I'm kind of confused with this system. My first question is. Is the Hamiltonian of one particle of this gas the following? H(x,y,z,px,py,pz)=12m(p2x+p2y+p2z)→12m(p2θR2+p2ϕR2sin2θ)
If so, then my second question is: in order to obtain the classical partition function of one particle, should I integrate over the 6-dimensional phase space (Considering in spherical coordinates r=constant, pr=0 and pϕ=constant)? i.e. Z=∫exp[−β2m(p2θR2+p2ϕR2sin2θ)]Jdrdθdϕdprdpθdpϕ(1)
or should I just integrate over the 4-D phase space, since r=constant, pr=0?
Z=∫exp[−β2m(p2θR2+p2ϕR2sin2θ)]Jdθdϕdpθdpϕ(2)
And my final question is: in the expressions (1) and (2), is J=1 or J=R2sinθ ?
Answer
Yes. The Hamiltonian is your H. pr=0 because ∂L∂˙r=0 with L:=m2(R2˙θ2+R2sin2θ˙ϕ2).
The integral is over the 4-dimensional phase space: (θ,ϕ,pθ,pϕ), because the particles just move over the 2D surface of the sphere.
J=1 regardless of the coordinate system you are expressing the Hamiltonian. This is because of the Liouville's theorem that states that the phase volume is conserved under canonical transformations. In particular a coordinate transformation is included. So, in 3 dimensions:
Z=1h3∫d3qd3pexp[−βH(ˉq,ˉp)](∗)
being ˉq=(x,y,z) or q=(r,θ,ϕ), etc.
*Note. The constant h introduced in the integral (*) in order to maintain Z dimensionless. So h is just a constant (yet unknown), with units of action, i.e. units of angular momentum.
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