# Insights A Brief on the Expansion of the Universe - Comments

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1. Mar 26, 2016

### nikkkom

2. Mar 27, 2016

### Staff: Mentor

Nice explanation of expansion, Nikkkom!

3. Mar 27, 2016

### Jorrie

Excellent intro to cosmology using so few words!

You wrote: "It’s a bit curious that in this scenario expansion seems to be very rapid at first (“density” of test particles falls very quickly), and millions of years later, it looks very gradual, but expansion speed is in fact constant."

This is well stated, but I just have a concern about the term "expansion speed" in that sentence, which has caused some confusion in the past. I think what you intended is normally referred to as the "recession rate" (between two particles) because the speed of expansion depends on the size of the area that one considers. If it will not lengthen the article too much, it may also be useful if you could bring in the expansion rate or Hubble value (H) somewhere.

4. Mar 27, 2016

### Ssnow

Good short but all cases has been examined ... very nice article...

5. Mar 27, 2016

### nikkkom

Yes, I think replacing "expansion speed" with "expansion rate" everywhere in the text is okay.

6. Mar 27, 2016

### Greg Bernhardt

Congrats nikkkom! I think you should be able to still make edits if you need to.

7. Mar 29, 2016

### Chronos

I object to this in the closing paragraph - “Λ going down” releases energy (in a form of appearance of new particles everywhere), you get a NON-empty symmetrical flat expanding Universe. As of 2016, this model fits observational data rather well."
What observational evidence supports this assertion?

8. Mar 29, 2016

### nikkkom

Observational data supports the theory that inflation did happen.
And observational data supports the theory that Universe is not empty.

9. Mar 29, 2016

### Chronos

10. Mar 29, 2016

### nikkkom

The text does not imply that we know that Lambda = inflaton field = dark energy.

In general, the text is meant as a FAQ for the new people who regularly pop in the Cosmology forum asking "what is that Big Bang thing? What does it mean that expansion is accelerating?"; it is not intended to be an in-depth discussion.

11. Mar 29, 2016

### Staff: Mentor

Your statement that "$\Lambda$ going down releases energy (in a form of appearance of new particles everywhere)" implies that the end of inflation involved "$\Lambda$ going down", which implies that Lambda = inflaton field = dark energy. If you don't intend that implication, then I think you need to reword the text.

12. Mar 29, 2016

### nikkkom

The text is "If “Λ (or dark energy) going down” releases energy...".
There is that "If" word. It is not a definitive statement.
Secondly, the statement does not say that we are sure it's Λ going down. It may be dark energy (whatever that is) going down.

13. Mar 29, 2016

### Staff: Mentor

Here is the actual text from your article:

"And finally, what if dark energy field is variable (e.g. it has several possible stable values) and one of these values is large (or there may be several such fields)? Alternatively, what if Λ can not only be larger than zero, but can be VERY MUCH larger than zero? Nothing unusual will happen, the Universe will behave as described above: grid of test particles grows faster with time. But very, very much faster. That’s inflation. (If you have a separate “large value dark energy field” for it, that’s “inflaton field”)."

This clearly implies that "inflation field" = "variable dark energy that can have a large value". But that's not the current mainstream model for inflation; in that model, the inflaton field is not the same as what we currently observe as dark energy, they just happen to have a similar property (that they cause accelerating expansion). If you want to mention the possibility that what we currently observe as "dark energy" is just the very small remnant of the inflaton field that's left over after the end of inflation, you should make clear that it's just a speculative hypothesis, since, as Chalnoth pointed out, it is clearly not the mainstream theory.

Also, the inflaton field is not the same as $\Lambda$, even a $\Lambda$ that is very much larger than zero. Nor can a variable dark energy be the same as $\Lambda$. Elsewhere in the article you are consistently using $\Lambda$ to refer specifically to the cosmological constant--i.e., it is by definition the same everywhere in spacetime. You should keep to the same usage at the end of the article as well, and in that usage, $\Lambda$ can't change, so obviously it can't be the inflaton field, since that has to change from a very large value to zero (in the current mainstream model) at the end of inflation, and it can't be a variable dark energy either. (You recognize this earlier where you say that a universe with $\Lambda = 0$ but some other field having similar properties could possibly explain the observed "dark energy".)

14. Mar 29, 2016

### PAllen

"Einstein’s General Relativity allows a solution (FLRW metric) where an empty Universe expands. Imagine that you have particles (say, hydrogen atoms) in a cubic grid with exactly 1 light year between nearest particles. (We assume that their mass is so tiny that this Universe essentially behaves as if it is empty). And after each second distance between each particle increases by 1 meter. Not because they move, but because space “grows”. That’s that solution."

This is very much subject to interpretation. Note the following:

1) The cosmologically interesting FLRW solutions are NOT empty in that the stress energy tensor is nowhere vanishing.
2) In the limit of actually empty, you have the Milne solution which is just flat SR spacetime in funny coordinates in which the 'hubble flow' bodies are inertial with invariant relative velocity accounting for all of the red shift. These relative velocities are all sub-luminal, but the expansion rate (which is a coordinate quantity, NOT an invariant) is super-luminal. Yet, in standard cosmological coordinates, this truly empty solution is the one with maximal expansion rate (of solutions with zero cosmological constant).
3) In the cosmologically meaningful solutions, if you set up Fermi-Normal coordinates, which are the closest analog if Minkowski coordinates in SR, you find that the Doppler of receding bodies is primarily due to relative velocity, and that this relative velocity is sub-luminal.
4) The growth of distance between bodies over time is wholly a coordinate dependent quantity. Distance growth over time requires a choice of foliation, and a choice of time coordinate, neither of which are invariant.
5) As is well known, there is actually no such thing as an unambiguous relative velocity over substantial distance in SR. Thus a claim that recession rate does not represent relative motion is meaningless (because relative motion is ambiguous). However, it is true that parallel transport of a 4-velocity from one event in any spactime to any other event in the spacetime, produces a path dependent relative velocity < c.

15. Apr 10, 2016

### Mordred

I find another aspect highly misleading. In a homogeneous and isotropic fluid there is no pressure gradient at any slice of time. Your correlations to positive and pressure matter suggests incorrectly that expansion is due to pressure.

In particular this section
"This all was about “normal” matter, with positive pressure. With “negative pressure matter” the effect is opposite – this was already described in the previous paragraph – that’s “dark energy”."

Reference https://www.physicsforums.com/insights/brief-expansion-universe/

In my opinion taking a lesson from Andrew Liddle's Introductory to Cosmology"

You should consider defining expansion in terms of potential vs kinetic energy.

There is some rewording here but this set of equations demonstrate the section

One of the key aspects to understand expansion is to understand the potential and kinetic energy aspects with conservation of energy.

For this we can detail using Newtons laws.

$$F=\frac{GMm}{r^2}$$

Mass density we will use $$\rho$$ which is the mass per unit volume.

Now assume a field of test particles. Motion and mass currently unimportant.

One of the aspects of the shell theorem in Newtons laws is the test particle will only notice a force from the center of mass.

In a homogeneous and isotropic distribution any test particle or CoM can be used.

As we're dealing with test particles we just need the mass relation.

$$M=\frac{4\pi\rho^3}{3}$$

So

$$E_p=-\frac{GMm}{r^2}=-\frac{4\pi G\rho^3 m}{3}$$

Kinetic energy is $$E_k=1/2m\dot{r}^2$$

$$U=E_k+E_p$$

U is just a dimensionless constant to equate total energy must be set as a constant value.

So the above translates to

$$U=\frac{1}{2}m\dot{r}^2-\frac{GMm}{r^2}=-\frac{4\pi G\rho^3 m}{3}$$

Now with the vector relation of the radius to length we can denote the scale factor.

$$\overrightarrow{r}=a(t)\overrightarrow{x}$$

Where a is a function of time. This leads to

$$U=\frac{1}{2}m\dot{a}^2x^2-\frac{4\pi}{3}G\rho a^2x^2 m$$

Multiply each side by $$2/ma^2x^2$$

$$(\frac{\dot{a}}{a})^2=\frac{8\pi G}{3}\rho-\frac{kc^2}{a^2}$$

$$kc^2=-2U/mx^2$$
$$k=-2U/mc^2x^2$$
K is the curvature constant.

Andrew Liddle covers this in better detail to the Einstein field equations later in the book.

Key note you don't require the cosmological constant for a matter only universe to expand.

A good book showing expansion of a matter only universe is Barbers Rydens "Introductory to Cosmology"

I prefer the solution above for the potential vs kinetic energy aspects.
The end result equation doesn't include the Cosmological constant.

With the cosmological constant the equation becomes.
$$(\frac{\dot{a}}{a})^2=\frac{8\pi G}{3}\rho-\frac{kc^2}{a^2}+\frac{\Lambda}{3}$$

This above better reflects the FLRW metric prior to the discovery of the cosmological constant. It shows you don't require the cosmological constant to have an expanding universe during any era

Last edited: Apr 10, 2016
16. Apr 10, 2016

### Mordred

The other consideration is if you look at the equation of state for matter p=0. However if you understand how the equation of state is calculated the value is given for temperature =0. (No kinetic energy) using the ideal gas laws. ( For matter EoS)
GENERAL RELATIVITY &
COSMOLOGY
Professor John W. Norbury

Has a good section on this.

Obviously I can't post Andrew Liddles textbook but this article covers the same principle in a different set of metrics.

It also does a better job showing the critical density relations.

Section 4.3 does the comparison to Newtonian aspects

Last edited: Apr 10, 2016
17. Apr 10, 2016

### Staff: Mentor

This is force, not potential energy; but you are using it later on as if it were potential energy. Potential energy would be

$$V = - \frac{G M m}{r}$$

No, this should be

$$M = \frac{4 \pi r^3}{3} \rho$$

You need to rework the rest of your post in the light of the above.

18. Apr 10, 2016

### Mordred

Yes that's why the minus sign here
$$E_p=-\frac{GMm}{r^2}=-\frac{4\pi G\rho^3 m}{3}$$

In that section I'm using the formulas in the sequence of the reference directly from Liddles textbook.

Oops there is a typo error there.

$$E_p=-\frac{GMm}{r^2}=-\frac{4\pi G\rho^2 m}{3}$$

Had the wrong exponent value for r, in the textbook he uses V instead of E_p but other that I tried not to make any changes, I'll check the rest to the book. Hopefully no other typo errors.

he set $$M=\frac{4\pi\rho r^3}{3}$$ on page 18.

Substitutes this into $$F=\frac{GMm}{r^2}=\frac{4\pi G\rho rm}{3}$$

In equation 3.2.

But then he ends up with
$$V=-\frac{4\pi G\rho r^2m}{3}$$
in equation 3.4

Lol

Last edited: Apr 10, 2016
19. Apr 10, 2016

### Staff: Mentor

And as you can see, there are evidently multiple typos.

In any case, this Newtonian reasoning about potential and kinetic energy implicitly assumes an underlying Newtonian absolute space and time, which makes other predictions that are not correct. So this "derivation" can't be anything more than a heuristic way of making the Friedmann equations seem intuitively plausible.

20. Apr 10, 2016

### Mordred

On that I agree, there are flaws on the Newtonian thinking. Yes there is a few typos.

I was just glancing at Nordbury and he gives
$$M=\frac{4}{3}\pi r^3\rho$$

I still don't agree with the quoted section from the insight article, for one thing the cosmological constant isn't required to have expansion.