Average Energy of Classial Oscillator and Planck's Oscillator

Average Energy of Classical Oscillator in Thermal Equilibrium

If p is the momentum and xis the displacement from equilibrium position of a classical oscillator of mass m vibrating under a force constant C , the frequency of the oscillator is

$$ v = \frac {1}{2\pi }\sqrt {\left (\frac Cm \right )} \dots (1) $$

And the energy of the oscillator is

$$E = \frac {p^2}{2m} + \frac 12 Cx^2 \dots (2)$$

If the oscillator is in thermal equilibrium with a heat reservoir at absolute temperature T, then the probability that the oscillator will have momentum and the displacement in the range p to p + dp and x and x + dx, according to Boltzmann’s law, E is proportional to \(e^{-E/KT}dpdx\).

The average energy of classical oscillator

\begin{align*} \vec {\varepsilon } &= \frac {\int \int \left (\frac {p^2}{2m}+\frac 12Cx^2 \right )e^{[(\frac {p^2}{2m})+\frac 12C^2]}/KT}{\int\int e^{-(\frac {p^2}{2m} + \frac 12Cx^2)/KT}dpdx}\end{align*}

\begin{align*} &=\frac {\int _{\infty}^{\infty}\frac{p^2}{2m}e^{(-P^2/2m)KT}dp\int _{\infty}^{\infty}e^{-Cx^2/2KT}dx}{\int _{\infty}^{\infty}e^{(-P^2/2m)KT}dp\int _{\infty}^{\infty}e^{-Cx^2/2KT}dx} \\ &+ \frac {\int _{\infty}^{\infty}\frac 12Cx^2e^{-Cx^2/2KT} dx\int _{\infty}^{\infty}e^{-p^2/2mKT}dp}{\int _{\infty}^{\infty}e^{-Cx^2/2mKT}dx\int _{\infty}^{\infty}e^{-p^2/2mKT}dx}\\ &= \frac {\int _{-\infty}^{\infty}\frac {P^2}{2m}e^{-P^2/2mKT}dp}{\int _{-\infty}^{\infty}e^{-p^2/2mKT}dp} + \frac {\int _{-\infty}^{\infty}\frac 12Cx^2e^{-x^2/2KT}dx}{\int _{-\infty}^{\infty}e^{-Cx^2/2KT}dx} \\ &= \frac 12KT + \frac 12KT \\ &= KT \end{align*} Thus the average energy of classical oscillator in thermal equation at absolute temperature T is KT as given by equipartition law. As for a simple harmonic oscillator average kinetic energy is equal to the average potential energy, if the average is taken over a complete period.

Average Energy of Planck’s Oscillator

If N is the total number of Planck’s resonators and E their total energy, then average energy per Plank’s oscillator is given by

$$\vec {\varepsilon} = \frac EN\dots (1)$$

According to Maxwell’s law for molecular motion if \(\varepsilon\) is a certain amount of energy, the probabilities that a system will have energies \(0,\varepsilon, 2\varepsilon, \dots n\varepsilon \dots \) are in the ratio

$$e^{0/KT}:e^{-\varepsilon/KT}:e^{-2\varepsilon/KT}\dots :e^{-n\varepsilon /KT}\dots \text {etc.}$$

Thus, if the number of resonators have energies \(0,\varepsilon, 2\varepsilon \dots r\varepsilon\) respectively \(N_0, N_1, N_2, \dots N_r, \dots , \) then from Boltzmann’s canonical distribution law

\begin{align*} N_0 &\propto e^{0/KT} \\ N_0 &= Ce^{-0} = C \\ N_1 &\propto e^{-\varepsilon /KT}\\ \text {or,}\: N_1 &= Ce^{-\varepsilon/KT} =N_0e^{-\varepsilon/KT} \\ N_2 &\propto e^{-2\varepsilon/KT} \\ \text {or}\: N_2 &= e^{-2\varepsilon /KT} = N_0e^{-2\varepsilon /KT} \\ \dots \dots \dots \\ \dots \dots \dots \\ \text {or}\: N_r &= e^{-r\varepsilon /KT} = N_0e^{-r\varepsilon /KT} \\ \dots \dots \dots \\ \dots \dots \dots \\ \end{align*}

Therefore, total number of resonators in the resonance chamber is as follows

\begin{align*} N &= N_0 + N_1 + N_2 + N_3 +\dots + N_r+ \dots \\ &= N_0 + N_0e^{-\varepsilon/KT} + N_0e^{-2\varepsilon/KT} + N_0e^{-3\varepsilon/KT}+ \\ \dots + N_0e^{-r\varepsilon/KT}+ \dots \\ &= N_0[1+ e^{-\varepsilon/KT} + e^{-2\varepsilon /KT} +\dots + \dots e^{-r\varepsilon/KT} + \dots ]\dots (2) \end{align*}

\begin{align*} \text {Let}\: e^{-\varepsilon/KT} = y \\ \therefore \: N &= N_0 [1 + y+ y^2 +\dots + y^r+ \dots ] \\ &= \frac {N_0}{1 - y} \\ \end{align*}

Total energy of these resonators will be

\begin{align*} E &= 0\times N_0 + E\times N_1 + 2E \times N_2 + \dots r\times N_r + \dots \\ E &= 0 + E\times N_0e^{-E/KT} + 2E \times N_0e^{-2E/KT}+\dots \\ &= N_0\varepsilon [e^{-\varepsilon/KT}+2e^{-2\varepsilon/KT}+ \dots + re^{r\varepsilon /KT+\dots }]\\ &= N_0\varepsilon y[1 + 2y + 3y^2+\dots + ry^{r-1}+\dots]\\ &= N_0\varepsilon yS \dots (4)\\ \text {where}\: S &=[1 + 2y + 3y^2+\dots +ry^{r-1}]\dots (5)\\ \therefore Sy &= [y +2y^2+\dots +(r - 1)y^{r-1}+ \dots]\dots {(6)} \end{align*}

\begin{align*} \text {subtracting}\:(6)\:\text {from}\:(5), \text {we get}\\ S(1-y)&= 1 + 2y + y^2 +\dots + y^r + = \frac {1}{1-y} \dots (7)\\ \therefore S &= \frac {1}{(1-y)^2}\dots (8)\\\text {subtracting this value in equation (4), we get}\\ E &= \frac{N_0\varepsilon y}{(1 -y)^2}\end{align*}

\begin{align*}\text {So, the average energy}\: \varepsilon \: \text {of Planck's resonator}\\ \vec {\varepsilon}&= \frac {E}{N}=\frac {N_0\varepsilon \frac {y}{(1-y)^2}}{\frac {N_0}{1 - y}} \\ &= \frac {\varepsilon y}{1-y}=\frac {\varepsilon \: \text {exp}(-\varepsilon /KT)}{1-\text {exp}(\varepsilon /KT)}\\ &= \frac {\varepsilon}{\text {exp}(\varepsilon /KT)-1}\\ \vec{\varepsilon} &= \frac {\varepsilon}{e^{\varepsilon}/KT -1}\end{align*}

Thus, the average energy of resonator is not KT, but \(\vec{\varepsilon} = \frac {\varepsilon}{e^{\varepsilon}/KT -1}\) according to quantum theory.

According to classical theory of energy of an oscillator may have any value, i.e, we must suppose \(\varepsilon \)to be infinitesimal. When \(\varepsilon \) tends to zero, the limiting value of \(\frac {\varepsilon }{\text {exp}(\varepsilon /KT) -1}\) is KT as given by equipartition law.

According to Planck’s hypothesis of quantum theory, the value of \(\varepsilon \) is not infinitesimal, but is equal to \(h\nu \)m therefore the average energy of Planck’s oscillator or resonator is given by

\begin{align*}\vec {\varepsilon}&= \frac {h\nu}{\text {exp}(h\nu /KT)-1}\dots (11)\\\text {This equation may be written as} \\ \vec {\varepsilon} &= \frac {h\nu}{[1 + \frac {h\nu}{KT}+ \frac {1}{2!}(\frac {h\nu }{KT})^2 + \frac {1}{3!}(\frac {h\nu }{KT})^3) + \dots -1]} \\ &= \frac {h\nu }{\frac {h\nu}{KT}[1 + \frac {1}{2!}\frac {h\nu}{KT}+ \frac {1}{3!}(\frac {h\nu }{KT})^2) + \dots]}\\ &= \frac {KT }{[1 + \frac {1}{2!}\frac {h\nu}{KT}+ \frac {1}{3!}(\frac {h\nu }{KT})^2) + \dots]}\\\text {Oviously if h were zero, then} \\ \vec {\varepsilon} &= KT\end{align*}

Thus is h were zero, then average energy of Planck’s oscillator would be the equipartition value KT.

Bibliography

S.S. Singhal, J.P. Agarwal, Satya Prakash. heat and thermodynamics and statistical physics. pragati prakashan, 2010.

—. Heat and Thermodynamics and Statistical Physics. Pragati Prakashan, 2010.

Vatsyayan, Dr. Rakesh Ranjan. Refresher Course in Physics. kathmandu: Surya Book Traders, 2015.