# How do you compute the radial coordinate in a hyperbolic universe using a metric?

In hyperbolic spacetime, we have this formula describing the surface: $$ds^2=-c^2dt^2+a^2(t)\frac{dr^2}{1-r^2}\tag 1$$ I'm leaving out the angular distances because I'm just interested in a line-of-sight measurement (e.g. distance to a supernova). This formula can be represented by the metric: $$g_{\mu\nu}=\begin{bmatrix}-c^2 & 0\\0 & \frac{a^2(t)}{1+r^2}\end{bmatrix}\tag 2$$ So now I want to test this metric out. I want to calculate the physical distance a photon travels between $$t1$$ and $$t0$$ (present day). So I employ this formula: $$s=\int_{t1}^{t0}\sqrt{dx^\mu\cdot g_{\mu\nu}\cdot dx^{\nu}}\space ds\tag 3$$ And this is where I'm stuck. The (2,2) term of the metric is $$\frac{a^2(t)}{1+r^2}$$, but $$r$$ is the answer! I'm trying to calculate the radial distance to an object, but the metric requires the radial coordinate in order to calculate the curvature.

What concept am I missing? Am I thinking about the metric incorrectly? Can it be employed like a machine to calculate the physical distance to an object (such as the distance to the SNe Ia in an expanding universe)? If so, what are the steps?

EDIT: The other problem I'm having with this concept is that the $$ds$$ line element is going to be zero, so the sum of the line elements should also be zero. So I understand how a metric can be used to find the distance in any number of spatial dimensions, but I'm confused how it can be employed in spacetime to find the radial coordinate between two events.

• have you forgotten the sign in the diagonal of $g_{\mu \nu}$? May 20 '20 at 21:01
• @NelsonVanegasA. - Yes, I did. thank you. May 20 '20 at 21:50

The metric is $$ds^2 = -c^2dt^2 + a^2[\frac{dr^2}{1-kr^2}]$$

for $$d\Omega = 0$$.

If you want to calculate the comoving distance taken by the photon you can use the above equation.

Note that the integral equation for the distance that you wrote is the same as the above metric. If you set $$ds=0$$, we have

$$c^2dt^2 = a^2[\frac{dr^2}{1-kr^2}]$$

which can be written as

$$\int_{t_e}^{t_0}c\frac{dt}{a} = -\int_r^0{\frac{dr}{\sqrt{1-kr^2}}}$$

We can define another comoving coordinate called $$\chi$$ and we claim that the distance traveled by the photon is equal to $$\chi$$ such that

$$\chi = \int_{t_e}^{t_0}c\frac{dt}{a} = \int_0^r{\frac{dr}{\sqrt{1-kr^2}}}$$

The R.H.S of the equation is useless because we cannot measure $$r$$. The proper way to do this is that calculate $$\int_{t_e}^{t_0}c\frac{dt}{a}$$ which can be written in terms of $$z$$. Then you can find $$\chi$$ which that's the answer we are looking for. So you don't need to $$r$$ to calculate the comoving distance. This can be directly seen if we write the metric in terms of $$\chi$$

$$ds^2 = -c^2dt^2 + a^2d\chi^2$$ for $$d\Omega = 0$$

for $$ds = 0$$ we have again

$$\chi = \int_{t_e}^{t_0}c\frac{dt}{a}$$ and we never used $$r$$.

If you particularly need to know $$r$$ just solve the integral

$$\chi = \int_0^r{\frac{dr}{\sqrt{1-kr^2}}}$$

and take the reverse so you'll obtain

$$r = \begin{cases} sinh(\chi), & k = -1 \\ \chi & k = 0 \\ sin(\chi) & k = 1 \end{cases}$$

• Would you please show me how to compute $a(t)$? I've been looking and can't find a reference anywhere. May 21 '20 at 13:08
• @GluonSoup What kind of universe are we talking about ? May 21 '20 at 14:11
• A universe governed by General Relativity having a constant Energy Density and Cold, Dark Matter. May 22 '20 at 13:49
• All you've done with this step $$\chi = \int_0^r\frac{dr}{\sqrt{1+r^2}}$$ is bury the radial coordinate in the scale factor. It doesn't seem to address my original question. You appear to need to know what the radial coordinate, $r$, is before you can solve the integration or, alternatively, you can use the absolute difference between the emitted time, $t_e$, and the observed time, $t_o$ as a proxy for the radial coordinate. To be clear, though, you can't integrate equation 1 using infinitesimals. You need to know the total distance (or the total time) and then solve a simple ODE. May 22 '20 at 14:35
• @GluonSoup I edited my answer. May 22 '20 at 16:15

You need to consider that photons travel along null geodesics $$(\mathrm{d}s)^2 = 0$$ and therefore the integral does not work this way. You asked what concept you are missing, that is an important one as you have already realized. Could it be that you want to calculate a distance then the integral should be performed over $$r$$ rather than $$s.$$ Or that you want a photo's traveled distance? for a photon, one has

$$ds^2=-c^2dt^2+a^2(t)\frac{dr^2}{1-r^2} = 0$$

and for an observer with proper time $$\mathrm{d}\tau$$

$$0 = -c^2 \dot{t}^2+a^2(t)\frac{\dot{r}^2}{1-r^2}$$ you need to work out the geodesics and the integral over $$r$$ from point $$r_a$$ to point $$r_b.$$

But you might be interested in the distance with $$\theta = \phi = t =$$ constant. In which case you just need the term involving $$r.$$ Then $$d = a(t) \int_{a}^{b} \frac{dr}{(1-r^2)^{1/2}}.$$ However you would need to know $$a,b$$ to work this out. So in cosmology people use the luminosity distance. Check D'inverno, chapter 22 section 22.11, for example.

• If I have $r_a$ and $r_b$, then why do I need to integrate? Can't I just take $r_a-r_b$? Seems like I have the same problem: the answer is one of the terms I need for the calculation. May 20 '20 at 21:48
• I can imagine using an ODE to solve for $r$, is that where you're going with this? Do I need to abandon the idea of using the metric as a kind of engine to compute the radial coordinate? May 20 '20 at 21:49
• In your question you wanted to know the distance traveled by a photon, then the null geodesic gives you the trajectory and you integrate the path for $dr$ along that curve. May 20 '20 at 22:15
• I'm asking for the radial coordinate. I'm pretty sure the distance travelled is 0 for the null geodesic. May 20 '20 at 23:34
• I appended my answer. May 20 '20 at 23:36