Z for classical particles in B-field

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SUMMARY

The discussion focuses on demonstrating that the free energy of classical particles without an internal magnetic moment is independent of the magnetic field. The partition function, defined as Z = (1/N!) ∫ exp(-βH) d^3p_1...d^3p_N d^3x_1...d^3x_N, is analyzed, where H includes the kinetic and potential energy terms. By changing variables to eliminate the vector potential from the integral, the magnetic field dependence can be effectively removed, confirming the independence of free energy from the magnetic field.

PREREQUISITES
  • Understanding of classical statistical mechanics
  • Familiarity with partition functions and free energy concepts
  • Knowledge of Hamiltonian mechanics
  • Basic concepts of electromagnetic potentials in classical physics
NEXT STEPS
  • Study the derivation of the partition function in classical systems
  • Learn about the implications of magnetic fields on particle dynamics
  • Explore the role of vector potentials in Hamiltonian mechanics
  • Investigate the relationship between free energy and thermodynamic potentials
USEFUL FOR

This discussion is beneficial for physics students, researchers in statistical mechanics, and anyone interested in the effects of magnetic fields on classical particle systems.

cscott
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Homework Statement



Show that the free energy of classical particles with no internal magnetic moment is always independent of magnetic field. Hint: Write down Z for N classical particles. Let the particles interact by U which depends only on the positions of the interacting particles. Show that the magnetic field dependence can be eliminated from the partition function by a change of variable.

Homework Equations



Partition function
Z = \frac{1}{N!} = \int \exp \left (-\beta H \right) d^3p_1..d^3p_Nd^3x_1..d^3x_N

Free energy:
\Omega = -\frac{\ln(Z)}{\beta}

The Attempt at a Solution



I would start by trying to write down H
H = \sum_n \left [ \frac{\vec{p_n}^2}{2m_n} + U(\vec{r_n}) \right]

but I don't know how to make any more use of that hint.
 
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If there's no magnetic moment, the particles couple to the magnetic field through the momentum term via

<br /> H = \sum_n \left [ \frac{(\vec{p_n}-q\vec{A})^2}{2m_n} + U(\vec{r_n}) \right]<br />

where \vec{A} is the scalar potential and q is the charge of a single particle.
 
So because U has nothing to do with the magnetic field, I need to eliminate the vector potential from,

\Pi_n \int \exp \left [-\frac{\beta}{2m_n} (\vec{p}_n - q \vec{A})^2 \right] d^3 p_n

and change variable,

\vec{u}_n = \vec{p_n}-q\vec{A}

thus eliminating vector potential?

Are the end points of the integral +/-infinity?
 

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