Box of dust

We will consider the simple example of dust in space to understand how particles transform in relativity.

Consider a box full of dust. The dust has no charge and doesn’t interact with itself. Initially we see the box at rest, i.e. we are in the box frame. From our perspective, the box has $N$ particles and a volume $V$ . We say the rest frame number density is

n_0 = \frac NV.

If we observe the box from some other frame in which the box appears to move at speed $v$ , the volume will look like $V/\gamma$ , so the density in the moving frame will be $n = \gamma N / V = \gamma n_0$ . The number of dust particles stays the same.

From our perspective the dust is moving. This means there are some volumes of space that the dust enters and some that the dust leaves, i.e. some dust flux through space. Assuming the box has a cross-sectional area $A$ in the direction of motion, in some time $\Delta t$ , the area at the front of the box gains $\Delta N = n A v \Delta t$ dust particles, and the area at the back loses $nAv\Delta t$ .

As a differential, we can write the flux as

\frac{\d N}{\d A \d t} = n v.

We define a number flux 3-vector as $\tv n = n \tv v = \gamma n_0 \tv v$ .

Now we can consider some other volume $V$ in space enclosed by a surface $S$ . The rate of change of dust in the volume is (applying the divergence theorem)

\begin{align*} \frac{\partial n}{\partial t} &= - \oint_S \tv n \cdot \d \tv A \\ \int_V \frac{\partial n}{\partial t} \d V &= - \int_V (\nabla \cdot \tv n) \d V \\ 0 &= \int_V \left(\frac{\partial n}{\partial t} + \nabla \cdot \tv n\right) \d V \end{align*}

The only way this holds for all volumes is if

\begin{align*} \frac{\partial n}{\partial t} + \nabla \cdot \tv n &= 0 \\ \frac{\partial n}{\partial t} + \frac{\partial n^x}{\partial x} + \frac{\partial n^y}{\partial y} + \frac{\partial n^z}{\partial z} &= 0. \end{align*}

Note that $n = \gamma n_0$ and $\tv n = \gamma n_0 \tv v$ . This almost looks like a 4-vector. In fact, we can create a 4-vector $\fv N$ by writing

\begin{align*} \fv N &\compeq \big( nc, nv^1, nv^2, nv^3 \big) \\ &\compeq \big( \gamma n_0 c, \gamma n_0v^1, \gamma n_0v^2, \gamma n_0v^3 \big) \\ &= n_0 \fv u. \end{align*}

This lets us write the continuity equation very nicely using the gradient.

\partial_\alpha N^\alpha = 0.

The magnitude is another invariant that follows from the 4-velocity.

N^\alpha N_\alpha = -n_0 c^2.

We can perform a similar analysis with charged dust. Assume every particle has charge $q$ . Then the charge density in space is $\rho_q = nq$ and the current $\tv J = q \tv n = \rho_q \tv v$ . We reach the continuity equation

\begin{align*} \frac{\partial \rho_q}{\partial t} + \nabla \cdot \tv J = 0. \end{align*}

And construct a 4-vector

\begin{align*} \fv J &\compeq \big(\rho_q c, J^1, J^2, J^3 \big) \\ \partial_\alpha J^\alpha &= 0. \end{align*}

If we further complicate the example by considering each dust particle to have a mass $m$ , we can consider the system’s energy. The energy density $\rho_0 = \frac NV mc^2$ .

We know that $N$ is invariant, but $V$ changes by a factor of $1/\gamma$ in a moving frame. Similarly, each particle’s energy picks up a $\gamma$ factor since it gains some kinetic energy in the moving frame. This gives us the energy density in another frame

\rho = \gamma^2 \frac NV mc^2 = \gamma^2 \rho_0.

Since there are two $\gamma$ factors, we assume this is a component of a rank-2 tensor. It turns out that $\rho = T^{00}$ , a component of the stress energy tensor.