# B How to do the calculations showing the Universe is flat?

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1. Jul 18, 2017

### Joshua P

I've been trying to understand how we know that the observable universe is flat, and I'm having difficulty finding any sources that explain exactly how the calculations were done. On this WMAP website (https://map.gsfc.nasa.gov/mission/sgoals_parameters_geom.html), it says:
"A central feature of the microwave background fluctuations are randomly placed spots with an apparent size ~1 degree across. These are produced by sound waves that travel through the hot ionized gas in the universe at a known speed (the speed of light divided by the square root of 3) for a known length of time (375,000 years). By using the relation: distance = rate * time, we can infer the distance the sound travels, and thus the actual size of a typical hot (compressed) or cold (rarefacted) spot. By comparing the apparent size of the spots to their known actual size, we can measure a combination of the distance to the last scattering surface and the curvature of the light path between us and this surface, which depends on the geometry of the universe. Then If we independently know the Hubble constant, we can determine the distance to the last scattering surface and thus use the spot size to determine the geometry uniquely."
I was wondering how the "the actual size of a typical hot (compressed) or cold (rarefacted) spot" was calculated? How was the Hubble constant used? How did they ultimately show that the spots should be 1 degree across in a Euclidean universe? I understand that it should be basic Euclidean geometry, but I'm not quite understanding the problem. Please keep it simple for me to understand.

2. Jul 18, 2017

### Bandersnatch

3. Jul 18, 2017

### Joshua P

Thank you so much for the reply. I'm getting really excited.
I've been trying to get my head around the paper for a while now. I'm not a physicist (yet), so this is quite difficult for me.

It looks like the calculated angle of 1 degree is based on the actual size of the sound horizon. Is that right?
To prove the universe is flat, the actual size of the sound horizon has to match the apparent size of the sound horizon. I know that the CMB shows that it appears to be 1 degree apart. I was looking for a source that showed why the fluctuations should be 1 degree apart in a flat universe. I'm just confirming, but that is what it's doing?

I was wondering what the value "a" represented in all the equations, I know it has something to do with the Hubble constant? Called... the scale factor?
Also, Cs, the speed of sound in the primordial plasma, was said to equal the square root of 3 in this paper. The WMAP website said that the speed of sound through this hot ionized gas was the speed of light divided by the square root of 3. What am I missing?

There are other things I don't understand, but they depend on me understanding these things I've mentioned. Thanks.

4. Jul 18, 2017

### Bandersnatch

I'll have to keep this short, since I've a deadline to meet that I just can't procrastinate (any more). So just a few points for now:

The scale factor is defined as $a(t)=r(t)/x_0$, where r(t) is any arbitrarily chosen distance (e.g., between some two test galaxies), and $x_0$ is the value of this distance at the present time.
Intuitively, the scale factor tells you how much the universe (=all distances) at some time t will be/was larger/smaller than now. a=1 today; it was equal to 1/1000 when the universe was 1/1000 the current size; it'll be equal to 2 when the universe will have grown to twice the current size.

If the scale factor grows $\dot a >0$, the universe expands, and if $\dot a<0$ then it contracts. If $\ddot a>0$, the universe accelerates, and vice versa.

Hubble parameter is the rate of expansion. It is defined as $H^2=(\frac{\dot a}{a})^2$. Its intuitive meaning is the instantaneous percentage growth of the universe at a given time. E.g., the current value of the Hubble parameter, $H_0$ i.e. the 'Hubble constant', translates to something like 1/144 % growth per million years.

If you have the time to watch Leonard Susskind's cosmology lectures (available on youtube), he covers it in the first one. It's generally a great resource if you're new to the topic.

The one degree is calculated under the assumption that the universe is flat - the Friedmann equation given in (19) and used to get the scale factor, does not include the curvature parameter k, which was assumed to be equal to 0 (=flat). The full equation is $$H^2=(\frac{\dot a}{a})^2=\frac{8piG}{3}\rho - \frac{kc^2}{a^2}$$
With $k=0$ it reduces to eq. (19).

So, again, that's the size you'd expect if the universe were flat.

You mean in that bit after the eq. 21? It's given as $c_s^{-1}=\sqrt{3}$, so it's actually $c_s=1/\sqrt{3}$, and it uses units where the speed of light is equal to 1 - so they're actually the same.

O.k., I really have to leave you here. Maybe somebody else will jump in while I'm gone, though. Heck, let me cast a bat-signal calling @bapowell (the author), so that you may get your info straight from the horse's mouth.

5. Jul 18, 2017

### kimbyd

Yes, that's a big part of it. The other way to look at it is to compare the average distances between relatively nearby galaxies to the distances expected from the CMB fluctuations. This large difference in distance between these galaxies and the surface of last scattering provides a long lever-arm with which to measure curvature.

6. Jul 20, 2017

### Joshua P

Thank you so much, @Bandersnatch. You've been an amazing help.
@kimbyd, that's interesting. Distances of what objects are expected from the CMB fluctuations? The surface of last scattering? How does the difference in distance help measure curvature?

I have a bunch more questions about the Insights article (https://www.physicsforums.com/insights/poor-mans-cmb-primer-part-4-cosmic-acoustics/), if anyone else would be willing to answer them. (No need to answer all of them at once, of course.)

How does equation 19, the Friedman equation (where we've assumed k = 0) show that density is proportional to a-3?
How did they then "find that in a matter dominated universe the scale factor grows as a power-law, a(t)∝t2/3"?
I was able to work out how they got from there to calculate ds, but I'm having difficulty doing the same thing to find xls from equation 20. Can anyone show me how its done?
And how do the units cancel out if cs is measured in terms of the speed of light? Are the distances measured in lightyears?

7. Jul 20, 2017

### George Jones

Staff Emeritus
A heuristic way to see this: consider a box of volume $V$ that contains $N$ particles that each have mass $m$, so that the density of the stuff in the box in $\rho = Nm/V$. If the box follows the expansion of the universe, and if the particles don't have peculiar velocities, then the number of particles in the box stays constant as the box expands. Without loss of generality (WLOG), assume that the box is $a \times a \times a$ (with $a$ the scale factor of the universe, so that $V = a^3$. At time $t_1$, density is $\rho_1 = Nm/a_1^3$; at time $t_2$, density is $\rho_2 = Nm/a_2^3$. Hence, $\rho_2/\rho_1 = a_1^3/a_2^3$, i.e., the density of non-relativistic matter is inversely proportional to the cube of the scale factor.

A less heuristic way to to see this uses local conservation of energy and the specific form of the connection coefficients for Friedmann-Lemaitre-Robertson-Walker (FLRW) universes to give an equation that resembles the first law of thermodynamics,
\begin{align} \frac{d}{dt} \left( \rho a^3 \right) &= -P \frac{d}{dt} \left( a^3 \right) \\ a^s d\rho + 3\rho a^2 da &= -3P a^2 da \\ a \frac{d\rho}{da} &= -3 \left( \rho + P \right) \end{align}
In the above $c = 1$, and $P$ is pressure, which can be taken to be zero, since for non-relativistic matter, $P$ is much smaller than $\rho$. This gives
\begin{align} \frac{d\rho}{\rho} &= -3 \frac{da}{a} \\ \int_{\rho_1}^{\rho_2} \frac{d\rho}{\rho} &= -3 \int_{a_1}^{a_2} \frac{da}{a} \\ \ln \frac{\rho_2}{\rho_1} &= \ln \frac{a_1^3}{a_2^3} \\ \frac{\rho_2}{\rho_1} &= \frac{a_1^3}{a_2^3} \end{align}
Substitute $H = \dot{a}/a$ and $\rho = K a^{-3}$ into (19). Here, $K$ is a constant of proportionality (and has nothing to do with spatial curvature).

8. Jul 20, 2017

### kimbyd

Think of it as the equivalent of summing the angles of a triangle.

In flat space, the angles of a triangle always add up to 180 degrees. In positively-curved space, the sum is greater. In negatively-curved space, the sum is smaller.

In principle, you could measure the curvature by just summing up the angles of a single triangle, and the larger the triangle the better.

The difficulty enters if there is some uncertainty as to precisely what the true length of the far-away side of the triangle is. We can't measure it directly: we have to infer from a model. We can resolve this degeneracy by comparing this triangle to another triangle measured in the more nearby universe. Relatively simple physics ties the triangle measured from the typical distances between galaxies and the triangle measured from distances between the temperature peaks on the CMB.

In detail, physicists don't actually set up triangles and measure them. This is just a heuristic device for explaining why it's helpful to look at both relatively near and relatively far objects for estimating curvature.

9. Jul 20, 2017

### kimbyd

Or, to put it more simply:
That the density of matter scales as $1/a^{-3}$ is a result of the conservation of stress-energy. Since matter doesn't experience pressure, this reduces to just the conservation of energy, which for non-relativistic matter is just the mass, whose density just decreases as $1/a^{-3}$ as the universe expands.

To get the scaling of other forms of matter, you have to use an argument more similar to George Jones above. But matter is easy.

10. Jul 20, 2017

### Joshua P

@George Jones thanks. How did you get your first equation? Does it come simply from equation 19?
Also, after you "Substitute H=(change in a)/a and p=Ka-3 into (19)", how do you simplify? I don't know what to do with the (change in a) part.
@kimbyd, how is that "relatively simple physics ties the triangle measured from the typical distances between galaxies and the triangle measured from distances between the temperature peaks on the CMB"?

11. Jul 20, 2017

### kimbyd

12. Jul 20, 2017

### George Jones

Staff Emeritus
What is the first law of thermodynamics?

\begin{align} H &= \sqrt{\frac{8 \pi G}{3} \rho} \\ \frac{1}{a} \frac{da}{dt} &= \sqrt{\frac{8 \pi G}{3} \frac{K}{a^3}} \\ a^{1/2} da &= \sqrt{\frac{8 \pi G}{3}} dt \\ \int_0^{a_1} a^{1/2} da &= \sqrt{\frac{8 \pi G}{3}} \int_0^{t_1} dt \\ \frac{2}{3} a_1^{3/2} &= \sqrt{\frac{8 \pi G}{3}} t_1 \end{align}

13. Jul 21, 2017

### Bandersnatch

That's one option. Other options include light-seconds (for c=1 light-second/second), light-fortnights (c=1 light-fortnight/fortnight), light-heartbeats (c=1 light-heartbeat/heartbeat), etc.
Doesn't matter which it is, as long as you're consistent with your units when you start evaluating numerical values.

14. Jul 21, 2017

### Joshua P

@kimbyd thanks, that was an interesting article.
@Bandersnatch thanks again.
@George Jones thanks for the explanation (and your time), I understood that last one.
I took a look at the first law of thermodynamics, and a closer look at your answers.
I believe that the first law states: dU = dQ - PdV
So dQ is change in energy from the surroundings, but given that we are talking about the whole universe, there is none. Right?
PdV was clear enough for me. As you said a3 is volume. dU is said to be the change in internal energy. From what you wrote, I take it also to be change in mass in the universe, from the d(density*a3)/dt. Why?
In the insights article Bandersnatch originally cited, they used the first Friedman equation to show what you've shown. Is what you're doing related to that?

Seperate question: I was looking at the Friedman equations, the first one also includes the cosmological constant? We assume that to be zero here, right? Why?

15. Jul 21, 2017

### George Jones

Staff Emeritus
@Joshua P : I just realized that you tagged this thread as B. Given some of the previous posts in this thread, I had assumed it was I. What background level of physics and maths should we assume?

16. Jul 21, 2017

### Joshua P

Well... I've made it most of the way through high school. I'm doing IB HL Maths and Physics, so I've done vectors and calculus (although only at a high school level). A lot of this physics is new to me. I've been doing research for about a month so far on this particular subject (curvature of the universe). That being said, I am looking to understand this subject in depth (and I want to know how the findings are made), though I know I have a ways to go. I apologise for asking such basic questions, and thanks to everyone for their patience. :)

17. Jul 21, 2017

### Bandersnatch

No, if anyone, it's me who should apologise for not paying attention to the thread level and leading you into deeper waters than I probably should.

Would you like to take a step back, towards a more conversational manner? Maybe with suggestions for reading to get you up to speed?

18. Jul 22, 2017

### Joshua P

@Bandersnatch You were simply answering my initial question; it was what I asked for, so thank you. The Insights article you gave me was exactly what I wanted.
It's quite difficult to explain what I'm looking for, but I'll try.
To give some context, I'm hoping to write a Maths paper on non-Euclidean geometry and what it has told us about the curvature of the universe. Since it's a Maths paper, I plan on simplifying the physics a lot. The entire essay aims to show that scientific theory tells us how far apart the fluctuations should be in a flat universe, and the apparant angle between the fluctuations on the CMB matches that, showing that the observable universe is more or less flat. Everything that everyone has given me so far is helping me write this essay.
I don't really need to fully understand any of the science (Eistein's field equations, the Friedman equations, or even the first law of thermodynamics) or justify it in my paper. I would just like to show that the science can be used to find the actual angle between fluctuations in a Euclidean universe. I'll more or less be quoting the middle of the Insights article, but first I want to understand how the Insights article reaches its conclusion. There are a bunch of places where I just didn't (or still don't) understand how it got from one point to the next. If I know how to explain the insights article (to a non-Physics-y person), I can do my essay. I would love some extra reading, I know I need it. But I don't think I need to understand all of the underlying physics in a whole lot of depth, just the main concepts?

19. Jul 22, 2017

### bapowell

Sorry to take so long to respond to @Bandersnatch 's bat signal, but I'm available to help if there are still questions!

20. Jul 22, 2017

### Joshua P

@bapowell Hi! Yes, I have more questions. There are still a couple questions that haven't been answered yet.

Last edited: Jul 22, 2017