- #36
Count Iblis
- 1,863
- 8
Art said:Isn't the definition of absolute zero when an ideal gas exerts zero pressure and so the temperature is derived indirectly and so would not be a violation of Heisenberg's uncertainty principle ?
Also AFAIK a gas at absolute zero still has quantum fluctuations. Isn't this motion the source of zero point energy?
At absolute zero the classical gas law breaks down (even for an ideal gas). To see this just imagine that you have N noninteracting molecules trapped in a box. What is the total ground state energy of this system? Well, you can just put all the molecules in the same singe particle ground state, so it is N times E0, where E0 is the ground state energy of a single molecule. What is E0? Well, the ground state wavefunction will have a single peak in the middle of the box and zero at the boundaries. If we put the boundaries at x = -L/2 and L/2 and similarly for y and z (so that the volume of the box is L^3), then
psi(x,y,z) = A cos(pi x/L)cos(pi y/L)cos(pi z/L)
The energy times the wavefunction is is - hbar^2 nabla^2/(2m) psi, where nabla^2 is the sum of the second derivatives w.r.t. x, y, and z, h-bar = h/(2 pi), and m is the mass of a single molecule. So we find that E0 is given by:
E0 = 3 pi^2 h-bar^2/(2 m L^2)
So, we see that the total ground state energy is:
E = 3/2 pi^2 h-bar^2 N /m V^(-2/3)
The system at absolute zero will be in the ground state and will have this energy. Since this energy depends on the volume, the pressure is nonzero. The pressure is minus the derivate of the energy w.r.t. volume. So we have:
P(absolute zero) = pi^2 h-bar^2 N /m V^(-5/3)