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Boson number density

  1. May 20, 2014 #1

    ChrisVer

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    Gold Member

    1. The problem statement, all variables and given/known data
    Show that the number density for bosons if T (temperature) >>μ (chemical energy) is:
    [itex] n= \frac{ζ(3)}{\pi^{2}} gT^{3}[/itex]

    (T>>m too)

    2. Relevant equations

    [itex] n= \frac{g}{2 \pi^{2}} \int_{m}^{∞} f(E) (E^{2}-m^{2})^{1/2} EdE[/itex]
    [itex] f(E)= \frac{1}{e^{\frac{E-μ}{T}} -1}[/itex]
    [itex]ζ(s)= \frac{1}{Γ(s)} \int_{0}^{∞} dx \frac{x^{s-1}}{e^{x}-1}[/itex]

    3. The attempt at a solution

    I am taking:
    [itex] n= \frac{g}{2 \pi^{2}} \int_{m}^{∞} \frac{1}{e^{\frac{E-μ}{T}} -1} (E^{2}-m^{2})^{1/2} EdE[/itex]

    Then use μ<<T to drop the part of it in the exponential

    [itex] n= \frac{g}{2 \pi^{2}} \int_{m}^{∞} \frac{1}{e^{\frac{E}{T}} -1} (E^{2}-m^{2})^{1/2} EdE[/itex]

    If now I make a change of variables, [itex] x= \frac{E}{T}[/itex] I think I'll get something similar to the definition of the zeta function... By this I have:
    [itex]x= \frac{E}{T}[/itex],
    [itex]dE= T dx[/itex],
    [itex] E^{2}= x^{2}T^{2}[/itex]
    and [itex]E= x T[/itex]
    The integration limits will be: [itex]x_{1}= \frac{m}{T}=0[/itex] and the upper one will be ∞...

    [itex] n= \frac{g}{2 \pi^{2}} \int_{0}^{∞} \frac{1}{e^{x} -1} (x^{2}T^{2}-m^{2})^{1/2} xT^{2}dx[/itex]

    Now I can see that dropping [itex]m^{2}[/itex] from the square root would give me the desired answer... However I don't know why can I?
    alright m<<T, but T also gets multiplied by the variable....
    If I'd drop it:

    [itex] n= \frac{g}{2 \pi^{2}} \int_{0}^{∞} \frac{1}{e^{x} -1} xT(1-\frac{m^{2}}{x^{2}T^{2}})^{1/2} xT^{2}dx[/itex]

    [itex] n= \frac{g}{2 \pi^{2}} \int_{0}^{∞} \frac{1}{e^{x} -1} x^{2} T^{3}dx[/itex]

    [itex] n= \frac{g}{2 \pi^{2}} ζ(3) Γ(3) T^{3}dx[/itex]

    [itex] Γ(3)= 2[/itex]
    and I get the right result...By the assumption I could drop m^2 from the square root...
     
  2. jcsd
  3. May 20, 2014 #2
    Do we have a question here?
     
  4. May 20, 2014 #3

    ChrisVer

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    Gold Member

    -haha that was a good one Dauto, by the fact that on paper I had done it wrong, while here I did it right-
    But yes, there's still a question on....
    "Now I can see that dropping m2 from the square root would give me the desired answer... However I don't know why can I?alright m<<T, but T also gets multiplied by the variable...."

    Could I instead do:
    [itex] (x^{2}T^{2}-m^{2})^{1/2}= T (x^{2}- m^{2}/T^{2})^{1/2}= T x[/itex]
    ?
     
  5. May 21, 2014 #4
    Since T>>m, the distribution has most of its weight on values E>>m. Therefore E^2 - m^2 ~ E^2 as far as that integral is concerned.
     
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