Contents

Problem set 5

Problem 5.2

  1. By definition, an expectation value is calculated by the sum (or integral) over all possible outcomes times their probability

    \[\langle E \rangle = \sum \limits_n E_n P(E_n).\]

    In this situation (an \(N,V,T\)-ensemble), the probability \(P(E_n)\) is given by the Boltzmann distribution

    \[P(E_n) = \frac{1}{Z}e^{-\beta E_n}\]

    so, one can substitute this to obtain

    (3)\[ \langle E \rangle = \frac{1}{Z} \sum \limits_n E_n e^{-\beta E_n} \]

    The factor \(Z\) does not depend on \(n\), so it can be taken out of the sum. Now, solving the right hand side of equation (5.2), one can use the chain rule

    \[\begin{split} \begin{aligned} -\frac{\partial}{\partial \beta} \ln{Z} &= -\frac{1}{Z}\frac{\partial Z}{\partial \beta}\\ &= -\frac{1}{Z}\frac{\partial}{\partial \beta} \sum \limits_n e^{-\beta E_n}\\ &= \frac{1}{Z} \sum \limits_n E_n e^{-\beta E_n} \end{aligned} \end{split}\]

    and the final expression is identical to equation (3), which demonstrates the equality in equation (5.2).

  2. The more difficult part of this exercise is to show

    \[\left \langle E^2 \right \rangle - \langle E \rangle^2= \frac{\partial^2}{\partial \beta^2} \ln {Z}\]

    which can be done in a similar way as in the previous question. The other equality follows immediately from the previous question, after adding another differential with respect to \(\beta\). Starting with the right hand side, and using the rules of differentiation:

    \[\begin{split} \begin{aligned} \frac{\partial^2}{\partial \beta^2} \ln{Z} &= \frac{\partial}{\partial \beta} \left( \frac{1}{Z}\frac{\partial Z}{\partial \beta} \right)\\ &= -\frac{1}{Z^2}\left(\frac{\partial Z}{\partial \beta}\right)^2 + \frac{1}{Z}\frac{\partial ^2 Z}{\partial \beta^2} \end{aligned} \end{split}\]

    The first term can be recognized as \(-\langle E \rangle^2\) (see previous question), and the second term results is equal to

    \[\begin{split} \begin{aligned} \frac{1}{Z}\frac{\partial ^2 Z}{\partial \beta^2} &= \frac{1}{Z} \frac{\partial^2}{\partial \beta^2}\sum \limits_n e^{-\beta E_n}\\ &= \frac{1}{Z} \sum \limits_n E_n^2 e^{-\beta E_n}\\ &= \sum \limits_n E_n^2 P(E_n)\\ &= \left \langle E_n^2 \right \rangle \end{aligned} \end{split}\]

    This completes the proof.

  3. One can use the result of previous question

    \[\sigma_E^2 = -\frac{\partial \langle E \rangle}{\partial \beta}\]

    and apply a substitution of variables, \(\beta = \frac{1}{k_\mathrm{B}T}\):

    \[\frac{\partial \langle E \rangle}{\partial \beta} = \frac{\partial \langle E \rangle}{\partial T} \frac{\partial T}{\partial \beta}\]

    and calculate that

    \[\frac{\partial T}{\partial \beta} = -k_\mathrm{B}T^2\]

    This equation is an example of the so-called fluctuation-dissipation theorem.

Problem 5.3

  1. This follows immediately from the rule that

    \[e^{a+b} = e^a e^b\]

  2. This statement follows immediately after using the previous question, and shuffling around some terms:

    \[\begin{split}\begin{aligned} Z &= \int e^{-\beta \sum \limits_{i=1}^N \left(\frac{p_n^2}{2m} + V(x_n)\right)}dx_1 dp_1 \cdots dx_N dp_N\\ &= \int \prod \limits_{n=1}^N e^{-\beta \left(\frac{p_n^2}{2m} + V(x_n)\right)} dx_1 dp_1 \cdots dx_N dp_N\\ &= \prod \limits_{n=1}^N \left ( \iint e^{-\beta \left(\frac{p_n^2}{2m} + V(x_n)\right)} dx_n dp_n \right) \\ &= \left( \int e^{-\beta V(x)} dx \int e^{-\beta \frac{p^2}{2m}} dp \right )^N \end{aligned}\end{split}\]

    as requested.

  3. The particles are identical, so there should not be a distinction between any of the particles. Even though the kinetic energies can be different for a given microstate, the expectation value should be the same for each particle. This can also be verified mathematically, by writing out the expression. It will be done below for particle 1. Writing out the full expression of the expectation value

    \[ \begin{aligned} \left \langle \frac{p_1^2}{2m} \right \rangle &= \frac { \int \frac{p_1^2}{2m} e^{-\beta \sum \limits_{i=1}^N \left(\frac{p_n^2}{2m} + V(x_n)\right)}dx_1 dp_1 \cdots dx_N dp_N } { \int e^{-\beta \sum \limits_{i=1}^N \left(\frac{p_n^2}{2m} + V(x_n)\right)}dx_1 dp_1 \cdots dx_N dp_N } \end{aligned} \]

    Using the result of previous question, one can factorise the integrals and obtain

    \[\begin{split} \begin{aligned} \left \langle \frac{p_1^2}{2m} \right \rangle &= \frac { \int \frac{p_1^2}{2m} e^{-\beta \frac{p_n^2}{2m}} dp_1 \int e^{-\beta V(x)} dx_1 \left( \int e^{-\beta V(x)} dx \int e^{-\beta \frac{p^2}{2m}} dp \right )^{N-1} } { \left( \int e^{-\beta V(x)} dx \int e^{-\beta \frac{p^2}{2m}} dp \right )^N } \\ &= \frac{ \int \frac{p_1^2}{2m} e^{-\beta \frac{p_1^2}{2m}} dp_1}{ \int e^{-\beta \frac{p_1^2}{2m}} dp_1} \end{aligned} \end{split}\]

  4. Direct evaluation of the integral, using the properties of a Gaussian integral

    \[\int \limits_{-\infty}^\infty e^{-a p^2} dp = \sqrt{\frac{\pi}{a}}\]

    and

    \[\int \limits_{-\infty}^\infty p^2 e^{-a p^2} dp = \frac{1}{2a} \sqrt{\frac{\pi}{a}}\]

    should yield

    \[\left \langle \frac{p^2}{2m} \right \rangle = \frac{1}{2}k_\mathrm{B}T\]

    In 3 dimensions, this value would be \(\frac{3}{2}k_\mathrm{B}T\).

    So, temperature relates directly to the kinetic energy of the particles, i.e. how fast they move and vibrate. This is independent of how the particles interact, or other properties of the system. It is a universal relation.

  5. …

Problem 5.4

  1. The probability of being in state \(n\) is given by the Boltzmann distribution, equation (5.1),

    \[P(n) = \frac{1}{Z}e^{-\beta E_n}\]

    The state is in this case determined by all the positions and momenta of all the particles,

    \[P(r_1,p_1,r_2,p_2,...,r_N,p_N)=\frac{1}{Z}e^{-\beta E(r_1,p_1,r_2,p_2,...,r_N,p_N)}\]

    and the energy is just a sum of terms, similar to the previous exercise

    \[E(r_1,p_1,r_2,p_2,...,r_N,p_N) = \sum \limits_{n=1}^{N} \frac{p_n^2}{2m} + mgh_n\]

    with \(h_n\) the height of particle \(n\) (the other coordinates do not change the energy, only the height matters). If one is only interested in a single particle, and not in all the other particles, one needs to integrate over all the other degrees of freedom, and then one obtains

    \[P(h) \propto e^{-\beta m g h}\]

    [Check why this only works if the particles do not interact with each other: in that case the energy would also depend on the relative positions of the particles, and contain terms of the form \(V(|r_i - r_j|)\), and that complicates the derivation of \(P(h)\) significantly]

  2. This would be

    \[\rho(h) \propto e^{-\beta m g h}\]

    The proportionality constant could be obtained in different ways, e.g. if the concentration at a given height were known, or if the total number of particles was given.

  3. In other words, find \(L\) for which

    \[e^{-\beta mg(h+L)} = e^{-1}e^{-\beta mgh}\]

    This length scale is

    \[L = \frac{k_\mathrm{B}T}{mg}\]

  4. Oxygen molecules have a mass \(2 \cdot 16\) g/mole, and there are \(6,022\cdot10^{23}\) particles per mole, so one oxygen molecule has a mass of \(5,314\cdot10^{-26}\) kg. The corresponding length scale \(L_\mathrm{O_2} = 7227\) m. This indicates why breathing at high altitude becomes more difficult, and why it can be dangerous to hike the Himalayas.

  5. The mass of a single nanoparticle is \(1,011\cdot10^{-20}\) kg, which is about a million times heavier than an oxygen molecule, so the length scale is also about a million times smaller, being around \(L_\mathrm{AuNP} = 3,80\) cm. Still, this length scale is enormous compared to the size of a living cell. Besides, the mass density of organic components is much lower than that of gold, so you can see that gravity is of no influence on a cellular scale.

Problem 5.5

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