- #1
Normandy
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(inspired partially by this blog post: http://johncarlosbaez.wordpress.com/2012/01/19/classical-mechanics-versus-thermodynamics-part-1/
To my understanding, the thermodynamic configuration space has a nice symplectic structure. For example, using the language of classical mechanics, starting from P-V coordinates, the 2-form \mathrm{d}P\wedge \mathrm{d}V is preserved under a canonical transformation to T-S coordinates generated by the internal energy U, i.e. \mathrm{d}U = T \mathrm{d}S - P \mathrm{d}V. The Maxwell relations follow from the symplectic structure:
\mathrm{d}^2 U = 0 = \mathrm{d}T \wedge \mathrm{d}S - \mathrm{d}P \wedge \mathrm{d}V
If T=T(P,S) and V=V(P,S) then (for brevity, with partials along P being constant in S and partials in S being constant in P)
\left(\frac{\partial T}{\partial P} dP + \frac{\partial T}{\partial S} dS\right) \wedge \mathrm{d}S - \mathrm{d}P \wedge \left(\frac{\partial V}{\partial P} dP + \frac{\partial V}{\partial S} dS\right) = \left(\frac{\partial T}{\partial P} + \frac{\partial V}{\partial S}\right) \mathrm{d}S \wedge \mathrm{d}P = 0
The other relations follow similarly.
The analogy to the symplectic structure of classical mechanics can be further extended by defining a poisson bracket (with partials in T being constant along S and vice versa unless otherwise stated):
\{f,g\}_{T,S} \equiv \frac{\partial f}{\partial T}\frac{\partial g}{\partial S} - \frac{\partial f}{\partial S}\frac{\partial g}{\partial T}
Then:
\{T,S\}_{T,S} = 1 \\<br /> \{P,V\}_{T,S} = 1 \\<br /> \{P,S\}_{T,S} = 0 \\<br /> \{T,V\}_{T,S} = 0
(The second one can be proved by setting P=P(T,S) and V=V(T,S) and using d2U = 0)
So my question is: does the symplectic structure of thermodynamics have any connection to the underlying symplectic structure of the hamiltonian of the microscopic variables when used with a statistical ensemble?
For concreteness, take the many free particle hamiltonian, which when applied with the microcanonical ensemble gives the ideal gas equation of state:
{\cal H}(\{\vec{x}_i,\vec{p}_i\}) = \sum\limits_{i=1}^N \frac{\vec{p}_i^2}{2m} \\<br /> \Rightarrow PV=NkT
In order to derive that equation, you hold N, V, and E constant and use S = S(N,V,E) = k ln(ω(N,V,E)) and dU = T dS - P dV. Holding N, V, and E constant can be interpreted as geometric constraints on the phase space and ω is the volume of this constraint hypersurface.
Could the microcanonical ensemble (or more generally, any statistical ensemble) then be interpreted as some sort of geometric projection or something? Doing a canonical transformation on the microscopic hamiltonian does not affect the equation of state or the resulting thermodynamic variables. This could be interpreted straightforwardly since canonical transformations do not change volume in phase space and the microcanonical ensemble uses phase space volumes, but I was just wondering if there were any deeper connections between the two geometries.
To my understanding, the thermodynamic configuration space has a nice symplectic structure. For example, using the language of classical mechanics, starting from P-V coordinates, the 2-form \mathrm{d}P\wedge \mathrm{d}V is preserved under a canonical transformation to T-S coordinates generated by the internal energy U, i.e. \mathrm{d}U = T \mathrm{d}S - P \mathrm{d}V. The Maxwell relations follow from the symplectic structure:
\mathrm{d}^2 U = 0 = \mathrm{d}T \wedge \mathrm{d}S - \mathrm{d}P \wedge \mathrm{d}V
If T=T(P,S) and V=V(P,S) then (for brevity, with partials along P being constant in S and partials in S being constant in P)
\left(\frac{\partial T}{\partial P} dP + \frac{\partial T}{\partial S} dS\right) \wedge \mathrm{d}S - \mathrm{d}P \wedge \left(\frac{\partial V}{\partial P} dP + \frac{\partial V}{\partial S} dS\right) = \left(\frac{\partial T}{\partial P} + \frac{\partial V}{\partial S}\right) \mathrm{d}S \wedge \mathrm{d}P = 0
The other relations follow similarly.
The analogy to the symplectic structure of classical mechanics can be further extended by defining a poisson bracket (with partials in T being constant along S and vice versa unless otherwise stated):
\{f,g\}_{T,S} \equiv \frac{\partial f}{\partial T}\frac{\partial g}{\partial S} - \frac{\partial f}{\partial S}\frac{\partial g}{\partial T}
Then:
\{T,S\}_{T,S} = 1 \\<br /> \{P,V\}_{T,S} = 1 \\<br /> \{P,S\}_{T,S} = 0 \\<br /> \{T,V\}_{T,S} = 0
(The second one can be proved by setting P=P(T,S) and V=V(T,S) and using d2U = 0)
So my question is: does the symplectic structure of thermodynamics have any connection to the underlying symplectic structure of the hamiltonian of the microscopic variables when used with a statistical ensemble?
For concreteness, take the many free particle hamiltonian, which when applied with the microcanonical ensemble gives the ideal gas equation of state:
{\cal H}(\{\vec{x}_i,\vec{p}_i\}) = \sum\limits_{i=1}^N \frac{\vec{p}_i^2}{2m} \\<br /> \Rightarrow PV=NkT
In order to derive that equation, you hold N, V, and E constant and use S = S(N,V,E) = k ln(ω(N,V,E)) and dU = T dS - P dV. Holding N, V, and E constant can be interpreted as geometric constraints on the phase space and ω is the volume of this constraint hypersurface.
Could the microcanonical ensemble (or more generally, any statistical ensemble) then be interpreted as some sort of geometric projection or something? Doing a canonical transformation on the microscopic hamiltonian does not affect the equation of state or the resulting thermodynamic variables. This could be interpreted straightforwardly since canonical transformations do not change volume in phase space and the microcanonical ensemble uses phase space volumes, but I was just wondering if there were any deeper connections between the two geometries.