1 What is a Category?

A category is just a bunch of dots with arrows going between them. The tricky part is interpreting what those dots and arrows mean.

In general, there are three conditions that these dots and arrows need to satisfy. Every dot needs a special arrow that goes from the dot to itself. We call this special arrow the identity.
Also, if we have an arrow from dot $A$ to dot $B$ and an arrow from dot $B$ to dot $C$, then we have to be able to combine these arrows to get an arrow from dot $A$ to dot $C$.
Finally, combining an arrow with the identity arrow shouldn't change it.

And that's all a category is. Just a collection of dots and arrows satisfying these three rules.

2 What Things Are Secretly Categories?

One simple category is Set, the category of sets. The dots in Set are, of course, sets. The arrows are functions between sets. Every set has an identity map to itself, functions between sets can be composed, and composing a function with the identity map does nothing. So the arrows in Set follow our rules.

A lot of categories follow this model, where the dots are some sort of collection and the arrows are functions between them. For example, the dots in the category Grp are groups and the arrows are group homomorphisms. Every group has an identity homomorphism to itself, the composition of two homomorphisms is a homomorphism, and composing a homomorphism with the identity does nothing. So Grp is also a category.

Along the same lines, we have Ring, the category of rings and homomorphism between them, and Top, the category of topological spaces and continuous maps between them. The list goes on.

But not every category is of this form. As a more exotic example, you can look at a group as a one-dot category. Each element of the group corresponds to an arrow in the category. The identity element of the group is the identity arrow. And combining two arrows corresponds to the group operation. This object is still a category, but in this case it doesn't make sense to view the dot as a set and the arrows as structure-preserving transformations.

3 Why Is This A Useful Notion?

To give some motivation, let's just consider Grp for the moment. The categorical viewpoint is that instead of studying how the individual elements of a group fit together, we should study structure-preserving transformations between groups. Often, when we look at a group, we want to know about its subgroups or quotient groups. But since every subgroup is the image of a homomorphism into the group, we can just study subgroups by studying homomorphisms into the group. And by the first isomorphism theorem, every quotient of a group is the image of a homomorphism out of the group. So we can really study subgroups and quotient groups just by studying group homomorphisms.

The idea of studying structured objects by studying maps that preserve that structure pops up all over in mathematics. Category theory just takes this idea to the extreme by studying only these maps and forgetting everything else about the objects.

The first image is from here

We see ice melting and water boiling every day. But why does it happen? If you think about it, it's kind of weird that the properties of water can change so suddenly. Today, I'm going to talk about why this happens. But first, we have to come up with some equations that we can use to describe the properties of gases.

The Van der Waals Equation

At some point in chemistry class, you might have seen the ideal gas law, $PV = nRT$. An ideal gas is a theoretical gas whose particles take up no volume and don't interact with each other. These properties make ideal gases easy to do math about, and give us nice equations like the ideal gas law. In fact, if we change our units, we can get an even nicer ideal gas law. If we let $N$ be the number of particles and $\tau$ be the temperature in joules, then instead of the normal chemistry ideal gas law, we just have $PV = N\tau$. It's nice and simple.

Unfortunately, if we want to study phase transitions, this ideal gas model is a little bit too simple. Many phase transitions happen because of interactions between molecules, which cannot happen in an ideal gas. If we want to understand why water boils, we'll need a more sophisticated model. We can add in some correction terms to make two of our assumptions a bit more realistic. First of all, instead of assuming our particles are point masses, we can instead give them each a little volume. This decreases the amount of empty space in the container of gas. Furthermore, how much the empty space decreases should depend linearly on the number of gas particles. So we can replace $V$ in our equation by $V - Nb$ for some positive constant $b$. Next, we add in an attractive force between the particles. For particles in the middle of the gas, this doesn't do much. They have particles surrounding them on all sides, so the attractive forces in all directions cancel each other out. But it does affect the particles near the edges of the container. They are pulled back towards the middle of the container since almost all of the other particles are in that direction. The magnitude of this force depends on the number density ($\frac N V$) of particles in the container. The number of particles at the edge of the container is also proportional to the number density of particles. This means that the pressure of our non-ideal gas is decreased by $a \frac {N^2}{V^2}$ for some positive constant $a$. These two modifications give us the Van der Waals equation \[ \left(p + \frac {N^2}{V^2} a\right) (V - Nb) = N \tau \] The Van der Waals equation is still a pretty crude approximation and still only works for dilute gases, but it will allow us to understand phase transitions qualitatively.

Phase Transitions

So, now we've got a fancy new equation to model non-ideal gases. Let's see what it tells us. We'll begin by looking at how pressure varies with volume. I picked some arbitrary $a, b$ and $N$ values and plotted $V$ vs $p$ for various temperatures. The plots look like this.

For $\tau$ below some critical temperature, we see that pressure first dips down, then goes up a bit, and then goes back down again as volume decreases. This is weird. Intuitively, if you squish a substance, the pressure should go up. But there's a region in the plot where decreasing the volume decreases the pressure. As you squish the material more, it resists you less. This seems pretty unrealistic. What's going on there?

To analyze this weird behavior, we need to think about the energy stored by squishing the gas. At some point in physics class, you might have seen that $W = \int F\;dx$. Work (energy) is a force applied over a distance. If we multiply through by $\frac{area}{area}$, we get another form of the equation that is often more useful with gases. \[ W = \int F \cdot \frac{area}{area}\;dx = \int \frac{F}{area} \; d(x \cdot area) = \int p\;dV \] So $\int p \; dV$ can be seen as the energy stored in the system. In fact, if our system is at a constant temperature, then $\int p \; dV$ is the Helmholtz free energy of the system. We denote the Helmholtz free energy by $A$.

Now, we'll try to use the Helmholtz free energy to understand what goes on in the weird region of the graph we identified above. We'll assume that our temperature is constant, so we have that the Helmholtz free energy when the system has volume $v$ is $A(v) = -\int^\infty_v p\;dV$ That weird dip in our $p-V$ graph makes the Helmholtz free energy flatten out a little bit in that area. This means that in that region, the Helmholtz free energy is not a convex function of $V$. In the following plot, I exaggerate the effect a bit, but it makes it a lot easier to see what happens next.

Systems don't like to have high energy, and gases are no exception. A gas tries to minimize its Helmholtz free energy however it can. This is what makes Helmholtz free energy a useful topic to study. And by using a neat trick, a gas can 'cheat' to get a lower free energy than our plot above would predict by taking advantage of all the weirdness that was confusing us earlier. Our plot describes the Helmholtz free energy of a homogenous substance. But a gas doesn't have to be homogenous. Some of it could be in one state, and some could be in another. Normally, this is not a helpful thing to do, so gases don't do it. But because of our weird graph, a gas can use this trick to lower its free energy.

If part of the gas is in the red state and part of it is in the blue state, then the Helmholtz free energy of the gas is a weighted average of the red energy and the blue energy. This means that by splitting itself into two states, the gas can follow the black line on the $A-V$ graph instead of the purple one, thus lowering its energy! This is only possible because of the unusual concavity of the graph, which in turn was caused by the weird dip in the $p-V$ graph. But what does this mean? When does a gas randomly split into two states? When it condenses into a liquid!

As a gas condenses, the liquid and gas form can coexist. This is precisely the gas becoming inhomogeneous as a means to lower its free energy.

So there you have it. Phase transitions occur because a material can decrease its energy by splitting into an inhomogeneous combination of states rather than smoothly changing its properties. The sudden, mysterious change from a gas to a liquid is a just trick that gases play to take advantage of little bumps in their energy curves!

The image of condensation came from here.

1 What is a Curried Function?

Currying is a way of taking a function of multiple arguments and returning a series of functions that take one argument. For example, consider the simple two-argument function

 function sum(x, y) {
    return x + y;
  }

If we write sum(1, 2), it gives us 3.

We could instead write it as

 function curried_sum(x) {
    return function (y) {
      return x + y;
    }
  }

Now, we have to call the function differently. Instead of writing sum(1, 2), you write curried_sum(1)(2), but it still gives you 3 as a result.

The obvious question to ask now is "Why on earth would I want to write functions that way?" It seems to add verbosity while still giving you the same end result. In fact, currying functions does give you a very useful tool - it allows you to partially apply your functions.

2 Partially Applied Functions

A partially applied function is what happens when you plug in some arguments to a function, but leave other arguments free. Using the example from above, we have that curried_sum(1) is the function that adds 1 to a number. It's what happens when you take the sum function, but fix 1 as the first argument.

This turns out to be a very useful way of defining new functions. The ++ increment operator present in many languages is essentially just curried_sum(1). But this idea of partially applying functions shows up in many other contexts.

2.1 Indefinite Integrals

In fact, we can actually look at the indefinite integral as a partially applied function. A definite integral is a function that takes two arguments: it takes a function to integrate, and an interval to integrate over. We usually write $\int_a^b f(x)\;dx$, but we could just as easily write definite_integral(f, [a, b]). Now, what happens if we partially apply definite_integral to just the first argument? definite_integral(f) should give us a function that takes in an interval and spits out the integral of f over that interval. And that's pretty much what the indefinite integral does! If $F(x) = \int f(x) \; dx$, then $\int_a^b f(x)\;dx = F(b) - F(a)$.

2.2 Dot Products and Dual Vectors

Now, we'll look at an example from linear algebra. Consider the real vector space $\mathbb{R}^3$. The dual vector space $\left(\mathbb{R}^3\right)^\star$ is the set of linear functions $f : \mathbb{R}^3 \to \mathbb{R}$. The Euclidean dot product on $\mathbb{R}^3$ gives us a bilinear function $g:\mathbb{R}^3 \times \mathbb{R}^3 \to \mathbb{R}^3$. For vectors $u = (u_1, u_2, u_3)$ and $v = (v_1, v_2, v_3)$, we define \[g(u, v) = u \cdot v = \sum_{i = 1}^3 u_i v_i \] Now, suppose we partially apply the dot product. $g(u, -)$ gives us a linear map $\mathbb{R}^3 \to \mathbb{R}$. This is an element of the dual space. So partial application of the dot product allows us to convert vectors to vectors in the dual space! This is a useful tool in tensor analysis (e.g. General Relativity).

2.3 Cross Products

As one last example, we'll look at the cross product of vectors in $\mathbb{R}^3$. Before we talk about the cross product, though, we have to talk about determinants. The determinant of a $3 \times 3$ matrix is signed the volume of the parallelepiped defined by its columns. We can also think of the determinant as taking in 3 vectors and returning the corresponding volume.

If you've seen cross products before, you might have been taught to take the cross product of 2 vectors by taking a determinant. \[ (a_1, a_2, a_3) \times (b_1, b_2, b_3) = \begin{vmatrix} \hat \imath & \hat \jmath & \hat k \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} \]

Now, how can we make this by partially applying functions? This looks similar to a partially applied determinant. $\det(-, A, B)$ is a function that takes in a vector and returns a scalar. So it lives in the dual vector space. But, similar to what we did above, we can convert the dual vector into a regular vector. And if we do that we get the standard formula for the cross product!

We can also look at this from a different geometric perspective. The triple product of three vectors is the quantity $A \cdot (B \times C)$, and is also equal to the signed volume of the parallelepiped that they define. If we write out the formula for the cross product, we find that \[ (a_1, a_2, a_3) \cdot \begin{vmatrix} \hat \imath & \hat \jmath & \hat k \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}\] So we apply our partially applied function by taking a dot product.

You might have heard the cross product of $A$ and $B$ defined as the vector whose magnitude is the area of the parallelogram defined by $A$ and $B$ and which points perpendicular to both $A$ and $B$. We can understand this geometric definition using what we just learned. The cross product is a partially applied volume function, and you finish applying the function by dotting it with the third vector. The volume of a parallelepiped is the volume of the base times the height. If the magnitude of the cross product is the area of the base, and it points in the direction of the height, then dotting that with the third vector gives base $\times$ height. Exactly the volume we wanted!

So, you googled Positive Semidefinite, hoping for someone to explain to you what a positive semidefinite matrix is. Maybe you are interested in doing some Principle Component Analysis, or maybe you want to solve a linear system with a Cholesky decomposition.
Well you came to the right place!

Positive Semidefinite

The positive in positive semidefinite is telling: the property of being positive semidefinite is a generalization of the property of being positive in real numbers to higher dimensional matrices. To further suggest this, we often denote a positive semidefinite matrix $M$ as \[ M \ge 0 \] and we say that \[ M \ge N \] if \[ M - N \ge 0 \] This partial order makes the ring of matrices into an ordered ring, i.e. we can show that if $M \ge N$, then for any $A$, \[ M+A \ge N + A \] and if $A \ge 0$, \[ AM \ge AN \]
TL;DR Positive semidefinite $\sim$ positive.

Bilinear Maps

One way to define positive semidefinite matrices is through their associated bilinear maps:
A bilinear map on a vector space $V$ over a field $K$ is a map \[ \langle\cdot, \cdot\rangle: V \times V \rightarrow K \] so that for every $v$, the maps \[ \langle \cdot, v\rangle \text{ and } \langle v, \cdot \rangle \] are both linear. Another way to look at a bilinear map is as a linear map from $V$ into the dual space $V^*$ of maps from $V$ to $K$ \[ f : V \rightarrow V^* \] where $f(v) = f_v$ is defined as a map in $V^*$ which maps $w$ to $f_v(w)= \langle v, w\rangle$. (If you are familiar with some computer science, this is the operation of currying the first element of the map.)

TL;DR A bilinear map as a linear map between every vectors and maps from vectors to the base field.

For a finite dimensional vector space, $V$ and $V^*$ are isomorphic, so by choosing a basis for each of them, we can identify each map from $V$ to $V^*$ with a square matrix $M$. In this description, a bilinear map is given by \[ vMw^{\intercal} = v\cdot (Mw^{\intercal}) \] where $\cdot$ is the usual dot product with respect to a given basis $\beta = \{\beta_i\}$, where if $v = \sum_i v_i\beta_i$ and $w = \sum_i w_i \beta_i$, then $v\cdot w = \sum_i w_i v_i$. (This can be derived from the definition of the dual basis).
Dually, given any matrix, we can define a bilinear map through the above definition.

TL;DR Bilinear maps and square matrices are isomorphic.

Inner Products

So far, this discussion has been entirely algebraic: these definitions hold over arbitrary fields, and can be entirely expressed through linear maps. To take a geometric point of view, we need to make a different definition. If we have a vector space over $\mathbb{R}$, then we can define an inner product as follows: A bilinear map $\langle \cdot, \cdot \rangle$ is an inner product if the following properties hold

Symmetry

$\langle v, w \rangle = \langle w, v \rangle$

Positive-Semidefiniteness

$\langle v, v \rangle \ge 0$, with $\langle v, v \rangle = 0$ iff $v=0$

An inner product is essentially a generalized way of measuring angles and lengths in a general vector space (see the formula $v\cdot w = |v||w|\cos(\theta)$ for the standard dot product).
Now, using the algebraic formalism, we know that associated to any inner product, there is a matrix $M$ such that $\langle \cdot, \cdot \rangle = vMw^{\intercal}$. The symmetric condition of an inner product corresponds to the symmetry of the matrix $M$, that is to say that $M_{ij} = M_{ji}$.

Perhaps not suprisingly, we define $M$ to be positive semidefinite if its associated bilinear map has the positive semidefiniteness condition.

TL;DR a square matrix $M$ is positive semidefinite iff \[ \forall v \in V,\; vMv^{\intercal} \ge 0 \]

For the rest of this article, we'll assume that these matrices are symmetric.

Eigenvalue characterization

It turns out that looking at the eigenvalues of positive semidefinite matrices give us a way of characterizing them. If we let $v$ be an eigenvalue of $M$, i.e. $Mv = \lambda v$, for some $\lambda$, then we can look at \[ v M v^{\intercal} = v \cdot (\lambda v) = \lambda (v \cdot v) \] Now, using the fact that the standard inner product satisfies the property that $v \cdot v \ge 0$ (you can see this from the fact that it is actually the sum of squares, which are all non-negative), we see that $vMv^{\intercal}$ has the same sign as $\lambda$. Therefore, we have the following result:

If $M$ has a basis of eigenvectors, then $M$ is positive semidefinite iff all of its eigenvalues are non-negative

We given an explicit proof: If $M$ is positive semidefinite, then in particular, we require that for all eigenvectors, \[ v M v^{\intercal} = \lambda (v \cdot v) \ge 0 \] which implies that $\lambda \ge 0$.
Now, if $M$ has the property that all of its eigenvalues are nonnegative, then for any $u, w$, we use the standard dot product defined on the basis of eigenvectors: $\beta = \{b_i\}$ to see that \[ u M u^{\intercal} = \sum_i \sum_j u_iu_j \lambda_i (v_i \cdot v_j) \] Now, for symmetric matrices we have that their eigenvalues are orthogonal, so that $v_i \cdot v_j = 0$ if $i \neq j$. This summation becomes \[ \sum_i u_i^2 \lambda_i \] which is nonnegative if all of the $\lambda_i$ are nonnegative, as the the squares of $u_i$ are positive.
TL;DR In linear algebra, we can generalize concepts about real numbers to concepts about matrices by simply applying those concepts to the eigenvalues of the matrices.Positive Semidefiniteness is just the generalization of non-negativity of the eigenvalues.

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.