To tackle the problem of a classical linear oscillator in thermal equilibrium, we can utilize concepts from statistical mechanics. The energy of the oscillator is given by the expression \( E = \frac{P^2}{2M} + bX^2 \), where \( P \) is the momentum, \( M \) is the mass, \( b \) is a constant related to the stiffness of the oscillator, and \( X \) is the displacement from equilibrium. Let's break down the calculations for the mean kinetic energy and the mean potential energy for an assembly of \( N \) such oscillators.
Mean Kinetic Energy Calculation
The mean kinetic energy \( \langle K \rangle \) of a single oscillator can be derived from the equipartition theorem, which states that each degree of freedom contributes \( \frac{1}{2} k_B T \) to the energy, where \( k_B \) is the Boltzmann constant and \( T \) is the temperature. For our oscillator, we have:
- One degree of freedom associated with the kinetic energy \( \frac{P^2}{2M} \).
Thus, the mean kinetic energy for one oscillator is:
Mean Kinetic Energy:
\[
\langle K \rangle = \frac{1}{2} k_B T
\]
For \( N \) oscillators, the total mean kinetic energy becomes:
Total Mean Kinetic Energy for N Oscillators:
\[
\langle K \rangle_{total} = N \cdot \langle K \rangle = N \cdot \frac{1}{2} k_B T
\]
Mean Potential Energy Calculation
Similarly, the mean potential energy \( \langle U \rangle \) can also be calculated using the equipartition theorem. The potential energy term \( bX^2 \) contributes another degree of freedom:
- One degree of freedom associated with the potential energy \( bX^2 \).
Thus, the mean potential energy for one oscillator is:
Mean Potential Energy:
\[
\langle U \rangle = \frac{1}{2} k_B T
\]
Again, for \( N \) oscillators, the total mean potential energy is:
Total Mean Potential Energy for N Oscillators:
\[
\langle U \rangle_{total} = N \cdot \langle U \rangle = N \cdot \frac{1}{2} k_B T
\]
Summary of Results
In summary, for an assembly of \( N \) classical linear oscillators in thermal equilibrium at temperature \( T \), both the mean kinetic energy and the mean potential energy can be expressed as:
- Mean Kinetic Energy: \( \langle K \rangle_{total} = N \cdot \frac{1}{2} k_B T \)
- Mean Potential Energy: \( \langle U \rangle_{total} = N \cdot \frac{1}{2} k_B T \)
This demonstrates how energy is distributed among the oscillators in thermal equilibrium, highlighting the fundamental principles of statistical mechanics and the equipartition theorem. Each oscillator contributes equally to the total energy, reflecting the system's equilibrium state.
