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diff --git a/content/know/concept/canonical-ensemble/index.pdc b/content/know/concept/canonical-ensemble/index.pdc new file mode 100644 index 0000000..9f75fc8 --- /dev/null +++ b/content/know/concept/canonical-ensemble/index.pdc @@ -0,0 +1,248 @@ +--- +title: "Canonical ensemble" +firstLetter: "C" +publishDate: 2021-07-10 +categories: +- Physics +- Thermodynamics +- Thermodynamic ensembles + +date: 2021-07-08T11:01:02+02:00 +draft: false +markup: pandoc +--- + +# Canonical ensemble + +The **canonical ensemble** or **NVT ensemble** builds on +the [microcanonical ensemble](/know/concept/microcanonical-ensemble/), +by allowing the system to exchange energy with a very large heat bath, +such that its temperature $T$ remains constant, +but internal energy $U$ does not. +The conserved state functions are +the temperature $T$, the volume $V$, and the particle count $N$. + +We refer to the system of interest as $A$, and the heat bath as $B$. +The combination $A\!+\!B$ forms a microcanonical ensemble, +i.e. it has a fixed total energy $U$, +and eventually reaches an equilibrium +with a uniform temperature $T$ in both $A$ and $B$. + +Assuming that this equilibrium has been reached, +we want to know which microstates $A$ prefers in that case. +Specifically, if $A$ has energy $U_A$, and $B$ has $U_B$, +which $U_A$ does $A$ prefer? + +Let $c_B(U_B)$ be the number of $B$-microstates with energy $U_B$. +Then the probability that $A$ is in a specific microstate $s_A$ is as follows, +where $U_A(s_A)$ is the resulting energy: + +$$\begin{aligned} + p(s_A) + = \frac{c_B(U - U_A(s_A))}{D} + \qquad \quad + D \equiv \sum_{s_A} c_B(U - U_A(s_A)) +\end{aligned}$$ + +In other words, we choose an $s_A$, +and count the number $c_B$ of compatible $B$-microstates. + +Since the heat bath is large, let us assume that $U_B \gg U_A$. +We thus approximate $\ln{p(s_A)}$ by +Taylor-expanding $\ln{c_B(U_B)}$ around $U_B = U$: + +$$\begin{aligned} + \ln{p(s_A)} + &= -\ln{D} + \ln\!\big(c_B(U - U_A(s_A))\big) + \\ + &\approx - \ln{D} + \ln{c_B(U)} - \bigg( \dv{(\ln{c_B})}{U_B} \bigg) \: U_A(s_A) +\end{aligned}$$ + +Here, we use the definition of entropy $S_B \equiv k \ln{c_B}$, +and that its $U_B$-derivative is $1/T$: + +$$\begin{aligned} + \ln{p(s_A)} + &\approx - \ln{D} + \ln{c_B(U)} - \frac{U_A(s_A)}{k} \Big( \pdv{S_B}{U_B} \Big) + \\ + &\approx - \ln{D} + \ln{c_B(U)} - \frac{U_A(s_A)}{k T} +\end{aligned}$$ + +We now define the **partition function** or **Zustandssumme** $Z$ as follows, +which will act as a normalization factor for the probability: + +$$\begin{aligned} + \boxed{ + Z + \equiv \sum_{s_A}^{} \exp\!(- \beta U_A(s_A)) + } + = \frac{D}{c_B(U)} +\end{aligned}$$ + +Where $\beta \equiv 1/ (k T)$. +The probability of finding $A$ in a microstate $s_A$ is thus given by: + +$$\begin{aligned} + \boxed{ + p(s_A) = \frac{1}{Z} \exp\!(- \beta U_A(s_A)) + } +\end{aligned}$$ + +This is the **Boltzmann distribution**, +which, as it turns out, maximizes the entropy $S_A$ +for a fixed value of the average energy $\expval{U_A}$, +i.e. a fixed $T$ and set of microstates $s_A$. + +Because $A\!+\!B$ is a microcanonical ensemble, +we know that its [thermodynamic potential](/know/concept/thermodynamic-potential/) +is the entropy $S$. +But what about the canonical ensemble, just $A$? + +The solution is a bit backwards. +Note that the partition function $Z$ is not a constant; +it depends on $T$ (via $\beta$), $V$ and $N$ (via $s_A$). +Using the same logic as for the microcanonical ensemble, +we define "equilibrium" as the set of microstates $s_A$ +that $A$ is most likely to occupy, +which must be the set (as a function of $T,V,N$) that maximizes $Z$. + +However, $T$, $V$ and $N$ are fixed, +so how can we maximize $Z$? +Well, as it turns out, +the Boltzmann distribution has already done it for us! +We will return to this point later. + +Still, $Z$ does not have a clear physical interpretation. +To find one, we start by showing that the ensemble averages +of the energy $U_A$, pressure $P_A$ and chemical potential $\mu_A$ +can be calculated by differentiating $Z$. +As preparation, note that: + +$$\begin{aligned} + \pdv{Z}{\beta} = - \sum_{s_A} U_A \exp\!(- \beta U_A) +\end{aligned}$$ + +With this, we can find the ensemble averages +$\expval{U_A}$, $\expval{P_A}$ and $\expval{\mu_A}$ of the system: + +$$\begin{aligned} + \expval{U_A} + &= \sum_{s_A} p(s_A) \: U_A + = \frac{1}{Z} \sum_{s_A} U_A \exp\!(- \beta U_A) + = - \frac{1}{Z} \pdv{Z}{\beta} + \\ + \expval{P_A} + &= - \sum_{s_A} p(s_A) \pdv{U_A}{V} + = - \frac{1}{Z} \sum_{s_A} \exp\!(- \beta U_A) \pdv{U_A}{V} + \\ + &= \frac{1}{Z \beta} \pdv{V} \sum_{s_A} \exp\!(- \beta U_A) + = \frac{1}{Z \beta} \pdv{Z}{V} + \\ + \expval{\mu_A} + &= \sum_{s_A} p(s_A) \pdv{U_A}{N} + = \frac{1}{Z} \pdv{N} \sum_{s_A} \exp\!(- \beta U_A) \pdv{U_A}{N} + \\ + &= - \frac{1}{Z \beta} \pdv{N} \sum_{s_A} \exp\!(- \beta U_A) + = - \frac{1}{Z \beta} \pdv{Z}{N} +\end{aligned}$$ + +It will turn out more convenient to use derivatives of $\ln{Z}$ instead, +in which case: + +$$\begin{aligned} + \expval{U_A} + = - \pdv{\ln{Z}}{\beta} + \qquad \quad + \expval{P_A} + = \frac{1}{\beta} \pdv{\ln{Z}}{V} + \qquad \quad + \expval{\mu_A} + = - \frac{1}{\beta} \pdv{\ln{Z}}{N} +\end{aligned}$$ + +Now, to find a physical interpretation for $Z$. +Consider the quantity $F$, in units of energy, +whose minimum corresponds to a maximum of $Z$: + +$$\begin{aligned} + F \equiv - k T \ln{Z} +\end{aligned}$$ + +We rearrange the equation to $\beta F = - \ln{Z}$ and take its differential element: + +$$\begin{aligned} + \dd{(\beta F)} + = - \dd{(\ln{Z})} + &= - \pdv{\ln{Z}}{\beta} \dd{\beta} - \pdv{\ln{Z}}{V} \dd{V} - \pdv{\ln{Z}}{N} \dd{N} + \\ + &= \expval{U_A} \dd{\beta} - \beta \expval{P_A} \dd{V} + \beta \expval{\mu_A} \dd{N} + \\ + &= \expval{U_A} \dd{\beta} + \beta \dd{\expval{U_A}} - \beta \dd{\expval{U_A}} - \beta \expval{P_A} \dd{V} + \beta \expval{\mu_A} \dd{N} + \\ + &= \dd{(\beta \expval{U_A})} - \beta \: \big( \dd{\expval{U_A}} + \expval{P_A} \dd{V} - \expval{\mu_A} \dd{N} \big) +\end{aligned}$$ + +Rearranging and substituting +the [fundamental thermodynamic relation](/know/concept/fundamental-thermodynamic-relation/) +then gives: + +$$\begin{aligned} + \dd{(\beta F - \beta \expval{U_A})} + &= - \beta \: \big( \dd{\expval{U_A}} + \expval{P_A} \dd{V} - \expval{\mu_A} \dd{N} \big) + = - \beta T \dd{S_A} +\end{aligned}$$ + +We integrate this and ignore the integration constant, +leading us to the desired result: + +$$\begin{aligned} + - \beta T S_A + &= \beta F - \beta \expval{U_A} + \quad \implies \quad + F = \expval{U_A} - T S_A +\end{aligned}$$ + +As was already suggested by our notation, +$F$ turns out to be the **Helmholtz free energy**: + +$$\begin{aligned} + \boxed{ + \begin{aligned} + F + &\equiv - k T \ln{Z} + \\ + &= \expval{U_A} - T S_A + \end{aligned} + } +\end{aligned}$$ + +We can therefore reinterpret +the partition function $Z$ and the Boltzmann distribution $p(s_A)$ +in the following "more physical" way: + +$$\begin{aligned} + Z + = \exp\!(- \beta F) + \qquad \quad + p(s_A) + = \exp\!\Big( \beta \big( F \!-\! U_A(s_A) \big) \Big) +\end{aligned}$$ + +Finally, by rearranging the expressions for $F$, +we find the entropy $S_A$ to be: + +$$\begin{aligned} + S_A + = k \ln{Z} + \frac{\expval{U_A}}{T} +\end{aligned}$$ + +This is why $Z$ is already maximized: +the Boltzmann distribution maximizes $S_A$ for fixed values of $T$ and $\expval{U_A}$, +leaving $Z$ as the only "variable". + + + +## References +1. H. Gould, J. Tobochnik, + *Statistical and thermal physics*, 2nd edition, + Princeton. diff --git a/content/know/concept/fermis-golden-rule/index.pdc b/content/know/concept/fermis-golden-rule/index.pdc new file mode 100644 index 0000000..5ed273e --- /dev/null +++ b/content/know/concept/fermis-golden-rule/index.pdc @@ -0,0 +1,91 @@ +--- +title: "Fermi's golden rule" +firstLetter: "F" +publishDate: 2021-07-10 +categories: +- Physics +- Quantum mechanics +- Optics + +date: 2021-07-03T14:41:11+02:00 +draft: false +markup: pandoc +--- + +# Fermi's golden rule + +In quantum mechanics, **Fermi's golden rule** expresses +the transition rate between two states of a system, +when a sinusoidal perturbation is applied +at the resonance frequency $\omega = E_g / \hbar$ of the +energy gap $E_g$. The main conclusion is that the rate is independent of +time. + +From [time-dependent perturbation theory](/know/concept/time-dependent-perturbation-theory/), +we know that the transition probability +for a particle in state $\ket{a}$ to go to $\ket{b}$ +is as follows for a periodic perturbation at frequency $\omega$: + +$$\begin{aligned} + P_{ab} + = \frac{|V_{ba}|^2}{\hbar^2} \frac{\sin^2\!\big((\omega_{ba} - \omega) t / 2\big)}{(\omega_{ba} - \omega)^2} +\end{aligned}$$ + +Where $\omega_{ba} \equiv (E_b - E_a) / \hbar$. +If we assume that $\ket{b}$ irreversibly absorbs an unlimited number of particles, +then we can interpret $P_{ab}$ as the "amount" of the current particle +that has transitioned since the last period $2 \pi n / (\omega_{ba} \!-\! \omega)$. + +For generality, let $E_b$ be the center +of a state continuum with width $\Delta E$. +In that case, $P_{ab}$ must be modified as follows, +where $\rho(E_x)$ is the destination's +[density of states](/know/concept/density-of-states/): + +$$\begin{aligned} + P_{ab} + &= \frac{|V_{ba}|^2}{\hbar^2} \int_{E_b - \Delta E / 2}^{E_b + \Delta E / 2} + \frac{\sin^2\!\big((\omega_{xa} - \omega) t / 2\big)}{(\omega_{xa} - \omega)^2} \:\rho(E_x) \dd{E_x} +\end{aligned}$$ + +If $E_b$ is not in a continuum, then $\rho(E_x) = \delta(E_x - E_b)$. +The integrand is a sharp sinc-function around $E_x$. +For large $t$, it is so sharp that we can take out $\rho(E_x)$. +In that case, we also simplify the integration limits. +Then we substitute $x \equiv (\omega_{xa}\!-\!\omega) / 2$ to get: + +$$\begin{aligned} + P_{ab} + &\approx \frac{2}{\hbar} |V_{ba}|^2 \rho(E_b) \int_{-\infty}^\infty \frac{\sin^2(x t)}{x^2} \:dx +\end{aligned}$$ + +This definite integral turns out to be $\pi |t|$, +so we find, because clearly $t > 0$: + +$$\begin{aligned} + P_{ab} + &= \frac{2 \pi}{\hbar} |V_{ba}|^2 \rho(E_b) \: t +\end{aligned}$$ + +The transition rate $R_{ab}$, +i.e. the number of particles per unit time, +then takes this form: + +$$\begin{aligned} + \boxed{ + R_{ab} + = \pdv{P_{ab}}{t} + = \frac{2 \pi}{\hbar} |V_{ba}|^2 \rho(E_b) + } +\end{aligned}$$ + +Note that the $t$-dependence has disappeared, +and all that remains is a constant factor involving $E_b = E_a \!+\! \hbar \omega$, +where $\omega$ is the resonance frequency. + + + +## References +1. D.J. Griffiths, D.F. Schroeter, + *Introduction to quantum mechanics*, 3rd edition, + Cambridge. diff --git a/content/know/concept/grand-canonical-ensemble/index.pdc b/content/know/concept/grand-canonical-ensemble/index.pdc new file mode 100644 index 0000000..5853a5f --- /dev/null +++ b/content/know/concept/grand-canonical-ensemble/index.pdc @@ -0,0 +1,83 @@ +--- +title: "Grand canonical ensemble" +firstLetter: "G" +publishDate: 2021-07-11 +categories: +- Physics +- Thermodynamics +- Thermodynamic ensembles + +date: 2021-07-08T11:01:11+02:00 +draft: false +markup: pandoc +--- + +# Grand canonical ensemble + +The **grand canonical ensemble** or **μVT ensemble** +extends the [canonical ensemble](/know/concept/canonical-ensemble/) +by allowing the exchange of both energy $U$ and particles $N$ +with an external reservoir, +so that the conserved state functions are +the temperature $T$, the volume $V$, and the chemical potential $\mu$. + +The derivation is practically identical to that of the canonical ensemble. +We refer to the system of interest as $A$, +and the reservoir as $B$. +In total, $A\!+\!B$ has energy $U$ and population $N$. + +Let $c_B(U_B)$ be the number of $B$-microstates with energy $U_B$. +Then the probability that $A$ is in a specific microstate $s_A$ is as follows: + +$$\begin{aligned} + p(s) + = \frac{c_B\big(U - U_A(s_A), N - N_A(s_A)\big)}{\sum_{s_A} c_B\big(U \!-\! U_A(s_A), N \!-\! N_A(s_A)\big)} +\end{aligned}$$ + +Then, as for the canonical ensemble, +we assume $U_B \gg U_A$ and $N_B \gg N_A$, +and approximate $\ln{p(s_A)}$ +by Taylor-expanding $\ln{c_B}$ around $U_B = U$ and $N_B = N$. +The resulting probability distribution is known as the **Gibbs distribution**, +with $\beta \equiv 1/(kT)$: + +$$\begin{aligned} + \boxed{ + p(s_A) = \frac{1}{\mathcal{Z}} \exp\!\Big(\!-\! \beta \: \big( U_A(s_A) \!-\! \mu N_A(s_A) \big) \Big) + } +\end{aligned}$$ + +Where the normalizing **grand partition function** $\mathcal{Z}(\mu, V, T)$ is defined as follows: + +$$\begin{aligned} + \boxed{ + \mathcal{Z} \equiv \sum_{s_A}^{} \exp\!\Big(\!-\! \beta \: \big( U_A(s_A) - \mu N_A(s_A) \big) \Big) + } +\end{aligned}$$ + +In contrast to the canonical ensemble, +whose [thermodynamic potential](/know/concept/thermodynamic-potential/) +was the Helmholtz free energy $F$, +the grand canonical ensemble instead +minimizes the **grand potential** $\Omega$: + +$$\begin{aligned} + \boxed{ + \begin{aligned} + \Omega(T, V, \mu) + &\equiv - k T \ln{\mathcal{Z}} + \\ + &= \expval{U_A} - T S_A - \mu \expval{N_A} + \end{aligned} + } +\end{aligned}$$ + +So $\mathcal{Z} = \exp\!(- \beta \Omega)$. +This is proven in the same way as for $F$ in the canonical ensemble. + + + +## References +1. H. Gould, J. Tobochnik, + *Statistical and thermal physics*, 2nd edition, + Princeton. diff --git a/content/know/concept/microcanonical-ensemble/index.pdc b/content/know/concept/microcanonical-ensemble/index.pdc index 89d114b..16628b8 100644 --- a/content/know/concept/microcanonical-ensemble/index.pdc +++ b/content/know/concept/microcanonical-ensemble/index.pdc @@ -34,8 +34,8 @@ $$\begin{aligned} = \sum_{U_A \le U} c_A(U_A) \: c_B(U - U_A) \end{aligned}$$ -Where $c_A$ and $c_B$ are the number of microstates of -the subsystems at the given energy levels. +Where $c_A$ and $c_B$ are the numbers of subsystem microstates +at the given energy levels. The core assumption of the microcanonical ensemble is that each of these microstates has the same probability $1 / c$. @@ -43,18 +43,20 @@ Consequently, the probability of finding an energy $U_A$ in $A$ is: $$\begin{aligned} p_A(U_A) - = \frac{c_A(U_A) \:c_B(U - U_A)}{c(U)} + = \frac{c_A(U_A) \: c_B(U - U_A)}{c(U)} \end{aligned}$$ If a certain $U_A$ has a higher probability, then there are more $A$-microstates with that energy, -meaning that $U_A$ is "easier to reach" or "more comfortable" for the system. -Note that $c(U)$ is a constant, because $U$ is given. +so, statistically, for an *ensemble* of many boxes, +we expect that $U_A$ is more common. -After some time, the system will reach equilibrium, -where both $A$ and $B$ have settled into a "comfortable" position. +The maximum of $p_A$ will be the most common in the ensemble. +Assuming that we have given the boxes enough time to settle, +we go one step further, +and refer to this maximum as "equilibrium". In other words, the subsystem microstates at equilibrium -must be maxima of their probability distributions $p_A$ and $p_B$. +are maxima of $p_A$ and $p_B$. We only need to look at $p_A$. Clearly, a maximum of $p_A$ is also a maximum of $\ln p_A$: @@ -66,13 +68,13 @@ $$\begin{aligned} Here, in the quantity $\ln{c_A}$, we recognize the definition of -the entropy $S_A \equiv k_B \ln{c_A}$, -where $k_B$ is Boltzmann's constant. -We thus multiply by $k_B$: +the entropy $S_A \equiv k \ln{c_A}$, +where $k$ is Boltzmann's constant. +We thus multiply by $k$: $$\begin{aligned} - k_B \ln p_A(U_A) - = S_A(U_A) + S_B(U - U_A) - k_B \ln{c(U)} + k \ln p_A(U_A) + = S_A(U_A) + S_B(U - U_A) - k \ln{c(U)} \end{aligned}$$ Since entropy is additive over subsystems, @@ -87,7 +89,7 @@ more concrete, equilibrium condition: $$\begin{aligned} 0 - = k_B \dv{(\ln{p_A})}{U_A} + = k \dv{(\ln{p_A})}{U_A} = \pdv{S_A}{U_A} + \pdv{S_B}{U_A} = \pdv{S_A}{U_A} - \pdv{S_B}{U_B} \end{aligned}$$ @@ -107,13 +109,14 @@ $$\begin{aligned} Recall that our partitioning into $A$ and $B$ was arbitrary, meaning that, in fact, the temperature $T$ must be uniform in the whole box. - We get this specific result because heat was the only thing that $A$ and $B$ could exchange. -The key point, however, -is that the total entropy $S$ must be maximized. -We also would have reached that conclusion if our imaginary wall -allowed changes in volume $V_A$ and particle count $N_A$. + +The point is that the most likely state of the box +maximizes the total entropy $S$. +We also would have reached that conclusion +if our imaginary wall was permeable and flexible, +i.e if it allowed changes in volume $V_A$ and particle count $N_A$. diff --git a/content/know/concept/thermodynamic-potential/index.pdc b/content/know/concept/thermodynamic-potential/index.pdc index b65891f..813f3b4 100644 --- a/content/know/concept/thermodynamic-potential/index.pdc +++ b/content/know/concept/thermodynamic-potential/index.pdc @@ -25,7 +25,7 @@ and is consequently assumed to be at the same temperature $T = T_{\mathrm{sur}}$ then $T$ must be a natural variable. The link from natural variables to potentials -is established by thermodynamic ensembles. +is established by [thermodynamic ensembles](/know/category/thermodynamic-ensembles/). Once enough natural variables have been found, the appropriate potential can be selected from the list below. |