields. In its full math-y glory, it is

\[ \iint_S\!\vec{E} \cdot d\vec{A} = \frac{1}{\epsilon_0}\iiint_V\! \rho\;dV \]

In more understandable terms, it says that the amount of electric field flowing out of an object just depends on the amount of charge inside the object.

\[ E\cdot A = \frac{Q_{enc}}{\epsilon_0} \]

where $E$ is the electric field flowing through your surface, $A$ is the surface area that electric field flows through, $Q_{enc}$ is the enclosed charge, and $\epsilon_0$ is the permittivity of free space. This is actually a really useful equation. For example, we can use it to find the electric field surrounding an infinitely long charged wire.

We note that the area of the side of this cylinder is $2\pi Rh$. We can tell by symmetry that the electric field from our infinitely long charged wire must go radially outwards. Suppose it pointed a little bit up. Then imagine flipping the wire upside down. The wire would still look the same, so the field should still point up. But since we flipped it upside down, the electric field that used to point up should now point down. The electric field can't both point up and point down! So it must not point up or down at all. This means that it just flows out the side of the cylinder. Thus, this is the only area we need. The charge enclosed by the cylinder is the charge density times the length, so $Q_{enc} = \lambda h$. Therefore,

\[\begin{aligned} E \cdot 2\pi Rh &= \frac{1}{\epsilon_0}\lambda h\\ E &= \frac{\lambda}{2\pi R \epsilon_0} \end{aligned}\]

3.4 Forces and Special Relativity

In order to understand electromagnetism in this sense we need to understand how forces transform under change of frame of reference. To do so, we need to slightly adjust our definition of force. At some point in physics class, you might have learned that $\vec{F} = m\vec{a}$. This is also not quite true. It is an approximation assuming that mass stays constant. Which is usually true. But the more general form is more useful in this case. Another random formula from physics is $\Delta \vec{p} = \vec{F}t$. A force applied over time changes momentum. This basically just means that forces push things, which is pretty intuitive. It turns out that this multiplication is actually an integral. $\Delta \vec{p} = \int \vec{F}\;dt$. We can get rid of the integral by taking the derivative of both sides. This tells us that \[\vec{F} = \frac{d\vec{p}}{dt}\] We can check this in the case that mass is constant. We get \[\vec{F} = \frac{d (m\vec{v})}{dt} = m\frac{d\vec{v}}{dt} = m\vec{a}\] So we can be pretty sure the equation is right. This is a useful equation because there is a momentum 4-vector, so we can use relativity! The momentum 4-vector has the form $(\frac{E}{c}, p_x, p_y, p_z)$ where the 'time' component is energy (with a factor of $c$ for units) and the 'space' components are momentum.

Now, we can look at how forces transform. First, we look at how the 4-momentum vector transforms under change of reference frame. We will use the Lorentz Transformation.

\[ \begin{pmatrix} \frac{E}{c}' \\ p_x'\\p_y'\\p_z' \end{pmatrix} = \begin{pmatrix} \gamma & -\beta\gamma & 0 & 0 \\ -\beta\gamma & \gamma & 0 & 0\\0 & 0 &1&0\\0&0&0&1 \end{pmatrix} \begin{pmatrix} \frac{E}{c} \\ p_x\\p_y\\p_z \end{pmatrix} \] \[ \begin{pmatrix} ct'\\x'\\y'\\z' \end{pmatrix} = \begin{pmatrix} \gamma & -\beta\gamma & 0 & 0 \\ -\beta\gamma & \gamma & 0 & 0\\0 & 0 &1&0\\0&0&0&1 \end{pmatrix} \begin{pmatrix} ct\\x\\y\\z \end{pmatrix} \]

We now need to look at $\frac{dp_x}{dt}$ and $\frac{dp_y}{dt}$ ($\frac{dp_z}{dt}$ will be pretty much the same as $y$, so it's not important to look at). Now watch:

\[\begin{aligned}\frac{dp_y'}{dt'} &= \frac{dp_y}{dt'} \quad \{\text{since }p_y' = p_y\\[4ex] &= \frac{dp_y}{\gamma dt - \frac{1}{c}\beta\gamma dx} \quad \{\text{since }ct' = c\gamma t - \beta\gamma x\\[4ex] &= \frac{\frac{d p_y}{dt}}{\gamma - \frac{1}{c}\beta\gamma \frac{dx}{dt}}\quad \{ \text{dividing above and below by}\; dt\\[4ex] &= \frac{F_y}{\gamma(1-\frac{\beta}{c}\vec{u})}\quad \{\text{let }\vec{u} = \frac{dx}{dt} \end{aligned}\]

$\vec{u}$ is the velocity of the particle in the original frame. If we have a particle originally at rest, then $F_\perp'$ (which applies to $F_y'$ and $F_z'$) $ = \frac{1}{\gamma}F_\perp$.

Similarly, we can do as follows for $F_x$:

\[\begin{aligned} \frac{dp_x'}{dt'} &= \frac{-\gamma\beta\frac{dE}{c} + \gamma dp_x}{dt'} \quad \{\text{ since }p_x' = -\gamma\beta\frac{E}{c} + \gamma p_x\\[4ex] &= \frac{\gamma(dp_x - \frac{\beta}{c}dE)}{\gamma dt - \frac{1}{c}\beta\gamma dx} \quad \{\text{ since }ct' = c\gamma t - \beta\gamma x\\[4ex] &= \frac{dp_x - \frac{\beta}{c}dE}{dt - \frac{\beta}{c}dx}\quad \{\text {dividing above and below by}\; \gamma \\[4ex] &= \frac{\frac{dp_x}{dt} - \frac{\beta}{c}\frac{dE}{dt}}{1 - \frac{\beta}{c}\frac{dx}{dt}}\quad \{\text {dividing above and below by}\; dt \\[4ex] &= \frac{\frac{dp_x}{dt} - \frac{\beta}{c}\frac{dE}{dt}}{1 - \frac{\beta}{c}\vec{u}}\quad \{\text {Let }\vec{u} = \frac{dx}{dt}\\[4ex] &= \frac{\frac{dp_x}{dt} - \frac{\beta}{c}\vec{F}\cdot\vec{u}}{1 - \frac{\beta}{c}\vec{u}}\quad \{\text {Random identity :} \frac{dE}{dt} = F\cdot u\\[4ex] &= \frac{F_x - \frac{\beta}{c}\vec{F}\cdot\vec{u}}{1 - \frac{\beta}{c}\vec{u}}\quad \{\text {since } \frac{dp_x}{dt} = F_x\\ \end{aligned}\]

Note that for a particle at rest initially, $\vec{u} = 0$. Thus, $F_\parallel' = F_\parallel$

>Now, we have two equations for transforming forces on particles at rest.

\[\begin{aligned} F_\perp' &= \frac{1}{\gamma} F_\perp\\ F_\parallel' &= F_\parallel \end{aligned}\]

3.5 Why Currents Push Things (Using Math)

Let us say that the positive charges are a distance $d_+$ apart with charge density $\lambda_0$ and the negative charges are a distance $d_-$ apart with charge density $-\lambda_0$.

Now, consider a frame moving at speed $v$ to the right. In this frame, charge $q$ is stationary. The distance between positive charges is contracted to $d_+' = \frac{d_+}{\gamma}$. Note that charge density is proportional to 1 over distance, so the positive charge density is $\lambda_+' = \gamma \lambda_0$.

Now, we have to look at the electron charge density. To do so, we need to find the spacing between the electrons in their rest frame. To do so, we just use length contraction again. The distance at rest is $d_{0-} = \gamma_0d_-$ where $\gamma_0 = \frac{1}{\sqrt{1-\frac{v_0^2}{c^2}}}$\par

Now, we need to boost into the frame moving to the right at $v$. The velocity of this frame relative to the electron rest frame is $v_0' = v - v_0$. Using the relativistic velocity addition formula, we get

\[ v_0' = \frac{ v - v_0}{1 - \frac{v_0v}{c^2}} \]

Dividing by $c$, we get

\[ \beta_0' = \frac{\beta - \beta_0}{1 - \beta\beta_0} \]

Now, to boost from the electron rest frame to the frame moving at $v$, we use a factor of $\gamma_0' = \frac{1}{\sqrt{1 - (\beta_0')^2}}$

Therefore, we can find $d_-' = \frac{1}{\gamma_0'}d_{0-} = \frac{1}{\gamma_0'}\gamma_0d_-$. This means that the electron charge density is $\lambda_0 \frac{\gamma_0'}{\gamma_0}$. We mess around with $\gamma_0'$ and find that

\[\begin{aligned} \gamma_0' &= \frac{1}{\sqrt{1 - (\beta_0')^2}}\\[4ex] &= \frac{1}{\sqrt{1 - (\frac{\beta - \beta_0}{1 - \beta\beta_0})^2}}\\[4ex] &= \frac{1 - \beta\beta_0}{\sqrt{(1 - \beta\beta_0)^2 - (\beta - \beta_0)^2}}\\[4ex] &= \frac{1 - \beta\beta_0}{\sqrt{1 - 2\beta\beta_0 + \beta^2\beta_0^2 - \beta^2 + 2\beta\beta_0 + \beta_0^2}}\\[4ex] &= \frac{1 - \beta\beta_0}{\sqrt{1 + \beta^2\beta_0^2 - \beta^2 - \beta_0^2}}\\[4ex] &= \frac{1 - \beta\beta_0}{\sqrt{(1 - \beta^2)( 1- \beta_0^2)}}\\[4ex] &= \gamma\gamma_0(1 - \beta\beta_0)\quad\text{since } \gamma = \frac{1}{\sqrt{1 - \beta^2}}\text{ and }\gamma_0\frac{1}{\sqrt{1 - \beta_0^2}} \end{aligned}\]

This formula is pretty nice. Using it, we find that the electron density is

\[ \lambda_-' = \lambda_0 \frac{\gamma_0'}{\gamma_0} = \lambda_0\gamma(1 - \beta\beta_0) \]

Now, we find the overall charge density by adding together the positive and negative (electron and nucleus) densities. We get

\[\begin{aligned} \lambda_0' &= \lambda_+' - \lambda_-' \\[2ex] &= \gamma \lambda_0 - \gamma\lambda_0(1 - \beta\beta_0) \\[2ex] &= \lambda_0\gamma\beta\beta_0 \end{aligned}\]

This tells us that

\[ \lambda_0' = \lambda_0\gamma\beta\beta_0 \]

which is greater than 0! So we started with a neutral wire. But by changing our frame of reference, the wire has become charged! This is crazy. Now, we can figure out the force from this charged wire using Gauss' Law. We derived earlier that the force due to a charged wire is

\[ E = \frac{\lambda_0'}{2\pi r \epsilon_0} \]

Therefore, since $F = qE$

\[ F = \frac{q\lambda_0 \gamma \beta \beta_0}{2\pi r \epsilon_0} \]

To transform back into the frame where the particle is moving, we use the fact that $F_\perp' = \frac{1}{\gamma}F_\perp$. We know that the force is pointing radially away from the wire, so we know it is perpendicular.

\[ F_\perp' = -\frac{q\lambda_0\beta\beta_0}{2\pi r \epsilon_0} \]

The negative comes from the fact that stuff is moving left. Now, we consider the $\lambda_0\beta_0$ factor. We know that $\beta_0 = \frac{v}{c}\lambda_0$. A velocity times a charge density gives the amount of charge moving per unit time, which is $I$. But remember, we are using electron flow, which is opposite conventional current. So it's actually $-I$. Thus, $\lambda_0\beta_0 = \frac{-1}{c}I$. Recalling that $\beta = \frac{v}{c}$, we can thus rewrite our equation for $F'_\perp$ as

\[ F'_\perp = \frac{q}{2\pi r \epsilon_0}\frac{I}{c}\frac{v}{c} \]

Now, we use a weird fact. It turns out that $\mu_0$, the permeability of free space, has a value of $\frac{1}{\epsilon_0 c^2}$. This lets us write our equation as

\[ F'_\perp = qv \frac{\mu_0I}{2\pi r} \]

But wait! $\frac{\mu_0 I}{2\pi r}$ is the electric field around a wire! And this is a force that's perpendicular to the velocity of the particle! Which means we can write it as $F = q\vec{v} \times \vec{B}$. Which is exactly the Lorentz Force Law we wanted to understand!

Now, we can imagine wires that generate any magnetic field and this relationship explains the magnetic force.