0
$\begingroup$

What precaution needs to be observed in writing down an expression for the total number of bosons $N$ valid at low temperatures?

$\endgroup$
  • $\begingroup$ You mean in second quantization? $\endgroup$ – Giuseppe Mar 30 '18 at 11:23
0
$\begingroup$

For a gas of $N$ bosons, the chemical potential $\mu_b$ is negative at high temperatures. As the temperature decreases towards a critical temperature $T_c$, $|\mu_b|$ also decreases until it reaches zero, and then remains stuck at zero for lower temperatures. The average number of bosons in the ground state is: $$N_0=\frac{1}{\exp\Big(\frac{|\mu_b|}{k_B T}\Big)-1}$$ Since the argument of the exponential is very small as $T\rightarrow T_c^+$ because $|\mu_b|\rightarrow 0$, you can taylor expand the exponential to first order to get: $$N_0=\frac{k_B T}{|\mu_b|}$$

In the case of $T\rightarrow T_c^-$, since the expression for $N_0$ would diverge, it is convenient to find an expression for the number of bosons which is not in the ground state, where $\mu_b=0$. $$N-N_0=\int_0^{\infty}\frac{g(\epsilon)d\epsilon}{\exp\Big(\frac{\epsilon}{k_B T}\Big)-1}$$ where $g(\epsilon)$ is the density of energy states. This was a specific example for the Bose-Einstein condensation, where you can see that the ground state is populated by a macroscopic number of particles. It is the most relevant example when considering a gas of bosons at low temperatures.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.