How to move from the space of moments to the space of energies?

[damarkk]

Homework Statement

Grand Canonical Ensemble of bosons: how to move from the space of moments to the space of energies?

Relevant Equations

I denote with #\epsilon# the energy of the state particle, and we have $2m\epsilon = p^2$ where
$p^2 = p_{x}^2+p_{y}^2+p_{z}^2$

Suppose we have a gas of bosons with spin 0 and the grand potential is

$\Phi =\frac{kTV}{h^3} \int ln(1-e^{-\beta(p^2/2m -\mu})d^3p$

we already integrated the function in the coordinate space and the result is the factor V (volume). Now, we know that $\epsilon = p^2/2m$ and $d^3p = 4\pi p^2dp$, and if we change variable because we want to integrate in energies states we have of course $p^2 = 2m\epsilon$ and we obtain $dp = (2m)^{1/2} \epsilon^{-1/2}d\epsilon$. The expression of $\Phi$ changes and become

$\Phi =\frac{kTV}{h^3} \int ln(1-e^{-\beta(p^2/2m -\mu})4\pi (2m)^{3/2}\epsilon^{1/2}d\epsilon$

or in a better manner

$\Phi =\frac{4\pi kTV(2m)^{3/2}}{h^3} \int ln(1-e^{-\beta(p^2/2m -\mu})\epsilon^{1/2}d\epsilon$ (1)


Starting from this expression of the grand potential we can obtain the average number of bosons of the gas system

$N = -\frac{\partial{\Phi}}{\partial{\mu}} = \frac{4\pi V (2m)^{3/2}}{h^3} \int \frac{\epsilon^{1/2}}{e^{\beta(\epsilon-\mu)}-1} d\epsilon$ (2)

the expressions (1) and (2) are different from the same that we can read on books like Greiner (Thermodynamics Statistical Mechanics) or Amit (Statistical Physics: and introductory course):

$N = -\frac{\partial{\Phi}}{\partial{\mu}} = \frac{2\pi V (2m)^{3/2}}{h^3} \int \frac{\epsilon^{1/2}}{e^{\beta(\epsilon-\mu)}-1} d\epsilon$ (3)

for a factor 2.

And I know this is ridicolous and of course there are some misunderstanding very basic, but I don't understand why my result is double compared with the (3).

 

[anuttarasammyak]

we already integrated the function in the coordinate space and the result is the factor V (volume). Now, we know that ϵ=p2/2m and d3p=4πp2dp, and if we change variable because we want to integrate in energies states we have of course p2=2mϵ and we obtain dp=(2m)1/2ϵ−1/2dϵ. T

$p^2=2m\epsilon$
$2pdp=2md\epsilon$
$dp=\frac{m}{p} d\epsilon=\frac{\sqrt{m} }{\sqrt{2}\sqrt{\epsilon}} d\epsilon$
it differs factor 1/2 from your result.

 

[damarkk]

I'm sorry because this is very a stupid question :D (if you want delete it). Of course the mistake occurs because

$N(k)= \frac{V}{(2\pi)^3}\frac{4}{3}\pi k^3$ and $\frac{d N}{d k} = \frac{V}{(2\pi)^3}=4\pi k^2$

I substitute $k^2$ with $2m\epsilon/\hbar^2$ and this is the error.

The substitution must be done before the derivation $N(\epsilon)=\frac{V}{(2\pi)^3}\frac{4}{3}\pi (\frac{2m\epsilon}{\hbar^2})^{3/2}$ and therefore we have

$\frac{d N}{d \epsilon}= \frac{2\pi (2m)^{3/2} \epsilon^{1/2}}{h^3}$

and this is correct.

Alternatively we can compute in this manner:


$\frac{d N}{d \epsilon}=\frac{d N}{d k}\frac{d k}{d \epsilon} =\frac{V}{(2\pi)^{3}} 4\pi (\frac{2m\epsilon}{\hbar^2}) \frac{m}{\hbar\sqrt{2m\epsilon}}$

that is equal to

$\frac{2\pi (2m)^{3/2} \epsilon^{1/2}}{h^3}$

 

[anuttarasammyak]

I substitute k2 with 2mϵ/ℏ2 and this is the error.

You may do this substitution anywhere. Just please be careful for taking derivatives as I posted #2.

 

[damarkk]

You're right.