I Q re the matter term in Friedmann’s equation

1. Feb 8, 2017

Buzz Bloom

It has occurred to me since May 2015 that for certain purposes the matter term in the Friedman equation might benefit from a modification.
The Friedmann equation
shows the matter term to be based on the assumption that the particles of matter never acquires any significant kinetic energy. One example for which this assumption is inadequate is calculating the affect on the solution t(a) by assuming DM consists of particles with small mass. In 1998 the idea that neutrinos might be the stuff of DM was an active possibility.
Apparently it is still active at MIT.
However, I have been unable to find any articles discussing the effect on t(a) based on modifying the mass term, even though the corresponding change in t(a) could have a significant impart on the primordial nucleosynthesis of deuterium.
To evaluate this possibility, the first step would seem to be to find the appropriate revised function F(a) to replace
F(a) = a-3​
as the coefficient of ΩM. I would much appreciate if one or more participants of PF would check my math and post descriptions of any errors they find, or to acknowledge that they found none.
The mass term in the Friedmann equation that I derived is:
It is clear that for a = ε << 1:
Q02 << 1, and F(a) ≈ a-3.
Also, for 1-a = e << 1:
Q02 ≈ 1, and F(a) ≈ a-4.

The following is the derivation of the modified F(a).

Last edited: Feb 8, 2017
2. Feb 8, 2017

Chalnoth

Neutrinos as a significant component of the dark matter are definitely out, as they are way too light, and wouldn't be able to form structures early enough in the universe to explain observations.

As for the possibility of kinetic energy, yes, it's absolutely conceivable that the dark matter has a temperature in our early universe that is measurably different from zero. This is contained in the proposed "warm dark matter" models. These models don't currently have much in the way of supporting evidence, but many are plausible.

3. Feb 8, 2017

Buzz Bloom

Hi @Chalnoth:
Thanks very much for responding. Also, thank you for making it clear to me that neutrinos are no longer considered to be a significant kind of DM. I have been unable to find anything definitive on this point searching for the past two years. Can you cite a reference for me?

I have the idea the using the modified F(t) it would be possible to estimate a lower bound for the mass of a DM particle that would make up a significant portion of DM. To make this calculation I would need an estimate for a temperature at which these particles would be in equilibrium with photons. Can you help me with that?

Regards,
Buzz

4. Feb 9, 2017

Chalnoth

For the known neutrino background, see here: https://en.wikipedia.org/wiki/Cosmic_neutrino_background

Attempts to use an additional neutrino species as the dark matter particle also exist, but the experimental evidence is strongly against this. See here for a summary of the Planck data on the matter, for example:
http://www.math.columbia.edu/~woit/wordpress/?p=5685

Basically, dark matter cannot be light neutrinos. There's not enough of them and they're too fast. If there's another, much heavier neutrino, then that could definitely be the dark matter particle, but it'd also have a much lower temperature.

5. Feb 9, 2017

Chalnoth

By the way, if you want something more in-depth, this paper provides a good overview of the cosmological constraints on dark matter:
https://arxiv.org/abs/1604.05701

6. Feb 9, 2017

Staff: Mentor

This isn't an assumption, it's a definition of the matter term. Particles that have significant kinetic energy compared to their rest mass appear in the radiation term. That's because the dependence on the scale factor for such particles is $a^{-4}$, not $a^{-3}$; the scale factor dependence is determined by the kinetic energy relative to the rest mass (or, equivalently, the pressure relative to the rest energy).

7. Feb 9, 2017

Buzz Bloom

Hi Peter:
I think I may have a misunderstanding about terminology.

My concept of "matter" as it pertains to the matter term in the Friedmann equation (FE), is that matter consists of particles that have some kinetic energy V, and typically V << mc2, so it is ignored in the FE. However, for very small values of a, and corresponding high temperature K, the particles becomes relativistic and then it behaves like radiation. When solving the FE to calculate numerical values for H(a) and a(t) when t0 >> t something must be done to deal with the transition regarding the particles which comprise the matter term when it is in between being a non-relativistic particle and a relativistic particle.

A somewhat simple approximate way to do this would be to use an intermediate value of t, say ti, such that for the particle at a(ti)
V = mc2.​
For
t=ti
set
revised Ωr ⋅ a-4 = Ωr ⋅ a-4 + Ωmp ⋅ a-3
and
revised Ωm ⋅ a-3 = Ωm ⋅ a-3 - Ωmp ⋅ a-3 .​
Ωmp
is that portion of Ωm corresponding to the particles with mass m.

However, I think a cleaner way to make the adjustment is to calculate Q0 for the particle and use the form for F(a) I derived in post #1. If there are several different kinds of particles with different masses, each would have it's own value for Q0, and in the FE each different kind of matter would have its own term. Do you see anything wrong with using this approach to get a more accurate calculation of t(a) for very small values of a?

Regards,
Buzz

Regards,
Buzz

Last edited: Feb 9, 2017
8. Feb 9, 2017

Staff: Mentor

As far as the Friedmann Equation is concerned, there is no such thing as "kinetic energy"; there is just energy density and pressure. The definition of "matter" is that pressure is negligible compared to energy density (mathematically the pressure of "matter" is zero). If you work out what this means in terms of the individual particles in a fluid, it means that kinetic energy is small compared to rest energy, yes. But the FE has no concept of such a split in the energy density.

Yes, that is why the universe is considered to be radiation dominated at early times. But again, in terms of the FE, "radiation" means $p = \rho / 3$, i.e., the pressure is one third of the energy density. If you look at what this implies for individual particles in a fluid, it means that kinetic energy is much greater than rest energy; but the FE doesn't know anything about kinetic vs. rest energy, it just knows about energy density (total) and pressure.

No, this is not necessary. All you have to do is deal with the transition from radiation dominated to matter dominated. The FE does not track individual particles; it just tracks energy density and pressure.

Instead of making this up on the fly, you should look in the literature to see how it is done in the current models. I think you will find that, by the time you have gone back to times early enough that the objects currently counted as "matter" are relativistic, the entire universe is already strongly radiation dominated anyway.

9. Feb 9, 2017

Chalnoth

If you want to be technical, the Friedmann equations say nothing at all about matter. They just relate energy density and pressure to expansion. The Friedmann equations can just as easily be used to determine the expansion from matter with a significant kinetic energy component as not. It's just that usually we make use of a matter term that assumes no pressure (i.e., negligible kinetic energy per particle).

10. Feb 9, 2017

Staff: Mentor

Yes, strictly speaking, that's true of the general version of the FE. But the version of the FE that is in the OP puts terms on the RHS that embody various assumptions about the pressure-energy density relationship, and shows how those different assumptions affect the scale factor dependence. The "matter" term under this definition is the one that embodies the assumption of zero pressure, and has the consequent $a^{-3}$ scale factor dependence. This version of the FE does not have a name for "matter with significant kinetic energy" because it assumes it away--there is just "matter" (zero pressure) and "radiation" (pressure 1/3 energy density), with no in between. (You could add another term for "in between", with pressure non-negligible but less than 1/3 energy density, but in practical terms there is no point since there does not seem to be significant energy density in such a component.)

11. Feb 10, 2017

Buzz Bloom

Hi Peter:

What I am interested in exploring is just how much the calculations of, for example t(a), would be affected by assuming DM consisting entirely of a particle type with a specific relatively small mass and corresponding small but non-negligible kinetic energy. It may well be that no value for this mass would lead to any significant changes, but I would like to make the calculation to understand why this is so. The purpose of this thread is to seek help to confirm that the form of F(a) I derived would be a mathematically correct way of adjusting the FE for this purpose. If the math is OK, but the physics is wrong, what other form of the FE would be better to use for this purpose?

Regards,
Buzz

12. Feb 10, 2017

Staff: Mentor

In other words, not zero pressure ("matter"), not $p = \rho / 3$ (radiation), but something in between?

This is verging on personal speculation, which is off limits. But heuristically, such a hypothetical "in-between" substance would, I would think, appear as a separate term in the equation, with its own $\Omega$ (and picking a new subscript), multiplied by a power of $a$ somewhere between $-3$ and $-4$. I would suggest looking at the literature to see if anyone has tried such a model.

13. Feb 10, 2017

Staff: Mentor

This thread is closed as it appears to be verging on personal speculation.