## Step 1: Find the Dispersion Relation and Density of States The **dispersion relation** for short-wavelength surface tension waves is given as: \[ \nu^2 = \frac{2\pi \sigma}{\rho \lambda^3} \] With \( \nu = \frac{\omega}{2\pi} \) and \( \lambda = \frac{2\pi}{k} \), we have: \[ \omega = 2\pi \nu = 2\pi \sqrt{\frac{2\pi \sigma}{\rho \lambda^3}} \] Substitute \( \lambda = \frac{2\pi}{k} \): \[ \omega = 2\pi \sqrt{\frac{2\pi \sigma}{\rho} \left(\frac{k}{2\pi}\right)^3} = 2\pi \sqrt{\frac{2\pi \sigma}{\rho} \frac{k^3}{(2\pi)^3}} = 2\pi \sqrt{\frac{\sigma}{4\pi^2 \rho} k^3 } \] \[ \omega \propto k^{3/2} \] For a 2D system of area \(A\), the number of modes with wavenumber less than \(k\) is: \[ N(k) = \frac{A}{(2\pi)^2} \cdot \pi k^2 \] The **density of states** per unit frequency: \[ g(\omega) = \frac{dN}{d\omega} \] From \( \omega \propto k^{3/2} \implies k \propto \omega^{2/3} \): \[ N(\omega) \propto k^2 \propto \omega^{4/3} \implies g(\omega) = \frac{dN}{d\omega} \propto \omega^{1/3} \] **Explanation:** We calculated the dispersion relation for capillary waves and expressed the density of states as a function of frequency, which is essential for calculating energy. --- ## Step 2: Calculate the Thermal Energy Using Bose-Einstein Statistics The total thermal energy is: \[ E = \int_{}^{\infty} d\omega \; g(\omega) \frac{\hbar \omega}{e^{\beta \hbar \omega} - 1} \] With \(g(\omega) \propto \omega^{1/3}\), we have: \[ E \propto \int_{}^{\infty} d\omega \; \omega^{4/3} \frac{1}{e^{\beta \hbar \omega} - 1} \] Let \(x = \beta \hbar \omega\): \[ E \propto (k_B T)^{7/3} \int_{}^{\infty} dx \; \frac{x^{4/3}}{e^x - 1} \] Thus, \[ E = C \cdot A \cdot (k_B T)^{7/3} \] where \(C\) involves physical constants and the area \(A\). **Low temperature limit:** At low \(T\), the energy is dominated by the lowest-frequency modes, leading to: \[ E \sim T^{7/3} \] (analogous to Wien's law). **High temperature limit:** At high \(T\), equipartition applies (\(k_B T\) per mode), so: \[ E \sim (k_B T) \times \text{number of modes} \sim T \] (analogous to the Rayleigh-Jeans law). **Explanation:** We used the density of states and Bose-Einstein statistics to derive the temperature dependence of the thermal energy. At low \(T\), energy scales as \(T^{7/3}\); at high \(T\), it scales linearly with \(T\). --- ## **Final Answer** The thermal energy of surface waves on water (with surface tension as the restoring force) is: \[ \boxed{ E = C \cdot A \cdot (k_B T)^{7/3} } \] where \(C\) is a constant involving physical parameters of the system. - **At low temperature:** \(E \sim T^{7/3}\) (Wien-like law) - **At high temperature:** \(E \sim T\) (Rayleigh-Jeans-like law) This parallels the behavior of photon energy in a cavity, with modified exponents due to the different dispersion relation.