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OPhO 2024, Теория — Задача 4. Hot Solids

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2024 Online Physics Olympiad: Invitational Contest — Theoretical Examination

Условие

In this problem, we investigate a one-dimensional model of atoms in a solid. Assume the atoms are point masses of mass $m$ connected by springs with spring constant $\kappa$ and rest length $a$ , and the total rest length of the chain is $L$ .

4.1

First, assume that the mass is spread continuously throughout the chain (in other words, $a$ is very small). Here, longitudinal waves have the same speed for all values of the angular frequency $\omega$ and wavenumber $k$ .

a. Find this speed of sound in the solid, $v$ , up to a dimensionless constant.

Now, we get rid of this assumption and solve fully.

b. Find a dispersion relation (a relationship between $\omega$ and $k$ ) for the chain of atoms if $a$ is not required to be small. Use this result to find the dimensionless constant from part (a).

4.2

We can use the above results to find the heat capacity of the chain. To do so, treat each possible frequency $\omega$ as its own quantum harmonic oscillator (QHO) with a particle of mass $m$ moving in a potential defined by $V(x)=\frac12m\omega^2x^2$ . Each of these harmonic oscillators is at thermal equilibrium, and the total energy of the chain is the sum of the contributions from each frequency. You may find the following integrals useful:

\int_0^\infty\frac x{e^x-1}\,dx=\frac{\pi^2}6,\qquad\int_0^\infty\frac{x^3}{e^x-1}\,dx=\frac{\pi^4}{15}

c. First, derive the energy levels of a quantum harmonic oscillator by using the WKB approximation:

\oint p(x)\,dx=2\pi\hbar(n+1/2)\qquad (1)

Here, $p(x)$ is the momentum of the particle as a function of position and the integral is across one classical period.

d. Using the model from part (a), derive the total energy and heat capacity as a function of the temperature $T$ . (Your result only needs to hold for $\beta\hbar\omega_{\mathrm{avg}}\gg1$ , with $\beta=1/k_BT$ .) Assume that the atoms at either end of the chain must remain fixed in place.

e. Using the dispersion relation from part (b), find the energy and heat capacity to the next order in $T$ .

f. Above we assumed $\beta\hbar\omega_{\mathrm{avg}}\gg1$ . Why do our results fail for high $T$ ?

4.3

When the mass-energy of a particle is small compared to its energy level, relativistic corrections are required. The relativistic energy levels of a particle in a harmonic oscillator potential are given by:

E_n=mc^2\left(-1+\sqrt{1+\frac{2\hbar\omega}{mc^2}\left(n+\frac12\right)}\right)\qquad (2)

g. Use the given energy levels to find the total energy and heat capacity of the chain where each particle is moving relativistically; you may assume that the dispersion relation is linear as in part (d). Give your answer to the lowest order in $\hbar\omega/mc^2$ .

Published organizer clarifications — separately quoted from the 2024 OPhO Clarifications document.

T4 - Hot Solids

$T$ is the temperature of the solid, and beta is the thermodynamic temperature $\beta=1/(k_BT)$ with $k_B$ the Boltzmann constant.
The integral is a line integral along the closed curve from $-A$ to $A$ and back.
(a) asks to find the speed up to a dimensionless constant (meaning we don’t care if you have a factor of 2 in front or not, for example). 4.1 (b) asks to find the value of this dimensionless constant (it could be 1). If you think you solved 4.1 (a) exactly, don’t worry about this part (we will give you the correct credit).
(c) Even if you already know the correct energy levels, please complete the derivation using the given WKB approximation.
(d) Don’t worry too much about $\omega_{\mathrm{av}}$ —this statement simply means that you can treat $\beta\hbar\omega$ as large for the entire problem (any relevant value of omega). (This previously said small, this was an error—we are finding heat capacity in the low temperature limit.)
Both ends are fixed
Assume the oscillations of the entire chain act like a single quantum harmonic oscillator (meaning the energy contributions for the entire chain from each frequency are that of a quantum harmonic oscillator of that frequency).
We are deriving the first order relativistic correction, so even though the mass energy is significant enough to need a relativistic correction we want the answer to first order in $\hbar\omega/mc^2$

Решение

Hot Solids – Solution

a.

By dimensional analysis,

v\sim a\sqrt{\frac\kappa m}\qquad (3)

b.

Let the displacement of the $n$ th mass be $\psi_n$ . Then

m\ddot\psi_n=\kappa((\psi_{n+1}-\psi_n)-(\psi_n-\psi_{n-1}))\qquad (4)

\Rightarrow m\ddot\psi_n=\kappa(\psi_{n+1}+psi_{n-1}-2\psi_n)\qquad (5)

We guess that the solution has the form $\psi(x,t)=A\exp(i(\omega t-kx))$ with $x$ being the equilibrium positions of the masses. Then

-m\omega^2Ae^{i(\omega t-kx)}=\kappa Ae^{i\omega t}(e^{-ik(x_n+a)}+e^{-ik(X_n-a)}-2e^{ikx_n})\qquad (6)

\Rightarrow m\omega^2=2\kappa[1-\cos(ka)]\qquad (7)

\Rightarrow\omega=2\sqrt{\frac\kappa m}\left|\sin\left(\frac{ka}2\right)\right|\qquad (8)

In the limit of small $a$ , we get that the dimensionless constant from (a) is 1.

c.

2\int_{-L}^Lm\omega\sqrt{L^2-x^2}\,dx=2\pi\hbar(n+1/2)\qquad (9)

\Rightarrow\pi L^2m\omega=2\pi\hbar(n+1/2)\qquad (10)

\Rightarrow E=\frac12m\omega^2L^2=\hbar\omega(n+1/2)\qquad (11)

d.

Each sound mode with wavenumber $k$ has energy in the associated modes with frequency $\omega$ . To derive the energy in a mode with frequency $\omega$ , we first find the partition function for the harmonic oscillator:

Z=\sum_{n=0}^\infty\exp(-\beta\hbar\omega(n+1/2))=\frac{\exp(-\frac12\beta\hbar\omega)}{1-\exp(-\beta\hbar\omega)}\qquad (12)

The total energy is given by $-\partial_\beta\ln Z$ , which gives

E_\omega=\frac{\hbar\omega}{\exp(\beta\hbar\omega)-1}\qquad (13)

The total energy, with $\omega(k)=ak\sqrt{\frac\kappa m}$

E=\sum E_\omega\qquad (14)

\Rightarrow=\frac L\pi\int_0^\infty dk\,\frac{\hbar\omega(k)}{\exp(\beta\hbar\omega(k))-1}\qquad (15)

\Rightarrow=\frac L\pi\int_0^\infty d\omega\,\frac1a\sqrt{\frac m\kappa}\frac{\hbar\omega}{\exp(\beta\hbar\omega)-1}\qquad (16)

\Rightarrow=\frac L{\pi a\beta^2\hbar}\sqrt{\frac m\kappa}\int_0^\infty\frac x{e^x-1}\,dx\qquad (17)

\Rightarrow=\frac{\pi L}{6a\beta^2\hbar}\sqrt{\frac m\kappa}=\frac{\pi Lk_B^2T^2}{6a\hbar}\sqrt{\frac m\kappa}\qquad (18)

Where we replace the sum with an integral,

\sum\longrightarrow\frac L\pi\int\qquad (19)

This is because the two ends are fixed, so we know that $n\lambda=2L$ for a mode of wavelength $\lambda$ . Since $\lambda=2\pi/k$ , we have that $k=n\pi/L$ and therefore the spacing between $k$ values is $\pi/L$ . An integral is approximated as this spacing times the value of the function for each $k$ value, so we can make the replacement as above. Finally, we compute

C_V=\frac{\pi Lk_B^2T}{3a\hbar}\sqrt{\frac m\kappa}\qquad (20)

e.

Let $v_0=a\sqrt{\frac\kappa m}$ . Then

\omega=\frac{2v_0}a\sin\left(\frac{ka}2\right)\qquad (21)

\Rightarrow\arcsin\left(\frac{a\omega}{2v_0}\right)=\frac{ka}2\qquad (22)

\Rightarrow k\approx\frac a{v_0}\left(1+\frac{a^2\omega^2}{24v_0^2}\right)\qquad (23)

\Rightarrow dk=\frac{d\omega}{v_0}\left(1+\frac{\omega^2a^2}{8v_0^2}\right)\qquad (24)

As above, we sum up $E_\omega$ as an integral:

E=\frac L{\pi v_0}\int_0^\infty d\omega\,\frac{\hbar\omega(1+\frac{\omega^2a^2}{8v_0^2})}{e^{\beta\hbar\omega}-1}\qquad (25)

\Rightarrow E=\frac{\pi L}{6a\beta^2\hbar}\sqrt{\frac m\kappa}+\frac L{\pi v_0}\int_0^\infty d\omega\,\frac{\frac{\hbar\omega^3a^2}{8v_0^2}}{e^{\beta\hbar\omega}-1}\qquad (26)

\Rightarrow E=\frac{\pi L}{6a\beta^2\hbar}\sqrt{\frac m\kappa}+\frac{La^2}{8\pi\beta^4v_0^3\hbar^3}\int_0^\infty\frac{x^3}{e^x-1}\,dx\qquad (27)

\Rightarrow E=\frac{\pi L}{6a\beta^2\hbar}\sqrt{\frac m\kappa}\left(1+\frac{\pi^2m}{20\beta^2\hbar^2\kappa}\right)\qquad (28)

\Rightarrow E=\frac{\pi Lk_B^2T^2}{6a\hbar}\sqrt{\frac m\kappa}\left(1+\frac{\pi^2mk_B^2T^2}{20\hbar^2\kappa}\right)\qquad (29)

And finally, we get $C_V=\partial_\beta E$ :

C_V=\frac{\pi Lk_B^2T}{3a\hbar}\sqrt{\frac m\kappa}\left(1+\frac{\pi^2mk_B^2T^2}{10\hbar^2\kappa}\right)\qquad (30)

f.

Our integral bounds go to infinity, but there are a finite number of of states. This is important at high temperatures, but for low temperatures the occupation of higher states is so low that we can make this approximation.

g.

Credit to SSHS for their awesome solution to this part! It was better than the current official solution so we present their approach here.

We find the partition function using energy values to first order in the relativistic correction:

Z\approx\sum_{n=0}^\infty\exp\left(-\beta\hbar\omega\left(n+\frac12\right)+\frac{\beta\hbar^2\omega^2}{2mc^2}\left(n+\frac12\right)^2\right)\qquad (31)

\approx\sum_{n=0}^\infty\exp\left(-\beta\hbar\omega\left(n+\frac12\right)\right)\left(1+\frac{\beta\hbar^2\omega^2}{2mc^2}\left(n+\frac12\right)^2\right)\qquad (32)

=\left(1+\frac\beta{2mc^2}\frac{\partial^2}{\partial\beta^2}\right)\sum_{n=0}^\infty\exp\left(-\beta\hbar\omega\left(n+\frac12\right)\right)\qquad (33)

Since we know that the right sum is the partition function for the classical case as above, we can go ahead and compute

Z=\frac{e^{\frac12\beta\hbar\omega}}{e^{\beta\hbar\omega}-1}+\frac{\hbar^2\omega^2}{8mc^2}\frac{\beta e^{\frac12\beta\hbar\omega}(6e^{\beta\hbar\omega}+e^{2\beta\hbar\omega}+1)}{(e^{\beta\hbar\omega}-1)^3}\qquad (34)

E_\omega=-\frac\partial{\partial\beta}\ln Z\qquad (35)

=-\frac\partial{\partial\beta}\ln\left[\left(\frac{e^{\frac12\beta\hbar\omega}}{e^{\beta\hbar\omega}-1}\right)\left(1+\frac{\hbar^2\omega^2}{8mc^2}\frac{\beta(6e^{\beta\hbar\omega}+e^{2\beta\hbar\omega}+1)}{(e^{\beta\hbar\omega}-1)^2}\right)\right]\qquad (36)

=-\frac\partial{\partial\beta}\left[\ln\left(\frac{e^{\frac12\beta\hbar\omega}}{e^{\beta\hbar\omega}-1}\right)+\ln\left(1+\frac{\hbar^2\omega^2}{8mc^2}\frac{\beta(6e^{\beta\hbar\omega}+e^{2\beta\hbar\omega}+1)}{(e^{\beta\hbar\omega}-1)^2}\right)\right]\qquad (37)

\approx-\frac\partial{\partial\beta}\left[\ln\left(\frac{e^{\frac12\beta\hbar\omega}}{e^{\beta\hbar\omega}-1}\right)+\frac{\hbar^2\omega^2}{8mc^2}\frac{\beta(6e^{\beta\hbar\omega}+e^{2\beta\hbar\omega}+1)}{(e^{\beta\hbar\omega}-1)^2}\right]\qquad (38)

=\frac{\hbar\omega}2+\frac{\hbar\omega}{e^{\beta\hbar\omega}-1}+\frac{\hbar^2\omega^2}{8mc^2}\left[\frac{16\beta\hbar\omega}{(e^{\beta\hbar\omega}-1)^3}+\frac{8(\beta\hbar\omega-1)}{e^{\beta\hbar\omega}-1}+\frac{8(3\beta\hbar\omega-1)}{(e^{\beta\hbar\omega}-1)^2}-1\right]\qquad (39)

Now, we integrate to find the total energy (dropping the zero point energy, of course):

E=\sum E_\omega\qquad (40)

=\frac{\pi L}{6a\beta^2\hbar}\sqrt{\frac m\kappa}\qquad (41)

+\frac{\hbar^2L}{8\pi mc^2a}\sqrt{\frac m\kappa}\int d\omega\,\omega^2\left[\frac{16\beta\hbar\omega}{(e^{\beta\hbar\omega}-1)^3}+\frac{8(\beta\hbar\omega-1)}{e^{\beta\hbar\omega}-1}+\frac{8(3\beta\hbar\omega-1)}{(e^{\beta\hbar\omega}-1)^2}-1\right]\qquad (42)

=\frac{\pi L}{6a\beta^2\hbar}\sqrt{\frac m\kappa}\qquad (43)

+\frac L{\pi mc^2\beta^3a\hbar}\sqrt{\frac m\kappa}\left[2\int_0^\infty dx\,\frac{x^3}{e^x-1}+\int_0^\infty dx\,\frac{x^3}{e^x-1}-\int_0^\infty dx\,\frac{x^2}{e^x-1}\right.\qquad (44)

\left.{}+3\int_0^\infty dx\,\frac{x^3}{(e^x-1)^2}-\int_0^\infty dx\,\frac{x^2}{(e^x-1)^2}\right]\qquad (45)

=\frac{\pi L}{6a\beta^2\hbar}\sqrt{\frac m\kappa}+\frac L{\pi mc^2\beta^3a\hbar}\sqrt{\frac m\kappa}[\pi^2-2\zeta(3)-0.88574]\qquad (46)

Which, finally, gives us

C_V=\frac{L\pi k_B^2T}{3\hbar a}\sqrt{\frac m\kappa}\left[1+\frac{9k_BT}{\pi^2mc^2}(\pi^2-2\zeta(3)-0.88574)\right]\qquad (47)

There are more ways to do this that give slightly different results due to the approximations made. Some ways are replacing the sum in the partition function with an integral or computing the total energy directly as an integral (without finding the partition function first and taking a derivative).

Оригинал в Архива: theory.pdf · официални решения: theory_sol.pdf · допълнителни данни: Clarifications.pdf