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| -rw-r--r-- | source/know/concept/canonical-ensemble/index.md | 2 | ||||
| -rw-r--r-- | source/know/concept/electromagnetic-wave-equation/index.md | 347 | ||||
| -rw-r--r-- | source/know/concept/euler-equations/index.md | 2 | ||||
| -rw-r--r-- | source/know/concept/fundamental-relation-of-thermodynamics/index.md | 326 | ||||
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| -rw-r--r-- | source/know/concept/laws-of-thermodynamics/index.md | 104 | ||||
| -rw-r--r-- | source/know/concept/martingale/index.md | 2 | ||||
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| -rw-r--r-- | source/know/concept/triple-product-rule/index.md | 97 |
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@@ -3,6 +3,7 @@ source "https://rubygems.org" gem "jekyll" gem "webrick" gem "json" +gem "erb" # https://github.com/rubyjs/mini_racer gem "mini_racer" @@ -14,9 +14,13 @@ all its extensions needed by this website as listed in the `Gemfile`: ```sh $ bundle config set --local path .bundle +$ bundle config build.posix-spawn --with-cflags="-Wno-incompatible-pointer-types" $ bundle install ``` +First try to run `bundle install` without changing any `CFLAGS`, +it used to work like that, but not anymore as of mid-2024. + Now you should be able to run a local development version with: ```sh diff --git a/_config.yml b/_config.yml index 5023f33..250ed0c 100644 --- a/_config.yml +++ b/_config.yml @@ -16,7 +16,7 @@ kramdown: macros: "\\Real": "\\mathop{\\mathrm{Re}}" "\\Imag": "\\mathop{\\mathrm{Im}}" - "\\pv": "\\:\\mathop{\\mathcal{P}}" + "\\pv": "\\:\\mathcal{P}\\!" "\\cross": "\\times" "\\va": "\\vec{\\mathbf{#1}}" "\\vb": "\\mathbf{#1}" diff --git a/source/know/concept/canonical-ensemble/index.md b/source/know/concept/canonical-ensemble/index.md index 8a96e91..da7d436 100644 --- a/source/know/concept/canonical-ensemble/index.md +++ b/source/know/concept/canonical-ensemble/index.md @@ -178,7 +178,7 @@ $$\begin{aligned} \end{aligned}$$ Rearranging and substituting -the [fundamental thermodynamic relation](/know/concept/fundamental-thermodynamic-relation/) +the [fundamental thermodynamic relation](/know/concept/fundamental-relation-of-thermodynamics/) then gives: $$\begin{aligned} diff --git a/source/know/concept/electromagnetic-wave-equation/index.md b/source/know/concept/electromagnetic-wave-equation/index.md index a27fe6f..559d943 100644 --- a/source/know/concept/electromagnetic-wave-equation/index.md +++ b/source/know/concept/electromagnetic-wave-equation/index.md @@ -1,7 +1,7 @@ --- title: "Electromagnetic wave equation" sort_title: "Electromagnetic wave equation" -date: 2021-09-09 +date: 2024-09-08 # Originally 2021-09-09, major rewrite categories: - Physics - Electromagnetism @@ -9,236 +9,281 @@ categories: layout: "concept" --- -The electromagnetic wave equation describes -the propagation of light through various media. -Since an electromagnetic (light) wave consists of +Light, i.e. **electromagnetic waves**, consist of an [electric field](/know/concept/electric-field/) and a [magnetic field](/know/concept/magnetic-field/), -we need [Maxwell's equations](/know/concept/maxwells-equations/) -in order to derive the wave equation. +one inducing the other and vice versa. +The existence and classical behavior of such waves +can be derived using only [Maxwell's equations](/know/concept/maxwells-equations/), +as we will demonstrate here. - -## Uniform medium - -We will use all of Maxwell's equations, -but we start with Ampère's circuital law for the "free" fields $$\vb{H}$$ and $$\vb{D}$$, -in the absence of a free current $$\vb{J}_\mathrm{free} = 0$$: - -$$\begin{aligned} - \nabla \cross \vb{H} - = \pdv{\vb{D}}{t} -\end{aligned}$$ - -We assume that the medium is isotropic, linear, -and uniform in all of space, such that: +We start from Faraday's law of induction, +where we assume that the system consists of materials +with well-known (linear) relative magnetic permeabilities $$\mu_r(\vb{r})$$, +such that $$\vb{B} = \mu_0 \mu_r \vb{H}$$: $$\begin{aligned} - \vb{D} = \varepsilon_0 \varepsilon_r \vb{E} - \qquad \quad - \vb{H} = \frac{1}{\mu_0 \mu_r} \vb{B} + \nabla \cross \vb{E} + = - \pdv{\vb{B}}{t} + = - \mu_0 \mu_r \pdv{\vb{H}}{t} \end{aligned}$$ -Which, upon insertion into Ampère's law, -yields an equation relating $$\vb{B}$$ and $$\vb{E}$$. -This may seem to contradict Ampère's "total" law, -but keep in mind that $$\vb{J}_\mathrm{bound} \neq 0$$ here: +We move $$\mu_r(\vb{r})$$ to the other side, +take the curl, and insert Ampère's circuital law: $$\begin{aligned} - \nabla \cross \vb{B} - = \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdv{\vb{E}}{t} + \nabla \cross \bigg( \frac{1}{\mu_r} \nabla \cross \vb{E} \bigg) + &= - \mu_0 \pdv{}{t} \big( \nabla \cross \vb{H} \big) + \\ + &= - \mu_0 \bigg( \pdv{\vb{J}_\mathrm{free}}{t} + \pdvn{2}{\vb{D}}{t} \bigg) \end{aligned}$$ -Now we take the curl, rearrange, -and substitute $$\nabla \cross \vb{E}$$ according to Faraday's law: +For simplicity, we only consider insulating materials, +since light propagation in conductors is a complex beast. +We thus assume that there are no free currents $$\vb{J}_\mathrm{free} = 0$$, leaving: $$\begin{aligned} - \nabla \cross (\nabla \cross \vb{B}) - = \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdv{}{t}(\nabla \cross \vb{E}) - = - \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{B}}{t} + \nabla \cross \bigg( \frac{1}{\mu_r} \nabla \cross \vb{E} \bigg) + &= - \mu_0 \pdvn{2}{\vb{D}}{t} \end{aligned}$$ -Using a vector identity, we rewrite the leftmost expression, -which can then be reduced thanks to Gauss' law for magnetism $$\nabla \cdot \vb{B} = 0$$: +Having $$\vb{E}$$ and $$\vb{D}$$ in the same equation is not ideal, +so we should make a choice: +do we restrict ourselves to linear media +(so $$\vb{D} = \varepsilon_0 \varepsilon_r \vb{E}$$), +or do we allow materials with more complicated responses +(so $$\vb{D} = \varepsilon_0 \vb{E} + \vb{P}$$, with $$\vb{P}$$ unspecified)? +The former is usually sufficient: $$\begin{aligned} - - \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{B}}{t} - &= \nabla (\nabla \cdot \vb{B}) - \nabla^2 \vb{B} - = - \nabla^2 \vb{B} + \boxed{ + \nabla \cross \bigg( \frac{1}{\mu_r} \nabla \cross \vb{E} \bigg) + = - \mu_0 \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{E}}{t} + } \end{aligned}$$ -This describes $$\vb{B}$$. -Next, we repeat the process for $$\vb{E}$$: -taking the curl of Faraday's law yields: +This is the general linear form of the **electromagnetic wave equation**, +where $$\mu_r$$ and $$\varepsilon_r$$ +both depend on $$\vb{r}$$ in order to describe the structure of the system. +We can obtain a similar equation for $$\vb{H}$$, +by starting from Ampère's law under the same assumptions: $$\begin{aligned} - \nabla \cross (\nabla \cross \vb{E}) - = - \pdv{}{t}(\nabla \cross \vb{B}) - = - \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{E}}{t} + \nabla \cross \vb{H} + = \pdv{\vb{D}}{t} + = \varepsilon_0 \varepsilon_r \pdv{\vb{E}}{t} \end{aligned}$$ -Which can be rewritten using same vector identity as before, -and then reduced by assuming that there is no net charge density $$\rho = 0$$ -in Gauss' law, such that $$\nabla \cdot \vb{E} = 0$$: +Taking the curl and substituting Faraday's law on the right yields: $$\begin{aligned} - - \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{E}}{t} - &= \nabla (\nabla \cdot \vb{E}) - \nabla^2 \vb{E} - = - \nabla^2 \vb{E} + \nabla \cross \bigg( \frac{1}{\varepsilon_r} \nabla \cross \vb{H} \bigg) + &= \varepsilon_0 \pdv{}{t} \big( \nabla \cross \vb{E} \big) + = - \varepsilon_0 \pdvn{2}{\vb{B}}{t} \end{aligned}$$ -We thus arrive at the following two (implicitly coupled) -wave equations for $$\vb{E}$$ and $$\vb{B}$$, -where we have defined the phase velocity $$v \equiv 1 / \sqrt{\mu_0 \mu_r \varepsilon_0 \varepsilon_r}$$: +And then we insert $$\vb{B} = \mu_0 \mu_r \vb{H}$$ to get the analogous +electromagnetic wave equation for $$\vb{H}$$: $$\begin{aligned} \boxed{ - \pdvn{2}{\vb{E}}{t} - \frac{1}{v^2} \nabla^2 \vb{E} - = 0 - } - \qquad \quad - \boxed{ - \pdvn{2}{\vb{B}}{t} - \frac{1}{v^2} \nabla^2 \vb{B} - = 0 + \nabla \cross \bigg( \frac{1}{\varepsilon_r} \nabla \cross \vb{H} \bigg) + = - \mu_0 \varepsilon_0 \mu_r \pdvn{2}{\vb{H}}{t} } \end{aligned}$$ -Traditionally, it is said that the solutions are as follows, -where the wavenumber $$|\vb{k}| = \omega / v$$: - -$$\begin{aligned} - \vb{E}(\vb{r}, t) - &= \vb{E}_0 \exp(i \vb{k} \cdot \vb{r} - i \omega t) - \\ - \vb{B}(\vb{r}, t) - &= \vb{B}_0 \exp(i \vb{k} \cdot \vb{r} - i \omega t) -\end{aligned}$$ - -In fact, thanks to linearity, these **plane waves** can be treated as -terms in a Fourier series, meaning that virtually -*any* function $$f(\vb{k} \cdot \vb{r} - \omega t)$$ is a valid solution. +This is equivalent to the problem for $$\vb{E}$$, +since they are coupled by Maxwell's equations. +By solving either, subject to Gauss's laws +$$\nabla \cdot (\varepsilon_r \vb{E}) = 0$$ and $$\nabla \cdot (\mu_r \vb{H}) = 0$$, +the behavior of light in a given system can be deduced. +Note that Gauss's laws enforce that the wave's fields are transverse, +i.e. they must be perpendicular to the propagation direction. -Keep in mind that in reality $$\vb{E}$$ and $$\vb{B}$$ are real, -so although it is mathematically convenient to use plane waves, -in the end you will need to take the real part. -## Non-uniform medium +## Homogeneous linear media -A useful generalization is to allow spatial change -in the relative permittivity $$\varepsilon_r(\vb{r})$$ -and the relative permeability $$\mu_r(\vb{r})$$. -We still assume that the medium is linear and isotropic, so: +In the special case where the medium is completely uniform, +$$\mu_r$$ and $$\varepsilon_r$$ no longer depend on $$\vb{r}$$, +so they can be moved to the other side: $$\begin{aligned} - \vb{D} - = \varepsilon_0 \varepsilon_r(\vb{r}) \vb{E} - \qquad \quad - \vb{B} - = \mu_0 \mu_r(\vb{r}) \vb{H} + \nabla \cross \big( \nabla \cross \vb{E} \big) + &= - \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{E}}{t} + \\ + \nabla \cross \big( \nabla \cross \vb{H} \big) + &= - \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{H}}{t} \end{aligned}$$ -Inserting these expressions into Faraday's and Ampère's laws -respectively yields: +This can be rewritten using the vector identity +$$\nabla \cross (\nabla \cross \vb{V}) = \nabla (\nabla \cdot \vb{V}) - \nabla^2 \vb{V}$$: $$\begin{aligned} - \nabla \cross \vb{E} - = - \mu_0 \mu_r(\vb{r}) \pdv{\vb{H}}{t} - \qquad \quad - \nabla \cross \vb{H} - = \varepsilon_0 \varepsilon_r(\vb{r}) \pdv{\vb{E}}{t} + \nabla (\nabla \cdot \vb{E}) - \nabla^2 \vb{E} + &= - \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{E}}{t} + \\ + \nabla (\nabla \cdot \vb{H}) - \nabla^2 \vb{H} + &= - \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{H}}{t} \end{aligned}$$ -We then divide Ampère's law by $$\varepsilon_r(\vb{r})$$, -take the curl, and substitute Faraday's law, giving: +Which can be reduced using Gauss's laws +$$\nabla \cdot \vb{E} = 0$$ and $$\nabla \cdot \vb{H} = 0$$ +thanks to the fact that $$\varepsilon_r$$ and $$\mu_r$$ are constants in this case. +We therefore arrive at: $$\begin{aligned} - \nabla \cross \Big( \frac{1}{\varepsilon_r} \nabla \cross \vb{H} \Big) - = \varepsilon_0 \pdv{}{t}(\nabla \cross \vb{E}) - = - \mu_0 \mu_r \varepsilon_0 \pdvn{2}{\vb{H}}{t} + \boxed{ + \nabla^2 \vb{E} - \frac{n^2}{c^2} \pdvn{2}{\vb{E}}{t} + = 0 + } \end{aligned}$$ -Next, we exploit linearity by decomposing $$\vb{H}$$ and $$\vb{E}$$ -into Fourier series, with terms given by: - $$\begin{aligned} - \vb{H}(\vb{r}, t) - = \vb{H}(\vb{r}) \exp(- i \omega t) - \qquad \quad - \vb{E}(\vb{r}, t) - = \vb{E}(\vb{r}) \exp(- i \omega t) + \boxed{ + \nabla^2 \vb{H} - \frac{n^2}{c^2} \pdvn{2}{\vb{H}}{t} + = 0 + } \end{aligned}$$ -By inserting this ansatz into the equation, -we can remove the explicit time dependence: +Where $$c = 1 / \sqrt{\mu_0 \varepsilon_0}$$ is the speed of light in a vacuum, +and $$n = \sqrt{\mu_0 \varepsilon_0}$$ is the refractive index of the medium. +Note that most authors write the magnetic equation with $$\vb{B}$$ instead of $$\vb{H}$$; +both are correct thanks to linearity. + +In a vacuum, where $$n = 1$$, these equations are sometimes written as +$$\square \vb{E} = 0$$ and $$\square \vb{H} = 0$$, +where $$\square$$ is the **d'Alembert operator**, defined as follows: $$\begin{aligned} - \nabla \cross \Big( \frac{1}{\varepsilon_r} \nabla \cross \vb{H} \Big) \exp(- i \omega t) - = \mu_0 \varepsilon_0 \omega^2 \mu_r \vb{H} \exp(- i \omega t) + \boxed{ + \square + \equiv \nabla^2 - \frac{1}{c^2} \pdvn{2}{}{t} + } \end{aligned}$$ -Dividing out $$\exp(- i \omega t)$$, -we arrive at an eigenvalue problem for $$\omega^2$$, -with $$c = 1 / \sqrt{\mu_0 \varepsilon_0}$$: +Note that some authors define it with the opposite sign. +In any case, the d'Alembert operator is important for special relativity. + +The solution to the homogeneous electromagnetic wave equation +are traditionally said to be the so-called **plane waves** given by: $$\begin{aligned} - \boxed{ - \nabla \cross \Big( \frac{1}{\varepsilon_r(\vb{r})} \nabla \cross \vb{H}(\vb{r}) \Big) - = \Big( \frac{\omega}{c} \Big)^2 \mu_r(\vb{r}) \vb{H}(\vb{r}) - } + \vb{E}(\vb{r}, t) + &= \vb{E}_0 e^{i \vb{k} \cdot \vb{r} - i \omega t} + \\ + \vb{B}(\vb{r}, t) + &= \vb{B}_0 e^{i \vb{k} \cdot \vb{r} - i \omega t} \end{aligned}$$ -Compared to a uniform medium, $$\omega$$ is often not arbitrary here: -there are discrete eigenvalues $$\omega$$, -corresponding to discrete **modes** $$\vb{H}(\vb{r})$$. +Where the wavevector $$\vb{k}$$ is arbitrary, +and the angular frequency $$\omega = c |\vb{k}| / n$$. +We also often talk about the wavelength, which is $$\lambda = 2 \pi / |\vb{k}|$$. +The appearance of $$\vb{k}$$ in the exponent +tells us that these waves are propagating through space, +as you would expect. + +In fact, because the wave equations are linear, +any superposition of plane waves, +i.e. any function of the form $$f(\vb{k} \cdot \vb{r} - \omega t)$$, +is in fact a valid solution. +Just remember that $$\vb{E}$$ and $$\vb{H}$$ are real-valued, +so it may be necessary to take the real part at the end of a calculation. -Next, we go through the same process to find an equation for $$\vb{E}$$. -Starting from Faraday's law, we divide by $$\mu_r(\vb{r})$$, -take the curl, and insert Ampère's law: + + +## Inhomogeneous linear media + +But suppose the medium is not uniform, i.e. it contains structures +described by $$\varepsilon_r(\vb{r})$$ and $$\mu_r(\vb{r})$$. +If the structures are much larger than the light's wavelength, +the homogeneous equation is still a very good approximation +away from any material boundaries; +anywhere else, however, they will break down. +Recall the general equations from before we assumed homogeneity: $$\begin{aligned} - \nabla \cross \Big( \frac{1}{\mu_r} \nabla \cross \vb{E} \Big) - = - \mu_0 \pdv{}{t}(\nabla \cross \vb{H}) - = - \mu_0 \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{E}}{t} + \nabla \cross \bigg( \frac{1}{\mu_r} \nabla \cross \vb{E} \bigg) + &= - \frac{\varepsilon_r}{c^2} \pdvn{2}{\vb{E}}{t} + \\ + \nabla \cross \bigg( \frac{1}{\varepsilon_r} \nabla \cross \vb{H} \bigg) + &= - \frac{\mu_r}{c^2} \pdvn{2}{\vb{H}}{t} \end{aligned}$$ -Then, by replacing $$\vb{E}(\vb{r}, t)$$ with our plane-wave ansatz, -we remove the time dependence: +In theory, this is everything we need, +but in most cases a better approach is possible: +the trick is that we only rarely need to explicitly calculate +the $$t$$-dependence of $$\vb{E}$$ or $$\vb{H}$$. +Instead, we can first solve an easier time-independent version +of this problem, and then approximate the dynamics +with [coupled mode theory](/know/concept/coupled-mode-theory/) later. + +To eliminate $$t$$, we make an ansatz for $$\vb{E}$$ and $$\vb{H}$$, shown below. +No generality is lost by doing this; +this is effectively a kind of [Fourier transform](/know/concept/fourier-transform/): $$\begin{aligned} - \nabla \cross \Big( \frac{1}{\mu_r} \nabla \cross \vb{E} \Big) \exp(- i \omega t) - = - \mu_0 \varepsilon_0 \omega^2 \varepsilon_r \vb{E} \exp(- i \omega t) + \vb{E}(\vb{r}, t) + &= \vb{E}(\vb{r}) e^{- i \omega t} + \\ + \vb{H}(\vb{r}, t) + &= \vb{H}(\vb{r}) e^{- i \omega t} \end{aligned}$$ -Which, after dividing out $$\exp(- i \omega t)$$, -yields an analogous eigenvalue problem with $$\vb{E}(r)$$: +Inserting this ansatz and dividing out $$e^{-i \omega t}$$ +yields the time-independent forms: $$\begin{aligned} \boxed{ - \nabla \cross \Big( \frac{1}{\mu_r(\vb{r})} \nabla \cross \vb{E}(\vb{r}) \Big) - = \Big( \frac{\omega}{c} \Big)^2 \varepsilon_r(\vb{r}) \vb{E}(\vb{r}) + \nabla \cross \bigg( \frac{1}{\mu_r} \nabla \cross \vb{E} \bigg) + = \Big( \frac{\omega}{c} \Big)^2 \varepsilon_r \vb{E} } \end{aligned}$$ -Usually, it is a reasonable approximation -to say $$\mu_r(\vb{r}) = 1$$, -in which case the equation for $$\vb{H}(\vb{r})$$ -becomes a Hermitian eigenvalue problem, -and is thus easier to solve than for $$\vb{E}(\vb{r})$$. +$$\begin{aligned} + \boxed{ + \nabla \cross \bigg( \frac{1}{\varepsilon_r} \nabla \cross \vb{H} \bigg) + = \Big( \frac{\omega}{c} \Big)^2 \mu_r \vb{H} + } +\end{aligned}$$ -Keep in mind, however, that in any case, -the solutions $$\vb{H}(\vb{r})$$ and/or $$\vb{E}(\vb{r})$$ -must satisfy the two Maxwell's equations that were not explicitly used: +These are eigenvalue problems for $$\omega^2$$, +which can be solved subject to Gauss's laws and suitable boundary conditions. +The resulting allowed values of $$\omega$$ may consist of +continuous ranges and/or discrete resonances, +analogous to *scattering* and *bound* quantum states, respectively. +It can be shown that the operators on both sides of each equation +are Hermitian, meaning these are well-behaved problems +yielding real eigenvalues and orthogonal eigenfields. + +Both equations are still equivalent: +we only need to solve one. But which one? +In practice, one is usually easier than the other, +due to the common approximation that $$\mu_r \approx 1$$ for many dielectric materials, +in which case the equations reduce to: $$\begin{aligned} - \nabla \cdot (\varepsilon_r \vb{E}) = 0 - \qquad \quad - \nabla \cdot (\mu_r \vb{H}) = 0 + \nabla \cross \big( \nabla \cross \vb{E} \big) + &= \Big( \frac{\omega}{c} \Big)^2 \varepsilon_r \vb{E} + \\ + \nabla \cross \bigg( \frac{1}{\varepsilon_r} \nabla \cross \vb{H} \bigg) + &= \Big( \frac{\omega}{c} \Big)^2 \vb{H} \end{aligned}$$ -This is equivalent to demanding that the resulting waves are *transverse*, -or in other words, -the wavevector $$\vb{k}$$ must be perpendicular to -the amplitudes $$\vb{H}_0$$ and $$\vb{E}_0$$. +Now the equation for $$\vb{H}$$ is starting to look simpler, +because it only has an operator on *one* side. +We could "fix" the equation for $$\vb{E}$$ by dividing it by $$\varepsilon_r$$, +but the resulting operator would no longer be Hermitian, +and hence not well-behaved. +To get an idea of how to handle $$\varepsilon_r$$ in the $$\vb{E}$$-equation, +notice its similarity to the weight function $$w$$ +in [Sturm-Liouville theory](/know/concept/sturm-liouville-theory/). + +Gauss's magnetic law $$\nabla \cdot \vb{H} = 0$$ +is also significantly easier for numerical calculations +than its electric counterpart $$\nabla \cdot (\varepsilon_r \vb{E}) = 0$$, +so we usually prefer to solve the equation for $$\vb{H}$$. + ## References diff --git a/source/know/concept/euler-equations/index.md b/source/know/concept/euler-equations/index.md index 2654d2b..415e2f1 100644 --- a/source/know/concept/euler-equations/index.md +++ b/source/know/concept/euler-equations/index.md @@ -146,7 +146,7 @@ When the fluid gets compressed in a certain location, thermodynamics states that the pressure, temperature and/or entropy must increase there. For simplicity, let us assume an *isothermal* and *isentropic* fluid, such that only $$p$$ is affected by compression, and the -[fundamental thermodynamic relation](/know/concept/fundamental-thermodynamic-relation/) +[fundamental thermodynamic relation](/know/concept/fundamental-relation-of-thermodynamics/) reduces to $$\dd{E} = - p \dd{V}$$. Then the pressure is given by a thermodynamic equation of state $$p(\rho, T)$$, diff --git a/source/know/concept/fundamental-relation-of-thermodynamics/index.md b/source/know/concept/fundamental-relation-of-thermodynamics/index.md new file mode 100644 index 0000000..a51c231 --- /dev/null +++ b/source/know/concept/fundamental-relation-of-thermodynamics/index.md @@ -0,0 +1,326 @@ +--- +title: "Fundamental relation of thermodynamics" +sort_title: "Fundamental relation of thermodynamics" +date: 2024-07-21 # Originally 2021-07-07, major rewrite +categories: +- Physics +- Thermodynamics +layout: "concept" +--- + +In most areas of physics, +we observe and analyze the behaviour +of physical systems that have been "disturbed" some way, +i.e. we try to understand what is *happening*. +In thermodynamics, however, +we start paying attention once the disturbance has ended, +and the system has had some time to settle down: +when nothing seems to be happening anymore. + +Then a common observation is that the system "forgets" what happened earlier, +and settles into a so-called **equilibrium state** +that appears to be independent of its history. +No matter in what way you stir your tea, once you finish, +eventually the liquid stops moving, cools down, +and just... sits there, doing nothing. +But how does it "choose" this equilibrium state? + + + +## Thermodynamic equilibrium + +This history-independence suggests that equilibrium +is determined by only a few parameters of the system. +Prime candidates are the **mole numbers** $$N_1, N_2, ..., N_n$$ +of each of the $$n$$ different types of particles in the system, +and its **volume** $$V$$. +Furthermore, the microscopic dynamics +are driven by energy differences between components, +and obey the universal principle of energy conservation, +so it also sounds reasonable to define a total +**internal energy** $$U$$. + +Thanks to many decades of empirical confirmations, +we now know that the above arguments can be combined into a postulate: +the equilibrium state of a closed system with fixed $$U$$, $$V$$ and $$N_i$$ +is completely determined by those parameters. +The system then "finds" the equilibrium +by varying its microscopic degrees of freedom +such that the **entropy** $$S$$ is maximized +subject to the given values of $$U$$, $$V$$ and $$N_i$$. +This statement serves as a definition of $$S$$, +and explains the **second law of thermodynamics**: +the total entropy never decreases. + +We do not care about those microscopic degrees of freedom, +but we do care about how $$U$$, $$V$$ and $$N_i$$ influence the equilibrium. +For a given system, we want a formula $$S(U, V, N_1, ..., N_n)$$, +which contains all thermodynamic information about the system +and is therefore known as its **fundamental relation**. + +The next part of our definition of $$S$$ +is that it must be invertible with respect to $$U$$, +meaning we can rearrange the fundamental relation +to $$U(S, V, N_1, ... N_n)$$ without losing any information. +Specifically, this means that $$S$$ must be continuous, +differentiable, and monotonically increasing with $$U$$, +such that $$S(U)$$ can be inverted to $$U(S)$$ and vice versa. + +The idea here is that maximizing $$S$$ at fixed $$U$$ +should be equivalent to minimizing $$U$$ for a given $$S$$ +(we prove this later). +Often it is mathematically more convenient +to choose one over the other, +but by definition both approaches are equally valid. +And because $$S$$ is rather abstract, +it may be preferable to treat it as a parameter +for a more intuitive quantity like $$U$$. + +Next, we demand that $$S$$ is additive over subsystems, +so $$S = S_1 + S_2 + ...$$, with $$S_1$$ being the entropy of subsystem 1, etc. +Consequently, $$S$$ is an **extensive** quantity of the system, +just like $$U$$ (and $$V$$ and $$N_i$$), +meaning they satisfy for any constant $$\lambda$$: + +$$\begin{aligned} + S(\lambda U, \lambda V, \lambda N_1, ..., \lambda N_n) + &= \lambda S(U, V, N_1, ..., N_n) + \\ + U(\lambda S, \lambda V, \lambda N_1, ..., \lambda N_n) + &= \lambda U(S, V, N_1, ..., N_n) +\end{aligned}$$ + +For $$U$$, this makes intuitive sense: +the total energy in two identical systems +is double the energy of a single of those systems. +Actually, reality is a bit hazier than this: +dynamics are governed by energy *differences* only, +so an offset $$U_0$$ can be added without a consequence. +We should choose an offset and a way to split the system into subsystems +such that the above relation holds for our convenience. +Fortunately, this choice often makes itself. + +$$S$$ does not suffer from this ambiguity, +since the **third law of thermodynamics** clearly defines +where $$S = 0$$ should occur: at a temperature of absolute zero. +In this article we will not explore the reason for this requirement, +which is also known as the **Nernst postulate**. +Furthermore, in most situations this law can simply be ignored. + +Since $$U$$, $$S$$, $$V$$ and $$N_i$$ are all extensive, +the partial derivatives of the fundamental relation are **intensive** quantities, +meaning they do not depend on the size of the system. +Those derivatives are very important, +since they are usually the equilibrium properties we want to find. + + + +## Energy representation + +When we have a fundamental relation of the form $$U(S, V, N_1, ..., N_n)$$, +we say we are treating the system's thermodynamics +in the **energy representation**. + +The following derivatives of $$U$$ are used as the thermodynamic *definitions* +of the **temperature** $$T$$, the **pressure** $$P$$, +and the **chemical potential** $$\mu_k$$ of the $$k$$th particle species: + +$$\begin{aligned} + \boxed{ + \begin{aligned} + T + &\equiv \bigg( \pdv{U}{S} \bigg)_{V, N_i} + \\ + P + &\equiv - \bigg( \pdv{U}{V} \bigg)_{S, N_i} + \\ + \mu_k + &\equiv \bigg( \pdv{U}{N_k} \bigg)_{S, V, N_{i \neq k}} + \end{aligned} + } +\end{aligned}$$ + +The resulting expressions of the form $$T(S, V, N_1, ..., N_n)$$ etc. +are known as the **equations of state** of the system. +Unlike the fundamental relation, a single equation of state +is not a complete thermodynamic description of the system. +However, if *all* equations of state are known +(for $$T$$, $$P$$, and all $$\mu_k$$), +then the fundamental relation can be reconstructed. + +As explained above, physical dynamics are driven by energy differences only, +so we expand an infinitesimal difference $$\dd{U}$$ as: + +$$\begin{aligned} + \dd{U} + = \bigg( \pdv{U}{S} \bigg)_{V, N_i} \!\dd{S} + \:\:+\:\: \bigg( \pdv{U}{V} \bigg)_{S, N_i} \!\dd{V} + \:\:+\:\: \sum_{k}^{} \bigg( \pdv{U}{N_k} \bigg)_{S, V, N_{i \neq k}} \!\dd{N_k} +\end{aligned}$$ + +Those partial derivatives look familiar. +Substituting $$T$$, $$P$$ and $$\mu_k$$ gives a result +that is also called the **fundamental relation of thermodynamics** +(as opposed to the fundamental relation of the system only, +just to make things confusing): + +$$\begin{aligned} + \boxed{ + \dd{U} + = T \dd{S} - P \dd{V} + \sum_{k}^{} \mu_k \dd{N_k} + } +\end{aligned}$$ + +Where the first term represents heating/cooling (also written as $$\dd{Q}$$), +and the second is physical work done on the system +by compression/expansion (also written as $$\dd{W}$$). +The third term is the energy change due to matter transfer and is often neglected. +Hence this relation can be treated as a form +of the **first law of thermodynamics** $$\Delta U = \Delta Q + \Delta W$$. + +Because $$T$$, $$P$$ and $$\mu_k$$ generally depend on $$S$$, $$V$$ and $$N_k$$, +integrating the fundamental relation can be tricky. +Fortunately, the fact that $$U$$ is extensive offers a shortcut. +Recall that: + +$$\begin{aligned} + \lambda U(S, V, N_1, ..., N_n) + &= U(\lambda S, \lambda V, \lambda N_1, ..., \lambda N_n) +\end{aligned}$$ + +For any $$\lambda$$. +Let us differentiate this equation with respect to $$\lambda$$, yielding: + +$$\begin{aligned} + U + &= \pdv{}{\lambda} U(\lambda S, \lambda V, \lambda N_1, ..., \lambda N_n) + \\ + &= \pdv{U(\lambda S)}{(\lambda S)} \pdv{(\lambda S)}{\lambda} + + \pdv{U(\lambda V)}{(\lambda V)} \pdv{(\lambda V)}{\lambda} + + \sum_{k} \pdv{U(\lambda N_k)}{(\lambda N_k)} \pdv{(\lambda N_k)}{\lambda} + \\ + &= \pdv{U(S)}{S} S + + \pdv{U(V)}{V} V + + \sum_{k} \pdv{U(N_k)}{N_k} N_k +\end{aligned}$$ + +Where we once again recognize the derivatives. +The resulting equation is known as the **Euler form** +of the fundamental relation of thermodynamics: + +$$\begin{aligned} + \boxed{ + U + = T S - P V + \sum_{k} \mu_k N_k + } +\end{aligned}$$ + +Plus a constant $$U_0$$ of course, +although $$U_0 = 0$$ is the most straightforward choice. + + + +## Entropy representation + +If the system's fundamental relation +instead has the form $$S(U, V, N_1, ..., N_i)$$, +we are treating it in the **entropy representation**. +Isolating the above fundamental relation of thermodynamics +for $$\dd{S}$$ yields its equivalent form in this representation: + +$$\begin{aligned} + \boxed{ + \dd{S} + = \frac{1}{T} \dd{U} + \frac{P}{T} \dd{V} - \sum_{k}^{} \frac{\mu_k}{T} \dd{N_k} + } +\end{aligned}$$ + +From which we can then read off the standard partial derivatives of $$S(U, V, N_1, ..., N_n)$$: + +$$\begin{aligned} + \boxed{ + \begin{aligned} + \frac{1}{T} + &= \bigg( \pdv{S}{U} \bigg)_{V, N_i} + \\ + \frac{P}{T} + &= \bigg( \pdv{S}{V} \bigg)_{U, N_i} + \\ + \frac{\mu_k}{T} + &= - \bigg( \pdv{S}{N_k} \bigg)_{U, V, N_{i \neq k}} + \end{aligned} + } +\end{aligned}$$ + +Note the signs: the parameters $$U$$, $$V$$ and $$N_i$$ are implicitly related +by our requirement that $$S$$ is stationary at a maximum, +so the [triple product rule](/know/concept/triple-product-rule/) +must be used, which brings some perhaps surprising sign changes. +Reading them off in this way is easier. + +And of course, since $$S$$ is defined to be an extensive quantity, +it also has an Euler form: + +$$\begin{aligned} + \boxed{ + S + = \frac{1}{T} U + \frac{P}{T} V - \sum_{k} \frac{\mu_k}{T} N_k + } +\end{aligned}$$ + +Finally, it is worth proving that minimizing $$U$$ +is indeed equivalent to maximizing $$S$$. +For simplicity, we consider a system +where only the volume $$V$$ can change +in order to reach an equilibrium; +the proof is analogous for all other parameters. +Clearly, $$S$$ is stationary at its maximum: + +$$\begin{aligned} + 0 + &= \bigg( \pdv{S}{V} \bigg)_{U, N_i} + = - \frac{ \bigg( \displaystyle\pdv{U}{V} \bigg)_{S, N_i} }{ \bigg( \displaystyle\pdv{U}{S} \bigg)_{V, N_i} } + = - \frac{1}{T} \bigg( \pdv{U}{V} \bigg)_{S, N_i} +\end{aligned}$$ + +Where we have used the triple product rule. +This can only hold if $$(\ipdv{U}{S})_{S, N_i} = 0$$, +meaning $$U$$ is also at an extremum. +But $$S$$ is not just at any extremum: it is at a *maximum*, so: + +$$\begin{aligned} + 0 + > \bigg( \pdvn{2}{S}{V} \bigg)_{U, N_i} + &= \bigg( \pdv{}{V} \Big( \frac{P}{T} \Big) \bigg)_{U, N_i} + \\ + &= \bigg( \pdv{}{V} \Big( \frac{P}{T} \Big) \bigg)_{S, N_i} + + \bigg( \pdv{}{S} \Big( \frac{P}{T} \Big) \bigg)_{V, N_i} \bigg( \pdv{S}{V} \bigg)_{U, N_i} + \\ + &= \bigg( \pdv{}{V} \Big( \frac{P}{T} \Big) \bigg)_{S, N_i} + + \frac{P}{T} \bigg( \pdv{}{S} \Big( \frac{P}{T} \Big) \bigg)_{V, N_i} + \\ + &= \frac{1}{T} \bigg( \pdv{P}{V} \bigg)_{S, N_i} + - \frac{P}{T^2} \bigg( \pdv{T}{V} \bigg)_{S, N_i} + + \frac{P}{T} \bigg( \pdv{}{S} \Big( \frac{P}{T} \Big) \bigg)_{V, N_i} + \\ + &= - \frac{1}{T} \bigg( \pdvn{2}{U}{V} \bigg)_{S, N_i} + + \frac{P}{T} \bigg[ \bigg( \pdv{}{S} \Big( \frac{P}{T} \Big) \bigg)_{V, N_i} + - \frac{1}{T} \bigg( \pdv{T}{V} \bigg)_{S, N_i} \bigg] +\end{aligned}$$ + +Because $$S$$ is at a maximum, we know that $$P/T = 0$$, +and $$T$$ is always above absolute zero +(since we defined $$S$$ to be monotonically increasing with $$U$$), +which leaves $$(\ipdvn{2}{U}{V})_{S, N_i} > 0$$ +as the only way to satisfy this inequality. +In other words, $$U$$ is at a minimum, as expected. + + + +## References +1. H.B. Callen, + *Thermodynamics and an introduction to thermostatistics*, 2nd edition, + Wiley. +2. H. Gould, J. Tobochnik, + *Statistical and thermal physics*, 2nd edition, + Princeton. diff --git a/source/know/concept/fundamental-thermodynamic-relation/index.md b/source/know/concept/fundamental-thermodynamic-relation/index.md deleted file mode 100644 index 0d945fa..0000000 --- a/source/know/concept/fundamental-thermodynamic-relation/index.md +++ /dev/null @@ -1,54 +0,0 @@ ---- -title: "Fundamental thermodynamic relation" -sort_title: "Fundamental thermodynamic relation" -date: 2021-07-07 -categories: -- Physics -- Thermodynamics -layout: "concept" ---- - -The **fundamental thermodynamic relation** combines the first two -[laws of thermodynamics](/know/concept/laws-of-thermodynamics/), -and gives the change of the internal energy $$U$$, -which is a [thermodynamic potential](/know/concept/thermodynamic-potential/), -in terms of the change in |