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GATE 2024 PH – Question 23

Thermodynamics and Statistical Mechanics · phase space · 1 mark · Multiple choice

Let $\rho(\mathbf p,\mathbf q,t)$ be ensemble phase-space density and $H(\mathbf p,\mathbf q)$ the Hamiltonian. With Poisson bracket $\{A,B\}$, $d\rho/dt=0$ implies

  1. $\partial_t\rho=0$
  2. $\partial_t\rho\propto\{\rho,H\}$
  3. $\partial_t\rho\propto\{\rho,\mathbf p\cdot\mathbf q/2\}$
  4. $\partial_t\rho\propto\{\rho,\mathbf q\cdot\mathbf q/2\}$

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Correct answer: (B) $\partial_t\rho\propto\{\rho,H\}$

Explanation

**Liouville's theorem.** The phase-space density $\rho$ of an ensemble is constant along a trajectory in phase space (the flow of the ensemble in phase space is incompressible):
$$\frac{d\rho}{dt}=0 .$$

**Expand the total derivative:**
$$\frac{d\rho}{dt}=\frac{\partial\rho}{\partial t}+\sum_i\left(\frac{\partial\rho}{\partial q_i}\dot q_i+\frac{\partial\rho}{\partial p_i}\dot p_i\right).$$

Using Hamilton's equations $\dot q_i=\dfrac{\partial H}{\partial p_i}$ and $\dot p_i=-\dfrac{\partial H}{\partial q_i}$, the sum is the Poisson bracket $\{\rho,H\}$:
$$\frac{\partial\rho}{\partial t}+\{\rho,H\}=0\;\Rightarrow\;\frac{\partial\rho}{\partial t}=-\{\rho,H\}.$$

So $\partial_t\rho\propto\{\rho,H\}$ (option B). It is zero only for a stationary (equilibrium) ensemble.