A bit of everything

Bound systems in the FRLW metric

A common question regarding the standard cosmological model is why matter seems to remain as it is while faraway galaxies receede.

The electron system in FRLW cosmology

The simplest case to consider is the hydrogen atom, in the limit of the nucleus as a fixed charge, in the FRLW metric. For simplicity, let's consider here the case of a flat cosmology, so that we have

$$ds^2 = -dt^2 + a(t) (dr^2 + r^2 d\theta + r^2 \sin^2 \theta d\phi^2)$$

This has a very simple tetrad field of the form

\begin{eqnarray} e_t &=& e^{\hat{t}}\\ e_r &=& \sqrt{a(t)} e^{\hat{r}}\\ e_\theta &=& \sqrt{a(t)} r e^{\hat{\theta}}\\ e_\phi &=& \sqrt{a(t)} r \sin(\theta) e^{\hat{\phi}} \end{eqnarray}

which indeed gives us

$$g_{\mu\nu} = \langle e_\mu, e_\nu \rangle$$

\begin{eqnarray} e^t &=& e_{\hat{t}}\\ e^r &=& \frac{1}{\sqrt{a(t)}} e_{\hat{r}}\\ e^\theta &=& \frac{1}{r \sqrt{a(t)}} e_{\hat{\theta}}\\ e^\phi &=& \frac{1}{r \sin \theta \sqrt{a(t)}} e_{\hat{\phi}} \end{eqnarray}

Fortunately, thanks to the symmetries of our metric, it can be shown that the Dirac equation we'll be interested in will be independent of the spin connection. Our action here simply reduces to

$$S[\psi, \bar{\psi}] = \int_\Omega \bar{\psi} \left[ i \gamma^a {e^\mu}_a \left( \partial_\mu \psi - i e A_\mu \psi \right) - m \psi \right] d\mu[g]$$

, and we simply get the following equation to solve :

$$i \gamma^a {e^\mu}_a \left( \partial_\mu \psi - i e A_\mu \psi \right) - m \psi= 0$$

We're also going to need the Hamiltonian. As the spacetime metric and electromagnetic potential are set quantities here, this should be fairly simple.

$$H =$$

The hydrogen atom in flat space

First, let's consider the problem in flat space, so that ${e^\mu}_a = {\delta^\mu}_a$. We can say without worrying too much here that

$$A_\mu(x) = (\frac{Ze}{r}, 0)$$

The Dirac equation here is

$$i \gamma^\mu \partial_\mu \psi + \gamma^0 \frac{Ze}{r} \psi - m \psi= 0$$

And our Hamiltonian is

The Dirac equation in the FRW metric

On the other side of the simplification, let's consider the Dirac equation in curved spacetime without any electromagnetic field. This will both help us out to work out the electron later on as well as find out what would describe the proton best in some limit.

Classical Coulomb field in the FRW metric

To simulate the electromagnetic field of a nucleus without any gravitational backreaction (otherwise we'd have to deal with the Reissner-Nordstrom-FRW metric, but the mass of a nucleus probably doesn't make this worth it), let's consider a static point charge in the Lorenz gauge :

$$\Box_{LB} A^\mu = - \mu_0 J^\mu + {R^\mu}_\nu A^\nu$$

The four-current $J$ is going to be some manner of Dirac distribution, with a total charge of $Ze$. First we need to establish what the total charge is. The local conservation law is

$$\nabla_\mu J^\mu = \frac{1}{\sqrt{g}} \partial_\mu (\sqrt{g} J^\mu) = 0$$

Fortunately, as an FRW spacetime, there is an obvious split of time and space, so that we can attempt to talk about our current on our spacelike hypersurfaces. First, let's expand this a bit

\begin{eqnarray} \frac{1}{\sqrt{g}} \partial_\mu (\sqrt{g} J^\mu) &=& \partial_\mu J^{\mu} + \frac{\partial_\mu \sqrt{g}}{\sqrt{g}} J^\mu\\ &=& \partial_t \rho + \vec{\nabla} \cdot \vec{J} + \frac{\partial_t a^{\frac{3}{2}}}{a^{\frac{3}{2}}} \rho + \frac{\partial_a r^2 \sin(\theta)}{r^2 \sin(\theta)} J^a \end{eqnarray}

We have two choices

$$J^\mu = (Ze \delta(r), 0)$$

Bibliography

1. Solution of the Dirac Equation for Hydrogen
2. V. M. Villalba, U. Percoco, Separation of variables and exact solution to Dirac and Weyl equations in Robertson–Walker space‐times
3. Chang Jun Gao, Shuang Nan Zhang, Reissner-Nordström Metric in the Friedman-Robertson-Walker Universe