# Using the RW Line element

Radiohannah
Hello! :-)

I've been thinking about the RW line element (c^2 d tau^2), and I understand that there is a term which describes the geometry of the Universe (d sigma^2) and then a term corresponding to the cosmic time (c^2 dt^2). The resulting (c^2 d tau^2 ) term would be the line measured by some fundamental observer.

I am confused as to how you may relate this line element to the density of the Universe, as I have read that there is an important relationship between the geometry of the Universe and its density.

I can see how the 'Friedmann equation' may give you a value for density, but I really cannot see the relationship between the Friedmann equation and the RW line element equation? :-S

Any help would be very much appreciated :-)

Thanks

## Answers and Replies

Science Advisor
The relationship between the geometry and density of the universe is given by the Einstein field equations. You have to take the matter density you want to talk about, solve the Einstein field equations to get the space-time (described by a metric), then use that to get the geodesic you want.

For a homogeneous, isotropic universe we use the metric:

$$ds^2 = c^2 dt^2 - a(t)^2d\Sigma^2$$
$$d\Sigma^2 = \frac{dr^2}{1-kr^2} + r^2\left(d\theta^2 + \mathrm{sin}^2\theta d\phi^2\right)$$

This metric has two parameters: a time-dependent scale factor and the overall spatial curvature (which is constant). If you work through the Einstein field equations using the same assumptions of homogeneity and isotropy used for the above metric, you find two equations for these two unknowns. First, for simplicity, we define:

$$H(t) = \frac{\dot{a}}{a}$$

One way of writing the two equations is as:

$$H^2 = \frac{8\pi G}{3}\rho - \frac{kc^2}{a^2}$$
$$\dot{\rho} = -3H\left(\rho + \frac{p}{c^2}\right)$$

Here $\rho$ is the overall energy density of the universe (and is a function of time), and $p$ is the total pressure of the various contents of the universe.

The first equation shows how the expansion rate is related to the overall density, while the second equation expresses conservation of the stress-energy tensor. Anyway, now that you have a metric, and can calculate how its parameters depend upon the density, you can compute any line element you want in a uniform, expanding universe. If you want to use a different distribution of matter, you have to use a different metric.

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Staff Emeritus
Science Advisor
Gold Member
In the link that Chalnoth gave for Einstein's equation, $G_{\mu \nu}$ is a geometrical tensor that is derived from the metric $g_{\mu \nu}$, and $T_{\mu \nu}$ is tensor that represents the distribution and flow of energy and momentum in the universe.

As Chalnoth mentioned spatial homogeneity and isotropy is assumed (backed on large scales by observation), and, in order to solve Einstein's equation, a specific form, that of a perfect fluid, for $T_{\mu \nu}$ is assumed. In a perfect fluid, there are no forces between particles (clusters of galaxies), and, in a local inertial reference frame, there is no heat conduction or viscosity.

Science Advisor
As Chalnoth mentioned spatial homogeneity and isotropy is assumed (backed on large scales by observation),
I'd just like to point out one little caveat that homogeneity and isotropy aren't always assumed. Our universe has galaxies, after all, and those are pretty darned big departures from homogeneity.

Radiohannah
Wowee! Thankyou :-) That's cleared it up significantly.

May I ask one last thing?

When you said : 'If you want to use a different distribution of matter'

By different distribution of matter, is that regarding the open, closed and flat cases?

And for the pressure 'P'; this is composed from the matter, radiation and vacuum contributions?

Science Advisor
By different distribution of matter, is that regarding the open, closed and flat cases?
No, no, the parameter "k" takes care of that. For other distributions, you could compute, for example, how light bends when it travels near galaxies or galaxy clusters. Or you could look at how galaxies and galaxy clusters interact with one another. Or you could just consider what happens near a black hole. There are all sorts of things you could deal with besides the simple approximation of our universe as homogeneous and isotropic, and they require you use a rather different metric than the FRW metric.

And for the pressure 'P'; this is composed from the matter, radiation and vacuum contributions?
Right, the pressure depends upon the type of matter you're talking about. With normal matter, the pressure is effectively zero on cosmological scales (e.g. galaxies have no pressure with respect to one another). Radiation has a pressure equal to one third its energy density. Dark energy has a pressure close to or equal its energy density, but negative.

Radiohannah
Woops (!!) yes, I see. k does account for those cases. Thanks again! :-)

(Sorry), but for the pressure....

These dark/normal matter and radiation contributions to the pressure are then denoted Omega_m and Omega_R?

I promise that's my last question! :-D (Thankyou so much).

Science Advisor
These dark/normal matter and radiation contributions to the pressure are then denoted Omega_m and Omega_R?
Those are the contributions to the energy density. They're the energy density of the particular form of matter divided by the so-called "critical density".

The critical density is a function of the current expansion rate: it's the energy density required for k=0 at the current expansion rate. For example, take the first Friedmann equation:

$$H^2 = \frac{8\pi G}{3} \rho - \frac{k c^2}{a^2}$$

In this case, the curvature k is dependent upon the relationship between the expansion rate and matter density. Consider, for instance, if we set the matter density as:

$$\rho(now) = \frac{3 H\left(\mathrm{now}\right)^2}{8\pi G}$$

In this particular situation, the current matter density and the current expansion rate are perfectly balanced, so k is forced to be zero. This is what is known as the "critical density". We can divide the entire equation by the critical density, and define the current expansion rate as $H_0$ to rewrite the Friedmann equations in a simpler form:

$$H^2 = H_0^2 \left(\frac{\Omega_m}{a^3} + \frac{\Omega_r}{a^4} + \Omega_\Lambda + \frac{\Omega_k}{a^2}\right)$$

This largely allows for simpler calculations, and means we don't have to deal with messy constants and unit conversions.

I promise that's my last question! :-D (Thankyou so much).
That's fine! Wouldn't be here if I didn't enjoy good questions!