# I Q re Wiki Equation for t x T^2

1. Feb 14, 2017

### Buzz Bloom

The following is from

The era began at temperatures of around 10 MeV (116 gigakelvin) and ended at temperatures below 100 keV (1.16 gigakelvin). The corresponding time interval was from a few tenths of a second to up to 103 seconds. The temperature/time relation in this era can be given by the equation:​

tT2 = (0.74 s MeV2) × ( 10.75 / g∗ )1/2

where t is time in seconds, T is temperature in MeV and g* is the effective number of particle species. (g* includes contributions of 2 from photons, 7/2 from electron-positron pairs and 7/4 from each neutrino flavor. In the standard model g* is 10.75).​

Assuming g* = 10.75 simplifies this to
tT2 = 0.74 s MeV2

I assume that what this means is that the product of time in seconds (t) and temperature squared (T2) in MeV2 will equal 0.74. However if I multiply the corresponding t and T2 values, I fail to get aything close to 0.74.

(1) 10 MeV corresponds to "a few tenths of a second",
and
(2) 100 keV = 0.1 MeV corresponds to 1000 seconds.

(1) is a bit ambiguous, so I arbitrarily substitute 0.3 seconds for "a few tenths of a second". Multiplying by 102 yields the product 30, not close to 0.74.
For (2), multiplying 0.12 = 0.01 by 1000 yields the product 10, again not close to 0.74.

What am I misunderstanding?

2. Feb 15, 2017

### Staff: Mentor

Let's do the reverse: Use the equation to find initial time and final temperature. 0.74s/102 = 7.4 ms for 10 MeV and $\sqrt{\frac{0.74MeV^2}{1000}} = 27 keV$ for 1000s. "a few tenths of a second" is a bit of a stretch, but "below 100 keV" certainly fits.

Those steps are never sharp lines, so there is some ambiguity when BBN starts. Different authors might use different points in time for the start. Reference 2, used for the time, says "At times from around 10-2 seconds through the first several minutes after the Big Bang, the temperature passed through the range from around 10 to below 10-1 MeV", which fits better to the equation but not to the article citing it. I changed it in the article, it is now "from 10 ms to 1000 seconds".

For high energies, the formula doesn't work because g* depends on the energy, see reference 3 in the article, Figure 21.3, but that effect is small below 100 MeV.

3. Feb 18, 2017

### Buzz Bloom

Hi @mfb:
I think I get what you are saying about the vagueness and variability of the T and t values, and also that the 0.74 value is more reliable. I am I interpreting you correctly?

I also understand that during this radiation dominated era when all of the matter is highly relativistic
(1) t ∝ a2
and
(2) T ∝ 1/a.​
Therefore t × T2 is a constant.

It will be convenient to chose a character, say τns, to represent the nucleosynthesis value of tT2 assuming g*=10.75.
τns = 0.74 MeV2 s = 1.00 × 1020 s K2 .​

The derivation of τ in K s units is based on
1 electron-volt (eV) = 11604.5250061657 kelvin (K),
and
1 MeV2 = 134665000618725000000 K2

Converting 0.74 MeV2 to K2 and rounding to two digits yields the above result for τns .
What a coincidence to get a power of 10 with two decimal places of precision!
I am making a suspicious guess that this precision of the value of τns was estimated, by the authors of the source of the Wikipedia article, only to an order of magnitude rather than to two digits. What do you think?

Next I derive a value for τns by numerical integration of the Friedmann equation. The first step is to derive a value for a corresponding to the nucleosynthesis era.
ans = T0 / Tns = 4.256 × 10-9
where T0 is current Temperature = 2.72548 K.

The Friedmann equation can be integrated to derive
tns = t(ans).​
I used the following online tool to do the following integration.

The following shows the results of the online tool calculating the integral for τns/H0 .

Multiplying this result for
H0 × τns by 1/H0 = 4.555179 ×1017 s​
yields
τns= 433 s = 7.21 min.​
This is within the range of “a few tenths of a second to up to 103 seconds” but not close to the harmonic mean. Multiplying
(11.6 GK)2 = 135 GK2 by 433 s​
yields
5.83 × 1022 K2 s .​
This is 2.7 orders of magnitude more than the 1020 value for τns K s equivalent of the 0.74 GeV s given in the Wikeipedia article for tT2.

I do not know what to make of this discrepancy.

BTW, using the tool to integrate the above Friedmann integrand from 0 to ∞ gives the result
t0 × H0 = 0.95583187680499,​
which is quite close to the value using the published values for t0 and H0 yielding the product
t0 × H0 = 0.95595.​

Regards,
Buzz

Last edited: Feb 18, 2017
4. Feb 18, 2017

### Staff: Mentor

Nuclei were not relativistic any more, but the universe was completely dominated by radiation.

I don't think Kelvin values were involved anywhere. The reference gives 2.4/sqrt(N(t)), that is equivalent to the Wikipedia formula: 2.4/sqrt(10.75)=0.732. The small rounding error is probably on my side, from using (21.43) instead of (21.42). This is all, everything you try to interpret into it later is not part of the reference of the Wikipedia article.

Where does that value come from, and why do you expect it to correspond to something special?

5. Feb 18, 2017

### Buzz Bloom

Hi @mfb:
I made a mistake. Thanks for catching it. (I am a terrible proof reader of my own work no matter how hard I try.)

The following is a correction.

The calculation of a value for ans comes from:
(1) the assumption that T∝ 1/a which implies that a1/a2 = T2/T1;
(2) choosing the corresponding current temperature, T0, as T1, and
(3) choosing a representative temperature of the nucleosynthesis era,Tns, as T2,
(4) choosing the current value of the universe scale, a1 = 1, and the scale, ans, as a2.
The Tns value I chose, 11.6 GK was the harmonic mean between the upper and lower temperatures given in the Wikipedia article, 116 KG and 1.16 KG.

Therefore
ans = 2.72548 / 11.6 × 109 = 2.35 × 10-10.
xns = ln (ans) = -24.474.
The value of the integral from 24.474 to ∞ of f(x) dx is
Xns = 2.92 × 10-20 .

tns = Xns / H0
= 2.92 × 10-20 / 0.21953 × 10-17 s
= 1.33 x 10-2 s = 0.0133 s.

This seems to be too small, so I have probably made another mistake. I will check it more thoroughly tomorrow.

Regards,
Buzz

BTW, I am puzzled by:
Nuclei were not relativistic any more, but the universe was completely dominated by radiation.​

Last edited: Feb 18, 2017
6. Feb 21, 2017

### Buzz Bloom

Hi @mfb:

I found an error. I have been thinking about the problem for several days now, and I had two ideas about what the error might be. One was that the online integration tool might have hidden limitations with the integral I was calculating. I have since confirmed that that was not the problem. While determining that I decided I must have made a typo using a calculator tool to calculated the log of ans. I discovered that I had accidentally entered an extra zero, so the logarithm was off by a factor of ln 10.

Here are the new recalculated results for three temperatures.
T = 11.6 GK -> t = 0.01333 s
T = 116 GK -> t = 1.333 s
T = 1.16 GK -> t = 133.3 s
The product tT2 = 1.793 × 1020 GK2for all three value of T. This is reasonably close (within a power of 2) to the value
1020 GK2 which corresponds to the Wikipedia value 0.74 MeV2.

I am now guessing that this relatively small discrepancy might be due to the fact that during the annihilation of antiparticles, and later the annihilation of positrons, the photon temperature rose without any changes to the scale factor a and the corresponding value of t(a). Do you think this is a plausible explanation?

Regarding the non-relativity o protons during the nucleosynthesis era, I calculated that that the average
Q = v/c = 0.22753398.
v = 83129569 km/hr.
Although not relativistic, protons are moving along at quite a good clip.

Regards,
Buzz

7. Feb 21, 2017

### Staff: Mentor

Might be.

0.22 c is 19 MeV per nucleon, still a bit too hot for nucleosynthesis.

8. Feb 22, 2017

### Buzz Bloom

Hi @mfb:

I am puzzled by the 19 MeV. Perhaps it's an indication another of my errors.

I assumed a temperature of 116 GK = 10 MeV, which the Wikipedia article gives as the temperature at the start of the nucleosynthesis era.
The average photon energy at 116 GK is E = 6.14698161990616E-12
I assume that this will be the average kinetic energy V of the protons.
The mass energy (mc2) of the proton is P = 1.50327759289611E-10 J.
The ratio R = V/P = 0.0408905291 .
Q2 = (v/c)2 = 1- 1/(1+R)^2 = 0.0770251099177389
Therefore Q = 0.2775339798.

The equation
Q2 = 1- 1/(1+R)^2​
is derived from
mc2 + V = mc2 / (1-Q2) (1/2).
(1 + R)2 = 1 / (1-Q2)​

Please explain how you derived the 19 MeV.

Regards,
Buzz

Last edited: Feb 22, 2017
9. Feb 22, 2017

### Staff: Mentor

The nucleon mass is 940 MeV, and 1/2 (0.22)^2 * 940 MeV = 22.8 MeV. Note that the speed of light cancels if we use MeV as mass. In the previous post I forgot the second "2" and used 0.2 c. 0.22c leads to a higher energy.

We can also use the relativistic formula, but that leads to an even higher energy (23.8 MeV).

10. Feb 22, 2017

### Buzz Bloom

Hi @mfb:
Thanks for your post. You seem to have uncovered another error of mine I will need to investigate.

I don't get where the Q = 0.22 came from. It should be 0.2775. Also I am using mp = 638 MeV.

I started with a guess that the approximation you used was not precise enough.
You multiply the proton mass-energy by (1/2) Q2 to get a kinetic energy in MeV units. This calculation assumes that
V = (1/2) mv2.​
This is an approximation of
1/(1-Q2)1/2 ≈ 1+(1/2)Q2 .​
For Q = 0.2775, this approximate equation becomes
1.038512555 ≈ 1.0408905291​
This is much closer than I expected, so there is still a big discrepancy to be found.
0.0408905291 * 938 = 38.4 MeV.​
I don't understand how the 10 MeV I stared with ends up as 38 MeV.
I will have to do some more checking.

Regards,
Buzz

11. Feb 22, 2017

### Staff: Mentor

The proton mass is 938 MeV, not 638 MeV. That should increase your estimates by 50% already.
I edited the relativistic result in my post, it is 23.8 MeV (for 0.22c).

0.2775 c leads to 38.3 MeV (relativistic).

12. Feb 24, 2017

### Buzz Bloom

Hi mfb:

I feel pretty confident that I found the cause of the discrepancy between my interpretation of Q = 0.2775 corresponding to 10 MeV and your interpretation that this Q corresponds to 38.3 MeV.

The first point is that I calculate the average the average photon energy at 10 MeV to be
hμ = 6.14698161990617e-12 J. (Please see below for more details related to the derivation.)

Using the conversion factor
1 J = 6.2415093433e+18 eV = 6.2415093433e+12 MeV .
http://www.rapidtables.com/convert/energy/joule-to-ev.htm
results in
6.14698162e-12 J = 38.3664432137 MeV .​

The direct conversion of temperature units to energy units seems to be based on the assumption
x = hμ/kT = 1 exactly, or some value very close to 1.​
From the Planck equation for the distribution of photon energies, hμ,
B(μ,T) = (2 h μ3 / c2) × 1 / (ex - 1) .​

The mean energy corresponds to xmean = 3.83813983744828 .
This is calculated as the quotient of two integrals.

http://solvemymath.com/online_math_calculator/calculus/definite_integral/index.php

Note that the xmean value is very close to the ratio between the two temperatures
10 MeV and 38.366. (I am not sure why these two values are not much closer.)

The fraction of photons at temperature 10 MeV with less than the mean energy is 56.5%.
The fraction of photons at temperature 10 MeV with less than the energy corresponding to
x=1 is 3.47%. It seems to me to be unreasonable to use x=1 as the basis for calculating average particle kinetic energy.

Regards,
Buzz

13. Feb 24, 2017

### Staff: Mentor

I just calculated the energy of particles moving at a given speed.
For a distribution that does not consist of only a single speed, and ignoring relativity, the average kinetic energy has to correspond to a speed that is above the average speed.

14. Feb 24, 2017

### Buzz Bloom

Hi @mfb:

I have not done any calculations regarding the distribution of speeds.

I calculated the average photon energy Emean based on Planck's law. I assumed that matter particles in equilibrium with photons would have an average kinetic energy equal to the average photon energy. Is this wrong?

The assignment of an average energy to particles by just converting the temperature units to energy units seems to not be based on the average photon energy for which
xmean = 3.83813983744828 ,​
but instead seems to be based on x=1 which produces an energy
E (nearly?) = emean / xmean .​

Regards,
Buzz

15. Feb 24, 2017

### Staff: Mentor

The average photon energy for a given temperature is $\frac{E}{N} = \frac{\pi^4 }{30 \zeta(3)} k T \approx 2.7 kT$ (calculated from here). That is not equivalent to the average kinetic energy in a gas of massive particles, $\frac 3 2 kT$.

16. Feb 24, 2017

### Buzz Bloom

Hi @mfb:

Thank you very much for the link. It will be very helpful to me in rechecking my calculations.

Regards,
Buzz

17. Feb 25, 2017

### Buzz Bloom

Hi @mfb:

I now understand the difference between the two calculations of "average energy". I calculated the average value of hμ in the Planck distribution. The calculation you used produces the average energy per photon. I do not understand the physics of "equilibrium" sufficiently to grasp why one or the other calculations gives the correct average kinetic energy per matter particle. I would much appreciate any reference to such an explanation that you can post.

I have not yet done all of the calculations based on the average energy per photon, but I hope to get it done during this weekend.

Buzz

18. Feb 25, 2017

### Staff: Mentor

I can see where your value of 3.83 comes from, and I could reproduce it, and I don't know where the discrepancy to the other value comes from. But that gets off-topic I think.

None of them is the average kinetic energy of a nonrelativistic massive particle.

19. Feb 25, 2017

### Buzz Bloom

Hi @mfb:

I have been assuming that the concept of equilibrium at a given temperature, with respect to a mixture of particles, including photons and matter particles, means that for each kind of matter particle, the average kinetic energy of those particles is the same as the average photon energy. Is this wrong? If this is wrong, can you suggest where I can find out what the correct relationship is?

Regards,
Buzz

20. Feb 25, 2017

### Staff: Mentor

It is wrong. Classically (and the protons behave like classical particles here), you have the same average energy for every degree of freedom. The degrees of freedom of the electromagnetic field are a bit more complex, they do not correspond to photons moving around. In addition, you get a Bose-Einstein statistics due to quantum mechanics, which differs from the classical statistics.