# Question re equilibrium abundance of a protons and electrons

1. Nov 27, 2015

### Buzz Bloom

I am thinking of a time period between two events:
1. After the annihilation of most protons/neutrons with all anti-protons/anti-neutrons, and
2. Before the annihilation of electrons and positrons.​
During this period the temperature decreases from T1 to T2. T1 is the highest temperature at which (almost?) no photons have the (approximate?) energy equivalent of two protons. T2 is the lowest temperature at which (almost?) all photons have the (approximate?) energy equivalent of two electrons. Also during this period there are no anti-protons/anti-neutrons, and a great many electrons and positrons.

I would much appreciate some help in calculating how the ratio between the number of electrons and the number of protons decreases as a function of the decreasing temperature.

2. Nov 27, 2015

### Staff: Mentor

Why do you think it would change at all? By hypothesis, between $T_1$ and $T_2$ there is no annihilation of either protons/antiprotons (because that annihilation is already done) or electrons/positrons (because that annihilation hasn't started yet) going on, so the number of all of them would be constant.

3. Nov 27, 2015

### Buzz Bloom

Hi Peter:

I am assuming that electron-positron pairs are continuously being created and destroyed, maintaining a equilibrium with the photons. As the temperature falls, the distribution of photon energies changes. Consequently, the equilibrium point between destruction and creation of the pairs changes. If this idea is correct, what I am interested in is:
In what way does the equilibrium point, as a function of temperature and volume, change the density of electron-positron pairs?​

Regards,
Buzz

4. Nov 27, 2015

### Staff: Mentor

Yes, this is correct.

As long as the temperature is high enough that almost all of the photons have enough energy for pair production, the equilibrium point does not change significantly. It only starts changing when a significant number of photons start having insufficient energy for pair production. Then the equilibrium point shifts, fairly rapidly, from "lots of electron-positron pairs" to "virtually no electron-positron pairs, with a small residue of electrons left over". That rapid shift in the equilibrium point is what you are calling the "electron-positron annihilation". (Similar remarks apply to the proton-antiproton annihilation that happens earlier.)

In the regime where almost all the photons have enough energy for pair production, what changes as the photon temperature falls is not the density of electron-positron pairs but their average kinetic energy, i.e., their temperature. When the photon temperature is much, much higher than the threshold for pair production, the electron-positron pairs have lots of kinetic energy, i.e., they have a high temperature. When the photon temperature is only a little bit higher than the threshold for pair production, the electron-positron pairs have very little kinetic energy, i.e., they have a low temperature. This is because the kinetic energy of a given electron-positron pair is whatever energy is left over from the photon energy after the pair is produced, i.e., after the rest energy of the pair is supplied.

5. Nov 27, 2015

### Buzz Bloom

H i Peter:

Thanks you very much for the explanation. I now feel I have learned what I wanted to learn about this question.

Regards,
Buzz

6. Nov 29, 2015

### Buzz Bloom

Hi @PeterDonis:

I have been thinking about our discussion over the last few posts, and I see a new consequence that I find confusing. I hope you can help me understand this.

The following is Planks Law with respect to frequency:

Each photon of frequency ν has energy hν. So if I divide the RHS by hν I get the number of photons of frequency ν per unit area, that is
2 × F(ν,T) where F =
Therefore the number of photons per unit area increases with T. My confusion about this is that this relationship is independent of scale factor effects. Therefore if is the energy equivalent of a electron-positron pair, at an earlier time with a higher temperature there would be more photons capable of creating such pairs. If there were fewer pairs than the increase of photons capable of creating them, then this would not be an equilibrium condition, and the equilibrium number of pairs would be higher than number at a lower temperature.

Since the above thought process seems to contradict the conclusion we discussed previously, that is:
the equilibrium number of pairs would remain the same, with their kinetic energy being greater at a higher temperature.​
I hope you will be able to help find the error in my thinking.

Regards,
Buzz

7. Nov 29, 2015

### Staff: Mentor

This doesn't look correct. There should be a factor of $\nu^2 / c^2$ in front.

That is the number of photons per unit area of the body emitting them. It is not the number of photons per unit area being measured. But in the case of the early universe, what is the "area" of the emitting body? Photons are everywhere, and the universe is expanding; photons are not being emitted from a body with a fixed area.

Also, you are focusing on photons of one particular frequency; but the photons are distributed over the entire range of frequencies. The total number of photons capable of pair production is an integral over all frequencies greater than or equal to the threshold frequency (which is the frequency for which $h \nu$ is the threshold energy for pair production).

That's because it's the wrong relationship. As above, the formula you are using assumes a body with a fixed area emitting photons. That does not describe the early universe, which was expanding rapidly. What you need is a formula in terms of the photon density, not photons per unit area. Any such formula will involve the scale factor.

8. Nov 30, 2015

### Buzz Bloom

Hi @PeterDonis:

Thanks for continuing this discussion with me.

Quite right. I apologize for my carelessness. Rather than fix that error, I would like to start with a different equation from the Wikipedia article on Planks Law in the section: Spectral energy density form".

In the article, the equation has subscripts on u and B because the equation applies to all the forms of the Spectral Radiance B. The text accompanying this equation says:
These distributions have units of energy per volume per spectral unit.​
I understand this to mean that if the equation is multiplied by a range of frequencies dν, then u(T)dν gives the energy per unit volume in the range {ν, ν+dν}. Now If we use the form of B given in my Post #6. and divide by hν and multiply by dν we get
where
The units for both sides of the equation are:
the number per unit volume of photons whose frequencies are in the range {ν, ν+dν}.​
To get the number density of photons with sufficient energy, Eep, to create a electron and positron we would integrate n(ν,T) from νep to ∞, where
νep = Eep/h.​
However, the confusion I have is not about the value of this integral, but only about whether or not the integral increases with temperature T. The factor to the left of N in the equation does not vary with T. Only N(ν,T) varies with T. The exponential gets smaller as T gets bigger. Therefore the denominator gets smaller, and the fraction gets bigger. Therefore the integral bets bigger because for any value of ν the integrand gets bigger.

If all the above is correct, then is the following also correct.
Time t1 has a higher temperature than time t2.
Therefore at time t1 there are more photons that can create electron-positron pairs than at time t2.

Question: Isn't it also then true that more pairs will be created at time t1 than at time t2, and the equilibrium number of pairs will be larger at time t1 than at time t2?

If the answer to this is "No," please explain how the dynamics of equilibrium fails to make this happen.

Regards,
Buzz

9. Nov 30, 2015

### Staff: Mentor

"More", yes; but how many more?

Go back and read my post #4 again. I didn't say the rate of pair production doesn't change at all with $T$; I said it doesn't change significantly until the temperature gets into a range where a significant number of photons don't have enough energy to create a pair. In terms of the function $N$ you wrote down, the key is not just how it varies with $T$, but how it varies with $\nu$ for different values of $T$.

I recommend actually running some numbers: try evaluating $N$ at various frequencies for a range of values of $T$, from $k_B T$ well above the threshold energy for pair production, to $k_B T$ only a little above that energy, to $k_B T$ a little below that energy, to $k_B T$ well below that energy. Further, try integrating $n$ (i.e., the number density with the factor in front of $N$ included) from the threshold value $\nu_e$ (defined such that $h \nu_e$ equals the threshold energy for pair production) up to arbitrarily large values of $\nu$, and see how the integral varies for different values of $T$. (This integral, of course, gives the total number of photons capable of producing a pair, as a function of $T$.)

10. Dec 1, 2015

### Buzz Bloom

Hi Peter:

Thanks for your post. I apologize again for reading carelessly. I will do as you suggest and run some numbers.

I tried integrating n by parts, and lost my way. After one integration I get the following form:
x2 (log(eax - 1) - 2 ∫ x log(eax - 1) dx​

Any suggestions of how to make progress from there?

Regards,
Buzz

11. Dec 1, 2015

### Staff: Mentor

I'm not sure there's a closed-form integral for $n$; you might have to do it numerically.

12. Dec 1, 2015

### Buzz Bloom

Hi Peter:

OK. I can do numerical integration.

Regards,
Buzz

13. Dec 3, 2015

### Buzz Bloom

Hi Peter:

Now I understand the point you were making.
I substituted
x = hν / kB T​
into n(ν,T) and I got
n(x,T) = (8 π/c3) (kB3) T3 f(x)​
where
f(x) = x2/(ex - 1)
I did the numerical integration of f(x) to get the normalize integral
F(x) = ∫0x f(x) dx / ∫0 f(x) dx
I found the value xb = 0.022 produces F(x) = 0.01%. This seemed like a reasonable value for x to correspond to the an equilibrium at which 99.99% of the photons would be productive, that is, able to create an electron-positron pair. Thus 0.022 is a reasonable value for x at which the destroying of the pairs begins.

I solved for temperature
Tb = hν / xb kB
where
hν = 2 me/c2 = 1.637×10-13 J​
giving
Tb = 1.637×10-13 / 0.022 × 1.38x10-23 = 5.39×1011 K​