# Charge and current

### Charge density

We can imagine space being filled with particles of different kinds, some negatively charged, some positively charged and ssome neutral.

We could imagine so many that it makes sense to think of charge being continuously (smoothly) distributed across space. Let $\mathrm{d}V$ be a small region (volume) of space which contains a small quantity of charge, $\mathrm{d}q$.

Then, the charge density, $\rho$, is defined to be the net charge per unit volume, $$\rho = \frac{\mathrm{d} q}{\mathrm{d} V}$$ which can vary over space and time, $$\rho = \rho(t,x,y,z)$$

We can define the total charge in space, $Q$, as the integral of the charge density over all space, $$Q = \int_{\text{space}}\mathrm{d}q = \int_{\text{space}} \rho \mathrm{d}V$$

Similarly, we can define the total charge, $Q_V$, in a volume of space, as the integral of the charge density over that region, $$Q_V = \int_{V}\mathrm{d}q = \int_{V} \rho \mathrm{d}V$$

Charge density can change, but only in a particular way. First of we have the empirical law of conservation of charge, which says that the total amount of charge in the universe remains constant. But even more than that, charge cannot just disappear in one place and re-appear instantaneously somewhere else. This leads us to the concept of currents.

### Current density

If charge is disappearing from region of space and appearing in another, then there must be some flow of charge between them.

The current along a wire, say, is defined to be the amount of charge passing through the wire per unit time. That is, the current is the rate of change of charge over time, $$I = \frac{\mathrm{d} q}{\mathrm{d} t}$$

We can generalise this by asking how much charge passes into, or out of, a given, fixed, volume of space, $$I_V = \frac{\partial Q_v}{\partial t} = \frac{\partial}{\partial t} \int_{V} \rho \mathrm{d}V = \int_{V} \frac{\partial \rho}{\partial t} \mathrm{d}V$$

The only way that the amount of charge, $Q_v$ inside the volume, $V$, can change is if charge is passing through the surface, $S$ say, of the volume, so we should consider the current through $S$, which we might call $I_S$.

Let $\mathrm{d}\sigma$ is a small surface element of $S$ and let $\hat{n}_\sigma$ be the associated unit normal vector, pointing out of the volume. For ease of notation, let the surface element vector be $$\mathrm{d}\vec{\sigma} = \hat{n}_\sigma \mathrm{d}\sigma$$

Then, we define the current density to be a vector $\vec{j}$, such that the dot product of the current density with the surface element vector is the current passing through the surface element of $S$. That is, $$\vec{j} \cdot \mathrm{d}\vec{\sigma} = \mathrm{d} I_S$$

Then, the total current through the surface is $$I_S = \int_{S} \vec{j} \cdot \mathrm{d}\vec{\sigma}$$

### Continuity equation

Now, if $I_V \lt 0$ so that charge is decreasing inside the volume, then charge must be passing out of the surface, so $I_S \gt 0$. On the other hand, if $I_V \gt 0$ so that charge is increasing inside the volume, then charge must be passing into the surface, $I_S \lt 0$. Of course, the magnitude of $I_V$ and $I_S$ must be the same due to charge conservation, so the only way to satisfy these three conditions is if $$I_V = - I_S$$

Thus, we can write an expression relating the time-rate of change of the charge density inside a volume to the current density on the volume's surface, $$\int_{V} \frac{\partial \rho}{\partial t} \mathrm{d}V = - \int_{S} \vec{j} \cdot \mathrm{d}\vec{\sigma}$$

Now, we use Gauss' theorem - also called the divergence theorem - to convert the surface integral on the right-hand side to a volume integral. Gauss' theorem states that, given a volume $V$, bounded by surface $S$, then for any vector field $\vec{a}$, $$\int_{S} \vec{a} \cdot \mathrm{d}\vec{\sigma} = \int_{V} \nabla \cdot \vec{a} \mathrm{d}V$$

Thus, the relationship between charge density and current density becomes $$\int_{V} \frac{\partial \rho}{\partial t} \mathrm{d}V = - \int_{V} \nabla \cdot \vec{j} \mathrm{d}V$$

We can re-write this as $$\int_{V} \frac{\partial \rho}{\partial t} + \nabla \cdot \vec{j} \mathrm{d}V = 0$$ and since we chose the volume $V$ completely arbitrarily, we arrive at the continuity equation, $$\frac{\partial \rho}{\partial t} + \nabla \cdot \vec{j} = 0$$ which holds over all space.

### Four-current

In the derivation of the continuity equation, we never specified a reference frame at all, so it must be true for all reference frames and we should be able to convert it to tensor notation.

In full, the continuity equation states that $$\frac{\partial \rho}{\partial t} + \frac{\partial j_x}{\partial x} + \frac{\partial j_y}{\partial y} + \frac{\partial j_z}{\partial z} = 0$$

So, define a new four-vector, $j^\mu$, called the four-current, which has the charge density as its time component and $\vec{j}$ as its spatial components, $$j^\mu \rightarrow (\rho, j_x, j_y, j_z)$$ and then we can write the continuity equation succinctly as $$\partial_\mu j^\mu = 0$$

Nb. We can compare the four-current to the four-momentum. The time component of the four-momentum relates to mass which is analogous to charge, related to the time component of the four-current. The spatial components of the four-momentum relate to the motion of mass and the spatial components of the four-current relate to the motion of charge.

### Covariant version of Maxwell's equations

The classical (reduced) set of Maxwell's equations are $$\begin{matrix}\nabla \times \vec{E} &= -\frac{\partial \vec{B}}{\partial t}&\nabla \times \vec{B} &= \frac{\partial \vec{E}}{\partial t} + \vec{j}\\\nabla \cdot \vec{E} &= \rho&\nabla \cdot \vec{B} &= 0\end{matrix}$$

The change from the vacuum equations - by including charges and currents - doesn't involve the two equations on the left-hand side hence, in their covariant form, we still have $$\partial_\mu \tilde{F}^{\mu \nu} = 0$$

But, the equations on the right do change when we introduce charge and current densities. The covariant version becomes $$\partial_\mu F^{\mu \nu} = j^\nu$$ where $j^\nu$ are the components of the four-current, where it is straightforward to show that \begin{align*}\partial_\mu F^{\mu 0} = j^0 &\Rightarrow & \nabla \cdot \vec{E} = \rho\\\partial_\mu F^{\mu m} = j^m &\Rightarrow & \nabla \times \vec{B} = \frac{\partial \vec{E}}{\partial t} + \vec{j}\end{align*}

In other words, the source of electromagnetic fields are produced by electric charges and electric currents.

The lack of magnetic monopoles - that is, the lack of either positive or negative magnetic ‘charge’ - means that only electric charges produce electromagnetic forces.