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-rw-r--r--content/know/concept/fundamental-thermodynamic-relation/index.pdc59
-rw-r--r--content/know/concept/laws-of-thermodynamics/index.pdc109
-rw-r--r--content/know/concept/legendre-transform/index.pdc4
-rw-r--r--content/know/concept/maxwell-boltzmann-distribution/index.pdc1
-rw-r--r--content/know/concept/reduced-mass/index.pdc140
-rw-r--r--content/know/concept/thermodynamic-potential/index.pdc279
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diff --git a/content/know/concept/fundamental-thermodynamic-relation/index.pdc b/content/know/concept/fundamental-thermodynamic-relation/index.pdc
new file mode 100644
index 0000000..8f3a742
--- /dev/null
+++ b/content/know/concept/fundamental-thermodynamic-relation/index.pdc
@@ -0,0 +1,59 @@
+---
+title: "Fundamental thermodynamic relation"
+firstLetter: "F"
+publishDate: 2021-07-07
+categories:
+- Physics
+- Thermodynamics
+
+date: 2021-07-05T17:39:57+02:00
+draft: false
+markup: pandoc
+---
+
+# Fundamental thermodynamic relation
+
+The **fundamental thermodynamic relation** combines the first two
+[laws of thermodynamics](/know/concept/laws-of-thermodynamics/),
+and gives the change of the internal energy $U$,
+which is a [thermodynamic potential](/know/concept/thermodynamic-potential/),
+in terms of the change in
+entropy $S$, volume $V$, and the number of particles $N$.
+
+Starting from the first law of thermodynamics,
+we write an infinitesimal change in energy $\dd{U}$ as follows,
+where $T$ is the temperature and $P$ is the pressure:
+
+$$\begin{aligned}
+ \dd{U} &= \dd{Q} + \dd{W} = T \dd{S} - P \dd{V}
+\end{aligned}$$
+
+The term $T \dd{S}$ comes from the second law of thermodynamics,
+and represents the transfer of thermal energy,
+while $P \dd{V}$ represents physical work.
+
+However, we are missing a term, namely matter transfer.
+If particles can enter/leave the system (i.e. the population $N$ is variable),
+then each such particle costs an amount $\mu$ of energy,
+where $\mu$ is known as the **chemical potential**:
+
+$$\begin{aligned}
+ \dd{U} = T \dd{S} - P \dd{V} + \mu \dd{N}
+\end{aligned}$$
+
+To generalize even further, there may be multiple species of particle,
+which each have a chemical potential $\mu_i$.
+In that case, we sum over all species $i$:
+
+$$\begin{aligned}
+ \boxed{
+ \dd{U} = T \dd{S} - P \dd{V} + \sum_{i}^{} \mu_i \dd{N_i}
+ }
+\end{aligned}$$
+
+
+
+## References
+1. H. Gould, J. Tobochnik,
+ *Statistical and thermal physics*, 2nd edition,
+ Princeton.
diff --git a/content/know/concept/laws-of-thermodynamics/index.pdc b/content/know/concept/laws-of-thermodynamics/index.pdc
new file mode 100644
index 0000000..190f0fd
--- /dev/null
+++ b/content/know/concept/laws-of-thermodynamics/index.pdc
@@ -0,0 +1,109 @@
+---
+title: "Laws of thermodynamics"
+firstLetter: "L"
+publishDate: 2021-07-07
+categories:
+- Physics
+- Thermodynamics
+
+date: 2021-07-05T17:44:53+02:00
+draft: false
+markup: pandoc
+---
+
+# Laws of thermodynamics
+
+The **laws of thermodynamics** are of great importance
+to physics, chemistry and engineering,
+since they restrict what a device or process can physically achieve.
+For example, the impossibility of *perpetual motion*
+is a consequence of these laws.
+
+
+## First law
+
+The **first law of thermodynamics** states that energy is conserved.
+When a system goes from one equilibrium to another,
+the change $\Delta U$ of its energy $U$ is equal to
+the work $\Delta W$ done by external forces,
+plus the energy transferred by heating ($\Delta Q > 0$) or cooling ($\Delta Q < 0$):
+
+$$\begin{aligned}
+ \boxed{
+ \Delta U = \Delta W + \Delta Q
+ }
+\end{aligned}$$
+
+The internal energy $U$ is a state variable,
+so is independent of the path taken between equilibria.
+However, the work $\Delta W$ and heating $\Delta Q$ do depend on the path,
+so the first law means that
+the act of transferring energy is path-dependent,
+but the result has no "memory" of that path.
+
+
+## Second law
+
+The **second law of thermodynamics** states that
+the total entropy never decreases.
+An important consequence is that
+no machine can convert energy into work with 100% efficiency.
+
+It is possible for the local entropy $S_{\mathrm{loc}}$
+of a system to decrease, but doing so requires work,
+and therefore the entropy of the surroundings $S_{\mathrm{sur}}$
+must increase accordingly, such that:
+
+$$\begin{aligned}
+ \boxed{
+ \Delta S_{\mathrm{tot}} = \Delta S_{\mathrm{loc}} + \Delta S_{\mathrm{sur}} \ge 0
+ }
+\end{aligned}$$
+
+Since the total entropy never decreases,
+the equilibrium state of a system must be a maximum
+of its entropy $S$, and therefore $S$ can be used as
+a [thermodynamic "potential"](/know/concept/thermodynamic-potential/).
+
+The only situation where $\Delta S = 0$ is a reversible process,
+since then it must be possible to return to
+the previous equilibrium state by doing the same work in the opposite direction.
+
+According to the first law,
+if a process is reversible, or if it is only heating/cooling,
+then (after one reversible cycle) the energy change
+is simply the heat transfer $\dd{U} = \dd{Q}$.
+An entropy change $\dd{S}$ is then expressed as follows
+(since $\pdv*{S}{U} = 1 / T$ by definition):
+
+$$\begin{aligned}
+ \boxed{
+ \dd{S}
+ = \Big( \pdv{S}{U} \Big)_{V, N} \dd{U}
+ = \frac{\dd{Q}}{T}
+ }
+\end{aligned}$$
+
+Confusingly, this equation is sometimes also called the second law of thermodynamics.
+
+
+## Third law
+
+The **third law of thermodynamics** states that
+the entropy $S$ of a system goes to zero when the temperature reaches absolute zero:
+
+$$\begin{aligned}
+ \boxed{
+ \lim_{T \to 0} S = 0
+ }
+\end{aligned}$$
+
+From this, the absolute quantity of $S$ is defined, otherwise we would
+only be able to speak of entropy differences $\Delta S$.
+
+
+
+## References
+1. H. Gould, J. Tobochnik,
+ *Statistical and thermal physics*, 2nd edition,
+ Princeton.
diff --git a/content/know/concept/legendre-transform/index.pdc b/content/know/concept/legendre-transform/index.pdc
index 9cb6824..732c6b6 100644
--- a/content/know/concept/legendre-transform/index.pdc
+++ b/content/know/concept/legendre-transform/index.pdc
@@ -86,8 +86,8 @@ $$\begin{aligned}
Legendre transformation is important in physics,
since it connects [Lagrangian](/know/concept/lagrangian-mechanics/)
-and Hamiltonian mechanics to each other.
-It is also used to convert between thermodynamic potentials.
+and [Hamiltonian](/know/concept/hamiltonian-mechanics/) mechanics to each other.
+It is also used to convert between [thermodynamic potentials](/know/concept/thermodynamic-potential/).
diff --git a/content/know/concept/maxwell-boltzmann-distribution/index.pdc b/content/know/concept/maxwell-boltzmann-distribution/index.pdc
index 082d2db..38b56fd 100644
--- a/content/know/concept/maxwell-boltzmann-distribution/index.pdc
+++ b/content/know/concept/maxwell-boltzmann-distribution/index.pdc
@@ -5,6 +5,7 @@ publishDate: 2021-05-08
categories:
- Physics
- Statistics
+- Thermodynamics
date: 2021-05-08T18:35:37+02:00
draft: false
diff --git a/content/know/concept/reduced-mass/index.pdc b/content/know/concept/reduced-mass/index.pdc
new file mode 100644
index 0000000..feaddd6
--- /dev/null
+++ b/content/know/concept/reduced-mass/index.pdc
@@ -0,0 +1,140 @@
+---
+title: "Reduced mass"
+firstLetter: "R"
+publishDate: 2021-07-05
+categories:
+- Physics
+
+date: 2021-07-03T11:39:03+02:00
+draft: false
+markup: pandoc
+---
+
+# Reduced mass
+
+Problems with two interacting objects can be simplified
+by combining them into a pseudo-object with **reduced mass** $\mu$,
+whose position equals the relative position of the objects.
+For bodies 1 and 2 with respective masses $m_1$ and $m_2$:
+
+$$\begin{aligned}
+ \boxed{
+ \mu \equiv \frac{m_1 m_2}{m_1 + m_2}
+ }
+\end{aligned}$$
+
+If $\vec{x}_1$ and $\vec{x}_2$ are the objects' respective positions,
+then we define
+the relative position $\vec{x}_r$,
+the relative velocity $\vec{v}_r$,
+and the relative acceleration $\vec{a}_r$:
+
+$$\begin{aligned}
+ \vec{x}_r
+ \equiv \vec{x}_1 - \vec{x}_2
+ \qquad
+ \vec{v}_r
+ \equiv \vec{v}_1 - \vec{v}_2
+ = \dv{\vec{x}_r}{t}
+ \qquad \quad
+ \vec{a}_r
+ \equiv \vec{a}_1 - \vec{a}_2
+ = \dv[2]{\vec{x}_r}{t}
+\end{aligned}$$
+
+We now choose the coordinate system's origin
+to be the center of mass of both objects:
+
+$$\begin{aligned}
+ m_1 \vec{x}_1 + m_2 \vec{x}_2 = \vec{0}
+\end{aligned}$$
+
+Rearranging and differentiating then yields the following useful equations:
+
+$$\begin{aligned}
+ \vec{x}_2 = - \frac{m_1}{m_2} \vec{x}_1
+ \qquad \quad
+ \vec{v}_2 = - \frac{m_1}{m_2} \vec{v}_1
+ \qquad \quad
+ \vec{a}_2 = - \frac{m_1}{m_2} \vec{a}_1
+\end{aligned}$$
+
+Using these relations, we can rewrite the relative quantities we defined earlier:
+
+$$\begin{aligned}
+ \vec{x}_r
+ = \Big( 1 + \frac{m_1}{m_2} \Big) \vec{x}_1
+ = \frac{m_1 + m_2}{m_2} \vec{x}_1
+ \qquad
+ \vec{v}_r
+ = \frac{m_1 + m_2}{m_2} \vec{v}_1
+ \qquad
+ \vec{a}_r
+ = \frac{m_1 + m_2}{m_2} \vec{a}_1
+\end{aligned}$$
+
+Meanwhile, Newton's third law states that
+if object 1 experiences a force $\vec{F}_1 = m_1 \vec{a}_1$ caused by object 2,
+then object 2 experiences an opposite and equal force $\vec{F}_2 = - \vec{F}_1$.
+In fact, our earlier relation between $\vec{a}_1$ and $\vec{a}_1$
+boils down to Newton's third law:
+
+$$\begin{aligned}
+ \vec{F}_2 = m_2 \vec{a}_2 = - m_1 \vec{a}_1 = - \vec{F}_1
+ \quad \implies \quad
+ \vec{a}_2 = - \frac{m_1}{m_2} \vec{a}_1
+\end{aligned}$$
+
+With all that in mind, let us take a closer look at the relative acceleration $\vec{a}_r$:
+
+$$\begin{aligned}
+ \vec{a}_r
+ = \frac{m_1 + m_2}{m_2} \Big( \frac{m_1}{m_1} \Big) \vec{a}_1
+ = \frac{m_1 + m_2}{m_1 m_2} \big( m_1 \vec{a}_1 \big)
+ = \frac{\vec{F}_1}{\mu}
+ = - \frac{\vec{F}_2}{\mu}
+\end{aligned}$$
+
+Where $\mu$ is the reduced mass, as defined above.
+In other words, the relative acceleration $\vec{a}_r$
+is just $\vec{a}_1 = \vec{F}_1 / m_1$ multiplied by $m_1 / \mu$.
+This can be regarded as focusing on the dynamics of body 1,
+while correcting for the effects of body 2.
+
+This also suggests the following way
+to recover the original positions $\vec{x}_1$ and $\vec{x}_2$
+from $\vec{x}_r$, which you can easily verify for yourself:
+
+$$\begin{aligned}
+ \vec{x}_1
+ = \frac{\mu}{m_1} \vec{x}_r
+ = \frac{m_2}{m_1 + m_2} \vec{x}_r
+ \qquad \quad
+ \vec{x}_2
+ = \frac{\mu}{m_2} \vec{x}_r
+ = - \frac{m_1}{m_1 + m_2} \vec{x}_r
+\end{aligned}$$
+
+With this, we can rewrite the total kinetic energy $T$ in an elegant way:
+
+$$\begin{aligned}
+ T
+ &= \frac{1}{2} m_1 \vec{v}_1^2 + \frac{1}{2} m_2 \vec{v}_2^2
+ = \frac{1}{2} m_1 \Big( \frac{\mu}{m_1} \vec{v}_r \Big)^2 + \frac{1}{2} m_2 \Big( \frac{\mu}{m_2} \vec{v}_r \Big)^2
+ \\
+ &= \frac{1}{2} \frac{\mu^2}{m_1} \vec{v}_r^2 + \frac{1}{2} \frac{\mu^2}{m_2} \vec{v}_r^2
+ = \frac{1}{2} \Big( \frac{m_2 \mu^2}{m_1 m_2} + \frac{m_1 \mu^2}{m_1 m_2} \Big) \vec{v}_r^2
+ \\
+ &= \frac{1}{2} \frac{(m_1 + m_2) \mu^2}{m_1 m_2} \vec{v}_r^2
+ = \frac{1}{2} \frac{\mu^2}{\mu} \vec{v}_r^2
+ = \frac{1}{2} \mu \vec{v}_r^2
+\end{aligned}$$
+
+Then, assuming that the system's potential energy $V$
+only depends on the distance between the two objects,
+i.e. $V = V(|\vec{x}_1 - \vec{x}_2|) = V(|\vec{x}_r|)$,
+we just showed that we can rewrite both $T$ and $V$
+to contain only $\mu$ and relative quantities.
+This is relevant for both [Lagrangian mechanics](/know/concept/lagrangian-mechanics/)
+and [Hamiltonian mechanics](/know/concept/hamiltonian-mechanics/),
+where $L = T - V$ and $H = T + V$ respectively.
diff --git a/content/know/concept/thermodynamic-potential/index.pdc b/content/know/concept/thermodynamic-potential/index.pdc
new file mode 100644
index 0000000..5d154d5
--- /dev/null
+++ b/content/know/concept/thermodynamic-potential/index.pdc
@@ -0,0 +1,279 @@
+---
+title: "Thermodynamic potential"
+firstLetter: "T"
+publishDate: 2021-07-07
+categories:
+- Physics
+- Thermodynamics
+
+date: 2021-07-03T14:40:22+02:00
+draft: false
+markup: pandoc
+---
+
+# Thermodynamic potential
+
+**Thermodynamic potentials** are state functions
+whose minima or maxima represent equilibrium states of a system.
+Such functions are either energies (hence *potential*) or entropies.
+
+Which potential (of many) decides the equilibrium states for a given system?
+That depends which variables are assumed to already be in automatic equilibrium.
+Such variables are known as the **natural variables** of that potential.
+For example, if a system can freely exchange heat with its surroundings,
+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.
+
+Once enough natural variables have been found,
+the appropriate potential can be selected from the list below.
+All non-natural variables can then be calculated
+by taking partial derivatives of the potential
+with respect to the natural variables.
+
+Mathematically, the potentials are related to each other
+by [Legendre transformation](/know/concept/legendre-transform/).
+
+
+## Internal energy
+
+The **internal energy** $U$ represents
+the capacity to do both mechanical and non-mechanical work,
+and to release heat.
+It is simply the integral
+of the [fundamental thermodynamic relation](/know/concept/fundamental-thermodynamic-relation/):
+
+$$\begin{aligned}
+ \boxed{
+ U(S, V, N) \equiv T S - P V + \mu N
+ }
+\end{aligned}$$
+
+It is a function of the entropy $S$, volume $V$, and particle count $N$:
+these are its natural variables.
+An infinitesimal change $\dd{U}$ is as follows:
+
+$$\begin{aligned}
+ \boxed{
+ \dd{U} = T \dd{S} - P \dd{V} + \mu \dd{N}
+ }
+\end{aligned}$$
+
+The non-natural variables are
+temperature $T$, pressure $P$, and chemical potential $\mu$.
+They can be recovered by differentiating $U$
+with respect to the natural variables $S$, $V$, and $N$:
+
+$$\begin{aligned}
+ \boxed{
+ T = \Big( \pdv{U}{S} \Big)_{V,N}
+ \qquad
+ P = - \Big( \pdv{U}{V} \Big)_{S,N}
+ \qquad
+ \mu = \Big( \pdv{U}{N} \Big)_{S,V}
+ }
+\end{aligned}$$
+
+It is convention to write those subscripts,
+to help keep track of which function depends on which variables.
+They are meaningless; these are normal partial derivatives.
+
+
+## Enthalpy
+
+The **enthalpy** $H$ of a system, in units of energy,
+represents its capacity to do non-mechanical work,
+plus its capacity to release heat.
+It is given by:
+
+$$\begin{aligned}
+ \boxed{
+ H(S, P, N) \equiv U + P V
+ }
+\end{aligned}$$
+
+It is a function of the entropy $S$, pressure $P$, and particle count $N$:
+these are its natural variables.
+An infinitesimal change $\dd{H}$ is as follows:
+
+$$\begin{aligned}
+ \boxed{
+ \dd{H} = T \dd{S} + V \dd{P} + \mu \dd{N}
+ }
+\end{aligned}$$
+
+The non-natural variables are
+temperature $T$, volume $V$, and chemical potential $\mu$.
+They can be recovered by differentiating $H$
+with respect to the natural variables $S$, $P$, and $N$:
+
+$$\begin{aligned}
+ \boxed{
+ T = \Big( \pdv{H}{S} \Big)_{P,N}
+ \qquad
+ V = \Big( \pdv{H}{P} \Big)_{S,N}
+ \qquad
+ \mu = \Big( \pdv{H}{N} \Big)_{S,P}
+ }
+\end{aligned}$$
+
+
+## Helmholtz free energy
+
+The **Helmholtz free energy** $F$ represents
+the capacity of a system to
+do both mechanical and non-mechanical work,
+and is given by:
+
+$$\begin{aligned}
+ \boxed{
+ F(T, V, N) \equiv U - T S
+ }
+\end{aligned}$$
+
+It depends on the temperature $T$, volume $V$, and particle count $N$:
+these are natural variables.
+An infinitesimal change $\dd{H}$ is as follows:
+
+$$\begin{aligned}
+ \boxed{
+ \dd{F} = - P \dd{V} - S \dd{T} + \mu \dd{N}
+ }
+\end{aligned}$$
+
+The non-natural variables are
+entropy $S$, pressure $P$, and chemical potential $\mu$.
+They can be recovered by differentiating $F$
+with respect to the natural variables $T$, $V$, and $N$:
+
+$$\begin{aligned}
+ \boxed{
+ S = - \Big( \pdv{F}{T} \Big)_{V,N}
+ \qquad
+ P = - \Big( \pdv{F}{V} \Big)_{T,N}
+ \qquad
+ \mu = \Big( \pdv{F}{N} \Big)_{T,V}
+ }
+\end{aligned}$$
+
+
+## Gibbs free energy
+
+The **Gibbs free energy** $G$ represents
+the capacity of a system to do non-mechanical work:
+
+$$\begin{aligned}
+ \boxed{
+ G(T, P, N)
+ \equiv U + P V - T S
+ }
+\end{aligned}$$
+
+It depends on the temperature $T$, pressure $P$, and particle count $N$:
+they are natural variables.
+An infinitesimal change $\dd{G}$ is as follows:
+
+$$\begin{aligned}
+ \boxed{
+ \dd{G} = V \dd{P} - S \dd{T} + \mu \dd{N}
+ }
+\end{aligned}$$
+
+The non-natural variables are
+entropy $S$, volume $V$, and chemical potential $\mu$.
+These can be recovered by differentiating $G$
+with respect to the natural variables $T$, $P$, and $N$:
+
+$$\begin{aligned}
+ \boxed{
+ S = - \Big( \pdv{G}{T} \Big)_{P,N}
+ \qquad
+ V = \Big( \pdv{G}{P} \Big)_{T,N}
+ \qquad
+ \mu = \Big( \pdv{G}{N} \Big)_{T,P}
+ }
+\end{aligned}$$
+
+
+## Landau potential
+
+The **Landau potential** or **grand potential** $\Omega$, in units of energy,
+represents the capacity of a system to do mechanical work,
+and is given by:
+
+$$\begin{aligned}
+ \boxed{
+ \Omega(T, V, \mu) \equiv U - T S - \mu N
+ }
+\end{aligned}$$
+
+It depends on temperature $T$, volume $V$, and chemical potential $\mu$:
+these are natural variables.
+An infinitesimal change $\dd{\Omega}$ is as follows:
+
+$$\begin{aligned}
+ \boxed{
+ \dd{\Omega} = - P \dd{V} - S \dd{T} - N \dd{\mu}
+ }
+\end{aligned}$$
+
+The non-natural variables are
+entropy $S$, pressure $P$, and particle count $N$.
+These can be recovered by differentiating $\Omega$
+with respect to the natural variables $T$, $V$, and $\mu$:
+
+$$\begin{aligned}
+ \boxed{
+ S = \Big( \pdv{\Omega}{T} \Big)_{V,\mu}
+ \qquad
+ P = - \Big( \pdv{\Omega}{V} \Big)_{T,\mu}
+ \qquad
+ N = - \Big( \pdv{\Omega}{\mu} \Big)_{T,V}
+ }
+\end{aligned}$$
+
+
+## Entropy
+
+The **entropy** $S$, in units of energy over temperature,
+is an odd duck, but nevertheless used as a thermodynamic potential.
+It is given by:
+
+$$\begin{aligned}
+ \boxed{
+ S(U, V, N) \equiv \frac{1}{T} U + \frac{P}{T} V - \frac{\mu}{T} N
+ }
+\end{aligned}$$
+
+It depends on the internal energy $U$, volume $V$, and particle count $N$:
+they are natural variables.
+An infinitesimal change $\dd{S}$ is as follows:
+
+$$\begin{aligned}
+ \boxed{
+ \dd{S} = \frac{1}{T} \dd{U} + \frac{P}{T} \dd{V} - \frac{\mu}{T} \dd{N}
+ }
+\end{aligned}$$
+
+The non-natural variables are $1/T$, $P/T$, and $\mu/T$.
+These can be recovered by differentiating $S$
+with respect to the natural variables $U$, $V$, and $N$:
+
+$$\begin{aligned}
+ \boxed{
+ \frac{1}{T} = \Big( \pdv{S}{U} \Big)_{V,N}
+ \qquad
+ \frac{P}{T} = \Big( \pdv{S}{V} \Big)_{U,N}
+ \qquad
+ \frac{\mu}{T} = - \Big( \pdv{S}{N} \Big)_{U,V}
+ }
+\end{aligned}$$
+
+
+
+## References
+1. H. Gould, J. Tobochnik,
+ *Statistical and thermal physics*, 2nd edition,
+ Princeton.