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| -rw-r--r-- | content/know/concept/blasius-boundary-layer/index.pdc | 119 | ||||
| -rw-r--r-- | content/know/concept/fredholm-alternative/index.pdc | 67 | ||||
| -rw-r--r-- | content/know/concept/optical-wave-breaking/index.pdc | 54 | ||||
| -rw-r--r-- | content/know/concept/prandtl-equations/index.pdc | 211 | ||||
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diff --git a/content/know/concept/blasius-boundary-layer/index.pdc b/content/know/concept/blasius-boundary-layer/index.pdc new file mode 100644 index 0000000..d9563c2 --- /dev/null +++ b/content/know/concept/blasius-boundary-layer/index.pdc @@ -0,0 +1,119 @@ +--- +title: "Blasius boundary layer" +firstLetter: "B" +publishDate: 2021-05-29 +categories: +- Physics +- Fluid mechanics +- Fluid dynamics + +date: 2021-05-10T18:41:28+02:00 +draft: false +markup: pandoc +--- + +# Blasius boundary layer + +In fluid dynamics, the **Blasius boundary layer** is an application of +the [Prandtl equations](/know/concept/prandtl-equations/), +which govern the flow of a fluid +at large Reynolds number $\mathrm{Re} \gg 1$ +close to a surface. +Specifically, the Blasius layer is the solution +for a half-plane approached from the edge by a fluid. + +A fluid with velocity field $\va{v} = U \vu{e}_x$ flows to the plane, +which starts at $y = 0$ and exists for $x \ge 0$. +To describe this, we make an ansatz +for the *slip-flow* region's $x$-velocity $v_x(x, y)$: + +$$\begin{aligned} + v_x + = U f'(s) + \qquad \quad + s + \equiv \frac{y}{\delta(x)} +\end{aligned}$$ + +Note that $f'(s)$ is the derivative of an unknown $f(s)$, +and that it obeys the boundary conditions $f'(0) = 0$ and $f'(\infty) = 1$. +Furthermore, $\delta(x)$ is the thickness of the stationary boundary layer at the surface. +To derive the Prandtl equations, +the estimate $\delta(x) = \sqrt{\nu x / U}$ was used, +which we will stick with. +For later use, it is worth writing the derivatives of $s$: + +$$\begin{aligned} + \pdv{s}{x} + = - y \frac{\delta'}{\delta^2} + = - s \frac{\delta'}{\delta} + \qquad \quad + \pdv{s}{y} + = \frac{1}{\delta} +\end{aligned}$$ + +Inserting the ansatz for $v_x$ into the incompressibility condition then yields: + +$$\begin{aligned} + \pdv{v_y}{y} + = - \pdv{v_x}{x} + = U s f'' \frac{\delta'}{\delta} +\end{aligned}$$ + +Which we integrate to get an expression for the $y$-velocity $v_y$, namely: + +$$\begin{aligned} + v_y + = U \frac{\delta'}{\delta} \int s f'' \dd{y} + = U \delta' \: (s f' - f) +\end{aligned}$$ + +Now, consider the main Prandtl equation, +assuming that the attack velocity $U$ is constant: + +$$\begin{aligned} + v_x \pdv{v_x}{x} + v_y \pdv{v_x}{y} + = \nu \pdv[2]{v_x}{y} +\end{aligned}$$ + +Inserting our expressions for $v_x$ and $v_y$ into this leads us to: + +$$\begin{aligned} + - U^2 \frac{\delta'}{\delta} s f'' f' + U^2 \frac{\delta'}{\delta} f'' (s f' - f) + = \nu U \frac{1}{\delta^2} f''' +\end{aligned}$$ + +After multiplying it by $\delta^2 / U$ and cancelling out some terms, +it reduces to: + +$$\begin{aligned} + \nu f''' + U \delta' \delta f'' f + = 0 +\end{aligned}$$ + +Then, substituting $\delta(x) = \sqrt{\nu x / U}$ and $\delta'(x) = (1/2) \sqrt{\nu / (U x)}$ yields: + +$$\begin{aligned} + \nu f''' + U \frac{\nu}{2 U} f'' f + = 0 +\end{aligned}$$ + +Simplifying this leads us to the **Blasius equation**, +which is a nonlinear ODE for $f(s)$: + +$$\begin{aligned} + \boxed{ + 2 f''' + f'' f = 0 + } +\end{aligned}$$ + +Unfortunately, this cannot be solved analytically, only numerically. +Nevertheless, the result shows a boundary layer $\delta(x)$ +exhibiting the expected downstream thickening. + + + +## References +1. B. Lautrup, + *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition, + CRC Press. diff --git a/content/know/concept/fredholm-alternative/index.pdc b/content/know/concept/fredholm-alternative/index.pdc new file mode 100644 index 0000000..8f20e57 --- /dev/null +++ b/content/know/concept/fredholm-alternative/index.pdc @@ -0,0 +1,67 @@ +--- +title: "Fredholm alternative" +firstLetter: "F" +publishDate: 2021-05-29 +categories: +- Mathematics + +date: 2021-03-09T20:36:01+01:00 +draft: false +markup: pandoc +--- + +# Fredholm alternative + +The **Fredholm alternative** is a theorem regarding equations involving +a linear operator $\hat{L}$ on a [Hilbert space](/know/concept/hilbert-space/), +and is useful in the context of multiple-scale perturbation theory. +It is an *alternative* because it gives two mutually exclusive options, +given here in [Dirac notation](/know/concept/dirac-notation/): + +1. $\hat{L} \ket{u} = \ket{f}$ has a unique solution $\ket{u}$ for every $\ket{f}$. +2. $\hat{L}^\dagger \ket{w} = 0$ has non-zero solutions. + Then regarding $\hat{L} \ket{u} = \ket{f}$: + 1. If $\braket{w}{f} = 0$ for all $\ket{w}$, then it has infinitely many solutions $\ket{u}$. + 2. If $\braket{w}{f} \neq 0$ for any $\ket{w}$, then it has no solutions $\ket{u}$. + +Where $\hat{L}^\dagger$ is the adjoint of $\hat{L}$. +In other words, $\hat{L} \ket{u} = \ket{f}$ has non-trivial solutions if +and only if for all $\ket{w}$ (including the trivial case $\ket{w} = 0$) +it holds that $\braket{w}{f} = 0$. + +As a specific example, +if $\hat{L}$ is a matrix and the kets are vectors, +this theorem can alternatively be stated as follows using the determinant: + +1. If $\mathrm{det}(\hat{L}) \neq 0$, then $\hat{L} \vec{u} = \vec{f}$ + has a unique solution $\vec{u}$ for every $\vec{f}$. +2. If $\mathrm{det}(\hat{L}) = 0$, + then $\hat{L}^\dagger \vec{w} = \vec{0}$ has non-zero solutions. + Then regarding $\hat{L} \vec{u} = \vec{f}$: + 1. If $\vec{w} \cdot \vec{f} = 0$ for all $\vec{w}$, then it has + infinitely many solutions $\vec{u}$. + 2. If $\vec{w} \cdot \vec{f} \neq 0$ for any $\vec{w}$, then it has + no solutions $\vec{u}$. + +Consequently, the Fredholm alternative is also brought up +in the context of eigenvalue problems. +Define $\hat{M} = (\hat{L} - \lambda \hat{I})$, +where $\lambda$ is an eigenvalue of $\hat{L}$ +if and only if $\mathrm{det}(\hat{M}) = 0$. +Then for the equation $\hat{M} \ket{u} = \ket{f}$, we can say that: + +1. If $\lambda$ is *not* an eigenvalue, + then there is a unique solution $\ket{u}$ for each $\ket{f}$. +2. If $\lambda$ is an eigenvalue, then $\hat{M}^\dagger \ket{w} = 0$ + has non-zero solutions. Then: + 1. If $\braket{w}{f} = 0$ for all $\ket{w}$, then there are + infinitely many solutions $\ket{u}$. + 2. If $\braket{w}{f} \neq 0$ for any $\ket{w}$, then there are no + solutions $\ket{u}$. + + + +## References +1. O. Bang, + *Nonlinear mathematical physics: lecture notes*, 2020, + unpublished. diff --git a/content/know/concept/optical-wave-breaking/index.pdc b/content/know/concept/optical-wave-breaking/index.pdc index 757a633..2ab3ff1 100644 --- a/content/know/concept/optical-wave-breaking/index.pdc +++ b/content/know/concept/optical-wave-breaking/index.pdc @@ -75,11 +75,10 @@ the instantaneous frequencies for these separate effects: $$\begin{aligned} \omega_i(z,t) &\approx \omega_\mathrm{GVD}(z,t) + \omega_\mathrm{SPM}(z,t) - \\ % &= \frac{\beta_2 z / T_0^2}{1 + \beta_2^2 z^2 / T_0^4} \frac{t}{T_0^2} % + \frac{2\gamma P_0 z}{T_0^2} t \exp\!\Big(\!-\frac{t^2}{T_0^2}\Big) % \\ - &= \frac{tz}{T_0^2} \bigg( \frac{\beta_2 / T_0^2}{1 + \beta_2^2 z^2 / T_0^4} + = \frac{tz}{T_0^2} \bigg( \frac{\beta_2 / T_0^2}{1 + \beta_2^2 z^2 / T_0^4} + 2\gamma P_0 \exp\!\Big(\!-\!\frac{t^2}{T_0^2}\Big) \bigg) \end{aligned}$$ @@ -97,14 +96,14 @@ and $N_\mathrm{sol}$ is the **soliton number**, which is defined as: $$\begin{aligned} - N_\mathrm{sol}^2 = \frac{L_D}{L_N} = \frac{\gamma P_0 T_0^2}{|\beta_2|} + N_\mathrm{sol}^2 \equiv \frac{L_D}{L_N} = \frac{\gamma P_0 T_0^2}{|\beta_2|} \end{aligned}$$ This quantity is very important in anomalous dispersion, but even in normal dispesion, it is still a useful measure of the relative strengths of GVD and SPM. As was illustrated earlier, $\omega_i$ overtakes itself at the edges, -so OWB only occurs when $\omega_i$ is not monotonic, -which is when its $t$-derivative, +so OWB occurs when $\omega_i$ oscillates there, +which starts when its $t$-derivative, the **instantaneous chirpyness** $\xi_i$, has *two* real roots for $t^2$: @@ -122,11 +121,17 @@ leading to the following exact minimum value $N_\mathrm{min}^2$ for $N_\mathrm{s such that OWB can only occur when $N_\mathrm{sol}^2 > N_\mathrm{min}^2$: $$\begin{aligned} - N_\mathrm{min}^2 = \frac{1}{4} \exp\!\Big(\frac{3}{2}\Big) \approx 1.12 + \boxed{ + N_\mathrm{min}^2 = \frac{1}{4} \exp\!\Big(\frac{3}{2}\Big) \approx 1.12 + } \end{aligned}$$ -Now, consider two times $t_1$ and $t_2$ in the pulse, separated by -a small initial interval $(t_2 - t_1)$. +If this condition $N_\mathrm{sol}^2 > N_\mathrm{min}^2$ is not satisfied, +$\xi_i$ cannot have two roots for $t^2$, meaning $\omega_i$ cannot overtake itself. +GVD is unable to keep up with SPM, so OWB will not occur. + +Next, consider two points at $t_1$ and $t_2$ in the pulse, +separated by a small initial interval $(t_2 - t_1)$. The frequency difference between these points due to $\omega_i$ will cause them to displace relative to each other after a short distance $z$ by some amount $\Delta t$, @@ -136,21 +141,21 @@ $$\begin{aligned} \Delta t &\approx z \Delta\beta_1 \qquad - &&\Delta\beta_1 = \beta_1(\omega_i(z,t_2)) - \beta_1(\omega_i(z,t_1)) + &&\Delta\beta_1 \equiv \beta_1(\omega_i(z,t_2)) - \beta_1(\omega_i(z,t_1)) \\ &\approx z \beta_2 \Delta\omega_i \qquad - &&\Delta\omega_i = \omega_i(z,t_2) - \omega_i(z,t_1) + &&\Delta\omega_i \equiv \omega_i(z,t_2) - \omega_i(z,t_1) \\ &\approx z \beta_2 \Delta\xi_i \,(t_2 - t_1) \qquad \quad - &&\Delta\xi_i = \xi_i(z,t_2) - \xi_i(z,t_1) + &&\Delta\xi_i \equiv \xi_i(z,t_2) - \xi_i(z,t_1) \end{aligned}$$ Where $\beta_1(\omega)$ is the inverse of the group velocity. OWB takes place when $t_2$ and $t_1$ catch up to each other, which is when $-\Delta t = (t_2 - t_1)$. -The distance where this happens, $z = L_\mathrm{WB}$, +The distance where this happens first, $z = L_\mathrm{WB}$, must therefore satisfy the following condition for a particular value of $t$: @@ -161,7 +166,7 @@ $$\begin{aligned} \end{aligned}$$ The time $t$ of OWB must be where $\omega_i(t)$ has its steepest slope, -which is at the minimum value of $\xi_i(t)$, and, by extension $f(x)$. +which is at the minimum value of $\xi_i(t)$, and by extension $f(x)$. This turns out to be $f(3/2)$: $$\begin{aligned} @@ -170,16 +175,17 @@ $$\begin{aligned} = 1 - N_\mathrm{sol}^2 / N_\mathrm{min}^2 \end{aligned}$$ -Clearly, $f_\mathrm{min} \ge 0$ when -$N_\mathrm{sol}^2 \le N_\mathrm{min}^2$, which, when inserted in the -condition above, confirms that OWB cannot occur in that case. Otherwise, -if $N_\mathrm{sol}^2 > N_\mathrm{min}^2$, then: +Clearly, $f_\mathrm{min} \ge 0$ when $N_\mathrm{sol}^2 \le N_\mathrm{min}^2$, +which, when inserted above, leads to an imaginary $L_\mathrm{WB}$, +confirming that OWB cannot occur in that case. +Otherwise, if $N_\mathrm{sol}^2 > N_\mathrm{min}^2$, then: $$\begin{aligned} - L_\mathrm{WB} - = - \frac{T_0^2}{\beta_2 \, \sqrt{f_\mathrm{min}}} - = \frac{L_D}{\sqrt{N_\mathrm{sol}^2 / N_\mathrm{min}^2 - 1}} - = \frac{L_D}{\sqrt{4 N_\mathrm{sol}^2 \exp(-3/2) - 1}} + \boxed{ + L_\mathrm{WB} + = \frac{T_0^2}{\beta_2 \, \sqrt{- f_\mathrm{min}}} + = \frac{L_D}{\sqrt{N_\mathrm{sol}^2 / N_\mathrm{min}^2 - 1}} + } \end{aligned}$$ This prediction for $L_\mathrm{WB}$ appears to agree well @@ -196,7 +202,7 @@ Filling $L_\mathrm{WB}$ in into $\omega_\mathrm{SPM}$ gives: $$\begin{aligned} \omega_{\mathrm{SPM}}(L_\mathrm{WB},t) - = \frac{2 \gamma P_0 t}{\beta_2 \sqrt{4 N_\mathrm{sol}^2 \exp(-3/2) - 1}} \exp\!\Big(\!-\frac{t^2}{T_0^2}\Big) + = \frac{2 \gamma P_0 t}{\beta_2 \sqrt{4 N_\mathrm{sol}^2 \exp(-3/2) - 1}} \exp\!\Big(\!-\!\frac{t^2}{T_0^2}\Big) \end{aligned}$$ Assuming that $N_\mathrm{sol}^2$ is large in the denominator, this can @@ -205,8 +211,8 @@ be approximately reduced to: $$\begin{aligned} \omega_\mathrm{SPM}(L_\mathrm{WB}, t) % = \frac{2 \gamma P_0 t \exp(-t^2 / T_0^2)}{\beta_2 \sqrt{N_\mathrm{sol}^2 / N_\mathrm{min}^2 - 1}} - \approx \frac{2 \gamma P_0 t}{\beta_2 N_\mathrm{sol}} \exp\!\Big(\!-\frac{t^2}{T_0^2}\Big) - = 2 \sqrt{\frac{\gamma P_0}{\beta_2}} \frac{t}{T_0} \exp\!\Big(\!-\frac{t^2}{T_0^2}\Big) + \approx \frac{2 \gamma P_0 t}{\beta_2 N_\mathrm{sol}} \exp\!\Big(\!-\!\frac{t^2}{T_0^2}\Big) + = 2 \sqrt{\frac{\gamma P_0}{\beta_2}} \frac{t}{T_0} \exp\!\Big(\!-\!\frac{t^2}{T_0^2}\Big) \end{aligned}$$ The expression $x \exp(-x^2)$ has its global extrema diff --git a/content/know/concept/prandtl-equations/index.pdc b/content/know/concept/prandtl-equations/index.pdc new file mode 100644 index 0000000..00f7773 --- /dev/null +++ b/content/know/concept/prandtl-equations/index.pdc @@ -0,0 +1,211 @@ +--- +title: "Prandtl equations" +firstLetter: "P" +publishDate: 2021-05-29 +categories: +- Physics +- Fluid mechanics +- Fluid dynamics + +date: 2021-05-10T18:41:20+02:00 +draft: false +markup: pandoc +--- + +# Prandtl equations + +In fluid dynamics, the **Prandtl equations** or **boundary layer equations** +describe the movement of a [viscous](/know/concept/viscosity/) fluid +with a large [Reynolds number](/know/concept/reynolds-number/) $\mathrm{Re} \gg 1$ +close to a solid surface. + +Fluids with a large Reynolds number +are often approximated as having zero viscosity, +since the simpler [Euler equations](/know/concept/euler-equations) +can then be used instead of the [Navier-Stokes equations](/know/concept/navier-stokes-equations/). + +However, in reality, a viscous fluid obeys the *no-slip* boundary condition: +at every solid surface the local velocity must be zero. +This implies the existence of a **boundary layer**: +a thin layer of fluid "stuck" to solid objects in the flow, +where viscosity plays an important role. +This is in contrast to the ideal flow far away from the surface. + +We consider a simple theoretical case in 2D: +a large flat surface located at $y = 0$ for all $x \in \mathbb{R}$, +with a fluid *trying* to flow parallel to it at $U$. +The 2D treatment can be justified by assuming that everything is constant in the $z$-direction. +We will not solve this case, +but instead derive general equations +to describe the flow close to a flat surface. + +At the wall, there is a very thin boundary layer of thickness $\delta$, +where the fluid is assumed to be completely stationary $\va{v} = 0$. +We are mainly interested in the region $\delta < y \ll L$, +where $L$ is the distance at which the fluid becomes practically ideal. +This the so-called **slip-flow** region, +in which the fluid is not stationary, +but still viscosity-dominated. + +In 2D, the steady Navier-Stokes equations are as follows, +where the flow $\va{v} = (v_x, v_y)$: + +$$\begin{aligned} + v_x \pdv{v_x}{x} + v_y \pdv{v_x}{y} + &= - \frac{1}{\rho} \pdv{p}{x} + \nu \Big( \pdv[2]{v_x}{x} + \pdv[2]{v_x}{y} \Big) + \\ + v_x \pdv{v_y}{x} + v_y \pdv{v_y}{y} + &= - \frac{1}{\rho} \pdv{p}{y} + \nu \Big( \pdv[2]{v_y}{x} + \pdv[2]{v_y}{y} \Big) + \\ + \pdv{v_x}{x} + \pdv{v_y}{y} + &= 0 +\end{aligned}$$ + +The latter represents the fluid's incompressibility. +We non-dimensionalize these equations, +and assume that changes along the $y$-axis +happen on a short scale (say, $\delta$), +and along the $x$-axis on a longer scale (say, $L$). +Let $\tilde{x}$ and $\tilde{y}$ be dimenionless variables of order $1$: + +$$\begin{aligned} + x + = L \tilde{x} + \qquad \quad + y + = \delta \tilde{x} + \qquad \quad + \pdv{x} + = \frac{1}{L} \pdv{\tilde{x}} + \qquad \quad + \pdv{y} + = \frac{1}{\delta} \pdv{\tilde{y}} +\end{aligned}$$ + +Furthermore, we choose velocity scales +to be consistent with the incompressibility condition, +and a pressure scale inspired +by [Bernoulli's theorem](/know/concept/bernoullis-theorem/): + +$$\begin{aligned} + v_x + = U \tilde{v}_x + \qquad \quad + v_y + = \frac{U \delta}{L} \tilde{v}_y + \qquad \quad + p + = \rho U^2 \tilde{p} +\end{aligned}$$ + +We insert these scalings into the Navier-Stokes equations, yielding: + +$$\begin{aligned} + \frac{U^2}{L} \tilde{v}_x \pdv{\tilde{v}_x}{\tilde{x}} + \frac{U^2}{L} \tilde{v}_y \pdv{\tilde{v}_x}{\tilde{y}} + &= - \frac{U^2}{L} \pdv{\tilde{p}}{\tilde{x}} + + \nu \Big( \frac{U}{L^2} \pdv[2]{\tilde{v}_x}{\tilde{x}} + \frac{U}{\delta^2} \pdv[2]{\tilde{v}_x}{\tilde{y}} \Big) + \\ + \frac{U^2 \delta}{L^2} \tilde{v}_x \pdv{\tilde{v}_y}{\tilde{x}} + \frac{U^2 \delta}{L^2} \tilde{v}_y \pdv{\tilde{v}_y}{\tilde{y}} + &= - \frac{U^2}{\delta} \pdv{\tilde{p}}{\tilde{y}} + + \nu \Big( \frac{U \delta}{L^3} \pdv[2]{\tilde{v}_y}{\tilde{x}} + \frac{U}{L \delta} \pdv[2]{\tilde{v}_y}{\tilde{y}} \Big) +\end{aligned}$$ + +For future convenience, +we multiply the former equation by $L / U^2$, and the latter by $\delta / U^2$: + +$$\begin{aligned} + \tilde{v}_x \pdv{\tilde{v}_x}{\tilde{x}} + \tilde{v}_y \pdv{\tilde{v}_x}{\tilde{y}} + &= - \pdv{\tilde{p}}{\tilde{x}} + + \nu \Big( \frac{1}{U L} \pdv[2]{\tilde{v}_x}{\tilde{x}} + \frac{L}{U \delta^2} \pdv[2]{\tilde{v}_x}{\tilde{y}} \Big) + \\ + \frac{\delta^2}{L^2} \tilde{v}_x \pdv{\tilde{v}_y}{\tilde{x}} + \frac{\delta^2}{L^2} \tilde{v}_y \pdv{\tilde{v}_y}{\tilde{y}} + &= - \pdv{\tilde{p}}{\tilde{y}} + + \nu \Big( \frac{\delta^2}{U L^3} \pdv[2]{\tilde{v}_y}{\tilde{x}} + \frac{1}{U L} \pdv[2]{\tilde{v}_y}{\tilde{y}} \Big) +\end{aligned}$$ + +We would like to estimate $\delta$. +Intuitively, we expect that higher viscosities $\nu$ give thicker layers, +and that faster velocities $U$ give thinner layers. +Furthermore, we expect *downstream thickening*: +with distance $x$, viscous stresses slow down the slip-flow, +leading to a gradual increase of $\delta(x)$. +Some dimensional analysis thus yields the following estimate: + +$$\begin{aligned} + \delta + \approx \sqrt{\frac{\nu x}{U}} + \sim \sqrt{\frac{\nu L}{U}} +\end{aligned}$$ + +We thus insert $\delta = \sqrt{\nu L / U}$ into the Navier-Stokes equations, giving us: + +$$\begin{aligned} + \tilde{v}_x \pdv{\tilde{v}_x}{\tilde{x}} + \tilde{v}_y \pdv{\tilde{v}_x}{\tilde{y}} + &= - \pdv{\tilde{p}}{\tilde{x}} + + \nu \Big( \frac{1}{U L} \pdv[2]{\tilde{v}_x}{\tilde{x}} + \frac{1}{\nu} \pdv[2]{\tilde{v}_x}{\tilde{y}} \Big) + \\ + \frac{\nu}{U L} \tilde{v}_x \pdv{\tilde{v}_y}{\tilde{x}} + \frac{\nu}{U L} \tilde{v}_y \pdv{\tilde{v}_y}{\tilde{y}} + &= - \pdv{\tilde{p}}{\tilde{y}} + + \nu \Big( \frac{\nu}{U^2 L^2} \pdv[2]{\tilde{v}_y}{\tilde{x}} + \frac{1}{U L} \pdv[2]{\tilde{v}_y}{\tilde{y}} \Big) +\end{aligned}$$ + +Here, we recognize the definition of the Reynolds number $\mathrm{Re} = U L / \nu$: + +$$\begin{aligned} + \tilde{v}_x \pdv{\tilde{v}_x}{\tilde{x}} + \tilde{v}_y \pdv{\tilde{v}_x}{\tilde{y}} + &= - \pdv{\tilde{p}}{\tilde{x}} + + \frac{1}{\mathrm{Re}} \pdv[2]{\tilde{v}_x}{\tilde{x}} + \pdv[2]{\tilde{v}_x}{\tilde{y}} + \\ + \frac{1}{\mathrm{Re}} \tilde{v}_x \pdv{\tilde{v}_y}{\tilde{x}} + \frac{1}{\mathrm{Re}} \tilde{v}_y \pdv{\tilde{v}_y}{\tilde{y}} + &= - \pdv{\tilde{p}}{\tilde{y}} + + \frac{1}{\mathrm{Re}^2} \pdv[2]{\tilde{v}_y}{\tilde{x}} + \frac{1}{\mathrm{Re}} \pdv[2]{\tilde{v}_y}{\tilde{y}} +\end{aligned}$$ + +Recall that we are only considering large Reynolds numbers $\mathrm{Re} \gg 1$, +in which case $\mathrm{Re}^{-1} \ll 1$, +so we can drop many terms, leaving us with these redimensionalized equations: + +$$\begin{aligned} + v_x \pdv{v_x}{x} + v_y \pdv{v_x}{y} + = - \frac{1}{\rho} \pdv{p}{x} + \nu \pdv[2]{v_x}{y} + \qquad \quad + \pdv{p}{y} + = 0 +\end{aligned}$$ + +The second one tells us that for a given $x$-value, +the pressure is the same at the surface +as in the main flow $y > L$, where the fluid is ideal. +In the latter regime, we apply Bernoulli's theorem to rewrite $p$, +using the *Bernoulli head* $H$ and the mainstream velocity $U(x)$: + +$$\begin{aligned} + p + = \rho H - \frac{1}{2} \rho U^2 + = p_0 - \frac{1}{2} \rho U^2 +\end{aligned}$$ + +Inserting this into the reduced Navier-Stokes equations, +we arrive at the Prandtl equations: + +$$\begin{aligned} + \boxed{ + v_x \pdv{v_x}{x} + v_y \pdv{v_x}{y} + = U \dv{U}{x} + \nu \pdv[2]{v_x}{y} + \qquad \quad + \pdv{v_x}{x} + \pdv{v_y}{y} + = 0 + } +\end{aligned}$$ + +A notable application of these equations is +the [Blasius boundary layer](/know/concept/blasius-boundary-layer/), +where the surface in question +is a semi-infinite plane. + + + +## References +1. B. Lautrup, + *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition, + CRC Press. diff --git a/content/know/concept/wicks-theorem/index.pdc b/content/know/concept/wicks-theorem/index.pdc new file mode 100644 index 0000000..824885a --- /dev/null +++ b/content/know/concept/wicks-theorem/index.pdc @@ -0,0 +1,194 @@ +--- +title: "Wick's theorem" +firstLetter: "W" +publishDate: 2021-05-29 +categories: +- Physics +- Quantum mechanics + +date: 2021-05-29T14:41:55+02:00 +draft: false +markup: pandoc +--- + +# Wick's theorem + +In the [second quantization](/know/concept/second-quantization/) formalism, +**Wick's theorem** helps to evaluate products +of creation and annihilation operators by +breaking them down into smaller products. + +Firstly, let us define the **normal product** or **normal order** as +a product of second quantization operators +reordered such that +all creation operators are on the left of +all annihilation operators. +For two operators this is written as follows, +at least in the case of bosons: + +$$\begin{aligned} + \underline{\hat{b}_\alpha \hat{b}_\beta^\dagger} + \equiv \hat{b}_\beta^\dagger \hat{b}_\alpha +\end{aligned}$$ + +For fermions, the result must be negated for each swapping of adjacent operators +(and every reordering of operators can be treated as a sequence of such swaps): + +$$\begin{aligned} + \underline{\hat{f}_\alpha \hat{f}_\beta^\dagger} + \equiv - \hat{f}_\beta^\dagger \hat{f}_\alpha +\end{aligned}$$ + +The normal product of three or more operators works in the same way, +but might not be unique depending, +on how many of each type there are. + +Next, the **contraction** of the operators $A$ and $B$ +is defined as the vacuum matrix element, +i.e. the expectation value of $\ket{0}$: + +$$\begin{aligned} + \expval{A B}_0 + \equiv \matrixel{0}{A B}{0} +\end{aligned}$$ + +Unsurprisingly, a contraction can only be nonzero if +$A = \hat{c}_\alpha$ is an annihilation and $B = \hat{c}_\alpha^\dagger$ +a creation for the same state $\alpha$. + +Wick's theorem states: +**any product of second quantization operators can be +rewritten as a sum of normal products, +from which 0, 1, 2, etc. contractions have been removed +in every possible way.** +For fermions, the sign of a term must also be swapped +every time two adjacent operators are swapped. +As an example, for four operators: + +$$\begin{aligned} + A B C D + = \underline{A B C D} + &+ \underline{A B} \expval{C D}_0 \pm \underline{A C} \expval{B D}_0 + \underline{A D} \expval{B C}_0 + \\ + &+ \underline{B C} \expval{A D}_0 \pm \underline{B D} \expval{A C}_0 + \underline{C D} \expval{A B}_0 + \\ + &+ \expval{A B}_0 \expval{C D}_0 \pm \expval{A C}_0 \expval{B D}_0 + \expval{A D}_0 \expval{B C}_0 +\end{aligned}$$ + +Where the negative signs apply to fermions only. +We take the normal product with 0 contractions removed ($\underline{ABCD}$), +then with 1 contraction removed in every possible way (first two lines), +then with 2 contractions removed in every possible way (last line), and so on. + + +## Proof + +We will prove this by induction, with the base case being two operators, +where Wick's theorem becomes as follows: + +$$\begin{aligned} + A B + = \underline{AB} + \expval{A B}_0 +\end{aligned}$$ + +This must be proven separately for fermions and bosons. +For fermions, a general consequence of the definition of the anticommutator is: + +$$\begin{aligned} + \hat{f}_\alpha \hat{f}_\beta^\dagger + = - \hat{f}_\beta^\dagger \hat{f}_\alpha + \{\hat{f}_\alpha, \hat{f}_\beta^\dagger\} +\end{aligned}$$ + +This anticommutator is known to be $\delta_{\alpha\beta}$, +so we can inconsequentially take +its inner product with the vacuum state $\ket{0}$: + +$$\begin{aligned} + \hat{f}_\alpha \hat{f}_\beta^\dagger + &= - \hat{f}_\beta^\dagger \hat{f}_\alpha + \matrixel{0}{\{\hat{f}_\alpha, \hat{f}_\beta^\dagger\}}{0} + = - \hat{f}_\beta^\dagger \hat{f}_\alpha + \matrixel{0}{\hat{f}_\alpha \hat{f}_\beta^\dagger + \hat{f}_\beta^\dagger \hat{f}_\alpha}{0} + \\ + &= - \hat{f}_\beta^\dagger \hat{f}_\alpha + \matrixel{0}{\hat{f}_\alpha \hat{f}_\beta^\dagger}{0} + = \underline{\hat{f}_\alpha \hat{f}_\beta^\dagger} + \expval*{\hat{f}_\alpha \hat{f}_\beta^\dagger}_0 +\end{aligned}$$ + +Which agrees with Wick's theorem. For bosons, we use the commutator: + +$$\begin{aligned} + \hat{b}_\alpha \hat{b}_\beta^\dagger + = \hat{b}_\beta^\dagger \hat{b}_\alpha + [\hat{b}_\alpha, \hat{b}_\beta^\dagger] +\end{aligned}$$ + +This commutator is known to be $\delta_{\alpha\beta}$, +so we take the inner product with $\ket{0}$, like before: + +$$\begin{aligned} + \hat{b}_\alpha \hat{b}_\beta^\dagger + &= \hat{b}_\beta^\dagger \hat{b}_\alpha + \matrixel{0}{[\hat{b}_\alpha, \hat{b}_\beta^\dagger]}{0} + = \hat{b}_\beta^\dagger \hat{b}_\alpha + \matrixel{0}{\hat{b}_\alpha \hat{b}_\beta^\dagger - \hat{b}_\beta^\dagger \hat{b}_\alpha}{0} + \\ + &= \hat{b}_\beta^\dagger \hat{b}_\alpha + \matrixel{0}{\hat{b}_\alpha \hat{b}_\beta^\dagger}{0} + = \underline{\hat{b}_\alpha \hat{b}_\beta^\dagger} + \expval*{\hat{b}_\alpha \hat{b}_\beta^\dagger}_0 +\end{aligned}$$ + +Which again agrees with Wick's theorem. +Next, we prove that if it holds for $N$ operators, then it also holds for $N + 1$. +To begin with, consider the following statement about right-multiplying +by an extra $A_{N+1}$, with $s = 1$ for bosons and $s = -1$ for fermions: + +$$\begin{aligned} + \underline{A_1 ... A_N} A_{N+1} + = \underline{A_1 ... A_N A_{N+1}} + + \sum_{n = 1}^N s^{n + N} \expval{A_n A_{N+1}}_0 \underline{A_1 ... A_{n-1} A_{n+1} ... A_N} +\end{aligned}$$ + +If $A_{N + 1}$ is an annihilation operator, then this is trivial: +appending it does not break the existing normal order, +and $\expval{A_n A_{N+1}}_0 = 0$ for all $A_n$. + +However, if $A_{N + 1}$ is a creation operator, +then to restore the normal order, +we move it to the front by swapping, +which introduces a bunch of (anti)commutators: + +$$\begin{aligned} + \underline{A_1 ... A_N} A_{N+1} + &= s^N A_{N+1} \underline{A_1 ... A_N} + + \sum_{n} s^{n + N} \{[A_n, A_{N+1}]\} \underline{A_1 ... A_{n-1} A_{n+1} ... A_N} + \\ + &= \underline{A_1 ... A_N A_{N+1}} + + \sum_{n} s^{n + N} \expval{A_n A_{N+1}}_0 \underline{A_1 ... A_{n-1} A_{n+1} ... A_N} +\end{aligned}$$ + +Where $\{[]\}$ is the anticommutator or commutator, +respectively for fermions or bosons. + +If we take Wick's theorem for $N$ operators $A_1 ... A_N$, +and right-multiply it by $A_{N + 1}$, +then each term will contain a product of the form $\underline{A_{v} ... A_{w}} A_{N+1}$. +Using the relation that we just proved, +each such product can be rewritten as follows: + +$$\begin{aligned} + \underline{A_v ... A_w} A_{N+1} + &= \underline{A_v ... A_w A_{N+1}} + + \sum_{n} s^{n + N} \expval{A_n A_{N+1}}_0 \underline{A_v ... A_{n-1} A_{n+1} ... A_w} +\end{aligned}$$ + +Inserting this back into Wick's theorem, +we get new terms with contractions of $A_{N+1}$. +After a lot of rearranging, +the result turns out to just be Wick's theorem for $N\!+\!1$ operators. +Therefore, +if Wick's theorem holds for $N$ operators, +it also holds for $N\!+\!1$. + +We showed that Wick's theorem holds for $N = 2$, +so, by induction, it holds for all $N \ge 2$. + + + +## References +1. L.E. Ballentine, + *Quantum mechanics: a modern development*, 2nd edition, + World Scientific. |