These notes, from the third week of Thermal and Statistical Physics cover the canonical ensemble and Helmholtz free energy. They include a number of small group activities.
The third week of PH 441(K&K 3, Schroeder 6)
This week we will be deriving the Boltzmann ratio and the Helmholtz free energy, continuing the microcanonical approach we started last week. Last week we saw that when two systems were considered together in a microcanonical picture, the energy of each system taken individually is not fixed. This provides our stepping stone to go from a microcanonical picture where all possible microstates have the same energy (and equal probability) to a canonical picture where all energies are possible and the probability of a given microstate being observed depends on the energy of that state.
We ended last week by finding that the following quantity is equal in two systems in thermal equalibrium \begin{align} \beta &= \left(\frac{\partial \ln g}{\partial E}\right)_{V} \end{align} where \(g(E)\) is the multiplicity in the microcanonical ensemble. To more definitively connect this with temperature, we will again consider two systems in thermal equilibrium using a microcanonical ensemble, but this time we will make one of those two systems huge. In fact, it will be so huge that we can treat it using classical thermodynamics, i.e. we can conclude that the above equation applies, and we can assume that the temperature of this huge system is unaffected by the small change in energy that could happen due to differences in the small system.
Let us now examine the multiplicity of our combined system, making \(B\) be our large system: \begin{align} g_{AB}(E) &= \sum_{E_A}g_A(E_A) g_B(E_{AB}-E_A) \end{align} We can further find the probability of any particular energy being observed from \begin{align} P_{A}(E_A|E_{AB}) &= \frac{g_A(E_A)g_B(E_{AB}-E_A)}{ \sum_{E_A'}g_A(E_A') g_B(E_{AB}-E_A') } \end{align} where we are counting how many microstatesstates of the combined system have this particular energy in system \(A\), and dividing by the total number of microstates of the combined system to create a probability. So far this is identical to what we had last week. The difference is that we are now claiming that system \(B\) is huge. This means that we can approximate \(g_B\). Doing so, however requires some care.
We might be tempted to simply Taylor expand \(g_B\) \begin{align} g_B(E_{AB}-E_A) &\approx g_B(E_{AB}) - \beta g_B(E_{AB}) E_A +\cdots \\ &\approx g_B(E_{AB})(1 - \beta E_A) \end{align} This however, would be wrong unless \(\beta E_A\ll 1\). One way to see that this expansion must have limited range is that if \(\beta E_a\ge 1\) then we will end up with a negative multiplicity, which is meaningless. The trouble is that we only assumed that \(E_A\) was small enough not to change the temperature (or \(\beta\)), which does not mean that \(\beta E_A<1\). Thus this expansion is guaranteed to fail.
When we run into this problem, we can consider that \(\ln g(E)\) is generally a smoother function than \(g(E)\). Based on the Central Limit Theorem, we expect \(g(E)\) to typically have a Gaussian shape, which is one of our analytic functions that is least well approximated by a polynomial. In contrast, \(\ln g\) will be parabolic (to the extent that \(g\) is Gaussian), which makes it a prime candidate for a Taylor expansion.
Now we can plug this into the probability equation above to find that \begin{align} P_{A}(E_A) &= \frac{g_A(E_A)\cancel{g_B(E_{AB})}e^{-\beta E_A} }{ \sum_{E_A'}g_A(E_A')\cancel{g_B(E_{AB})}e^{-\beta E_A'} } \\ &= \frac{g_A(E_A) e^{- \frac{E_A}{k_BT}}}{ \sum_{E_A'}g_A(E_A') e^{- \frac{E_A'}{k_BT}} } \end{align} Now this looks a bit different than the probabilities we saw previously (two weeks ago), because this is the probability that we see an energy \(E_A\), not the probability for a given microstate, and thus it has the factors of \(g_A\), and it sums over energies rather than microstates. To find the probability of a given microstate, we just need to divide the probability of its energy by the number of microstates at that energy, i.e. drop the factor of g: \begin{align} P_i^A &= \frac{e^{- \beta E_i}}{Z} \\ Z &= \sum_{E}^{\text{all energies}} g(E)e^{-\beta E} \\ &= \sum_i^{\text{all $\mu$states}} e^{- \beta E} \end{align} This is all there is to show the Boltzmann probability distribution from the microcanonical picture: Big system with little system, treat big system thermodynamically, count microstates.
Now that we have the set of probabilities expressed again in terms of \(\beta\), there are a few things we can solve for directly, namely any quantities that are directly defined from probabilities. Most specifically, the internal energy \begin{align} U &= \sum_i P_i E_i \\ &= \sum_i E_i \frac{e^{-\beta E_i}}{Z} \\ &= \frac1Z \sum_i E_i e^{-\beta E_i} \end{align} Now doing yet another summation will often feel tedious. There are a couple of ways to make this easier. The simplest is to examine the sum above and notice how very similar it is to the partition function itself. If you take a derivative of the partition function with respect to \(\beta\), you will find: \begin{align} \left(\frac{\partial Z}{\partial \beta}\right)_{V} &= \sum_i e^{-\beta E_i} (-E_i) \\ &= -UZ \\ U &= -\frac{1}{Z}\left(\frac{\partial Z}{\partial \beta}\right)_{V} \\ &= -\left(\frac{\partial \ln Z}{\partial \beta}\right)_{V} \end{align}
How do we compute pressure? So far, everything we have done has kept the volume fixed. Pressure tells us how the energy changes when we change the volume, i.e. how much work is done. From Energy and Entropy, we know that \begin{align} dU &= TdS - pdV \\ p &= -\left(\frac{\partial U}{\partial V}\right)_S \end{align} So how do we find the pressure? We need to find the change in internal energy when we change the volume at fixed entropy.
If we take a derivative of \(U\) with respect to volume while holding the probabilities fixed, we obtain the following result: \begin{align} p &= -\left(\frac{\partial U}{\partial V}\right)_S \\ &= -\left(\frac{\partial \sum_i E_i P_i}{\partial V}\right)_S \\ &= -\sum_i P_i \frac{d E_i}{d V} \\ &= -\sum_i \frac{e^{-\beta E_i}}{Z} \frac{d E_i}{d V} \end{align} So the pressure is just a weighted sum of derivatives of energy eigenvalues with respect to volume. We can apply the derivative trick to this also: \begin{align} p&= \frac1{\beta Z}\left(\frac{\partial Z}{\partial V}\right)_\beta \\ &= \frac1{\beta}\left(\frac{\partial \ln Z}{\partial V}\right)_\beta \end{align} Now we have an expression in terms of \(\ln Z\) and \(\beta\).
We saw a hint above that \(U\) somehow relates to \(\ln Z\), which hinted that \(\ln Z\) might be something special, and now \(\ln Z\) also turns out to relate to the pressure somehow. Let's put this into thermodynamics language.^{*}
\begin{align} U &= -\left(\frac{\partial \ln Z}{\partial \beta}\right)_{V} \\ d\ln Z &= -Ud\beta + \left(\frac{\partial \ln Z}{\partial V}\right)_{\beta} dV \\ d\ln Z &= -Ud\beta +\beta p dV \end{align} We can already see work (i.e. \(-pdV\)) showing up here. So now we're going to try a switch to a \(dU\) rather than a \(d\beta\), since we know something about \(dU\). \begin{align} d(\beta U) &= Ud\beta + \beta dU \\ d\ln Z &= -\left(d(\beta U) - \beta dU\right) +\beta p dV \\ &= \beta dU -d(\beta U) +\frac{p}\beta dV \\ \beta dU &= d\left(\ln Z + \beta U\right) -\beta p dV \\ dU &= \frac1\beta d\left(\ln Z + \beta U\right) - p dV \end{align} Comparing this result with the thermodynamic identity tells us that \begin{align} S &= k_B\ln Z + U/T \\ F &\equiv U-TS \\ &= U - T\left(k_B\ln Z + U/T\right) \\ &= U - k_BT\ln Z + U \\ &= - k_BT\ln Z \end{align} That was a bit of a differentials slog, but got us the same result for the Helmholtz free energy without assuming the Gibbs entropy \(S = -k\sum_i P_i\ln P_i\). It did, however, demonstrate a not-quite-contradiction, in that the expression we found for the entropy is not mathematically equal to the Boltmann entropy. It approaches the same thing for large systems, although I won't prove that now.In effect, this expression gives us a physical meaning for the partition function.
We begin by writing down the partition function \begin{align} Z &= \sum_i e^{-\beta E_i} \\ &= g e^{-\beta E_0} \end{align} Now we just need a log and we're done. \begin{align} F &= -kT\ln Z \\ &= -kT \ln\left(g e^{-\beta E_0}\right) \\ &= -kT \left(\ln g + \ln e^{-\beta E_0}\right) \\ &= E_0 - Tk\ln g \end{align} This is just what we would have concluded about the free energy if we had used the Boltzmann expression for the entropy in this microcanonical ensemble.
Waving our hands, we can understand \(F=-kT\ln Z\) in two ways:
Why the big deal with the free energy? One way to put it is that it is relatively easy to compute. The other is that once you have an analytic expression for the free energy, you can solve for pretty much anything else you want.
Recall: \begin{align} F &\equiv U-TS \\ dF &= dU - SdT - TdS \\ &= -SdT - pdV \\ -S &= \left(\frac{\partial F}{\partial T}\right)_V \\ -p &= \left(\frac{\partial F}{\partial V}\right)_T \end{align} Thus by taking partial derivatives of \(F\) we can find \(S\) and \(p\) as well as \(U\) with a little arithmetic. You have all seen the Helmholtz free energy before so this shouldn't be much of a surprise. Practically, the Helmholtz free energy is why finding an analytic expression for the partition function is so valuable.
In addition to the “fundamental” physical parameters, we can also find response functions, such as heat capacity or compressibility which are their derivatives. Of particular interest is the heat capacity at fixed volume. The heat capacity is vaguely defined as: \begin{align} C_V &\equiv \sim \left(\frac{\bar d Q}{\partial T}\right)_V \end{align} by which I mean the amount of heat required to change the temperature by a small amount, divided by that small amount, while holding the volume fixed. The First Law tells us that the heat is equal to the change in internal energy, provided no work is done (i.e. holding volume fixed), so \begin{align} C_V &= \left(\frac{\partial U}{\partial T}\right)_V \end{align} which is a nice equation, but can be a nuisance because we often don't know \(U\) as a function of \(T\), which is not one of its natural variables. We can also go back to our Energy and Entropy relationship between heat and entropy where \(\bar dQ = TdS\), and use that to find the ratio that defines the heat capacity: \begin{align} C_V &= T \left(\frac{\partial S}{\partial T}\right)_V. \end{align} Note that this could also have come from a manipulation of the previous derivative of the internal energy. However, the “heat” reasoning allows us to recognize that the heat capacity at constant pressure will have the same form when expressed as an entropy derivative. This expression is also convenient when we compute the entropy from the Helmholtz free energy, because we already know the entropy as a function of \(T\).
Let us work on the free energy of a particle in a 3D box.
From this point, we just need to sum over all states to find \(Z\), and from that the free energy and everything else! So how do we sum all these things up? \begin{align} Z &= \sum_{n_x=-\infty}^{\infty} \sum_{n_y=-\infty}^{\infty} \sum_{n_z=-\infty}^{\infty} e^{-\beta \frac{2\pi^2\hbar^2}{mL^2}\left(n_x^2+n_y^2+n_z^2\right) } \\ &= \sum_{n_x}\sum_{n_y} \sum_{n_z} e^{-\beta \frac{2\pi^2\hbar^2}{mL^2}n_x^2 } e^{-\beta \frac{2\pi^2\hbar^2}{mL^2}n_y^2 } e^{-\beta \frac{2\pi^2\hbar^2}{mL^2}n_z^2 } \\ &= \left(\sum_{n_x} e^{-\beta \frac{2\pi^2\hbar^2}{mL^2}n_x^2 }\right) \left(\sum_{n_y} e^{-\beta \frac{2\pi^2\hbar^2}{mL^2}n_y^2 }\right) \left(\sum_{n_z} e^{-\beta \frac{2\pi^2\hbar^2}{mL^2}n_z^2 }\right) \\ &= \left(\sum_{n=-\infty}^\infty e^{-\beta \frac{2\pi^2\hbar^2}{mL^2}n^2 }\right)^3 \end{align} The last bit here basically looks a lot like separation of variables. Our energy separates into a sum of \(x\), \(y\) and \(z\) portions (which is why we can use separation of variables for the quantum problem), but that also causes things to separate (into a product) when we compute the partition function.
This final sum here is now something we would like to approximate. If our box is reasonably big (and our temperature is not too low), we can assume that \(\frac{4\pi^2\hbar^2}{k_BTmL^2} \ll 1\), which is the classical limit. In this limit, the “thing in the exponential” hardly changes when we change \(n\) by 1, so we can reasonably replace this summation with an integral.
Extending from a single atom to several requires just a bit more subtlety. Naively, you could just argue that because we understand extensive and intensive quantities, we should be able to go from a single atom to \(N\) atoms by simply scaling all extensive quantities. That is almost completely correct (if done correctly).
The entropy has an extra term (the “entropy of mixing”), which also shows up in the free energy. Note that while we may think of this “extra term” as an abstract counting thing, it is physically observable, provided we do the right kind of experiment (which turns out to need to involve changing \(N\), so we won't discuss it in detail until we talk about changing \(N\) later).
There are a few different ways we could imagine putting \(N\) non-interacting atoms together. I will discuss a few here, starting from simplest, and moving up to the most complex.
One option is to consider a single box with volume \(V\) that holds \(N\) different atoms, each of a different kind, but all with the same mass. In this case, each microstate of the system will consist of a microstate for each atom. Quantum mechanically, the wave function for the entire state with \(N\) atoms will separate and will be a product of \(N\) single-particle states (or orbitals) \begin{align} \Psi_{\text{microstate}}(\vec r_1, \vec r_2, \cdots, \vec r_N) &= \prod_{i=1}^{N}\phi_{n_{xi},n_{yi},n_{zi}}(\vec r_i) \end{align} and the energy will just be a sum of different energies. The result of this will be that the partition function of the entire system will just be the product of the partition functions of all the separate non-interacting systems (which happen to all be equal). This is mathematically equivalent to what already happened with the three \(x\), \(y\) and \(z\) portions of the partition function. \begin{align} Z_N &= Z_1^N \\ F_N &= N F_1 \end{align} This results in simply scaling all of our extensive quantities by \(N\) except the volume, which didn't increase when we added more atoms.
This result sounds great, in that it seems to be perfectly extensive, but when we look more closely, we cna see that it is actually not extensive! \begin{align} F_N &= -NkT\ln(Vn_Q) \end{align} If we double the size of our system, so \(N\rightarrow 2N\) and \(V\rightarrow 2V\), you can see that the free energy does not simply double, because the \(V\) in the logarithm doubles while \(n_Q\) remains the same (because it is intensive). So there must be an error here, which turns out to be caused by having treated all the atoms as distinct. If each atom is a unique snowflake, then it doesn't quite make sense to expect the result to be extensive, since you aren't scaling up “interchangeable” things.
This is the picture for a real ideal gas. All of our atoms are the same, or perhaps some fraction are a different isotope, but who cares about that? Since they are all in the same box, we will want to write the many-atom wavefunction as a product of single-atom wavefunctions (sometimes called orbitals). Thus the wave function looks like our first option of “different atoms, same box”, but we have fewer distinct microstates, since swapping the quantum numbers of two atoms doesn't change the microstate.
How to remove this duplication, which is sort of a fundamental problem when our business is counting microstates? Firstly, we will consider it vanishingly unlikely for two atoms to be in the same orbital (when we study Bose condensation, we will see this assumption breaking down). Then we need to figure out exactly how many times we counted each orbital, so we can correct our number of microstates (and our partition function). That number is equal to the number of permutations of \(N\) distinct numbers, which is \(N!\), if we can assume that there is negligible probability that two atoms are in an identical state. Thus we have a corrected partition function \begin{align} Z_N &= \frac1{N!}Z_1^N \\ F_N &= N F_1 + k_BT \ln N! \\ &\approx N F_1 + Nk_BT(\ln N - 1) \\ &= -NkT\ln(Vn_Q) + NkT(\ln N - 1) \\ &= NkT\left(\ln\left(\frac{N}{Vn_Q}\right) - 1\right) \\ &= NkT\left(\ln\left(\frac{n}{n_Q}\right) - 1\right) \end{align} This answer is extensive, because now we have a ratio of \(V\) and \(N\) in the logarithm. So yay.
We now have the true free energy for an ideal gas at high enough temperature.
This is very similar to what we did with just one atom, but now it will give us the true answer for the monatomic ideal gas. \begin{align} S &= -\left(\frac{\partial F}{\partial T}\right)_V \\ &= -Nk(\ln\left(\frac{n}{n_Q}\right) - 1) - NkT\frac{\partial}{\partial T}\ln\left(\frac{n}{n_Q}\right) \\ &= -Nk\ln\left(\frac{n}{n_Q}\right) +Nk + NkT\frac{\partial}{\partial T}\ln n_Q \\ &= -Nk\ln\left(\frac{n}{n_Q}\right) +Nk + \frac32 Nk \\ &= -Nk\ln\left(\frac{n}{n_Q}\right) + \frac52 Nk \end{align} This is called the Sackur-Tetrode equation. The quantum mechanics shows up here (\(\hbar^2/m\)), even though we took the classical limit, because the entropy of a truly classical ideal gas has no minimum value. So quantum mechanics sets the zero of entropy. Note that the zero of entropy is a bit tricky to measure experimentally (albeit possible). The zero of entropy is in fact set by the Third Law of Thermodynamics, which you probably haven't heard of.
Now we can solve for the internal energy: \begin{align} U &= F + TS \\ &= NkT\left(\ln\left(\frac{n}{n_Q}\right) - 1\right) -NkT\ln\left(\frac{n}{n_Q}\right) + \frac52 NkT \\ &= \frac32 NkT \end{align} This is just the standard answer you're familiar with. You can notice that it doesn't have any quantum mechanics in it, because we took
The pressure is easier than the entropy, since the volume is only inside the log: \begin{align} p &= -\left(\frac{\partial F}{\partial V}\right)_T \\ &= \frac{NkT}{V} \end{align} This is the ideal gas law. Again, the quantum mechanics has vanished in the classical limit.
face Lecture
120 min.
paramagnet entropy temperature statistical mechanics
These lecture notes for the second week of Thermal and Statistical Physics involve relating entropy and temperature in the microcanonical ensemble, using a paramagnet as an example. These notes include a few small group activities.assignment Homework
Consider two noninteracting systems \(A\) and \(B\). We can either treat these systems as separate, or as a single combined system \(AB\). We can enumerate all states of the combined by enumerating all states of each separate system. The probability of the combined state \((i_A,j_B)\) is given by \(P_{ij}^{AB} = P_i^AP_j^B\). In other words, the probabilities combine in the same way as two dice rolls would, or the probabilities of any other uncorrelated events.
face Lecture
30 min.
assignment Homework
Suppose that a system of \(N\) atoms of type \(A\) is placed in diffusive contact with a system of \(N\) atoms of type \(B\) at the same temperature and volume.
Show that after diffusive equilibrium is reached the total entropy is increased by \(2Nk\ln 2\). The entropy increase \(2Nk\ln 2\) is known as the entropy of mixing.
If the atoms are identical (\(A=B\)), show that there is no increase in entropy when diffusive contact is established. The difference has been called the Gibbs paradox.
Since the Helmholtz free energy is lower for the mixed \(AB\) than for the separated \(A\) and \(B\), it should be possible to extract work from the mixing process. Construct a process that could extract work as the two gasses are mixed at fixed temperature. You will probably need to use walls that are permeable to one gas but not the other.
This course has not yet covered work, but it was covered in Energy and Entropy, so you may need to stretch your memory to finish part (c).
face Lecture
120 min.
ideal gas particle in a box grand canonical ensemble chemical potential statistical mechanics
These notes from week 6 of Thermal and Statistical Physics cover the ideal gas from a grand canonical standpoint starting with the solutions to a particle in a three-dimensional box. They include a number of small group activities.assignment Homework
\(\ln{x}+\ln{y}\)
\(\ln{a}-\ln{b}\)
\(2\ln{f}+3\ln{f}\)
\(e^{m}e^{k}\)
\(\frac{e^{c}}{e^{d}}\)
\(e^{(3h-j)}\)
\(e^{2(c+w)}\)
\(\ln{h/g}\)
\(\ln(kT)\)
\(\ln{\sqrt{\frac{q}{r}}}\)
face Lecture
120 min.
chemical potential Gibbs distribution grand canonical ensemble statistical mechanics
These notes from the fifth week of Thermal and Statistical Physics cover the grand canonical ensemble. They include several small group activities.assignment Homework
Nuclei of a particular isotope species contained in a crystal have spin \(I=1\), and thus, \(m = \{+1,0,-1\}\). The interaction between the nuclear quadrupole moment and the gradient of the crystalline electric field produces a situation where the nucleus has the same energy, \(E=\varepsilon\), in the state \(m=+1\) and the state \(m=-1\), compared with an energy \(E=0\) in the state \(m=0\), i.e. each nucleus can be in one of 3 states, two of which have energy \(E=\varepsilon\) and one has energy \(E=0\).
Find the Helmholtz free energy \(F = U-TS\) for a crystal containing \(N\) nuclei which do not interact with each other.
Find an expression for the entropy as a function of temperature for this system. (Hint: use results of part a.)
assignment Homework
Make sure that you have memorized the following identities and can use them in simple algebra problems: \begin{align} e^{u+v}&=e^u \, e^v\\ \ln{uv}&=\ln{u}+\ln{v}\\ u^v&=e^{v\ln{u}} \end{align}
assignment Homework
The goal of this problem is to show that once we have maximized the entropy and found the microstate probabilities in terms of a Lagrange multiplier \(\beta\), we can prove that \(\beta=\frac1{kT}\) based on the statistical definitions of energy and entropy and the thermodynamic definition of temperature embodied in the thermodynamic identity.
The internal energy and entropy are each defined as a weighted average over microstates: \begin{align} U &= \sum_i E_i P_i & S &= -k_B\sum_i P_i \ln P_i \end{align}: We saw in clase that the probability of each microstate can be given in terms of a Lagrange multiplier \(\beta\) as \begin{align} P_i &= \frac{e^{-\beta E_i}}{Z} & Z &= \sum_i e^{-\beta E_i} \end{align} Put these probabilities into the above weighted averages in order to relate \(U\) and \(S\) to \(\beta\). Then make use of the thermodynamic identity \begin{align} dU = TdS - pdV \end{align} to show that \(\beta = \frac1{kT}\).