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Diffstat (limited to 'content/know')
| -rw-r--r-- | content/know/concept/bernoullis-theorem/index.pdc | 9 | ||||
| -rw-r--r-- | content/know/concept/density-of-states/index.pdc | 156 | ||||
| -rw-r--r-- | content/know/concept/hydrostatic-pressure/index.pdc | 12 | ||||
| -rw-r--r-- | content/know/concept/maxwell-boltzmann-distribution/index.pdc | 219 | ||||
| -rw-r--r-- | content/know/concept/newtons-bucket/index.pdc | 97 | ||||
| -rw-r--r-- | content/know/concept/stokes-law/index.pdc | 6 | ||||
| -rw-r--r-- | content/know/concept/superdense-coding/index.pdc | 33 |
7 files changed, 511 insertions, 21 deletions
diff --git a/content/know/concept/bernoullis-theorem/index.pdc b/content/know/concept/bernoullis-theorem/index.pdc index 5ff5225..fbabff2 100644 --- a/content/know/concept/bernoullis-theorem/index.pdc +++ b/content/know/concept/bernoullis-theorem/index.pdc @@ -27,13 +27,8 @@ $$\begin{aligned} = \va{g} - \frac{\nabla p}{\rho} \end{aligned}$$ -Assuming that $\va{v}$ and $\va{g}$ are constant in $t$, -it becomes clear that a higher $\va{v}$ requires a lower $p$: - -$$\begin{aligned} - \frac{1}{2} \nabla \va{v}^2 - = \va{g} - \frac{\nabla p}{\rho} -\end{aligned}$$ +Assuming that $\va{v}$ is constant in $t$, +it becomes clear that a higher $\va{v}$ requires a lower $p$. ## Simple form diff --git a/content/know/concept/density-of-states/index.pdc b/content/know/concept/density-of-states/index.pdc new file mode 100644 index 0000000..195ac2a --- /dev/null +++ b/content/know/concept/density-of-states/index.pdc @@ -0,0 +1,156 @@ +--- +title: "Density of states" +firstLetter: "D" +publishDate: 2021-05-08 +categories: +- Physics +- Statistics + +date: 2021-05-08T18:35:46+02:00 +draft: false +markup: pandoc +--- + +# Density of states + +The **density of states** $g(E)$ of a physical system is defined such that +$g(E) \dd{E}$ is the number of states which could be occupied +with an energy in the interval $[E, E + \dd{E}]$. +In fact, $E$ need not be an energy; +it should just be something that effectively identifies the state. + +In its simplest form, the density of states is as follows, +where $\Gamma(E)$ is the number of states with energy +less than or equal to the argument $E$: + +$$\begin{aligned} + g(E) + = \dv{\Gamma}{E} +\end{aligned}$$ + +If the states can be treated as waves, +which is often the case, +then we can calculate the density of states $g(k)$ in +$k$-space, i.e. as a function of the wavenumber $k = |\vb{k}|$. +Once we have $g(k)$, we use the dispersion relation $E(k)$ to find $g(E)$, +by demanding that: + +$$\begin{aligned} + g(k) \dd{k} = g(E) \dd{E} + \quad \implies \quad + g(E) + = g(k) \dv{k}{E} +\end{aligned}$$ + +Inverting the dispersion relation $E(k)$ to get $k(E)$ might be difficult, +in which case the left-hand equation can be satisfied numerically. + + +Define $\Omega_n(k)$ as the number of states with +a $k$-value less than or equal to the argument, +or in other words, the volume of a hypersphere with radius $k$. +Then the $n$-dimensional density of states $g_n(k)$ +has the following general form: + +$$\begin{aligned} + \boxed{ + g_n(k) + = \frac{D}{2^n k_{\mathrm{min}}^n} \: \dv{\Omega_n}{k} + } +\end{aligned}$$ + +Where $D$ is each state's degeneracy (e.g. due to spin), +and $k_{\mathrm{min}}$ is the smallest allowed $k$-value, +according to the characteristic length $L$ of the system. +We divide by $2^n$ to limit ourselves to the sector where all axes are positive, +because we are only considering the magnitude of $k$. + +In one dimension $n = 1$, the number of states within a distance $k$ from the +origin is the distance from $k$ to $-k$ +(we let it run negative, since its meaning does not matter here), given by: + +$$\begin{aligned} + \Omega_1(k) + = 2 k +\end{aligned}$$ + +To get $k_{\mathrm{min}}$, we choose to look at a rod of length $L$, +across which the function is a standing wave, meaning that +the allowed values of $k$ must be as follows, where $m \in \mathbb{N}$: + +$$\begin{aligned} + \lambda = \frac{2 L}{m} + \quad \implies \quad + k = \frac{2 \pi}{\lambda} = \frac{m \pi}{L} +\end{aligned}$$ + +Take the smallest option $m = 1$, +such that $k_{\mathrm{min}} = \pi / L$, +the 1D density of states $g_1(k)$ is: + +$$\begin{aligned} + \boxed{ + g_1(k) + = \frac{D L}{2 \pi} \: 2 + = \frac{D L}{\pi} + } +\end{aligned}$$ + +In 2D, the number of states within a range $k$ of the +origin is the area of a circle with radius $k$: + +$$\begin{aligned} + \Omega_2(k) + = \pi k^2 +\end{aligned}$$ + +Analogously to the 1D case, +we take the system to be a square of side $L$, +so $k_{\mathrm{min}} = \pi / L$ again. +The density of states then becomes: + +$$\begin{aligned} + \boxed{ + g_2(k) + = \frac{D L^2}{4 \pi^2} \:2 \pi k + = \frac{D L^2 k}{2 \pi} + } +\end{aligned}$$ + +In 3D, the number of states is the volume of a sphere with radius $k$: + +$$\begin{aligned} + \Omega_3(k) + = \frac{4 \pi}{3} k^3 +\end{aligned}$$ + +For a cube with side $L$, we once again find $k_{\mathrm{min}} = \pi / L$. +We thus get: + +$$\begin{aligned} + \boxed{ + g_3(k) + = \frac{D L^3}{8 \pi^3} \:4 \pi k^2 + = \frac{D L^3 k^2}{2 \pi^2} + } +\end{aligned}$$ + +All these expressions contain the characteristic length/area/volume $L^n$, +and therefore give the number of states in that region only. +Keep in mind that $L$ is free to choose; +it need not be the physical size of the system. +In fact, we typically want the density of states +per unit length/area/volume, +so we can just set $L = 1$ in our preferred unit of distance. + +If the system is infinitely large, or if it has periodic boundaries, +then $k$ becomes a continuous variable and $k_\mathrm{min} \to 0$. +But again, $L$ is arbitrary, +so a finite value can be chosen. + + + +## References +1. H. Gould, J. Tobochnik, + *Statistical and thermal physics*, 2nd edition, + Princeton. diff --git a/content/know/concept/hydrostatic-pressure/index.pdc b/content/know/concept/hydrostatic-pressure/index.pdc index 90e57ce..001a198 100644 --- a/content/know/concept/hydrostatic-pressure/index.pdc +++ b/content/know/concept/hydrostatic-pressure/index.pdc @@ -141,8 +141,10 @@ $$\begin{aligned} With this, the equilibrium condition is turned into the following equation: $$\begin{aligned} - \nabla \Phi + \frac{\nabla p}{\rho} - = 0 + \boxed{ + \nabla \Phi + \frac{\nabla p}{\rho} + = 0 + } \end{aligned}$$ In practice, the density $\rho$ of the fluid @@ -156,7 +158,7 @@ the indefinite integral of the density: $$\begin{aligned} w(p) - = \int \frac{1}{\rho(p)} \dd{p} + \equiv \int \frac{1}{\rho(p)} \dd{p} \end{aligned}$$ Using this, we can rewrite the equilibrium condition as a single gradient like so: @@ -172,9 +174,7 @@ From this, let us now define the **effective gravitational potential** $\Phi^*$ as follows: $$\begin{aligned} - \boxed{ - \Phi^* = \Phi + w(p) - } + \Phi^* \equiv \Phi + w(p) \end{aligned}$$ This results in the cleanest form yet of the equilibrium condition, namely: diff --git a/content/know/concept/maxwell-boltzmann-distribution/index.pdc b/content/know/concept/maxwell-boltzmann-distribution/index.pdc new file mode 100644 index 0000000..082d2db --- /dev/null +++ b/content/know/concept/maxwell-boltzmann-distribution/index.pdc @@ -0,0 +1,219 @@ +--- +title: "Maxwell-Boltzmann distribution" +firstLetter: "M" +publishDate: 2021-05-08 +categories: +- Physics +- Statistics + +date: 2021-05-08T18:35:37+02:00 +draft: false +markup: pandoc +--- + +# Maxwell-Boltzmann distribution + +The **Maxwell-Boltzmann distributions** are a set of closely related +probability distributions with applications in classical statistical physics. + + +## Velocity vector distribution + +In the canonical ensemble +(where a fixed-size system can exchange energy with its environment), +the probability of a microstate with energy $E$ is given by the Boltzmann distribution: + +$$\begin{aligned} + f(E) + \:\propto\: \exp\!\big(\!-\! \beta E\big) +\end{aligned}$$ + +Where $\beta = 1 / k_B T$. +We split $E = K + U$, +where $K$ and $U$ are the total kinetic and potential energy contributions. +If there are $N$ particles in the system, +with positions $\tilde{r} = (\vec{r}_1, ..., \vec{r}_N)$ +and momenta $\tilde{p} = (\vec{p}_1, ..., \vec{p}_N)$, +then $K$ only depends on $\tilde{p}$, +and $U$ only depends on $\tilde{r}$, +so the probability of a specific microstate +$(\tilde{r}, \tilde{p})$ is as follows: + +$$\begin{aligned} + f(\tilde{r}, \tilde{p}) + \:\propto\: \exp\!\Big(\!-\! \beta \big( K(\tilde{p}) + U(\tilde{r}) \big) \Big) +\end{aligned}$$ + +Since this is classical physics, +we can split the exponential. +In quantum mechanics, +the canonical commutation relation would prevent that. +Anyway, splitting yields: + +$$\begin{aligned} + f(\tilde{r}, \tilde{p}) + \:\propto\: \exp\!\big(\!-\! \beta K(\tilde{p}) \big) \exp\!\big(\!-\! \beta U(\tilde{r}) \big) +\end{aligned}$$ + +Classically, the probability +distributions of the momenta and positions are independent: + +$$\begin{aligned} + f_K(\tilde{p}) + \:\propto\: \exp\!\big(\!-\! \beta K(\tilde{p}) \big) + \qquad + f_U(\tilde{r}) + \:\propto\: \exp\!\big(\!-\! \beta U(\tilde{r}) \big) +\end{aligned}$$ + +We cannot evaluate $f_U(\tilde{r})$ further without knowing $U(\tilde{r})$ for a system. +We thus turn to $f_K(\tilde{p})$, and see that the total kinetic +energy $K(\tilde{p})$ is simply the sum of the particles' individual +kinetic energies $K_n(\vec{p}_n)$, which are well-known: + +$$\begin{aligned} + K(\tilde{p}) + = \sum_{n = 1}^N K_n(\vec{p}_n) + \qquad \mathrm{where} \qquad + K_n(\vec{p}_n) + = \frac{|\vec{p}_n|^2}{2 m} +\end{aligned}$$ + +Consequently, the probability distribution $f(p_x, p_y, p_z)$ for the +momentum vector of a single particle is as follows, +after normalization: + +$$\begin{aligned} + f(p_x, p_y, p_z) + = \Big( \frac{1}{2 \pi m k_B T} \Big)^{3/2} \exp\!\Big( \!-\!\frac{(p_x^2 + p_y^2 + p_z^2)}{2 m k_B T} \Big) +\end{aligned}$$ + +We now rewrite this using the velocities $v_x = p_x / m$, +and update the normalization, giving: + +$$\begin{aligned} + \boxed{ + f(v_x, v_y, v_z) + = \Big( \frac{m}{2 \pi k_B T} \Big)^{3/2} \exp\!\Big( \!-\!\frac{m (v_x^2 + v_y^2 + v_z^2)}{2 k_B T} \Big) + } +\end{aligned}$$ + +This is the **Maxwell-Boltzmann velocity vector distribution**. +Clearly, this is a product of three exponentials, +so the velocity in each direction is independent of the others: + +$$\begin{aligned} + f(v_x) + = \sqrt{\frac{m}{2 \pi k_B T}} \exp\!\Big( \!-\!\frac{m v_x^2}{2 k_B T} \Big) +\end{aligned}$$ + +The distribution is thus an isotropic gaussian with standard deviations given by: + +$$\begin{aligned} + \sigma_x = \sigma_y = \sigma_z + = \sqrt{\frac{k_B T}{m}} +\end{aligned}$$ + + +## Speed distribution + +We know the distribution of the velocities along each axis, +but what about the speed $v = |\vec{v}|$? +Because we do not care about the direction of $\vec{v}$, only its magnitude, +the [density of states](/know/concept/density-of-states/) $g(v)$ is not constant: +it is the rate-of-change of the volume of a sphere of radius $v$: + +$$\begin{aligned} + g(v) + = \dv{v} \Big( \frac{4 \pi}{3} v^3 \Big) + = 4 \pi v^2 +\end{aligned}$$ + +Multiplying the velocity vector distribution by $g(v)$ +and substituting $v^2 = v_x^2 + v_y^2 + v_z^2$ +then gives us the **Maxwell-Boltzmann speed distribution**: + +$$\begin{aligned} + \boxed{ + f(v) + = 4 \pi \Big( \frac{m}{2 \pi k_B T} \Big)^{3/2} v^2 \exp\!\Big( \!-\!\frac{m v^2}{2 k_B T} \Big) + } +\end{aligned}$$ + +Some notable points on this distribution are +the most probable speed $v_{\mathrm{mode}}$, +the mean average speed $v_{\mathrm{mean}}$, +and the root-mean-square speed $v_{\mathrm{rms}}$: + +$$\begin{aligned} + f'(v_\mathrm{mode}) + = 0 + \qquad + v_\mathrm{mean} + = \int_0^\infty v \: f(v) \dd{v} + \qquad + v_\mathrm{rms} + = \bigg( \int_0^\infty v^2 \: f(v) \dd{v} \bigg)^{1/2} +\end{aligned}$$ + +Which can be calculated to have the following exact expressions: + +$$\begin{aligned} + \boxed{ + v_{\mathrm{mode}} + = \sqrt{\frac{2 k_B T}{m}} + } + \qquad + \boxed{ + v_{\mathrm{mean}} + = \sqrt{\frac{8 k_B T}{\pi m}} + } + \qquad + \boxed{ + v_{\mathrm{rms}} + = \sqrt{\frac{3 k_B T}{m}} + } +\end{aligned}$$ + + +## Kinetic energy distribution + +Using the speed distribution, +we can work out the kinetic energy distribution. +Because $K$ is not proportional to $v$, +we must do this by demanding that: + +$$\begin{aligned} + f(K) \dd{K} + = f(v) \dd{v} + \quad \implies \quad + f(K) + = f(v) \dv{v}{K} +\end{aligned}$$ + +We know that $K = m v^2 / 2$, +meaning $\dd{K} = m v \dd{v}$ +so the energy distribution $f(K)$ is: + +$$\begin{aligned} + f(K) + = \frac{f(v)}{m v} + = \sqrt{\frac{2 m}{\pi}} \: \bigg( \frac{1}{k_B T} \bigg)^{3/2} v \exp\!\Big( \!-\!\frac{m v^2}{2 k_B T} \Big) +\end{aligned}$$ + +Substituting $v = \sqrt{2 K/m}$ leads to +the **Maxwell-Boltzmann kinetic energy distribution**: + +$$\begin{aligned} + \boxed{ + f(K) + = 2 \sqrt{\frac{K}{\pi}} \: \bigg( \frac{1}{k_B T} \bigg)^{3/2} \exp\!\Big( \!-\!\frac{K}{k_B T} \Big) + } +\end{aligned}$$ + + + +## References +1. H. Gould, J. Tobochnik, + *Statistical and thermal physics*, 2nd edition, + Princeton. diff --git a/content/know/concept/newtons-bucket/index.pdc b/content/know/concept/newtons-bucket/index.pdc new file mode 100644 index 0000000..3f074f5 --- /dev/null +++ b/content/know/concept/newtons-bucket/index.pdc @@ -0,0 +1,97 @@ +--- +title: "Newton's bucket" +firstLetter: "N" +publishDate: 2021-05-13 +categories: +- Physics +- Fluid mechanics +- Fluid statics + +date: 2021-05-13T17:06:45+02:00 +draft: false +markup: pandoc +--- + +# Newton's bucket + +**Newton's bucket** is a cylindrical bucket +that rotates at angular velocity $\omega$. +Due to [viscosity](/know/concept/viscosity/), +any liquid in the bucket is affected by the rotation, +eventually achieving the exact same $\omega$. + +However, once in equilibrium, the liquid's surface is not flat, +but curved upwards from the center. +This is due to the centrifugal force $\va{F}_\mathrm{f} = m \va{f}$ on a molecule with mass $m$: + +$$\begin{aligned} + \va{f} + = \omega^2 \va{r} +\end{aligned}$$ + +Where $\va{r}$ is the molecule's position relative to the axis of rotation. +This (fictitious) force can be written as the gradient +of a potential $\Phi_\mathrm{f}$, such that $\va{f} = - \nabla \Phi_\mathrm{f}$: + +$$\begin{aligned} + \Phi_\mathrm{f} + = - \frac{\omega^2}{2} r^2 + = - \frac{\omega^2}{2} (x^2 + y^2) +\end{aligned}$$ + +In addition, each molecule feels a gravitational force $\va{F}_\mathrm{g} = m \va{g}$, +where $\va{g} = - \nabla \Phi_\mathrm{g}$: + +$$\begin{aligned} + \Phi_\mathrm{g} + = \mathrm{g} z +\end{aligned}$$ + +Overall, the molecule therefore feels an "effective" force +with a potential $\Phi$ given by: + +$$\begin{aligned} + \Phi + = \Phi_\mathrm{g} + \Phi_\mathrm{f} + = \mathrm{g} z - \frac{\omega^2}{2} (x^2 + y^2) +\end{aligned}$$ + +At equilibrium, the [hydrostatic pressure](/know/concept/hydrostatic-pressure/) $p$ +in the liquid is the one that satisfies: + +$$\begin{aligned} + \frac{\nabla p}{\rho} + = - \nabla \Phi +\end{aligned}$$ + +Removing the gradients gives integration constants $p_0$ and $\Phi_0$, +so the equilibrium equation is: + +$$\begin{aligned} + p - p_0 + = - \rho (\Phi - \Phi_0) +\end{aligned}$$ + +We isolate this for $p$ and rewrite $\Phi_0 = \mathrm{g} z_0$, +where $z_0$ is the liquid height at the center: + +$$\begin{aligned} + p + = p_0 - \rho \mathrm{g} (z - z_0) + \frac{\omega^2}{2} \rho (x^2 + y^2) +\end{aligned}$$ + +At the surface, we demand that $p = p_0$, where $p_0$ is the air pressure. +The $z$-coordinate at which this is satisfied is as follows, +telling us that the surface is parabolic: + +$$\begin{aligned} + z + = z_0 + \frac{\omega^2}{2 \mathrm{g}} (x^2 + y^2) +\end{aligned}$$ + + + +## References +1. B. Lautrup, + *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition, + CRC Press. diff --git a/content/know/concept/stokes-law/index.pdc b/content/know/concept/stokes-law/index.pdc index acc98af..dee40d6 100644 --- a/content/know/concept/stokes-law/index.pdc +++ b/content/know/concept/stokes-law/index.pdc @@ -356,8 +356,10 @@ This is an equation for the **terminal velocity** $U_t$, which we find to be as follows: $$\begin{aligned} - U_t - = \frac{2 a^2 (\rho_s - \rho_f) g_0}{9 \eta} + \boxed{ + U_t + = \frac{2 a^2 (\rho_s - \rho_f) g_0}{9 \eta} + } \end{aligned}$$ The falling sphere will accelerate until $U_t$, diff --git a/content/know/concept/superdense-coding/index.pdc b/content/know/concept/superdense-coding/index.pdc index e50be2b..1a9337c 100644 --- a/content/know/concept/superdense-coding/index.pdc +++ b/content/know/concept/superdense-coding/index.pdc @@ -32,12 +32,33 @@ Based on the values of the two classical bits $(a_1, a_2)$, Alice performs the following operations on her side $A$ of the Bell state: -| $(a_1, a_2)$ | Operator $\qquad$ | Result | -|:--:|:--|:---------| -| $00$ | $\hat{I}$ | $\ket*{\Phi^{+}} = \frac{1}{\sqrt{2}} \Big(\ket{0}_A \ket{0}_B + \ket{1}_A \ket{1}_B \Big)$ | -| $01$ | $\hat{\sigma}_z$ | $\ket*{\Phi^{-}} = \frac{1}{\sqrt{2}} \Big(\ket{0}_A \ket{0}_B - \ket{1}_A \ket{1}_B \Big)$ | -| $10$ | $\hat{\sigma}_x$ | $\ket*{\Psi^{+}} = \frac{1}{\sqrt{2}} \Big(\ket{0}_A \ket{1}_B + \ket{1}_A \ket{0}_B \Big)$ | -| $11$ | $\hat{\sigma}_x \hat{\sigma}_z$ | $\ket*{\Psi^{-}} = \frac{1}{\sqrt{2}} \Big(\ket{0}_A \ket{1}_B - \ket{1}_A \ket{0}_B \Big)$ | +<table style="width:70%;margin:auto;text-align:center;"> + <tr> + <th>$(a_1, a_2)$</th> + <th>Operator</th> + <th>Result</th> + </tr> + <tr> + <td>$00$</td> + <td>$\hat{I}$</td> + <td>$\ket*{\Phi^{+}} = \frac{1}{\sqrt{2}} \Big(\ket{0}_A \ket{0}_B + \ket{1}_A \ket{1}_B \Big)$</td> + </tr> + <tr> + <td>$01$</td> + <td>$\hat{\sigma}_z$</td> + <td>$\ket*{\Phi^{-}} = \frac{1}{\sqrt{2}} \Big(\ket{0}_A \ket{0}_B - \ket{1}_A \ket{1}_B \Big)$</td> + </tr> + <tr> + <td>$10$</td> + <td>$\hat{\sigma}_x$</td> + <td>$\ket*{\Psi^{+}} = \frac{1}{\sqrt{2}} \Big(\ket{0}_A \ket{1}_B + \ket{1}_A \ket{0}_B \Big)$</td> + </tr> + <tr> + <td>$11$</td> + <td>$\hat{\sigma}_x \hat{\sigma}_z$</td> + <td>$\ket*{\Psi^{-}} = \frac{1}{\sqrt{2}} \Big(\ket{0}_A \ket{1}_B - \ket{1}_A \ket{0}_B \Big)$</td> + </tr> +</table> Her actions affect the state on Bob's side $B$ due to entanglement. Alice then sends her qubit $A$ to Bob over the quantum channel, |