Thermodynamic equilibrium means that each object in thermal contact with each other have the same temperature
Ideal gases, as you cool them down, all converge to a single point, which defines absolute zero. For Kelvin,, the triple point of water defines the size of each step on the scale (273.16 K)
The change is energy can either come from work, or heat
$dE = E_{f}-E_{i}$
Energy is a function of state (path independent)
$\delta W$ and $\delta Q$ are path dependent
Can expand out $\delta W$ into more parts:
$dE = -p dV + \delta Q$, where the negative sign comes from the fact that the system does work on the outside environment
You don’t need to be restricted to pressure and volume. In general, you can have some “displacements” $\chi_{i}$ (ie. extensive quantities) and their conjugate forces $J_{i}$ (ie. intensive quantities) to get that $dW = \Sigma_{i} J_{i} d\chi_{i}$
Efficiency of an ideal heat engine is defined by $\eta = \frac{W}{Q_{H}} = \frac{Q_{H}-Q_{C}}{Q_{H}} \leq 1$
The figure of merit for an ideal refrigerator is $\omega = \frac{Q_{C}}{W} = \frac{Q_{C}}{Q_{H}-Q_{C}}$
Kelvin formulation of 2nd Law: No process is possible whose sole result is the complete conversion of heat into work
Clausius’s Statement: No process is possible whose sole result is the transfer of heat from a colder to a hotter body
The above two are equivalent: Can show that violation of one implies a violation of another via hooking up outputs of an ideal engine to an ideal refrigerator
Carnot’s Theorem: You can’t beat the Carnot engine in terms of efficiency
A Carnot cycle is defined by two isotherms at $T_{H}$ and $T_{C}$, and two adiabatic curves linking the isotherms
The efficiency of a Carnot engine is defined by $\eta = \frac{T_{H}-T_{C}}{T_{H}}$. This can be derived hooking up two different Carnot engines at 3 temperatures. The heat from one gets dumped to the next. The overall efficiency is then the product of the two, which implies that the efficiency is some ratio of temperatures
Clausius’s Theorem: For any cyclic transformation (reversible or not) $\oint \frac{dQ}{T} \leq 0$, where $dQ$ is the heat increment supplied to the system at temperature T
Imagine dividing the cycle into a series of infinitesimal portions, where the system receives energy from dQ and dW. Imagine that dQ gets funneled into some Carnot engine which is attached to some reservoir at a fixed temperature $T_{0}$
Since the sign of dQ is unspecified, the Carnot engine must be able to run in both directions. In order to do this, the engine must extract some heat $dQ_{R}$ from the fixed reservoir
This implies that $dQ_{R} = \frac{T_{0}}{T} dQ$
The net effect of this process is that some heat $Q_{R} = \oint dQ_{R}$ is extracted from the reservoir and converted to external work W. By the Kelvin formulation of the 2nd law, $Q_{R} = W \leq 0$ which implies $\oint \frac{dQ}{T} \leq 0$
For a reversible cycle, we know $\oint \frac{dQ_{rev}}{T} = 0$ This implies that for a reversible cycle, this integral is independent of the path. We can use this to define the entropy S: $S(B)-S(A) = \int_{A}^{B} \frac{dQ_{rev}}{T}$
Imagine that you make an irreversible change from A to B, but make a reversible regression from B to A: $\int_{A}^{B} \frac{dQ}{T}+ \int \frac{dQ_{rev}}{T} \leq 0$ which implies $\int_{A}^{B} \frac{dQ}{T} \leq S(B)-S(A)$
This is the statement that entropy always increases or stays the same
For a reversible (and/or quasi-static) process, we can write $dQ = TdS$, where T and S and conjugate variables
The energy gets extremized in equillibrium for an adiabatically isolated system
If you are in an out of equillibrium system that is not adiabatically isolated and subject to external work, then you can define other thermodynamic potentials which are extremized in equillibrium
If there is no heat exchange, and the system comes to mechanical equilibrium subject to a constant external force, then enthalpy is the appropriate potential
Can be though of as minimizing the energy, plut the work from the external agent
$H = E- \vec{J}\cdot \vec{x}$
$dH = dE -d(J\cdot x)$ (Just the legendre transform w.r.t. J and x)
What if the number of particles in the system changes? we can define the chemical work as $dW = \vec{\mu} \cdot d\vec{N}$, where each species has it’s own chemical potential
We can introduce this work by simply adding this work to a potential
Random variable x has a set of outcome S (can be continuous or discrete)
An event is a subset of outcomes from S, and is assigned a probability
Probabilities must obey the following rules
They must be greater than or equal to 0
They must be additive if the events subsets are disjoint
They must be normalizable (ie. probability of S should be 1)
a cumulative probability function (CPF) denoted P(x) is the probability of an outcome with any value less than x
P(x) must be a monotonically increasing function of x
$P(-\infty) = 0$ and $P(\infty) = 1$
a probability density function (PDF) is defined by $p(x) = \frac{dP(x)}{dx}$
This must satisfy $\int p(x) dx = 1$
The expectation value of any function F(x) of the random variable is $<F(x)> = \int_{-\infty}^{\infty} dx p(x) F(x)$
Easily extendable to multiple dimensions
The moments of a PDF are expectation values for power of the random variable
The nth moment is $m_{n} = <x^{n}> = \int dx p(x) x^{n}$
The characteristic function is the generator of moments of the distribution. It’s just the Fourier transform of the PDF
$\tilde{p}(k) = \int dx p(x) exp(-ikx)$
The PDF can be recovered from the characteristic function via the inverse Fourier transform $p(x) = \frac{1}{2\pi} \int dk \tilde{p}(k) exp(ikx)$
The moments can be calculated by expanding $\tilde{p}(k)$ in powers of k: $\tilde{p}(k) = \Sigma_{n=0}^{\infty} \frac{(-ik)^{n}}{n!}<x^{n}>$
Moments of the PDF around any $x_{0}$ can also be generated by substituting x for $x-x_{0}$ in the moment generation
The cumulant generation function is the logarithm of the characteristic function
You can generate cumulants of the distribution by expansions in powers of x: $ln \tilde{p}(k) = \Sigma_{n=1}^{\infty} \frac{(-ik)^{n}}{n!}<x^{n}>_{c}$
Moments can be related to cumulants by using $\ln(1+\epsilon) = \Sigma_{n=1}^{\infty} (-1)^{\epsilon+1} \frac{\epsilon^{n}}{n}$ to expand out $\ln\tilde{p}(k)$
A Joint PDF is the probability density of many variables. If all variables are independent, this is just the product of the individual pdfs
Stirling’s approximation for N! holds for very large N. This states that
$ln(N!) \approx N ln N - N + \frac{1}{2}ln(2\pi N)$
Characteristic function is $\tilde{p}(k) = (p_{A}e^{-ik}+p_{B})^{N}$
Poisson Distribution
This is the limit of the binomial distribution. We want to find the probability of observing N decays in time interval T. Subdivide the interval into $N = \frac{T}{dt} » 1$
In each interval, the chance of an event occuring os $p_{A} = \alpha dt$ and the chance of no event in the interval is $p_{B} = 1-\alpha dt$
This implies the characteristic function is $\tilde{p}(k) = (p_{A}exp(-ik)+p_{B})^{N} = lim_{dt\rightarrow 0}(1+\alpha dt (e^{-ik}-1))^{\frac{T}{dt}} = exp(\alpha(e^{-ik}-1)T)$
Taking the inverse Fourier transform yields the Poisson PDF: $p(x) = \int_{-\infty}^{\infty} \frac{dk}{2\pi} e^{ikx} \Sigma_{M=0}^{\infty} \frac{(\alpha T)^{M}}{M!}e^{-ikM}$ (using power series for the exponential)
Using the identity $\int_{-\infty}^{\infty} \frac{dk}{2\pi} exp(ik(x-M)) = \delta(x-M)$, which implies $p_{\alpha,T}(M) = e^{-\alpha T}\frac{(\alpha T)^{M}}{M!}$
Concisely, Liouville’s Theorem states that $\rho$ behaves like an incompressible fluid
The proof is as follows:
Imagine that you Taylor expand out all the q and p’s as a function of time
The volume of phase space before is simply $\Pi_{i} dq_{i}dp_{i}$
The volume of phase space after is then the dot product of the Taylor expanded momeenta and position. You can use Hamilton’s equations to eliminate the time dependent portion of the expansion, which implies that the phase space volume is unchanged
A consequence of this is that $\frac{d\rho}{dt} = \frac{\partial \rho}{\partial t}+\Sigma_{\alpha=1}^{3N} (\frac{\partial \rho}{\partial p_{\alpha}}\frac{dp_{\alpha}}{dt}+\frac{\partial \rho}{\partial q_{\alpha}}\frac{dq_{\alpha}}{dt}) = 0$
Alternatively: $\frac{\partial \rho }{\partial t} = -\{\rho, H\}$, where $\{\}$ is the Poisson bracket
Another consequence of this is that $\frac{d<O>}{dt} = <\{A,H\}>$
straightforward to prove (integration by parts, and using Hamilton equations gives it straightforwardly)
If a density has reached equillibrium, this implies that it’s independent of time. This implies that $\{\rho_{eq},H\}= 0$
If $\rho_{eq}$ is a function of the Hamiltonian, then this equation holds
This implies that $\rho$ is constant of surfaces of constant energy in phase space
This is the basic assumption of statistical mechanics! You have equal probabilities at constant energies
For multiple particle, you can generate the s-particle density via $f_{s}(p_{1}…q_{s},t) = \frac{N!}{(N-s)!}\int \Pi_{i=s+1}^{N} dV_{i} \rho(p,q,t) = \frac{N!}{(N-s)!}\rho_{s}(p_{1},…,q_{s},t)$
The normalization is $\frac{N!}{(N-s)!}$ and $p_{s}$ is the unconditional PDF for the coordinates of s particles
Can imagine a Hamiltonian which is $H = \Sigma_{i=1}^{N} (\frac{p_{i}^{2}}{2m}+U(q_{i}))+\frac{1}{2}\Sigma_{i,j=1}^{N} V(q_{i}-q_{j})$ where U is some external potential and V is a two-body interaction
You can write the time evolution of $\rho_{s}$ as $\frac{\partial \rho_{s}}{\partial t} = -\int \Pi_{i=s+1}^{N} dV_{i} \{\rho, H_{s}+H_{N-s}+H^{’}\}$
If you take the Poisson bracket of each term, you find that $\frac{\partial f_{s}}{\partial t} = -\{H_{s}, f_{s}\} = \Sigma_{n=1}^{s} \int dV_{s+1} \frac{\partial V(q_{n}-q_{s+1})}{\partial q_{n}}\frac{\partial f_{s+1}}{\partial p_{n}}$
Note the hierarchy here: $\frac{\partial f_{s}}{\partial t}$ depends on $\frac{\partial f_{s+1}}{\partial t}$
To terminate the hierarchy, we need to be able to neglect some terms. Since all terms have units of inverse time, we can seperate terms according to time scale.
$\frac{1}{\tau_{U}}\approx \frac{\partial U}{\partial q}\frac{\partial}{\partial p}$ is some extrinsic time scale, which can be made arbitrarily small by increasing the system size. On the order of 1E-5 sec for length scales of 1E-3 m ($\tau \approx \frac{d}{v}$)
$\frac{1}{\tau_{c}}\approx \frac{\partial V}{\partial q}\frac{\partial }{\partial p}$ has some collision time, on the order of 1E-12 sec
Longe range interactions are harder to get a collision time
There are also collision terms which depend on $f_{s+1}$
$\frac{1}{\tau_{x}} \approx \int dV \frac{\partial V}{\partial q}\frac{\partial}{\partial p}\frac{f_{s+1}}{f_{s}}$ where $\tau_{x}$ is the mean free time
With the dilute gas approximation, you can truncate the BBGKY hierarchy by setting the RHS of the 2nd equation to 0
It’s also reasonable to expect that at very large distances, the particles become independent (ie. $f_{2}(p_{1},q_{1},p_{2},q_{2},t) \rightarrow f_{1}(p_{1},q_{1},t)f_{1}(p_{2},q_{2},t)$)
The final closed form of Boltzmann’s equation is $(\frac{\partial}{\partial t} - \frac{\partial U}{\partial q} \frac{\partial}{\partial p }+\frac{p_{1}}{m}\frac{\partial}{\partial q}) f_{1} = -\int d^{3}\vec{p}d^{2}\Omega |\frac{\partial \sigma}{\partial \Omega}||v_{1}-v_{2}| (f_{1}(p_{1},q_{1},t)f_{1}(p_{2},q_{1},t)-f_{1}(p_{1}’,q_{1},t)f_{1}(p_{2}’,q_{1},t))$
The Theorem: If $f_{1}(\vec{p},\vec{q},t)$ satisfies the Boltzmann equation, then $\frac{dH}{dt} \leq 0$, where $H(t) \int d^{3}\vec{p}d^{3}\vec{q} f_{1} ln f_{1}$
H is very closely related to the information content of a one particle PDF
The proof is straight forward
Take the time derivative of H, and use approximation $ln (f_{1})+1 \approx ln f_{1}$
Use Boltzmann equation
Use some integration by parts to elimnate some streaming terms
Observe that the equation must if you swap particle 1 for particle 2. Do the swap, then average the two epxressions
Make the change of integration variables from the initiators of the collision $(p_{1},p_{2},b)$ to the products of the collision $(p_{1}’,p_{2}’,b’)$
We know that the Jacobian is unitary from time symmetry
We realize that we can swap the initiators and the products. Do this swap, then average the two equations again
The final results is something like $\frac{dH}{dt} = -\frac{1}{4}\int d\Gamma x$ where x is some strictly positive integrand. Hence $\frac{dH}{dt}\leq 0$
The microstates are defined by points in phase space, and their time evolution is governed by the Hamiltonian. Since the Hamiltonian is energy conserving, all micro-states are confined to a constant energy surface in phase space
This implies that all points on the surface are mutually accessible, which gives the central postulate of stat mech $p(E,x) (\mu) = \frac{1}{\Omega(E,x)}* \delta(H(\mu)-E)$ where $\mu$ enumerate the microstates
ie. on the surface, all microstates are equally likely. If you are off the surface, you can’t access it
The normalization $\Omega$ is the area of the surface of the constant energy E in phase space. Typically, we define $E-\Delta \leq H(\mu) \leq E+\Delta$, where $\Delta$ is uncertainty in the energy. This allows us to have the normalization be a spherical shell
The entropy if this uniform probability distribution is $S = k_{b} ln \Omega$
The overall allowed phase space is the product of individual ones (assuming independent systems)
$\Omega$ is the number of ways of choosing $N_{1}$ excited levels among the available N atoms. This is just ${N}\choose{N_{1}}$
Entropy is just $S = k_{b} \ln N$
You can use Stirling’s formula to approximate $\ln N! \approx N \ln N -N$
You can then get the temperature from $\frac{\partial S}{\partial E} = \frac{1}{T}$, and invert this to get the erngy as a function of temperature
You can figure out the occupancy of a particular level by imagining a a small subsystem that is decreased by the energy of that state, with particle number that is one less (ie. $p_{n_{1}} = \frac{\Omega(E-n_{1}\epsilon, N-1)}{\Omega(E,N)}$)
The Hamiltonian of this system is $H = \Sigma_{i=1}^{N} \frac{p_{i}^{2}}{2m}+U(q_{i})$ where U is some external potential. For now, assume that U=0
Assuming some fixed E, $\Omega$ is just
the spatial components ($V^{N}$, since each particle can be anywhere in the volume)
times momentum components (this is just the surface of a hypersphere of dimension 3N and radius $\Sigma_{i=1}^{N} p_{i}^{2} = \sqrt{2mE}$)
The area of a hypersphere is given by $A_{d} = S_{d}R^{d-1}$
Can calculate $S_{d}$ via $I_{d} =( \int_{\infty}^{\infty} dx e^{-x^{2}})^{d}$
This equals $ \pi^{\frac{d}{2}}$
This integral is the product of d one-dimensional gaussians. This is spherically symmetric. Making that change in variable ($dV_{d} = S_{d}R^{d-1}dR$), then make another change of variable $y=R^{2}$, then use the integral form of the Gamma function $\int_{0}^{\infty} dy y^{\frac{d}{2}-1}e^{-y} = (\frac{d}{2}-1)!$
This implies that $S_{d} = \frac{2\pi^{\frac{d}{2}}}{(\frac{d}{2}-1)!}$
Hence $\Omega = V^{N} \frac{2\pi^{\frac{d}{2}}}{(\frac{d}{2}-1)!} (2mE)^{\frac{3N-1}{2}}\Delta R$ where $\delta R$ is the thickness of the hypersphere
The entropy (after Sterling approximation) is then $S = Nk_{b} \ln (V (\frac{4\pi e m E}{3N})^{\frac{3}{2}})$
$\frac{1}{T} = \frac{3}{2}\frac{Nk_{b}}{E}$
Can recover ideal gas law from $\frac{P}{V} = \frac{\partial S}{\partial V}$
Can calculate the probability of finding a particle of momentum $\vec{p_{1}}$ via $p(\vec{p_{1}}) = \frac{V \Omega(E-\frac{p_{1}^{2}}{2m},V,N-1)}{\Omega(E,V,C)}$
After some algebra and Sterling’s approximation, you get the normalized Maxwell-Boltzmann distribution
$p(p_{1}) = (\frac{3N}{4\pi m E})^{\frac{3}{2}}exp(\frac{-3N}{2}\frac{p_{1}^{2}}{2mE})$, where $E = \frac{3Nk_{b}T}{2}$
After mixing $S_{f} = N_{1}k_{b} \ln(V_{1}+V_{2})+N_{2}k_{b} \ln(V_{1}+V_{2})$
Which means the change in entropy is $\Delta S = -N k_{b} (\frac{N_{1}}{N}\ln\frac{V_{1}}{V}+\frac{N_{2}}{N}\ln\frac{V_{2}}{V})$
One problem with this is that if you are “mixing” two gases. Imagine that you have a partition separating the gas. Removing the partition shouldn’t change the entropy, but the above argument shows that you have some entropy change
The resolution to this is that with indistinguishable particles, you are over counting the number of states by a factor of N!. The above entropy works for for distinguishable particles