Question 705041

1. Identify the Phase Space Variables For two classical particles moving in one dimension, the phase space is 4-dimensional. The variables are their positions $(x_1, x_2)$ and their momenta $(p_1, p_2)$. The total phase space volume $\Omega$ for a given energy $E$ is the integral over all allowed positions and momenta: $\Omega = \int \int \int \int dx_1 dx_2 dp_1 dp_2$

2. Evaluate the Spatial Integral Both particles are confined within a one-dimensional box of length $L$. The potential energy inside the box is zero, and the limits for $x_1$ and $x_2$ are from $0$ to $L$. $\int_0^L \int_0^L dx_1 dx_2 = \left( \int_0^L dx_1 \right) \left( \int_0^L dx_2 \right) = L \times L = L^2$

3. Set up the Momentum Condition The total energy $E$ of the system is the sum of the kinetic energies of the two non-interacting particles. The problem states that the total energy does not exceed $E$: $\frac{p_1^2}{2m_1} + \frac{p_2^2}{2m_2} \le E$

4. Evaluate the Momentum Integral The inequality $\frac{p_1^2}{2m_1} + \frac{p_2^2}{2m_2} \le E$ represents the interior region of an ellipse in the 2D momentum space $(p_1, p_2)$. We can rewrite this in the standard equation form of an ellipse ($\frac{x^2}{a^2} + \frac{y^2}{b^2} \le 1$): $\frac{p_1^2}{2m_1 E} + \frac{p_2^2}{2m_2 E} \le 1$

From this, the lengths of the semi-axes are: $a = \sqrt{2m_1 E}$ $b = \sqrt{2m_2 E}$

The area of an ellipse is $\pi a b$. Thus, the momentum volume (area in this 2D momentum space) is: $\iint dp_1 dp_2 = \pi \times (\sqrt{2m_1 E}) \times (\sqrt{2m_2 E})$ $\iint dp_1 dp_2 = \pi \sqrt{4 m_1 m_2 E^2}$ $\iint dp_1 dp_2 = 2 \pi E \sqrt{m_1 m_2}$

5. Calculate the Final Phase Space Volume To find the total phase space volume, multiply the spatial volume by the momentum volume: $\Omega = (\text{Spatial Integral}) \times (\text{Momentum Integral})$ $\Omega = L^2 \times 2 \pi E \sqrt{m_1 m_2}$ $\Omega = 2 \pi L^2 E \sqrt{m_1 m_2}$

Correct Option: 4