Friiedman Fun Facts to know and tell

1. Oct 20, 2007

jonmtkisco

Hi SpaceTiger, Pervect & Hellfire,

Here are some follow-up thoughts about the Friedmann equation for expansion. Pervect, thank you for using Noether's Theorem to demonstrate that normal momentum (of movement) is conserved. The next challenge is to demonstrate that the "momentum-like" continuation of the original expansion of space is conserved.

1. By playing around with a spreadsheet, I have determined that if mass is held constant, the "momentum-like" continuation of expansion is conserved, by the formula:

PP = $$\Delta$$volume$$^{2}$$ /volume

where PP is the "momentum-like" continuation of expansion. So my earlier suggestion that $$\Delta$$volume/ $$\Delta$$time might be the metric turns out to be wrong. It's a relief to find that this "momentum-like" quantity remains constant in the Friedmann equation when mass/energy is constant.

2. However, that finding merely leads to the next question, which I find to be of great concern. That is, that the Friedmann equation calculates that when the mass/energy of the universe increases (e.g., due to the cosmological constant), the expansion rate increases. And most relevant, when mass/energy decreases, the expansion rate decreases. This is an important scenario, because under the traditional $$\Lambda$$CDM model, the total mass/energy of the universe is held to have decreased dramatically during the radiation dominated era, as radiation gave up energy to expansionary redshift. Sure enough, the Friedmann equation dutifully calculates that the "momentum-like" continuation of expansion declines throughout the radiation-dominated era. Yet if mass/energy is held constant at its pre-decline value, Friedmann calculates that the expansion rate remains higher (and higher than we observe). You can convince yourself of this just by noting that mass is in the top line of the Friedmann equation, and R is not.

Now I really need a clear explanation as to how, in a GR-based model, a large decline in mass/energy can cause a large decline in the expansion rate. That certainly violates the expected behavior of a "momentum-like" quantity. Less gravity ought to cause faster expansion than more gravity. That's pretty basic.

The Friedmann equation was created in the 1920's when there was no observational evidence that the universe was expanding, and therefore the idea that the universe might not be purely adabiatic, because its mass/energy might actually change over time, was not incorporated in the formula. If anyone is aware of this specific question having been addressed subsequently by mathematicians, I would very much appreciate a reference.

Jon

2. Oct 21, 2007

pervect

Staff Emeritus
The detailed explanation is in terms of Einstein's field equations.

A short, terse, and somewhat popularized answer is that one has to consider not only energy, but pressure, as causing gravity.

Details can be found in Baez's paper:

http://www.math.ucr.edu/home/baez/einstein/einstein.html

which I'll quote from below. (I'll certainly encourage people to read the original paper in its entierty, though).

Note that "by the rate at which the ball shrinks", Baez means the second derivative of the volume divided by the volume, i.e (d^2 V / dt^2) / V. This is explained in the paper.

Here the volume V is measured in the co-moving frame of the particles (which are all at rest relative to each other).

So, while the cosmological constant causes empty space to have a positive energy, it also causes empty space to have a negative pressure. The gravitational effects of the positive energy are smaller than the gravitational effects of the negative pressure.

A small ball of particles will contain positive energy, but because gravity is driven by rho+Px+Py+Pz, which equals rho+3P when the pressure is isotropic, the net effect on a small ball of particles in a space-time with a cosmological constant will be that they will expand, i.e. that d^2V/ dt^2 will be positive, in spite of the fact that the ball of particles contains a positive amount of energy.

Note that there are no non-gravitational effects of the negative pressure because the pressure is the same outside the ball of particles as it is inside.

Last edited: Oct 21, 2007
3. Oct 21, 2007

jonmtkisco

Hi Pervect,

Thanks for the explanation, although I am very familiar with the concept of negative pressure. You may have notice that I explained the same concept to Holocene in a recent post about why the expansion rate is expanding.

Because we are also familiar with negative pressure and the cosmological constant, that's why I focused in my post on the opposite situation, when mass/energy was decreasing during the radiation-dominated period. I said:

"And most relevant, when mass/energy decreases, the expansion rate decreases. This is an important scenario, because under the traditional CDM model, the total mass/energy of the universe is held to have decreased dramatically during the radiation dominated era, as radiation gave up energy to expansionary redshift. Sure enough, the Friedmann equation dutifully calculates that the "momentum-like" continuation of expansion declines throughout the radiation-dominated era. Yet if mass/energy is held constant at its pre-decline value, Friedmann calculates that the expansion rate remains higher (and higher than we observe). You can convince yourself of this just by noting that mass is in the top line of the Friedmann equation, and R is not.

Now I really need a clear explanation as to how, in a GR-based model, a large decline in mass/energy can cause a large decline in the expansion rate. That certainly violates the expected behavior of a "momentum-like" quantity. Less gravity ought to cause faster expansion than more gravity. That's pretty basic."

Subsequent to writing the post, it occurred to me that maybe the decline in mass/energy coincided with the decline in the expansion rate because gravity was much, much higher than the simple mass/energy of the radiation at the time. Since radiation is supposed to have positive pressure, and the temperature was supposed to be very high, maybe these factors increased the gravity dramatically above what matter and radiation alone would have caused. And then maybe the forced expansion of the universe, (due to the supposed original expansion driven by the 'initial conditions') happened to drive gravity down dramatically at a rate which coincided with the startlingly abrupt deceleration rate.

But then it occurred to me that this could not be a valid case, because it would have caused total gravity during that period to exceed the expansion rate. And that would have meant that the universe was not flat. But it is believed that it was flat.

So, I have no explanation still, as to why a decrease in mass/energy would cause a decrease in the expansion rate. Any suggestions or referrences would be appreciated.

4. Oct 21, 2007

pervect

Staff Emeritus

OK, I'm going to have to get a bit "nitpicky" here, because there's some sort of communication issue here.

Einstein's equation doesn't say anything about "the expansion rate" being a function of mass energy.

Einstein's equation can be interpreted as saying that for a small volume, the second derivative of the volume of a sphere of comoving particles, is proportional to the mass-energy density at the center of the sphere plus three times the pressure (assuming isotropy of the pressure).

The "expansion rate" will be some function of the first derivative of the rate of change of the volume of the comoving sphere.

So if we define the scale factor of the universe as a(t) (the a(t) in the FRW metric), and the expansion rate as da/dt, then the volume of a sphere of particles will be V0*a(t)^3, the rate of change of the volume will be 3 v0 a(t)^2 da/dt, and the second derivative of the rate of change of the volume will be

d^2V/dt^2 = 6 v0 a(t) (da/dt)^2 + 3 v0 a(t)^2 d^2 a/ dt^2

This is what's proportional to energy density + 3*pressure (rho+3P for short).

So if by "expansion rate" we mean da/dt, then you can see that dV/dt is proportional to the expansion rate, and that d^2V/dt^2 depends on both the expansion rate and it's first derivative (the first derivative of da/dt, the second derivative d^2a/dt^2).

Thus, we can see that we cannot solve for da/dt knowing rho+3P, because it could have literally any value.

You've totally lost me here.

The "total mass-energy of the universe" is a phrase which is rather ill-defined. For example, see MTW's "Gravitation" which says

One might guess that you mean by "total mass energy of the universe" that you multiply the density, rho, by the comoving volume.

But we don't need to know the "total-mass energy of the universe", and I don't see why it's relevant, anyway.

Einstein's field equation, which gives us the Friedman equation, is a purely local equation, and the question of what the "mass-energy of the universe" is is irrelevant.

Furthermore, while there are tricky issues with regards to the global defintion of energy conservation, there aren't any issues with regards to local energy conservation, which is already built into Einstein's field equations.

So in short, you only need Einstein's field equations to get Friedman's equations, and the question of what "the total mass-energy of the universe" might be isn't relevant to the problem - i.e. we only need to know the density of mass energy, and the pressure.

Last edited: Oct 21, 2007
5. Oct 22, 2007

jonmtkisco

Hi Pervect,

Well you can feel free to analyze the problem using the Einstein Field Equations. I haven't gotten there yet, I'm still working with the 2nd Friedmann equation.

You must be kidding if you are suggesting that "total mass/energy" isn't relevant to the Friedmann equation. As you yourself pointed out, and Hellfire documented, and I subsequently accepted, the reference in the 2nd Friedmann equation to "mass/energy density" is nothing more than "total mass/energy" divided by volume. It couldn't be more straightforward.

Don't give me a bunch of mumbo-jumbo about how "total mass/energy" is a meaningless quantity. If that's so, then the 2nd Friedmann equation by your definition is invalid. Again, all the 2nd Friedmann equation does is to divide this supposedly fictitious "total mass/energy" by volume to calculate energy density, which is the cornerstone of the equation.

6. Oct 22, 2007

jonmtkisco

Hi Pervect,

Given that, for purposes of the 2nd Friedmann equation "total mass/energy" and "energy density" are essentially interchangeable (the former being the latter multiplied by volume), my original question is equally valid if asked using energy density terminology:

"When energy density decreases, the expansion rate decreases. This is an important scenario, because under the traditional LCDM model, the energy density of the universe is held to have decreased dramatically during the radiation dominated era, as radiation gave up energy to expansionary redshift. Sure enough, the Friedmann equation dutifully calculates that the "momentum-like" continuation of expansion declines throughout the radiation-dominated era. Yet if energy density is held constant at its pre-decline value, Friedmann calculates that the expansion rate remains higher (and higher than we observe). You can convince yourself of this just by noting that energy density is in the top line of the Friedmann equation, and R is not.

Now I really need a clear explanation as to how, in a GR-based model, a large decline in energy density can cause a large decline in the expansion rate. That certainly violates the expected behavior of a "momentum-like" quantity. Less gravity ought to cause faster expansion than more gravity. That's pretty basic."

7. Oct 22, 2007

hellfire

Although you seem to be able to understand the math, you rely on some wrong heuristic arguments to understand the physics. Such a scenario we had already in this thread where I was tring to put in math terms your words and questions, and at the end I think we got the right answer. So humbly I suggest to get a good reference about cosmology or general relativity and rethink carefully in math terms what you have written above; what are the conditions for this to hold and how it relates to the apparent (incorrect) "paradox" in your argument.

8. Oct 22, 2007

pervect

Staff Emeritus
Less gravity ought to cause a slower deacceleraton than more gravity. And it does. Less gravity doesn't have anything to do directly with the rate of expansion at all.

Gravity causes acceleration - it doesn't cause velocity.

The dynamic Friedman equation

$$3 \frac{a''}{a} = \Lambda - 4 \pi G \, \left(\rho+ \frac{3P}{c^2}\right)$$

(FromThe Wikipedia

should be a consequence of the initial value equation

$$\left(\frac{a'}{a}\right)^2 = \frac{8 \pi G}{3} \rho + \frac{\Lambda}{3} - K \frac{c^2}{a^2}$$

along with the local energy conservation principle which says that density and pressure must satisfy

$$\frac{d}{dt} \left(\rho V \right) = -P \frac{d V}{dt}$$

where V, the volume of a fluid element, can be taken as V0 * a(t)^3 for some constant V0.

This is discussed on MTW, pg 728-729, where it is also pointed out that the full version of Einstien's equation automatically gives you the the above conservation principle without having to assume it separately. (It would take some work to get this conservation principle out of Baez's approach, though Baez says it's possible).

(I was originally having some problem confirming this, but I think I'm now in a position to say that's what I get, too. My textbook uses geometric units, the Wiki uses standard units, just to make life interesting).

Note that with the re-defintion of variables suggested by the Wikipedia, and the assumption that a'=0 this dynamic equation is essentially what Baez derives - the main point is that

a'' / a is proportional to (rho+3P) when rho and P include contributions from Lambda rather than Lambda being a separate variable. Don't sweat the units overmuch, Baez is using some funky units, the proportionality is what matters.

I assume it should also be possible to work backwards from the dynamic equation, and the above relation between rho and P to get the initial value equation, one should be able to regard this as a "solution" to the second order differential equation given by the dynamic equation along with the energy conservation equation, since it's possible to work forwards from the initial value equation + the local energy conservation equation to the dynamic equation.

Last edited: Oct 22, 2007
9. Oct 22, 2007

pervect

Staff Emeritus
Add: if you like Newtonian analogies, the dynamic equation is a bit like

m d^2 x / dt^2 = force

the initial value equation is like

(m/2) (dx/dt)^2 = E^2 - V(x)

V(x) being a potential function.

note the formal similarities. I'm not sure how far this analogy can be pushed, though - that's why I say "a bit like".

Specifically, the dynamic friedman equation and the dynamic Newtonian equation both involve a second derivative with respect to time, while the IV versions involve only the square of the first derivative.

You also go from the initial value equation to the dynamic equation in much the same way - differentiate with respect to time. In the Newtonian example, we get

m (dx/dt) = -dV/dx = force.

This analogy may make it clearer as to what I mean when I say gravity causes acceleration (d^2 x/dt^2), and not velocity (dx/dt).

Last edited: Oct 22, 2007
10. Oct 22, 2007

jonmtkisco

Hi Pervect,

I agree with you that gravity causes acceleration, and that any subsequent change in expansion velocity is therefore a second order effect. If you interpreted me to say anything different, that's sloppy wording on my part. Sorry about that.

Thanks also for the reference to Baez's GR website. I was very encouraged that he describes Vol"/Vol to be a critical metric in the Einstein Equations, as I found it to be when fiddling with the 2nd Friedmann equation. As I said, I found that Vol"/Vol is a constant during the Friedmann expansion, if total mass/energy (density*volume) is held constant.

In any event, the "paradox" (as Hellfire describes it) that I'm trying understand here is so simpleminded that the communication gap obviously exists only because of my inability to explain myself clearly.

Maybe it will help to just walk through the steps I took. I have a spreadsheet that calculates radius, volume, mass/energy, radius', vol', and vol" at various intervals between the end of inflation and 1400GY in the future. I've run various "what if" scenarios.

Here is the "what if" scenario that confuses me. I asked, "how would the Friedmann expansion rate calculation change, if the total mass/energy (and density) had its historical value at the end of inflation, but if that mass/energy were comprised 100% of matter and 0% of radiation." To simplify the question, I omitted the cosmological constant from the calculation. My reason for using 100% matter is to create a scenario where mass/energy remains a constant through time. No other significance.

The result is that in my test case, at any given absolute volume, the calculated radius' (meters/second) is more than 1000 times faster than what is calculated using the historical original mix of matter and radiation.

I'm not suggesting that this difference in calculated expansion rates between the "what if" and "historical" scenarios is a surprise. I'm just trying to interpret what it means. The following interpretation seems reasonable to me:

"For any given initial expansion rate and energy density, the 2nd Friedmann equation will calculate lower total expansion over any given time interval, if total mass/energy declines during that time interval (due to redshift of free radiation), than if total mass/energy had remained constant during that time interval." (Again, "total mass/energy" = Friedmann energy density*volume).

It indeed seems paradoxical if, starting from the same initial energy density value, a more rapid decrease in density over time results in a smaller universe over time. What gives?

(As one possible "fix" to this problem, I understand that adding positive pressure (of radiation) to the "historical" calculation might eliminate the discrepency, but as I mentioned earlier, adding a pressure factor would increase gravity during the early universe. I think that a higher gravity might cause the universe to be "closed" rather than "flat".)

11. Oct 23, 2007

pervect

Staff Emeritus
Let me see if I understand this. You assumed that the denisity, rho, remains the same, and when you say that it "was composed 100% of matter", you've kept rho constant, but you've reduced P to zero. You have not kept K, the Gaussian curvature, constant.

Assuming that this is actually what you did, it's easiest to explain this from the dynamic equation.

Numerically, using the dynamic form of the equation, rho remains consant, and P goes to zero. Because P = 1/3 rho for radiation, rho+3P should drop to half its value when you convert from a radiation dominated universe to a matter dominated universe, meaning that the deaccleraton also drops to 1/2.

The consequence of this is that you reduce the deacceleration of the universe, meaning that if you project the universe forward to the current age, it will be a lot bigger.

I'm assuming that that's when you talk about the calculated radius, you talk about the calculated radius "now".

I think this is wrong, or perhaps you just lost me again. I suspect, however, that you are under the illusion that total mass energy causes gravity. This is not correct. Pressure also causes gravity, i.e. what is important is not rho, but rho+3P.

An modified version of an example I worked out and put in the wikipedia might help clarify what this means, though it's not cosmogical, it's related to the Komar mass.

The original article is at http://en.wikipedia.org/wiki/Mass_in_general_relativity with a different slant.

Suppose you have a heavy hollow spherical pressure vessel, and in the center you have an matter-antimatter bomb. You put accelerometers at various locations to measure the force of Newtonian gravity - the interesting points are on the outside and inside surface of the pressure vessel, and at some distant location.

You explode the bomb, converting matter into energy, and generating relativistically significant pressures in the interior of the pressure vessel in the process. What happens to the gravity at the various accelerometers (assuming that you wait until the system reaches some sort of equilibrium to take your readings?).

For the accelerometers far away from the pressure vessel, and even for the accelerometers on the outside surface of the pressure vessel, there is no change in the gravity readings on the accelerometers. Not so on the accelrometer mounted on the inside of the pressure vessel. It's reading essentially doubles when you set off the bomb.

What happened? Pressure causes gravity (not just total mass energy), and the extra pressure due to the explosion causes the extra gravity on the inside accelerometers. The gravity does not change in the exterior region, because there is a tension in the walls of the pressure vessel. This reduces the gravitational contribution of the walls of the pressure vessel, so that there is no change in the accelerometers measuring the gravity outside the pressure vessel.

So if you look only at the exterior of the pressure vessel, you might think that "total mass-energy causes gravity". But when you look at the accelerometers on the interior of the pressure vessel, (which are quite well engineered so that they don't melt :-)), you realize that pressure also causes gravity.

Last edited: Oct 23, 2007
12. Oct 23, 2007

jonmtkisco

Pervect,

Thanks for the answer. I agree that pressure causes gravity.

What befuddles me is why there are two separate Friedmann equations, and how to apply them both at the same time. Can one derive a correct expansion solution by using only one of the two equations?

In the 2nd Friedmann equation, why is there no pressure component? How can the expansion rate be modeled accurately without including pressure?

Is rho the same thing in both Friedmann equations? That is, is it always equal to mass/energy "without" the pressure component? Is it always "total" mass/energy, including both "matter energy" and "radiation energy"? If so, then why are the two equations not precisely interchangeable?

How can a"/a require a pressure component, when a'/a does not? Mathematically, the two figures should be directly related.

Jon

13. Oct 23, 2007

Jorrie

2nd Friedmann Equation

Jon, just for clarification: I know the first Friedmann equation as the 'energy equation' or the 'expansion rate equation' (which is often just called the 'Friedmann equation') and the second one as the 'acceleration equation', which includes a pressure term. It seems that you refer to them the other way round, perhaps causing some confusion.

[Edit: I see Wallace and Pervect have clarified the issue and the standard names of the Friedmann equations below.]

Last edited: Oct 24, 2007
14. Oct 24, 2007

Wallace

To throw a spanner in the works there are actually three equations that are at times referred to as the Friedmann equations. They are not independent however, any two can be combined to give the third, so you only ever need two of them.

In any case all three contain a pressure term, I'm not sure what has lead you to believe otherwise?

Here are the three equations in the most general way I can think to express them:

The energy conservation equation for each individual energy component (radiation, matter, dark energy, curavture etc etc):

$$\frac{d\rho_i}{dt} = -3H\rho_i(1+w_i)$$

where H is the Hubble parameter and $$w_i$$ is the equation of state of the energy component i, defined as $$w_i = \frac{p_i}{\rho_i}$$

The expansion equation

$$\frac{H}{H_0} = \sqrt{\frac{8 \pi G }{3}\Sigma_i [\rho_i]}$$

The acceleration equation

$$\frac{\ddot{a}}{a} = -\frac{4 \pi G}{3}\Sigma_i [ \rho_i(1+3w_i)]$$

As an example of how you can reduce this to only two equations, if the equation of state, w, of a component is constant for all time the energy equation for that component can be solved to give:

$$\rho_i(a) = \rho_i(a_0)a^{3(1+w_i)}$$

and therefore the expansion equation becomes

$$\frac{H}{H_0} = \sqrt{\frac{\pi G 8}{3}\Sigma_i [\rho_i(a_0)a^{3(1+w_i)}]}$$

As you can see, pressure plays a part in all three equations. You may have been reading something that was a simplified equation assuming there is only matter in the universe (which is pressure less).

Last edited: Oct 24, 2007
15. Oct 24, 2007

pervect

Staff Emeritus
I thought I wrote something about this before, but I've been a bit distracted recently.

Let me quote from my textbook, MTW's "Gravitation", pg 728, with a few modifications in notation to clarify this:

While MTW choses to ignore the acceleration equation, I utilized it in my reply, because I like it from an intuitive standpoint. The point is that you should get the same answer either way, because both equations are equivalent when combined with the "energy conservation equation" which I have numbered (3) for future reference.

So we've got three equations (F1), (F2), and (3).

(F1) is the so-called initial-value equation:
$$\left(\frac{a'}{a}\right)^2 = \frac{8 \pi G}{3} \rho + \frac{\Lambda}{3} - K \frac{c^2}{a^2}$$

(F2) is the so-called dynamic equation
$$3 \frac{a''}{a} = \Lambda - 4 \pi G \, \left(\rho+ \frac{3P}{c^2}\right)$$

and we have a third equation, which is the "energy conservation equation:
$$\frac{d}{dt} \left(\rho a^3\right) = -P \frac{d}{dt} \left(a^3 \right) \hspace{1 in}$$

Any two of the above equations implies the third, i.e.:

(F1) + (3) -> (F2)
(F2) + (3) -> (F1)
(F1) + (F2) ->(3)

though I have personally only verified that (F1)+3 -> (F2)

I don't know if telling you that (3) for the relationship between pressure,density and the scale factor comes from

$$\nabla \cdot T = 0$$

where T is the stress-energy tensor of the cosmological fluid, will give you any insight into why it gets called the "law of conservation of energy", because I'm not quite sure of your background. But I'll add it in case it helps. So there are good reasons to assume that the energy equation is true on its own, though it turns out to be automatically generated by Einstein's field equations. (The full field equation route generates (F1) and (F2), which implies (3).)

Now, given all of these equations, how do we solve them?

I'm still very hazy on exactly what *you* are doing, other than you've got some sort of spreadsheet, which doesn't help much. I guess the solution is to describe what *I* am doing a bit more clearly.

Basically, the equations of state based on your matter model are going to imply some relation between rho and P. You have to combine these with the Friedmann equations to get the final answer, which is an expression for a(t).

If you consider a universe of pure matter, things are simple because P=0, always. (Actually if you have hot matter this might not be strictly true, a more exact statement is that P is negligible rather than zero, because the "hot" matter is not hot enough to have a relativistically significant pressure).

You can use P=0 plus (3) to find that rho*a^3 = constant for a universe of pure matter, because (d/dt) rho*a^3 = 0 via the energy equation.

If you consdier a universe of pure radiation, then P = 1/3 rho. In a similar manner, you can find that rho*a^4 = constant for a universe of pure radiation.

You can then use whichever of the Friedmann equations you like to find the evolution of a(t). (F1) is computationally more convenient, but I think that it doesn't give you as good an insight as to what's going on as (F2) does because (F1) is rather involved, and (F2) is much simpler.

Last edited: Oct 24, 2007
16. Oct 24, 2007

jonmtkisco

Hi Pervect, Wallace and Jorrie,

First, thank you for your insights. I'll use Pervect's terminology to refer to the equation I've been using as the "Friedmann Initial Value" or "Friedmann IV" equation. I'm glad that the results work out the same with any of the 3 equations.

Second, Wallace, the version of the Friedmann IV equation you posted was the first time I've ever seen it containing an explicit pressure component. I'm confused, can you explain?

Third, Pervect, you said:

"Let me see if I understand this. You assumed that the denisity, rho, remains the same, and when you say that it "was composed 100% of matter", you've kept rho constant, but you've reduced P to zero. You have not kept K, the Gaussian curvature, constant.

"Assuming that this is actually what you did, it's easiest to explain this from the dynamic equation.

"Numerically, using the dynamic form of the equation, rho remains consant, and P goes to zero. Because P = 1/3 rho for radiation, rho+3P should drop to half its value when you convert from a radiation dominated universe to a matter dominated universe, meaning that the deaccleraton also drops to 1/2.

"The consequence of this is that you reduce the deacceleration of the universe, meaning that if you project the universe forward to the current age, it will be a lot bigger."

I want to be clear that I deleted both $$\Lambda$$ and gaussian curvature elements in order to simplify the interpretation of my calculations. Shouldn't curvature always be zero, since the universe is believed to have been flat throughout the original expansion (post-inflation)? As a purely mathematical perspective, I think that by forcing the curvature=0 over time, I forced the equation to generate a faster deceleration in the 100% matter scenario. Still, I don't see why doing so is in any way invalid.

I'm a bit unclear as to whether the effect of positive radiation pressure is most properly incorporated by doubling gravity in the historical case of the Friedmann IV equation, or on the contrary by halving the gravity in the "what if 100% matter" case of the same equation. Since doubling gravity in the historical case seems...er...non-historical, I will follow your suggested route of halving gravity in the 100% matter case.

Initially, as you say, this must cause the expansion rate to decelerate more slowly in the 100% matter case. However, the critical problem here is that in the "historic" case, both total mass/energy density of radiation and radiation pressure decline over time (due to redshift), at the rate of a$$^{2}$$. At that rate, the halving of gravity (in the 100% matter case) caused by the inclusion of pressure in the formula quickly becomes a tiny rounding error, with virtually no impact on the long-term deceleration of the expansion rate, as compared to the "historic" case. In fact, its effect is entirely washed out within the first tiny fraction of a second after inflation ends. In the "historic case", by 3.6 seconds into the expansion, total mass has declined by a factor of 10$$^{16}$$, which obviously has a far more significant impact on the calculation. By comparison, that enormous reduction does not occur at all in the 100% matter case.

So I continue to find that dR/R remains a healthy 10$$^{3}$$ slower in the "historical" case than in the 100% matter case, at every scale factor "a" higher than a*10$$^{-5}$$.

The seeming paradox remains: a faster decline in mass/energy density results in a slower expansion and smaller universe.

Last edited: Oct 24, 2007
17. Oct 24, 2007

pervect

Staff Emeritus
What you need to think about (I have to think about it too) is whether or not this assumption about K satisfies the energy-conservation relationship that I described as (3).

18. Oct 24, 2007

Jorrie

Hi Jon. My 2 cents for what it's worth.

If you refer to Wallace's

$$\frac{H}{H_0} = \sqrt{\frac{\pi G 8}{3}\Sigma_i [\rho_i(a_0)a^{3(1+w_i)}]}$$

as "Friedmann IV", I suggest you rather don't - it may just confuse the issue further. It's the same as the expansion (or initial values) equation, just expressed differently for the purpose of showing the pressure influence clearly.

If you force the curvature to zero for a matter only case, then $\Omega_m = 1$ and for a radiation only case, then $\Omega_r = 1$. I think you cannot just delete the other density components and leave the rest at the same value. If you put $\Omega_m = 1$ and $\Omega_r = 1$ separately into the initial value (F1) equation (leaving Ho the same), then yes, the initial expansion rate (a') is faster for a radiation only case, but the deceleration (a'' from F2) is also larger when compared to a matter only case.

I don't know what you mean by "historic" case, but doesn't radiation energy density decline by $a^{4}$? The effect of initial radiation pressure may quickly become insignificant in terms of deceleration, but its effect on the future expansion rate is significant for a very long time afterwards.

Not necessarily if the initial expansion rate is larger, as it must be for a flat universe with more radiation energy density (and the same Ho, one must add).

19. Oct 24, 2007

jonmtkisco

Hi Jorrie,

I'm not sure I understand all of your points. The "historic" case is not "radiation only", it's the actual mix of radiation and matter that the historic universe is believed to have had. The only modified case is the "100% matter" case -- in which radiation = 0, so I haven't departed from "omega"=1.

In the "historic" case, radiation density declines at a^4, but I was referring to total mass/energy of radiation, which declines at a^2.

As far as the possibility that in the "historic" case the universe may have expanded so much in the first tiny fraction of a second that it outweighs all of the dramatically faster expansion after that point in the "100% matter" case, I guess my calculations haven't ruled that out, although I'm skeptical. I would be grateful if you can help me figure that out. I know the expansion rate at every absolute volume, but in the 100% matter case I don't know what "t" corresponds with each "r" or "v, since I can't start with a Hubble value at the present time.

Jon

Last edited: Oct 24, 2007
20. Oct 25, 2007

Wallace

Let's compare the initial value equation I posted:

$$\frac{H}{H_0} = \sqrt{\frac{8\piG}{3}\Sigma_i[\rho_i]}$$

with the one Pervect posted which was (with c=1 and some slight extra notations to make things crystal clear) :

$$(\frac{H}{H_0})^2 = \frac{8\piG}{3}\rho_m + \frac{\Lambda}{3} - \frac{K}{a^2}$$

Now I'm sure you're familiar with Pervects version and indeed that is how it is commonly presented. Then you look at mine and are surprised that pressure seems to have appeared from nowhere! In fact the effect of pressure is included in both, it's just more explicit in the general form I posted. The reason it's not usually written this way is that it's more general than is usually needed and harder to 'get a feel for' but none the less they are equivalent. The energy component represented in Pervects version are matter, vaccum energy (or cosmological constant) and curvature. These have eqautions of state of w=0,-1, -1/3 respectively. Substituting those values into the equations I gave leads to the form of the equation that Pervect posted.

Note that the fact that the energy components have different equations of state is due to the different pressure they exert. Out of interest, radiation has an equation of state of w=1/3, but is usually not included since the energy density drops so rapidly as to be unimportant for the global dynamics except in the early universe.

21. Oct 25, 2007

Jorrie

I think I understand what you are trying to do (correct me if I'm wrong). It seems you want to keep the initial expansion rate a' (or H') after inflation the same, but want to omit the radiation energy density and then see what happens with the deceleration and expansion rate during the first moments after inflation. Correct?

I think the moment you do that, you deviate grossly from flatness when a is small, simply because there is then not enough energy density (gravity) to balance out the (now excessive) expansion rate and the cosmos would have been very open. Removing energy density at the beginning must be accompanied by an appropriate reduction in the initial expansion rate. Alternatively, you have to 'balance the books' by converting that (removed) radiation energy density into matter density and so keep the initial expansion rate the same.

I'm not sure how to calculate that and whether it's even valid to do that, but I'll think about it (and hope the mentors come up with an answer in the meantime).

Absolute volume is poorly defined, as Pervect said many times before. It's much better to use the normalized expansion factor a and its evolution over time, which is easily calculated from the initial values equation for K=0 (or any value of K, if you like). For your 100% matter scenario, all you need to decide on is the total matter-energy density for your starting a and K and then integrate expansion rate over time.

I may have taken a few 'engineering-liberties' here, but I do not think it's too far off the mark.

22. Oct 25, 2007

Jorrie

Jon is particularly interested in the scenario immediately after inflation. Is it correct to bring radiation energy density into your above equation as follows:

$$(\frac{H}{H_0})^2 = \frac{8\piG}{3}(\rho_m + \rho_r) + \frac{\Lambda}{3} - \frac{K}{a^2}$$

or must rho_r be divided by a?

Last edited: Oct 25, 2007
23. Oct 25, 2007

George Jones

Staff Emeritus
Yes, another a is needed. It's interesting to think of this from point view of the particle/wave nature of light. As the universe expands, the number density of photons decreases at the same rate as the number density of massive particles. However, the wavelength of light also expands, thus decreasing its energy density by another factor of a.

24. Oct 25, 2007

Jorrie

Thanks George, but what still bugs me is this: although not explicitly stated, I assume that in the equation:

$$(\frac{H}{H_0})^2 = \frac{8\piG}{3}(\rho_m + \rho_r) + \frac{\Lambda}{3} - \frac{K}{a^2}$$

$\rho_m$ and $\rho_r$ represent the energy densities when the expansion factor was $a$, so the dependency of the radiation energy density on $a$ has already been accounted for. So why divide it by $a$ again?

To clarify my uncertainty further, consider the radiation energy density shortly after inflation, at say $a = 10^{-25}$, which is in the order of $\rho_r=10^{-30}/10^{-100} = 10^{70}$ kg/m$^3$. It already includes the $1/a^4$ factor, so $1/a$ must not be introduced another time.

Or am I messing something up?

Last edited: Oct 25, 2007
25. Oct 25, 2007

George Jones

Staff Emeritus
No, I messed up. And I'm still a little confused.

I think that pervect posted ($c=G=1)$

$$H^2 = \frac{8\pi}{3}\rho + \frac{\Lambda}{3} - \frac{K}{a^2}$$

I see why Wallace's $\pi$'s disappeared (look at the latex), but I can't see why an $H_0$ appeared.

And, yes, $\rho = \rho_m + \rho_r$ when radiation is also included.

Last edited: Oct 25, 2007
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook