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Homework Help: Kosmologi problem

  1. Jun 4, 2005 #1
    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 [tex]^4 He[/tex] (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 [tex]He[/tex] to [tex]He^+[/tex] is [tex]Q_{He}=24,6eV[/tex]. At what temperature would the fractional ionization of helium be [tex]X=1/2[/tex]? Assume that [tex]\eta=5.5 \cdot 10^{-10}[/tex] and that the number density of [tex]He^{++}[/tex] is negligably small. [The relevatn statistical weight factor for the ionization of helium is [tex]g_{He}/(g_{e}g_{He^+}=1/4[/tex]]."

    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:

    [tex]\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}}[/tex]

    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:

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

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

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

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

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

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

    Define my constants:

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

    So i have:

    [tex]K_1=\frac{3x^2+K_2}{x} -> 0=3x^2 - x \cdot K_1 + K_2 [/tex]

    And solving for x:

    [tex]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} [/tex]

    From this i get two solutions [tex]x_{1}=0.75[/tex] and [tex]x_{2}=7,4[/tex]. 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 [tex]m_{e}c^2 = 511000ev[/tex]), so i'm not sure how i could get around taht either. Any suggestions?
     
    Last edited: Jun 4, 2005
  2. jcsd
  3. Jun 4, 2005 #2
    I actually didn't think this assignment was too bad, until i got around to having to solve
    [tex]
    \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)}}
    [/tex]
    which not even Maple liked to solve, i told me [tex]\math{1.440994055 \times 10^{16]}[/tex] which is opviously wrong, so i rearanged the equation a bit to give me
    [tex]\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)[/tex]
    Which Maple tells me gives [tex]1.440994053 \times 10^{16} K,\ 6702.288018 K[/tex] 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?
     
  4. Jun 4, 2005 #3
    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:

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

    And the final equation:

    [tex]K_1=x+\frac{K_2}{x} => 0=x^2-K_1x+K_2[/tex]

    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 :/
     
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