# Kosmologi problem

1. Jun 4, 2005

### Dracovich

Cosmology problem

Ok i've been trying to work on this assignment and it's driving me crazy agh, i'm pretty sure i should be able to do this with my hands and not need the use of maple or anything, although i do think the others used maple in the end and got a result of something like 7000K, but ok here's the question from the book:

"Imagine that at the time of recombination, the baryonic portion of the unvierse consisted entirely of $$^4 He$$ (that is, helium with two protons and two neutrons in its nucleus). The ionization energy of helium (that is the energy required to convert neutral $$He$$ to $$He^+$$ is $$Q_{He}=24,6eV$$. At what temperature would the fractional ionization of helium be $$X=1/2$$? Assume that $$\eta=5.5 \cdot 10^{-10}$$ and that the number density of $$He^{++}$$ is negligably small. [The relevatn statistical weight factor for the ionization of helium is $$g_{He}/(g_{e}g_{He^+}=1/4$$]."

Ok so here's what i've been doing so far:

The book already did the calculations on this assuming a Hydrogen only universe, so i redid the derivation with He and simply came to the same equation times 1/4, so the equation i'm assuming i should be working with is:

$$\frac{1-X}{X^2}=\frac{1}{4}3.84\eta \big( \frac{kT}{m_{e} c^2} \big)^{3/2} exp{\frac{Q}{kT}}$$

Where k is boltzmans constant. Then putting X=1/2, and putting it all to the power of 2/3 and throwing the constants on the left side i get:

$$\big( \frac{2}{0.96\eta} \big)^(2/3) = \frac{kT}{m_{e}c^2} exp{\frac{2Q}{3kT}}$$

And this is where i run into problems, i can't seem to isolate T properly. I at least can't do it without the help of maple or such as far as i can see, at least not in its unchanged form. So i tried to do some taylor expansions to see if i could solve it, but i've had no luck, i get a result but it's usually quite ludicrous numbers, first i tried expanding $$exp{\frac{2Q}{3kT}}$$ just to the second expansion, but i'm guessing the number becomes too big since it's divided by k, so then i tried to take the natural logarithm on both sides, and then trying to do a taylor expansion on $$ln(kT)$$ around the point kT=1 since that should be a relatively small number. I got a better result from that, i'll show my calculations so you can see if i go wrong somewhere:

ln on both sides
$$ln( \big( \frac{2}{0.96\eta} \big)^{2/3} )= ln( \frac{kT}{m_{e}c^2} exp{\frac{2Q}{3kT}})$$

Using a few logarithm rulse
$$\frac{2}{3} ln(\frac{2}{0.96\eta})= ln(kT)-ln(m_{e}c^2) +\frac{2Q}{3kT}$$

Putting in ln(x)=x-1 for a taylor expansion
$$\frac{2}{3} ln(\frac{2}{0.96\eta})= (kT-1)-ln(m_{e}c^2) +\frac{2Q}{3kT}$$

isolating the T's (and collecting the two ln's on the left side into one).
$$\frac{2}{3} ln(\frac{2m_{e}c^2}{0.96\eta})+1= kT+\frac{2Q}{3kT}$$

$$\frac{2}{3} ln(\frac{2m_{e}c^2}{0.96\eta})+1= \frac{3kT^2+2Q}{3kT}$$

Define my constants:

$$\frac{2}{3} ln(\frac{2m_{e}c^2}{0.96\eta})+1=24,46=K_1$$
$$2Q=49,6=K_2$$

So i have:

$$K_1=\frac{3x^2+K_2}{x} -> 0=3x^2 - x \cdot K_1 + K_2$$

And solving for x:

$$x=\frac{K_1 \pm \sqrt{K_1^2 - 4K_2}}{2 \cdot 3} = \frac{24,5 \pm \sqrt{598,6 - 198,4}}{6} =\frac{24,5 \pm 20}{6}$$

From this i get two solutions $$x_{1}=0.75$$ and $$x_{2}=7,4$$. Now the first one gives a somewhat good answer, around 8700K, which is decent but it still seems to be quite a big % off from what people got solving it with the computer, the other one is ridiculously high, but i'm not sure i would argue that i use the other one, other then intuitively i wouldn't expect it to be correct. Also in this all i did take the logarithm of a number with units (i used $$m_{e}c^2 = 511000ev$$), so i'm not sure how i could get around taht either. Any suggestions?

Last edited: Jun 4, 2005
2. Jun 4, 2005

### Whitey

I actually didn't think this assignment was too bad, until i got around to having to solve
$$\math{\frac{1-X}{X^2}=\frac{3.84 \eta}{4} \left(\frac{k T}{m_e c^2}\right)^{3/2} \exp{\left(\frac{Q}{k T}\right)}}$$
which not even Maple liked to solve, i told me $$\math{1.440994055 \times 10^{16]}$$ which is opviously wrong, so i rearanged the equation a bit to give me
$$\left(\frac{2}{3}\right) \ln\left(\frac{\left(\frac{1-X}{X^2}\right)}{\frac{3.84}{4} \eta)}\right)=\ln\left(\frac{k T}{m_e c^2}\right)+\left(\frac{2}{3}\right) \left(\frac{Q}{k T}\right)$$
Which Maple tells me gives $$1.440994053 \times 10^{16} K,\ 6702.288018 K$$ where i assume it is the second answer we are looking for.
ok, so we've gotten to a good answer, but i would much rather know how to get to that answer than just using Maple to find it. Can anyone help us, please?

3. Jun 4, 2005

### Dracovich

Caught quite a few calculations errors and got it now :) First off i was putting 2/3 in front of the ln and then joining the two, and i put my second degree equation up a little nicer, and got damn close this time. Ended up with this equation now:

$$ln((\frac{2}{0.96\eta})^{2/3}m_{e}c^2)+1=28.85=K_1$$
$$\frac{2}{3}Q=K_2$$

And the final equation:

$$K_1=x+\frac{K_2}{x} => 0=x^2-K_1x+K_2$$

And then two x's one crazy high and one was 0,58 which gives 6730K.

So now i got the right result, but i still took the logarithm of a number with units attatched to it (the rest energy of an electron), so i'm not quite sure how i can argue that :/