diff options
| -rw-r--r-- | content/know/concept/archimedes-principle/index.pdc | 22 | ||||
| -rw-r--r-- | content/know/concept/hagen-poiseuille-equation/index.pdc | 203 | ||||
| -rw-r--r-- | content/know/concept/navier-stokes-equations/index.pdc | 134 | ||||
| -rw-r--r-- | content/know/concept/renyi-entropy/index.pdc | 114 | ||||
| -rw-r--r-- | content/know/concept/viscosity/index.pdc | 100 |
5 files changed, 563 insertions, 10 deletions
diff --git a/content/know/concept/archimedes-principle/index.pdc b/content/know/concept/archimedes-principle/index.pdc index 6335a77..0837cc9 100644 --- a/content/know/concept/archimedes-principle/index.pdc +++ b/content/know/concept/archimedes-principle/index.pdc @@ -44,16 +44,19 @@ on the surface $S$ of $V$: $$\begin{aligned} \va{F}_p = - \oint_S p \dd{\va{S}} + = - \int_V \nabla p \dd{V} \end{aligned}$$ -We rewrite this using Gauss' theorem, -and replace $\nabla p$ by demanding -[hydrostatic equilibrium](/know/concept/hydrostatic-pressure/): +The last step follows from Gauss' theorem. +We replace $\nabla p$ by assuming +[hydrostatic equilibrium](/know/concept/hydrostatic-pressure/), +leading to the definition of the **buoyant force**: $$\begin{aligned} - \va{F}_p - = - \int_V \nabla p \dd{V} - = - \int_V \va{g} \rho_\mathrm{f} \dd{V} + \boxed{ + \va{F}_p + = - \int_V \va{g} \rho_\mathrm{f} \dd{V} + } \end{aligned}$$ For the body to be at rest, we require $\va{F}_g + \va{F}_p = 0$. @@ -66,10 +69,9 @@ $$\begin{aligned} } \end{aligned}$$ -It is commonly assumed that $\va{g}$ has a constant direction -and magnitude $\mathrm{g}$ everywhere. -If we also assume that $\rho_\mathrm{b}$ and $\rho_\mathrm{f}$ are constant, -and only integrate over the "submerged" part, we find: +It is commonly assumed that $\va{g}$ is constant everywhere, with magnitude $\mathrm{g}$. +If we also assume that $\rho_\mathrm{f}$ is constant on the "submerged" side, +and zero on the "non-submerged" side, we find: $$\begin{aligned} 0 diff --git a/content/know/concept/hagen-poiseuille-equation/index.pdc b/content/know/concept/hagen-poiseuille-equation/index.pdc new file mode 100644 index 0000000..b33e085 --- /dev/null +++ b/content/know/concept/hagen-poiseuille-equation/index.pdc @@ -0,0 +1,203 @@ +--- +title: "Hagen-Poiseuille equation" +firstLetter: "H" +publishDate: 2021-04-13 +categories: +- Physics +- Fluid mechanics +- Fluid dynamics + +date: 2021-04-13T10:42:46+02:00 +draft: false +markup: pandoc +--- + +# Hagen-Poiseuille equation + +The **Hagen-Poiseuille equation**, or simply the **Poiseuille equation**, +describes the flow of a fluid with nonzero [viscosity](/know/concept/viscosity/) +through a cylindrical pipe. +Due to its viscosity, the fluid clings to the sides, +limiting the amount that can pass through, for a pipe with radius $R$. + +Consider the [Navier-Stokes equations](/know/concept/navier-stokes-equations/) +of an incompressible fluid with spatially uniform density $\rho$. +Assuming that the flow is steady $\pdv*{\va{v}}{t} = 0$, +and that gravity is negligible $\va{g} = 0$, we get: + +$$\begin{aligned} + (\va{v} \cdot \nabla) \va{v} + = - \frac{\nabla p}{\rho} + \nu \nabla^2 \va{v} + \qquad \quad + \nabla \cdot \va{v} = 0 +\end{aligned}$$ + +Into this, we insert the ansatz $\va{v} = \vu{e}_z \: v_z(r)$, +where $\vu{e}_z$ is the $z$-axis' unit vector. +In other words, we assume that the flow velocity depends only on $r$; +not on $\phi$ or $z$. +Plugging this into the Navier-Stokes equations, +$\nabla \cdot \va{v}$ is trivially zero, +and in the other equation we multiply out $\rho$, yielding this, +where $\eta = \rho \nu$ is the dynamic viscosity: + +$$\begin{aligned} + \nabla p + = \vu{e}_z \: \eta \nabla^2 v_z +\end{aligned}$$ + +Because only $\vu{e}_z$ appears on the right-hand side, +only the $z$-component of $\nabla p$ can be nonzero. +However, $v_z(r)$ is a function of $r$, not $z$! +The left thus only depends on $z$, and the right only on $r$, +meaning that both sides must equal a constant, +which we call $-G$: + +$$\begin{aligned} + \dv{p}{z} + = -G + \qquad \quad + \eta \frac{1}{r} \dv{r} \Big( r \dv{v_z}{r} \Big) + = - G +\end{aligned}$$ + +The former equation, for $p(z)$, is easy to solve. +We get an integration constant $p(0)$: + +$$\begin{aligned} + p(z) + = p(0) - G z +\end{aligned}$$ + +This gives meaning to the **pressure gradient** $G$: +for a pipe of length $L$, +it describes the pressure difference $\Delta p = p(0) - p(L)$ +that is driving the fluid, +i.e. $G = \Delta p / L$ + +As for the latter equation, for $v_z(r)$, +we start by integrating it once, introducing a constant $A$: + +$$\begin{aligned} + \dv{r} \Big( r \dv{v_z}{r} \Big) + = - \frac{G}{\eta} r + \quad \implies \quad + \dv{v_z}{r} + = - \frac{G}{2 \eta} r + \frac{A}{r} +\end{aligned}$$ + +Integrating this one more time, +thereby introducing another constant $B$, +we arrive at: + +$$\begin{aligned} + v_z + = - \frac{G}{4 \eta} r^2 + A \ln{r} + B +\end{aligned}$$ + +The velocity must be finite at $r = 0$, so we set $A = 0$. +Furthermore, the Navier-Stokes equation's *no-slip* condition +demands that $v_z = 0$ at the boundary $r = R$, +so $B = G R^2 / (4 \eta)$. +This brings us to the **Poiseuille solution** for $v_z(r)$: + +$$\begin{aligned} + \boxed{ + v_z(r) + = \frac{G}{4 \eta} (R^2 - r^2) + } +\end{aligned}$$ + +How much fluid can pass through the pipe per unit time? +This is denoted by the **volumetric flow rate** $Q$, +which is the integral of $v_z$ over the circular cross-section: + +$$\begin{aligned} + Q + = 2 \pi \int_0^R v_z(r) \: r \dd{r} + = \frac{\pi G}{2 \eta} \int_0^R R^2 r - r^3 \dd{r} + = \frac{\pi G}{2 \eta} \bigg[ \frac{R^2 r^2}{2} - \frac{r^4}{4} \bigg]_0^R +\end{aligned}$$ + +We thus arrive at the main Hagen-Poiseuille equation, +which predicts $Q$ for a given setup: + +$$\begin{aligned} + \boxed{ + Q + = \frac{\pi G R^4}{8 \eta} + } +\end{aligned}$$ + +Consequently, the average flow velocity $\expval{v_z}$ +is simply $Q$ divided by the cross-sectional area: + +$$\begin{aligned} + \expval{v_z} + = \frac{Q}{\pi R^2} + = \frac{G R^2}{8 \eta} +\end{aligned}$$ + +The fluid's viscous stickiness means it exerts a drag force $D$ +on the pipe as it flows. For a pipe of length $L$ and radius $R$, +we calculate $D$ by multiplying the internal area $2 \pi R L$ +by the [shear stress](/know/concept/cauchy-stress-tensor/) +$-\sigma_{zr}$ on the wall +(i.e. the wall applies $\sigma_{zr}$, the fluid responds with $- \sigma_{zr}$): + +$$\begin{aligned} + D + = - 2 \pi R L \: \sigma_{zr} \big|_{r = R} + = - 2 \pi R L \eta \dv{v_z}{r}\Big|_{r = R} + = 2 \pi R L \eta \frac{G R}{2 \eta} + = \pi R^2 L G +\end{aligned}$$ + +We would like to get rid of $G$ for being impractical, +so we substitute $R^2 G = 8 \eta \expval{v_z}$, yielding: + +$$\begin{aligned} + \boxed{ + D + = 8 \pi \eta L \expval{v_z} + } +\end{aligned}$$ + +Due to this drag, the pressure difference $\Delta p = p(0) - p(L)$ +does work on the fluid, at a rate $P$, +since power equals force (i.e. pressure times area) times velocity: + +$$\begin{aligned} + P + = 2 \pi \int_0^R \Delta p \: v_z(r) \: r \dd{r} +\end{aligned}$$ + +Because $\Delta p$ is independent of $r$, +we get the same integral we used to calculate $Q$. +Then, thanks to the fact that $\Delta p = G L$ +and $Q = \pi R^2 \expval{v_z}$, it follows that: + +$$\begin{aligned} + P + = \Delta p \: Q + = G L \pi R^2 \expval{v_z} + = D \expval{v_z} +\end{aligned}$$ + +In conclusion, the power $P$, +needed to drive a fluid through the pipe at a rate $Q$, +is given by: + +$$\begin{aligned} + \boxed{ + P + = 8 \pi \eta L \expval{v_z}^2 + } +\end{aligned}$$ + + + +## References +1. B. Lautrup, + *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition, + CRC Press. diff --git a/content/know/concept/navier-stokes-equations/index.pdc b/content/know/concept/navier-stokes-equations/index.pdc new file mode 100644 index 0000000..7256b7e --- /dev/null +++ b/content/know/concept/navier-stokes-equations/index.pdc @@ -0,0 +1,134 @@ +--- +title: "Navier-Stokes equations" +firstLetter: "N" +publishDate: 2021-04-12 +categories: +- Physics +- Fluid mechanics +- Fluid dynamics + +date: 2021-04-12T13:14:09+02:00 +draft: false +markup: pandoc +--- + +# Navier-Stokes equations + +While the [Euler equations](/know/concept/euler-equations/) govern *ideal* "dry" fluids, +the **Navier-Stokes equations** govern *nonideal* "wet" fluids, +i.e. fluids with nonzero [viscosity](/know/concept/viscosity/). + + +## Incompressible fluid + +First of all, we can reuse the incompressibility condition for ideal fluids, without modifications: + +$$\begin{aligned} + \boxed{ + \nabla \cdot \va{v} = 0 + } +\end{aligned}$$ + +Furthermore, from the derivation of the Euler equations, +we know that Newton's second law can be written as follows, +for an infinitesimal particle of the fluid: + +$$\begin{aligned} + \rho \frac{\mathrm{D} \va{v}}{\mathrm{D} t} + = \va{f^*} +\end{aligned}$$ + +$\mathrm{D}/\mathrm{D}t$ is the [material derivative](/know/concept/material-derivative/), +$\rho$ is the density, and $\va{f^*}$ is the effective force density, +expressed in terms of an external body force $\va{f}$ (e.g. gravity) +and the [Cauchy stress tensor](/know/concept/cauchy-stress-tensor/) $\hat{\sigma}$: + +$$\begin{aligned} + \va{f^*} + = \va{f} + \nabla \cdot \hat{\sigma}^\top +\end{aligned}$$ + +From the definition of viscosity, +the stress tensor's elements are like so for a Newtonian fluid: + +$$\begin{aligned} + \sigma_{ij} + = - p \delta_{ij} + \eta (\nabla_i v_j + \nabla_j v_i) +\end{aligned}$$ + +Where $\eta$ is the dynamic viscosity. +Inserting this, we calculate $\nabla \cdot \hat{\sigma}^\top$ in index notation: + +$$\begin{aligned} + \big( \nabla \cdot \hat{\sigma}^\top \big)_i + = \sum_{j} \nabla_j \sigma_{ij} + &= \sum_{j} \Big( \!-\! \delta_{ij} \nabla_j p + \eta (\nabla_i \nabla_j v_j + \nabla_j^2 v_i) \Big) + \\ + &= - \nabla_i p + \eta \nabla_i \sum_{j} \nabla_j v_j + \eta \sum_{j} \nabla_j^2 v_i +\end{aligned}$$ + +Thanks to incompressibility $\nabla \cdot \va{v} = 0$, +the middle term vanishes, leaving us with: + +$$\begin{aligned} + \va{f^*} + = \va{f} - \nabla p + \eta \nabla^2 \va{v} +\end{aligned}$$ + +We assume that the only body force is gravity $\va{f} = \rho \va{g}$. +Newton's second law then becomes: + +$$\begin{aligned} + \rho \frac{\mathrm{D} \va{v}}{\mathrm{D} t} + = \rho \va{g} - \nabla p + \eta \nabla^2 \va{v} +\end{aligned}$$ + +Dividing by $\rho$, and replacing $\eta$ +with the kinematic viscosity $\nu = \eta/\rho$, +yields the main equation: + +$$\begin{aligned} + \boxed{ + \frac{\mathrm{D} \va{v}}{\mathrm{D} t} + = \va{g} - \frac{\nabla p}{\rho} + \nu \nabla^2 \va{v} + } +\end{aligned}$$ + +Finally, we can optionally allow incompressible fluids +with an inhomogeneous "lumpy" density $\rho$, +by demanding conservation of mass, +just like for the Euler equations: + +$$\begin{aligned} + \boxed{ + \frac{\mathrm{D} \rho}{\mathrm{D} t} + = 0 + } +\end{aligned}$$ + +Putting it all together, the Navier-Stokes equations for an incompressible fluid are given by: + +$$\begin{aligned} + \boxed{ + \frac{\mathrm{D} \va{v}}{\mathrm{D} t} + = \va{g} - \frac{\nabla p}{\rho} + \nu \nabla^2 \va{v} + \qquad + \nabla \cdot \va{v} = 0 + \qquad + \frac{\mathrm{D} \rho}{\mathrm{D} t} + = 0 + } +\end{aligned}$$ + +Due to the definition of viscosity $\nu$ as the molecular "stickiness", +we have boundary conditions for the velocity field $\va{v}$: +at any interface, $\va{v}$ must be continuous. +Likewise, Newton's third law demands that the normal component +of stress $\hat{\sigma} \cdot \vu{n}$ is continuous there. + + + +## References +1. B. Lautrup, + *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition, + CRC Press. diff --git a/content/know/concept/renyi-entropy/index.pdc b/content/know/concept/renyi-entropy/index.pdc new file mode 100644 index 0000000..073d773 --- /dev/null +++ b/content/know/concept/renyi-entropy/index.pdc @@ -0,0 +1,114 @@ +--- +title: "Rényi entropy" +firstLetter: "R" +publishDate: 2021-04-11 +categories: +- Cryptography + +date: 2021-04-11T15:13:07+02:00 +draft: false +markup: pandoc +--- + +# Rényi entropy + +In information theory, the **Rényi entropy** is a measure +(or family of measures) of the "suprise" or "information" +contained in a random variable $X$. +It is defined as follows: + +$$\begin{aligned} + \boxed{ + H_\alpha(X) + = \frac{1}{1 - \alpha} \log\!\bigg( \sum_{i = 1}^N p_i^\alpha \bigg) + } +\end{aligned}$$ + +Where $\alpha \ge 0$ is a free parameter. +The logarithm is usually base-2, but variations exist. + +The case $\alpha = 0$ is known as the **Hartley entropy** or **max-entropy**, +and quantifies the "surprise" of an event from $X$, +if $X$ is uniformly distributed: + +$$\begin{aligned} + \boxed{ + H_0(X) + = \log N + } +\end{aligned}$$ + +Where $N$ is the cardinality of $X$; the number of different possible events. +The most famous case, however, is $\alpha = 1$. +Since $H_\alpha$ is problematic for $\alpha \to 1$, we must take the limit: + +$$\begin{aligned} + H_1(X) + = \lim_{\alpha \to 1} H_\alpha(X) + = \lim_{\alpha \to 1} \frac{\log\!\left( \sum_i p_i^\alpha \right)}{1 - \alpha} +\end{aligned}$$ + +We then apply L'Hôpital's rule to evaluate this limit, +and use the fact that all $p_i$ sum to $1$: + +$$\begin{aligned} + H_1(X) + = \lim_{\alpha \to 1} \frac{\dv{\alpha} \log\!\left( \sum_i p_i^\alpha \right)}{\dv{\alpha} (1 - \alpha)} + = \lim_{\alpha \to 1} \frac{\sum_i p_i^\alpha \log p_i}{- \sum_i p_i^\alpha} + = - \sum_{i = 1}^N p_i \log p_i +\end{aligned}$$ + +This quantity is the **Shannon entropy**, +which is the most general measure of "surprise": + +$$\begin{aligned} + \boxed{ + H_1(X) + = \lim_{\alpha \to 1} H_\alpha(X) + = - \sum_{i = 1}^N p_i \log p_i + } +\end{aligned}$$ + +Next, for $\alpha = 2$, we get the **collision entropy**, which describes +the surprise of two independent and identically distributed variables +$X$ and $Y$ yielding the same event: + +$$\begin{aligned} + \boxed{ + H_2(X) + = - \log\!\bigg( \sum_{i = 1}^N p_i^2 \bigg) + = - \log P(X = Y) + } +\end{aligned}$$ + +Finally, in the limit $\alpha \to \infty$, +the largest probability dominates the sum, +leading to the definition of the **min-entropy** $H_\infty$, +describing the surprise of the most likely event: + +$$\begin{aligned} + \boxed{ + H_\infty(X) + = \lim_{\alpha \to \infty} H_\alpha(x) + = - \log\!\big( \max_{i} p_i \big) + } +\end{aligned}$$ + +It is straightforward to convince yourself that these entropies +are ordered in the following way: + +$$\begin{aligned} + H_0 \ge H_1 \ge H_2 \ge H_\infty +\end{aligned}$$ + +In other words, from left to right, +they go from permissive to conservative, roughly speaking. + + +## References +1. P.A. Bromiley, N.A. Thacker, E. Bouhova-Thacker, + [Shannon entropy, Rényi entropy, and information](https://www.researchgate.net/publication/253537416_Shannon_Entropy_Renyi_Entropy_and_Information), + 2010, University of Manchester. +2. J.B. Brask, + *Quantum information: lecture notes*, + 2021, unpublished. diff --git a/content/know/concept/viscosity/index.pdc b/content/know/concept/viscosity/index.pdc new file mode 100644 index 0000000..d7abd7a --- /dev/null +++ b/content/know/concept/viscosity/index.pdc @@ -0,0 +1,100 @@ +--- +title: "Viscosity" +firstLetter: "V" +publishDate: 2021-04-12 +categories: +- Physics +- Fluid mechanics +- Fluid dynamics + +date: 2021-04-12T13:14:16+02:00 +draft: false +markup: pandoc +--- + +# Viscosity + +The **viscosity** of a fluid describes how +"sticky" its constituent molecules are; +when one part of the fluid moves, it "drags" +neighbouring parts by an amount proportional to the viscosity. + +Imagine a liquid in a canal, +flowing in the $x$-direction at a velocity $v(z)$ +as a function of depth $z$. +Due to the liquid's viscosity, +its molecules are "stuck" to the bottom of the canal $z = 0$, +such that it is stationary there $v(0) = 0$. +However, at the surface $z = z_s$, there is a flow at $v(z_s) = v_s$. + +This difference in $v$ means that there is a velocity gradient across $z$. +Each infinitesimal layer of the liquid +is dragging on the layers above and below it, +meaning there is a nonzero shear stress $\sigma_{xz}$ +(see [Cauchy stress tensor](/know/concept/cauchy-stress-tensor/)). +Formally, the **dynamic viscosity** $\eta$ is defined as follows: + +$$\begin{aligned} + \boxed{ + \sigma_{xz} + = \eta \dv{v}{z} + } +\end{aligned}$$ + +This is **Newton's law of viscosity**, +and fluids obeying it are known as **Newtonian**. +In a Newtonian fluid *at rest*, there are no such shear stresses, +and the Cauchy stress tensor $\hat{\sigma}$ is diagonal: + +$$\begin{aligned} + \sigma_{ij} = - p \delta_{ij} +\end{aligned}$$ + +Where $p$ is the pressure, and $\delta_{ij}$ is the Kronecker delta. +If the fluid flows according to a velocity field $\va{v}$, +then a more general definition of $\eta$ is as follows, +in index notation with $\nabla_i \!=\! \pdv*{x_i}$: + +$$\begin{aligned} + \boxed{ + \sigma_{ij} + = - p \delta_{ij} + \eta (\nabla_i v_j + \nabla_j v_i) + } +\end{aligned}$$ + +The double term $\nabla_i v_j + \nabla_j v_i$ comes from the fact that +the stress tensor of a Newtonian fluid is always symmetric; +this definition of $\sigma_{ij}$ enforces that. + +Another quantity is the **kinematic viscosity** $\nu$, +which is simply $\eta$ divided by the density $\rho$: + +$$\begin{aligned} + \boxed{ + \nu + \equiv \frac{\eta}{\rho} + } +\end{aligned}$$ + +With this, Newton's law of viscosity is written +using the momentum density $P = \rho v$: + +$$\begin{aligned} + \sigma_{xz} + = \nu \dv{P}{z} +\end{aligned}$$ + +Because momentum is "more fundamental" than velocity, +is $\nu$ often more useful than $\eta$. +However, this comes at the cost of our intuition: +for example, as you would expect, $\eta_\mathrm{water} > \eta_\mathrm{air}$, +but you may be surprised that $\nu_\mathrm{water} < \nu_\mathrm{air}$. +Since air is less dense, it is easier to set in motion, +hence we expect it to be less viscous than water, +but in fact air's molecules are stickier than water's. + + +## References +1. B. Lautrup, + *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition, + CRC Press. |