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Unit 1
Equipartition Theorem
Equipartition Theorem: Law of Equipartition of Energy
State and Prove the Law of Equipartition of Energy
The kinetic energy of a dynamical system containing a large number of particles in thermal equilibrium is equally divided among all degrees of freedom. The average energy associated with each degree of freedom is:
\[
\frac{1}{2}kT
\]
Proof:
Since the molecules are similar and distinguishable:
\[
\begin{aligned}
n(E) &= e^{-\alpha - \beta E} \\
&= A e^{-\beta E} \\
\Rightarrow \langle E \rangle &= \dfrac{\int n(E) E \, d\Gamma}{\int n(E) \, d\Gamma}\\
&= \dfrac{\int A e^{-\beta E} E \, dq \, dp}{\int A e^{-\beta E} \, dq \, dp} \\
&= \dfrac{\iiint e^{-p^2 / 2mkT} \left( \dfrac{p^2}{2m} \right) \, dp}{\iiint e^{-p^2 / 2mkT} \, dp} \\
&= \dfrac{1}{2m} \cdot \dfrac{\int_{-\infty}^{\infty} p^2 e^{-a p^2} dp}{\int_{-\infty}^{\infty} e^{-a p^2} dp} \quad \text{where } a = \dfrac{1}{2mkT} \\
&= \dfrac{1}{2m} \cdot \dfrac{\dfrac{1}{2} \sqrt{\pi} a^{-3/2}}{\sqrt{\pi} a^{-1/2}} \\
&= \dfrac{1}{2m} \cdot \dfrac{1}{2a} \\
&= \dfrac{1}{2}kT
\end{aligned}
\]
Mean Energy of a Harmonic Oscillator (1D)
For a classical harmonic oscillator vibrating in one dimension (say \( x \)) with force constant \( k \), the energy is:
\[
E = \dfrac{p_x^2}{2m} + \dfrac{1}{2} k x^2
\]
The mean energy is:
\[
\begin{aligned}
\langle E \rangle &= \dfrac{\int E A e^{-E / kT} dx \, dp_x}{\int A e^{-E / kT} dx \, dp_x} \\
&= \dfrac{\int \left( \dfrac{p_x^2}{2m} + \dfrac{1}{2} k x^2 \right) e^{- \left( \dfrac{p_x^2}{2mkT} + \dfrac{k x^2}{2kT} \right)} dx \, dp_x}
{\int e^{- \left( \dfrac{p_x^2}{2mkT} + \dfrac{k x^2}{2kT} \right)} dx \, dp_x} \\
&= \underbrace{\dfrac{\int \dfrac{p_x^2}{2m} e^{-a p_x^2} dp_x}{\int e^{-a p_x^2} dp_x}}_{\text{Kinetic part}} + \underbrace{\dfrac{1}{2} k \cdot \dfrac{\int x^2 e^{-b x^2} dx}{\int e^{-b x^2} dx}}_{\text{Potential part}} \\
&= \dfrac{1}{4ma} + \dfrac{k}{4b} \quad \text{where } a = \dfrac{1}{2mkT}, \quad b = \dfrac{k}{2kT} \\
&= \dfrac{2mkT}{4m} + \dfrac{2kT}{4} \\
&= \dfrac{kT}{2} + \dfrac{kT}{2} \\
&= kT
\end{aligned}
\]
Hence, the mean energy of a harmonic oscillator in classical statistical mechanics is \( kT \), with \( \frac{1}{2}kT \) from each quadratic degree of freedom.
