How Does High Temperature Affect Boson Number Density in Quantum Statistics?

[ChrisVer]

Homework Statement

Show that the number density for bosons if T (temperature) >>μ (chemical energy) is:
$n= \frac{ζ(3)}{\pi^{2}} gT^{3}$

(T>>m too)

Homework Equations

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

The Attempt at a Solution

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

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

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

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

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

Now I can see that dropping $m^{2}$ 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:

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

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

$n= \frac{g}{2 \pi^{2}} ζ(3) Γ(3) T^{3}dx$

$Γ(3)= 2$
and I get the right result...By the assumption I could drop m^2 from the square root...

 

[dauto]

Do we have a question here?

 

[ChrisVer]

-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:
$(x^{2}T^{2}-m^{2})^{1/2}= T (x^{2}- m^{2}/T^{2})^{1/2}= T x$
?

 

[clamtrox]

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.

 

[paranormal]

But I don't know why...
I would say that the assumption of T>>m allows us to neglect the term m^2 in the square root because it is much smaller compared to the term x^2T^2. This is because m<<T, so m^2 is even smaller. Therefore, dropping m^2 from the square root will not significantly affect the overall result and we can make this simplification. This simplification also makes the integration easier and leads to the desired result.