# How does one measure the curvature $k$ in FLRW metric?

How does one measure the curvature parameter $k$ in the FLRW metric? $$ds^2=-c^2dt^2+a^2(t)[\frac{dr^2}{1-kr^2}+r^2d\theta^2+r^2\sin^2\theta d\phi^2]$$ In particular, what is the convenient equation (involving $k$) that is/can be used to measure $k$?

EDIT I'm looking for an answer that will explain the measurement of curvature $k$ with the same clarity as the measurement of spring constant $\kappa$ from the one-dimensional simple harmonic equation $F=-\kappa x$ i.e., having measured the applied force $F$ (can be done with a spring balance may be) and the corresponding displacement $x$ (by a meter rule), one can measure $\kappa$. Similarly, if the equation involving curvature $k$ contains non-trivial physical quantities (such as the components of Riemann curvature tensor etc), I would like to know how each of them is measured.

You simply measure the ratio of the circumference of a circle to its radius.

Take a spatial submanifold, and for convenience we'll take $a=1$ (the units of the radial distance can always be chosen to make $a=1$ as any chosen time). Then the spatial metric becomes:

$$d\ell^2=\frac{dr^2}{1-kr^2}+r^2d\theta^2+r^2\sin^2\theta d\phi^2$$

Draw a circle with yourself at the origin. To get the circumference of the circle we integrate around the equatorial angle $\phi$ while keeping $r$ fixed and $\theta$ fixed at $\pi/2$. Since $dr = d\theta = 0$ the metric becomes:

$$d\ell^2=r^2\sin^2\theta d\phi^2$$

The circumference is then:

$$C = \int_0^{2\pi}\,rd\phi = 2\pi r$$

which shouldn't surprise us unduly :-)

Now we take a measuring tape and measure out the distance to the circle. In this process we are keeping $\theta$ and $\phi$ fixed so $d\theta = d\phi = 0$ so our metric becomes:

$$d\ell^2=\frac{dr^2}{1-kr^2}$$

So the distance we measure is:

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

The integral depends on the sign of $k$. For positive $k$ (closed universe) we get:

$$R = \frac{\sin^{-1}(\sqrt{k}\,r)}{\sqrt{k}}$$

and for negative $k$ (open universe) we get:

$$R = \frac{\sinh^{-1}(\sqrt{|k|}\,r)}{\sqrt{|k|}}$$

To find $k$ simply substitute $r = C/2\pi$, where $C$ is our experimentally measured circumference, and solve the resulting equation for $k$.

• That's a nice and simple approach but I can't imagine this working in an experimental context, your circle would have to be incredible large for this to work. And you also forgot that k is rescaled such that k = 0 or +-1. So you final formula cannot differentiate positive and negative curvatures. – gertian Mar 16 '17 at 9:23
• @gertian: yes, this isn't a measurement you would be able to do in practice. I note that your answer explains how the measurement could be done using real cosmological data, and I think it's a very good answer. I think our answers are complimentary since mine is more of an in principle answer while yours is more practical. Which SRS prefers is going to depend on what his motivation for the question is. BTW while we usually rescale to set $k = 0,\pm 1$ we don't have to do that. – John Rennie Mar 16 '17 at 9:29

This is a very hard question to answer in detail as it requires several pages of mathematics to derive the required formulas (there is no easy fit like $F=-kx$ as you suggested)

I will not derive the formula (it can be found in e.g. Dodelson) but after some work you obtain:

$$\Delta(m-M) = 5\log\left\{ \left( \frac{c}{H_0}\sqrt{\frac{k}{\Omega_{total}-1}}(1+z) \right)S_k\left(\sqrt{\frac{\Omega_{total}-1}{k}}\int_0^z \frac{dz'}{(1+z')\sqrt{(1+z')(1-\Omega_\Lambda) + \Omega_\Lambda/(1+z')^2}} \right)\right\} \\ - 5\log\left(\frac{1}{2}((1+z)^2-1)\right)$$

Where $\Delta(m-M) = (m-M)_{'real'\ universe} - (m-M)_{empty\ universe}$. M can be obtained by using standard candles such as supernovae type Ia, it is than easy calculate $(m-M)_{empty\ universe}$ via an equation similar to the one above and $m_{'real'\ universe}$ is simply the magnitude that we measure. Therefore $\Delta(m-M)$

$S_k(...) = sinh(...), sin(...) or 1$ depending on the value of $k$

$\Omega_\Lambda, \Omega_{total}=\Omega_\lambda+\Omega_{matter}, H_0$ and k remain unknown.

The next step is to measure a lot of standard candles at various redshifts z and plot their $\Delta(m-M)$ relation as a function of z. This should obey the relation above. All that remains to be done is to run a fitting script that fits the above function to $\Delta(m-M)$ for various values of $k, \Omega_\Lambda,...$ the best fit gives us the observed cosmology.

In the figure below you can see such a fit from a project I made previous semester where we had to calculate k for some dataset.

Obviously the fits are hard to make due to degeneracies in the fit and uncertainty plots can be made as like this one:

My results for the above fit were: $\Omega_{matter} = 0.286 \pm 0.031$, $\Omega_{\Lambda} = 0.721 \pm 0.025$, $H_0(km/s/MPc) = 70.3 \pm 2.58$ which is consistent with k = 0.

I hope this helped :)

• Perhaps you could say what your best fit $k$ is and its uncertainty? Also, are you sure your statement shouldn't be $S_k(...) = \sin(...), \sinh(...)\text{ or }1$ rather than what you have? From the look of your first plot, you would seem to get a result that "$k$ can be no larger than x with uncertainty y" kind of result given the spreads in the high $z$ data. Is this so? Also is this how real astronomers would go about measuring $k$; I could imagine there are several different methods. – Selene Routley Mar 16 '17 at 9:23
• I edited in my results (can't seem to find my final result on k though). And indeed you are right $S_k = sin, sinh or 1$ I was mistaken... Well once you got your cosmological parameters it is just a question of getting k out of them and indeed you get a spread on the result but I seem to have lost it... As far as I know this is the most accepted method for "fitting" cosmologies to datasets as it allows you to keep into account what happened long ago if you find standard candles at those redshits. Current research goes towards finding those high redshifts objects. – gertian Mar 16 '17 at 9:35
• A shame you have lost it: it's a great, practical answer. Please post it if you find it. – Selene Routley Mar 16 '17 at 9:36
• Thanks, that's nice to hear :) If I happen to have some time soon I'll search for it and tag you in the comment. But my master thesis yells loud at me these days... – gertian Mar 16 '17 at 10:08