Discussion
Let \(\rho(\vec{p}, \vec{q}, t)\) be the phase space density of an ensemble of a system. The Hamiltonian of the system is \(H(\vec{p}, \vec{q})\). If \(\{A, B\}\) denotes the Poisson bracket of \(A\) and \(B\), then
\(\dfrac{d\rho}{dt} = 0\)
implies
Answer: B
The condition implies \(\dfrac{\partial\rho}{\partial t} \propto \{\rho, H\}\).
The total time derivative of a phase-space function along the motion is its explicit time derivative plus its Poisson bracket with the Hamiltonian:
\[ \begin{aligned} \frac{d\rho}{dt} &= \frac{\partial\rho}{\partial t} + \{\rho, H\} = 0 \\ \frac{\partial\rho}{\partial t} &= -\{\rho, H\} \end{aligned} \]
This is Liouville's theorem: the phase-space density is constant along trajectories, and its local rate of change is fixed by the bracket with H. ∂ρ/∂t = 0 holds only for a stationary ensemble, which dρ/dt = 0 alone does not guarantee. Brackets with p·q/2 or q·q/2 involve functions that are not the Hamiltonian of the system and do not generate its motion.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development