# What is the best current value for Omega_r?

• I
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## Summary:

I have been trying to track down the most current published value for Omega_r. The most recent (?) I have found is 0.824 x 10^-4. There is a list of references, but I am not sure which was the source of the value. My guess it might be the one dated 2017.
The source of the 0.824×10-4 value is
I also looked at the Planck 2018, paper I but could not find a value for Ωr. I searched the internet with:
but I had no success. I also looked at Planck 2013 and found the value 0.92364×10-4.

Any suggestions would be greatly appreciated.

BTW: The reason behind this search is I want to calculate the best value I can for the time at which the gravitational influences of radiation and matter are the same.

Last edited:
vanhees71

PeterDonis
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Any suggestions would be greatly appreciated.
I don't have any better data sources to suggest, but there is an easy way to estimate ##\Omega_r## for yourself. Just calculate the energy density of black-body radiation at the temperature of the CMB (which you can find a formula for, for example, at the end of this article), and divide by the critical density. This estimate won't include other radiation or relativistic particles (such as neutrinos), but it will give you the right rough order of magnitude.

vanhees71
Arman777
Gold Member
If you want to include neutrinos, you can use the temperature of the neutrino background density, which is related to the CMB temperature as

$$T_{\nu} = (4/13)^{1/3}T_{CMB}$$

Buzz Bloom
Gold Member
If you want to include neutrinos, . . .
Hi Arman:

I have some doubts about neutrinos. I think the formula you posted was the relationship in an early era which also assumed that either neutrinos were mass-less, or that the mass was so small that neutrinos would contintue to behave until the present as relativistic particles. In another thread I calculated the behavior of relativistic H atoms at decoupling, and the result was that the temperature would become TH(a=1) = 0.002959 K, compared with the CMB temperature of 2.725 K.
It was pointed out to me that this was wrong because of reionization. However, reionization would not affect neutrinos. I have not as yet used (but plan to use) the same math method to calculate the current average energy of neutrinos. I will post the result when the calculation is completed.

Regards,
Buzz

PeterDonis
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In another thread I calculated the behavior of relativistic H atoms at decoupling
No, you didn't. You calculated the behavior of non-relativistic H atoms at decoupling. The temperature of decoupling is way, way below the temperature at which H atoms would be relativistic. Your calculations in that thread would not be correct for particles that remained relativistic up until the present.

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Your calculations in that thread would not be correct for particles that remained relativistic up until the present.
Hi Peter:

Thank you for pointing out my mistake in thinking H atoms at about 3000 K were relativistic at decoupling. I would very much like to understand the math that shows that that neutrinos at about 3000 K were still relativistic at decoupling. Can you please recommend some reference?

Regards,
Buzz

PeterDonis
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I would very much like to understand the math that shows that that neutrinos at about 3000 K were still relativistic at decoupling.
It all depends on their masses. Roughly speaking, if the energy equivalent of the particle's mass is smaller than the energy equivalent of the temperature in question, then the particle is relativistic.

The energy equivalent of a temperature is given by that temperature times Boltzmann's constant, which is about ##8.62 \times 10^{-5}## eV per K. So the energy equivalent of 3000 K is about 0.26 eV. Our current best estimates of neutrino masses are much smaller than that, so neutrinos would indeed have been relativistic at decoupling.

In fact, our current best estimates of neutrino masses are even smaller than the energy equivalent of 3 K, which is the CMB temperature today, so those estimates indicate that primordial neutrinos would still be relativistic today.

Buzz Bloom
Arman777
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I have found much much better solution.

We know that matter radiation equality ##a(t=t_{rm})## corresponds to ##a_{rm}## and it can be written as

$$1=\frac{Ω_m}{Ω_r}=\frac{Ω_{m,0}a_{rm}}{Ω_{r,0}}$$

So $$a_{rm}=\frac{Ω_{r,0}}{Ω_{m,0}}$$

Let us use $$a_{rm}=(1+z_{rm})^{-1}$$ and write

But we know ##z_{rm}## and ##Ω_{m,0}## from the Planck 2018. (See table 2, https://arxiv.org/pdf/1807.06209.pdf)

So all we have to do this algebra and we find

$$\Omega_{r,0} = \frac{\Omega_{m,0}}{1+z_{rm}}$$

Thus we obtain $$\Omega_{r,0} = 9.2899 \times 10^{-5}$$

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Gold Member
But we know zrm and from Ωm,0 the Planck 2018.
Hi Arman:

I can use a little help.

I found a row in Table 2 (pg 16) for Ωm. Which of the 6 columns did you use to pick a value?

Regarding z, I found zre, z*, zdrag, and zeq. Which of these is zrm, and why?

Regards,
Buzz

Arman777
Gold Member
I found a row in Table 2 (pg 16) for Ωm. Which of the 6 columns did you use to pick a value?
You can use any of them
Regarding z, I found zre, z*, zdrag, and zeq. Which of these is zrm, and why?
Its the ##z_{eq}## which represents matter radiaiton equality. I just preferred to use ##z_{rm}##

Buzz Bloom
Gold Member
You can use any of them
Hi Arman:

Your quote above confuses me. When I do the calculations of the six possible values, I get a range from the minimum of 8.650 10-5 to the maximum 9.408 10-5. This seems like the best I can do is a single digit of precision: 9 10-5. If there is a rational reason for one of the six to be best, then perhaps I might find that choice to give me perhaps 3 or 4 digits of precision. Any suggestions?

Regards,
Buzz

PeterDonis
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We know that matter radiation equality at=trm corresponds to a(trm) and it can be written as

$$1=\frac{Ω_m}{Ω_r}=\frac{Ω_{m,0} \times a_rm}{Ω_{r,0}}$$

So $$a_{rm}=\frac{Ω_{r,0}}{Ω_{m,0}}$$
Where is all this coming from?

$$\Omega_{r,0} = \frac{\Omega_{m,0}}{1+z_{rm}}$$
This equation is correct, but I'm not sure you got it via a correct process.

The way we can see that it is obviously correct is that, since "matter radiation equality" means ##\Omega_m = \Omega_r = \Omega_{rm}##, and since ##\Omega_m## decreases as ##\left( 1 + z \right)^3## while ##\Omega_r## decreases as ##\left( 1 + z \right)^4##, we have

$$\Omega_{m, 0} = \frac{\Omega_{rm}}{\left( 1 + z_{rm} \right)^3}$$

$$\Omega_{r, 0} = \frac{\Omega_{rm}}{\left( 1 + z_{rm} \right)^4}$$

and therefore

$$\Omega_{m, 0} \left( 1 + z_{rm} \right)^3 = \Omega_{r, 0} \left( 1 + z_{rm} \right)^4$$

which leads to the equation I quoted above. But that doesn't seem to be the reasoning you are using.

Arman777
Gold Member
Hi Arman:

Your quote above confuses me. When I do the calculations of the six possible values, I get a range from the minimum of 8.650 10-5 to the maximum 9.408 10-5. This seems like the best I can do is a single digit of precision: 9 10-5. If there is a rational reason for one of the six to be best, then perhaps I might find that choice to give me perhaps 3 or 4 digits of precision. Any suggestions?

Regards,
Buzz
I would prefer TT,TE,EE+lowE+lensing for only CMB data, but if you want to include BAO, you can also use TT,TE,EE+lowE+lensing+BAO.

Arman777
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Where is all this coming from?
I already wrote, "matter radiation equality means" $$\frac{\Omega_r}{\Omega_m} = 1$$

PeterDonis
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I already wrote, "matter radiation equality means" $$\frac{\Omega_r}{\Omega_m} = 1$$
Yes. What about the rest of what you wrote?

Arman777
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Yes. What about the rest of what you wrote?
$$1=\frac{Ω_r}{Ω_m}=\frac{Ω_{r,0}a_{rm}^{-4}}{Ω_{m,0}a_{rm}^{-3}}$$

Its the same as yours.

PeterDonis
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$$1=\frac{Ω_r}{Ω_m}=\frac{Ω_{r,0}a_{rm}^{-4}}{Ω_{m,0}a_{rm}^{-3}}$$

Its the same as yours.
That's not what you wrote before.

Also, why are you writing ##a_{rm}## instead of ##1 + z_{rm}##?

Arman777
Gold Member
That's not what you wrote before.
Can you qoute that equation ?
Also, why are you writing arm instead of 1+zrm?
It does not matter which one I use. I just preferred that way.

PeterDonis
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Can you qoute that equation ?
It's what you wrote in post #8. It was not clear in that post what your reasoning was.

Arman777
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It's what you wrote in post #8. It was not clear in that post what your reasoning was.
Not clear ? I have wrote " We know that matter radiation equality " and showed it by

##1 = \frac{\Omega_m}{\Omega_r}##

I think its pretty clear.

PeterDonis
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I think its pretty clear.
I'm sure you do. Everyone always thinks their own posts are perfectly clear. But I don't think it's clear, and I've already explained why; in post #12 I quoted the exact equation you posted that I thought was problematic (which is not the equation ##1 = \Omega_m / \Omega_r##), and explained why in the very next sentence in that post after the quote. You haven't addressed any of that; all you did was rewrite the equation I wrote, and say that yours was the same, when it wasn't. Yours is mathematically equivalent to mine, but that doesn't mean it's "the same"; I explained exactly how mine was arrived at, and you didn't do that for yours, you just stated it without any explanation or argument.

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So the energy equivalent of 3000 K is about 0.26 eV. Our current best estimates of neutrino masses are much smaller than that, so neutrinos would indeed have been relativistic at decoupling.

In fact, our current best estimates of neutrino masses are even smaller than the energy equivalent of 3 K, which is the CMB temperature today, so those estimates indicate that primordial neutrinos would still be relativistic today.
Hi Peter:

I think I now more-or-less understand the concept of a relativistic particle remaining as such as it's energy reduces during the universe expansion. However, as I have tried to collect numbers, I seem to have a very much different answer than your post quoted above for the current average temperature of cosmological neutrinos, and for the neutrino's energy.

The following source gives a current temperature for a current neutrino.
(Eq 1) T = 1.96 K.

E is energy in eV units.
Mc2 is the energy equivalent, also in eV units, of one of the neutrinos' mass.
T is temperature in K units.

A recent report (below) gives the "mass" of the neutrino with the least mass as
(Eq 2) Mc2 = 0.086 eV.
This is the same as the number you quoted.
(I tried to find a more formal source, but I was unable to do so. I am also unfamiliar with the standards for the use of technical vocabulary, but I am guessing that this value actually is the energy equivalent Mc2 rather than the mass.)​
The upper bound sum of the three kinds of neutrinos is 0.26 eV. This means that the neutrino with the largest mass-energy E must be in the range
0.087 <= E <= 0.088.
The middle neutrino's mass-energy would need to be in the range.
0.086 <= E <= 0.087.
The following is a series of equations leading to a conclusion. Eqs 3 and 4 yield the average energy of a black body photon at temperature T.
(Eq 3) E = 3.832 kB T.
(Eq 4) kB = 8.617333 10-5 eV/K
(Eq 5) E = 3.3022 10-5 T.
Combining Eqs 1 and 5 yields the mass-energy of the current cosmological neutrino, assuming it is still relativistic.
(Eq 6) E = 0.00126 eV.
This is much less than the rest mass, so it is clear that the current cosmological neutrons are no longer relativistic.

What I think is needed is a criterion specifying a lowest acceptable value for velocity for which it is proper to call a particle relativistic. From such a value, it would be possible to calculate the temperature corresponding to this value. For non-relativistic particles, something similar to what I calculated for electrons (when I ignored reionization) would then be an appropriate way to calculate the energy of a non-relativistic particle as the universe expands.

Regards,
Buzz

PeterDonis
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A recent report (below) gives the "mass" of the neutrino with the least mass as
These are upper bounds, not actual values. The actual masses could be smaller.

I am guessing that this value actually is the energy equivalent Mc2 rather than the mass.
Most physicists in contexts like this will give "mass" in energy units, which means they are actually giving ##Mc^2##, yes.

it is clear that the current cosmological neutrons are no longer relativistic.
Yes, you're right, I had misread orders of magnitude. They were relativistic at neutrino decoupling, but, unless the actual neutrino masses are much, much smaller than the current upper bounds, they are not relativistic today.

What I think is needed is a criterion specifying a lowest acceptable value for velocity for which it is proper to call a particle relativistic.
The general rule of thumb is that if the ##\gamma## factor is close to ##1##, it's "non-relativistic"; if ##\gamma## is very large, it's "relativistic" (or sometimes "ultrarelativistic"). The gray area is if ##\gamma## is around ##2## (which happens for ##v / c = \sqrt{3} / 2##), so the kinetic energy is roughly equal to the rest energy. Some sources describe this as "relativistic", others don't; but no sources, to my knowledge, describe this as "non-relativistic".

Buzz Bloom