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diff --git a/content/know/concept/convolution-theorem/index.pdc b/content/know/concept/convolution-theorem/index.pdc index fc96f30..86543d8 100644 --- a/content/know/concept/convolution-theorem/index.pdc +++ b/content/know/concept/convolution-theorem/index.pdc @@ -68,7 +68,7 @@ $$\begin{aligned} \boxed{(f * g)(t) = \hat{\mathcal{L}}^{-1}\{\tilde{f}(s) \: \tilde{g}(s)\}} \end{aligned}$$ -Because the inverse Laplace transform $\hat{\mathcal{L}}^{-1}$ is quite +Because the inverse Laplace transform $\hat{\mathcal{L}}^{-1}$ is unpleasant, the theorem is often stated using the forward transform instead: diff --git a/content/know/concept/dirac-delta-function/index.pdc b/content/know/concept/dirac-delta-function/index.pdc index 6c9f2b0..97704d7 100644 --- a/content/know/concept/dirac-delta-function/index.pdc +++ b/content/know/concept/dirac-delta-function/index.pdc @@ -40,7 +40,7 @@ $$\begin{aligned} } \end{aligned}$$ -$\delta(x)$ is thus an effective weapon against integrals. This may not seem very +$\delta(x)$ is thus quite an effective weapon against integrals. This may not seem very useful due to its "unnatural" definition, but in fact it appears as the limit of several reasonable functions: diff --git a/content/know/concept/parsevals-theorem/index.pdc b/content/know/concept/parsevals-theorem/index.pdc index 8f653f8..ae34bda 100644 --- a/content/know/concept/parsevals-theorem/index.pdc +++ b/content/know/concept/parsevals-theorem/index.pdc @@ -13,8 +13,8 @@ markup: pandoc # Parseval's theorem -**Parseval's theorem** relates the inner product of two functions $f(x)$ and $g(x)$ to the -inner product of their [Fourier transforms](/know/concept/fourier-transform/) +**Parseval's theorem** is a relation between the inner product of two functions $f(x)$ and $g(x)$, +and the inner product of their [Fourier transforms](/know/concept/fourier-transform/) $\tilde{f}(k)$ and $\tilde{g}(k)$. There are two equivalent ways of stating it, where $A$, $B$, and $s$ are constants from the Fourier transform's definition: @@ -29,12 +29,11 @@ $$\begin{aligned} } \end{aligned}$$ -For this reason, physicists like to define their Fourier transform -with $A = B = 1 / \sqrt{2\pi}$ and $|s| = 1$, because then the FT nicely -conserves the total probability (quantum mechanics) or the total energy -(optics). +For this reason, physicists like to define the Fourier transform +with $A\!=\!B\!=\!1 / \sqrt{2\pi}$ and $|s|\!=\!1$, because then it nicely +conserves the functions' normalization. -To prove this, we insert the inverse FT into the inner product +To prove the theorem, we insert the inverse FT into the inner product definition: $$\begin{aligned} @@ -55,7 +54,7 @@ $$\begin{aligned} \end{aligned}$$ Where $\delta(k)$ is the [Dirac delta function](/know/concept/dirac-delta-function/). -Note that we can just as well do it in the opposite direction, +Note that we can equally well do the proof in the opposite direction, which yields an equivalent result: $$\begin{aligned} diff --git a/content/know/concept/repetition-code/bit-flip-detect.png b/content/know/concept/repetition-code/bit-flip-detect.png Binary files differnew file mode 100644 index 0000000..0532421 --- /dev/null +++ b/content/know/concept/repetition-code/bit-flip-detect.png diff --git a/content/know/concept/repetition-code/bit-flip-encode.png b/content/know/concept/repetition-code/bit-flip-encode.png Binary files differnew file mode 100644 index 0000000..7c78667 --- /dev/null +++ b/content/know/concept/repetition-code/bit-flip-encode.png diff --git a/content/know/concept/repetition-code/index.pdc b/content/know/concept/repetition-code/index.pdc new file mode 100644 index 0000000..7245cbc --- /dev/null +++ b/content/know/concept/repetition-code/index.pdc @@ -0,0 +1,349 @@ +--- +title: "Repetition code" +firstLetter: "R" +publishDate: 2021-05-07 +categories: +- Quantum information + +date: 2021-05-07T15:10:43+02:00 +draft: false +markup: pandoc +--- + +# Repetition code + +A **repetition code** is a simple approach to error correction: +to protect a bit $x$, make two copies: + +$$\begin{aligned} + 0 \to 000 + \qquad \quad + 1 \to 111 +\end{aligned}$$ + +If a single-bit error occurs, e.g. $000 \to 100$, +a majority vote resets the minority bit. +Clearly, this does not protect against multi-bit errors, +but that is usually not necessary. + +In quantum computing, where error correction is much more important, +repetition codes can also be used, +albeit with some complications, +as discussed below. + + +## Bit flip code + +Suppose that we want to detect errors in +the following arbitrary qubit state $\ket{\psi}$: + +$$\begin{aligned} + \ket{\psi} + = \alpha \ket{0} + \beta \ket{1} +\end{aligned}$$ + +For now, let us limit ourselves to detecting **bit flips**, +where $\alpha$ and $\beta$ get switched: + +$$\begin{aligned} + \alpha \ket{0} + \beta \ket{1} + \quad \to \boxed{\mathrm{Error}} \to \quad + \beta \ket{0} \!+\! \alpha \ket{1} +\end{aligned}$$ + +One way to defend against this is +the quantum version of a classical repetition code: + +$$\begin{aligned} + \ket{\psi} + \quad \to \boxed{\mathrm{Encoder}} \to \quad + \ket*{\overline{\psi}} + = \alpha \ket{000} + \beta \ket{111} +\end{aligned}$$ + +In other words, a *logical* $\ket{0}$ (written $\ket*{\overline{0}}$) +is represented by 3 *physical* qubits, and vice versa: + +$$\begin{aligned} + \boxed{ + \ket{0} + \to + \ket*{\overline{0}} + = \ket{000} + \qquad \quad + \ket{1} + \to + \ket*{\overline{1}} + = \ket{111} + } +\end{aligned}$$ + +Such a transformation is easy to achieve with the following sequence +of [quantum gates](/know/concept/quantum-gate/): + +<a href="bit-flip-encode.png"> +<img src="bit-flip-encode.png" style="width:32%;display:block;margin:auto;"> +</a> + +So, a little while after encoding the state $\ket{\psi}$ like that, +a bit flip occurs on the 2nd qubit: + +$$\begin{aligned} + \ket*{\overline{\psi}} + \quad \to \boxed{\mathrm{Error}} \to \quad + \alpha \ket{010} + \beta \ket{101} +\end{aligned}$$ + +But now there is a problem: how do we detect this error? +We could measure the state, but that would make it collapse, +which is probably not what we want. + +The trick is to use operators called **stabilizers**, +in this case for example $ZZI = Z_1 \otimes Z_2 \otimes I_3$, +where $I$ is identity and $Z$ is the Pauli-$Z$ gate. +The 3-qubit basis states are its eigenvectors: + +$$\begin{alignedat}{2} + ZZI \ket{000} + &= + \ket{000} + \qquad + ZZI \ket{001} + &&= + \ket{001} + \\ + ZZI \ket{010} + &= - \ket{010} + \qquad + ZZI \ket{011} + &&= - \ket{011} + \\ + ZZI \ket{100} + &= - \ket{100} + \qquad + ZZI \ket{101} + &&= - \ket{101} + \\ + ZZI \ket{110} + &= + \ket{110} + \qquad + ZZI \ket{111} + &&= + \ket{111} +\end{alignedat}$$ + +We could measure $ZZI$ for $\ket*{\overline{\psi}}$, +and if the eigenvalue is $-1$, +we know that a bit flip has occurred, +whereas if the eigenvalue is $+1$, +there is *maybe* no error ($\ket{001}$ and $\ket{110}$ are false negatives). + +These false negatives are fixed by including another stabilizer $IZZ$, +with these eigenvectors: + +$$\begin{alignedat}{2} + IZZ \ket{000} + &= + \ket{000} + \qquad + IZZ \ket{001} + &&= - \ket{001} + \\ + IZZ \ket{010} + &= - \ket{010} + \qquad + IZZ \ket{011} + &&= + \ket{011} + \\ + IZZ \ket{100} + &= + \ket{100} + \qquad + IZZ \ket{101} + &&= - \ket{101} + \\ + IZZ \ket{110} + &= - \ket{110} + \qquad + IZZ \ket{111} + &&= + \ket{111} +\end{alignedat}$$ + +In which case $\ket{100}$ and $\ket{011}$ are false negatives. +In other words, $IZZ$ cannot detect if the 1st qubit was flipped, +while $ZZI$ cannot protect the 3rd qubit. +But by using both, we know exactly which qubit was flipped +thanks to the eigenvalues: + +<table style="width:30%;margin:auto;text-align:center;"> + <tr> + <th>Error</th> + <th>$ZZI$</th> + <th>$IZZ$</th> + </tr> + <tr> + <td>$I$</td> + <td>$+1$</td> + <td>$+1$</td> + </tr> + <tr> + <td>$X_1$</td> + <td>$-1$</td> + <td>$+1$</td> + </tr> + <tr> + <td>$X_2$</td> + <td>$-1$</td> + <td>$-1$</td> + </tr> + <tr> + <td>$X_1$</td> + <td>$+1$</td> + <td>$-1$</td> + </tr> +</table> + +Where e.g. $X_3$ denotes that the 3rd qubit was flipped. +The measurement outcomes on the last three rows are called **error syndromes**, +and are obtained by a **syndrome measurement**. + +Fortunately, we can measure $ZZI$ and $IZZ$ +without affecting $\ket*{\overline{\psi}}$ itself, +by applying $\mathrm{CNOT}$s to some ancillary qubits +and then measuring those: + +<a href="bit-flip-detect.png"> +<img src="bit-flip-detect.png" style="width:62%;display:block;margin:auto;"> +</a> + +The two measurements, respectively representing $ZZI$ and $IZZ$, +yield $\ket{1}$ if a bit flip definitely occurred, +and $\ket{0}$ otherwise. +There is no entanglement, +so the input is untouched. + + +## Phase flip code + +The above system protects us against all single-qubit bit flips. +Unfortunately, that is not enough: +qubits can also experience a **phase flip**: + +$$\begin{aligned} + \alpha \ket{0} + \beta \ket{1} + \quad \to \boxed{\mathrm{Error}} \to \quad + \alpha \ket{0} - \beta \ket{1} +\end{aligned}$$ + +How to detect that? +If we want to protect against phase flips *instead of* bit flips, +we can simply do the same as before, +but along the $X$-axis intead of the $Z$-axis: + +$$\begin{aligned} + \boxed{ + \ket{0} + \to + \ket*{\overline{0}} + = \ket{+\!+\!+} + \qquad \quad + \ket{1} + \to + \ket*{\overline{1}} + = \ket{-\!-\!-} + } +\end{aligned}$$ + +Such that an arbitrary state $\ket{\psi}$ is encoded as follows, +by the circuit shown below: + +$$\begin{aligned} + \ket{\psi} + \quad \to \boxed{\mathrm{Encoder}} \to \quad + \ket*{\overline{\psi}} + = \alpha \ket{+\!+\!+} + \beta \ket{-\!-\!-} +\end{aligned}$$ + +<a href="phase-flip-encode.png"> +<img src="phase-flip-encode.png" style="width:40%;display:block;margin:auto;"> +</a> + +A phase flip along the $Z$-axis +corresponds to a bit flip along the $X$-axis $\ket{+} \to \ket{-}$. +In this case, the stabilizers are $XXI$ and $IXX$, +and the error detection circuit is as follows: + +<a href="phase-flip-detect.png"> +<img src="phase-flip-detect.png" style="width:70%;display:block;margin:auto;"> +</a> + +This system protects us against all single-qubit phase flips, +but not against bit flips. + + +## Shor code + +What kind of repetition code would we need +if we want to detect both bit flips *and* phase flips? +The most straightforward option is the **Shor code**. +Starting from a phase flip encoding: + +$$\begin{aligned} + \ket{0} \to + \ket*{\overline{0}} + &= \ket{+\!+\!+} + = \bigg( \frac{\ket{0} + \ket{1}}{\sqrt{2}} \bigg)^{\otimes 3} + \\ + \ket{1} \to + \ket*{\overline{1}} + &= \ket{-\!-\!-} + = \bigg( \frac{\ket{0} - \ket{1}}{\sqrt{2}} \bigg)^{\otimes 3} +\end{aligned}$$ + +We add protection against bit flips +by using a repetition code for each physical qubit: + +$$\begin{aligned} + \boxed{ + \ket*{\overline{0}} + = \bigg( \frac{\ket{000} + \ket{111}}{\sqrt{2}} \bigg)^{\otimes 3} + \qquad \quad + \ket*{\overline{1}} + = \bigg( \frac{\ket{000} - \ket{111}}{\sqrt{2}} \bigg)^{\otimes 3} + } +\end{aligned}$$ + +This encoding is achieved by the following quantum circuit, +which simply consists of the phase flip encoder, +followed by 3 copies of the bit flip encoder: + +<a href="shor-code-encode.png"> +<img src="shor-code-encode.png" style="width:55%;display:block;margin:auto;"> +</a> + +We thus use 9 physical qubits to store 1 logical qubit. +Fortunately, more efficient schemes exist. + +The bit flip stabilizers $ZZI$ and $IZZ$ +are applied on a per-block basis, like so: + +$$\begin{aligned} + ZZI \: III \: III \qquad\quad III \: ZZI \: III \qquad\quad III \: III \: ZZI + \\ + IZZ \: III \: III \qquad\quad III \: IZZ \: III \qquad\quad III \: III \: IZZ +\end{aligned}$$ + +Whereas the phase flip stabilizers $XXI$ and $IXX$ +are applied to entire blocks at once: + +$$\begin{aligned} + XXX \: XXX \: III + \qquad \quad + III \: XXX \: XXX +\end{aligned}$$ + + + +## References +1. J.S. Neergaard-Nielsen, + *Quantum information: lectures notes*, + 2021, unpublished. +2. S. Aaronson, + *Introduction to quantum information science: lecture notes*, + 2018, unpublished. + diff --git a/content/know/concept/repetition-code/phase-flip-detect.png b/content/know/concept/repetition-code/phase-flip-detect.png Binary files differnew file mode 100644 index 0000000..fe0243b --- /dev/null +++ b/content/know/concept/repetition-code/phase-flip-detect.png diff --git a/content/know/concept/repetition-code/phase-flip-encode.png b/content/know/concept/repetition-code/phase-flip-encode.png Binary files differnew file mode 100644 index 0000000..04de8ce --- /dev/null +++ b/content/know/concept/repetition-code/phase-flip-encode.png diff --git a/content/know/concept/repetition-code/shor-code-encode.png b/content/know/concept/repetition-code/shor-code-encode.png Binary files differnew file mode 100644 index 0000000..ab1a01e --- /dev/null +++ b/content/know/concept/repetition-code/shor-code-encode.png diff --git a/content/know/concept/reynolds-number/index.pdc b/content/know/concept/reynolds-number/index.pdc index bd18f2f..ff5e793 100644 --- a/content/know/concept/reynolds-number/index.pdc +++ b/content/know/concept/reynolds-number/index.pdc @@ -25,6 +25,12 @@ $$\begin{aligned} = - \frac{\nabla p}{\rho} + \nu \nabla^2 \va{v} \end{aligned}$$ +In this case, the gravity term $\va{g}$ +has been absorbed into the pressure term: +$p \to p\!+\!\rho \Phi$, +where $\Phi$ is the gravitational scalar potential, +i.e. $\va{g} = - \nabla \Phi$. + Let us introduce the dimensionless variables $\va{v}'$, $\va{r}'$, $t'$ and $p'$, where $U$ and $L$ are respectively a characteristic velocity and length of the system at hand: @@ -108,7 +114,7 @@ such that redimensionalizing yields: $$\begin{aligned} \pdv{\va{v}}{t} + (\va{v} \cdot \nabla) \va{v} - = - \nabla p + = - \frac{\nabla p}{\rho} \end{aligned}$$ Which is simply the main [Euler equation](/know/concept/euler-equations/) @@ -133,7 +139,8 @@ $$\begin{aligned} This equation is called the **unsteady Stokes equation**. Usually, however, such flows are assumed to be steady -(i.e. time-invariant), leading to the **steady Stokes equation**: +(i.e. time-invariant), leading to the **steady Stokes equation**, +with $\eta = \rho \nu$: $$\begin{aligned} \boxed{ diff --git a/content/know/concept/shors-algorithm/index.pdc b/content/know/concept/shors-algorithm/index.pdc index 0700408..3e3c05a 100644 --- a/content/know/concept/shors-algorithm/index.pdc +++ b/content/know/concept/shors-algorithm/index.pdc @@ -243,7 +243,7 @@ which is $\mathcal{O}(q^2 (\log{q}) \log{\log{q}})$ and therefore worse than the QFT, yielding a total complexity of $\mathcal{O}(q^2 (\log{q})^2 \log{\log{q}})$. -OK, but what does $s$ have to do factorizing integers? +OK, but what does $s$ have to do with factorizing integers? Well, recall that $f$ is given by: $$\begin{aligned} diff --git a/content/know/concept/stokes-law/index.pdc b/content/know/concept/stokes-law/index.pdc new file mode 100644 index 0000000..acc98af --- /dev/null +++ b/content/know/concept/stokes-law/index.pdc @@ -0,0 +1,371 @@ +--- +title: "Stokes' law" +firstLetter: "S" +publishDate: 2021-05-04 +categories: +- Physics +- Fluid mechanics +- Fluid dynamics + +date: 2021-05-04T19:28:30+02:00 +draft: false +markup: pandoc +--- + +# Stokes' law + +**Stokes' law** describes the size of the drag force $D$ +at low [Reynolds number](/know/concept/reynolds-number/) $\mathrm{Re} \ll 1$ +experienced by a spherical object in a steady, uniform flow at velocity $U$. + + +## Flow field + +Imagine a sphere with radius $a$ sinking in a viscous liquid. +To model this situation, let us pretend that the sphere is fixed instead, +and the fluid comes from infinity at velocity $U$ along the $z$-axis, +flows past the sphere, and continues to infinity at the same $U$. +The Reynolds number is: + +$$\begin{aligned} + \mathrm{Re} + = \frac{2 a U}{\nu} +\end{aligned}$$ + +We assume that $\mathrm{Re} \ll 1$, in which case +the incompressible [Navier-Stokes equations](/know/concept/navier-stokes-equations/) +are reduced to the **steady Stokes equations**: + +$$\begin{aligned} + \nabla p + = \eta \nabla^2 \va{v} + \qquad \quad + \nabla \cdot \va{v} + = 0 +\end{aligned}$$ + +The goal is to solve for $p$ and $\va{v}$. +We make the following ansatz in +[spherical coordinates](/know/concept/spherical-coordinates/) $(r, \theta, \phi)$, +where $q(r)$, $f(r)$ and $g(r)$ are unknown functions: + +$$\begin{gathered} + p + = \eta U q(r) \cos\theta + \\ + v_r + = U f(r) \cos\theta + \qquad + v_\theta + = - U g(r) \sin\theta + \qquad + v_\phi + = 0 +\end{gathered}$$ + +The fluid hits the sphere head on, +so the solution is taken to be $\phi$-independent due to symmetry. +Note that $\theta$ is the angle to the positive $z$-axis, +which is the direction of $\va{U} = U \vu{e}_z$. +Moreover, note that $\va{U} \cdot \vu{e}_r = U \cos\theta$ +and $\va{U} \cdot \vu{e}_\theta = - U \sin\theta$, +where $\vu{e}_r$ and $\vu{e}_\theta$ are basis vectors. + +To begin with, we insert this ansatz into the incompressibility condition, +yielding: + +$$\begin{aligned} + 0 + = \nabla \cdot \va{v} + &= \pdv{v_r}{r} + \frac{1}{r} \pdv{v_\theta}{\theta} + \frac{2 v_r}{r} + \frac{v_\theta}{r \tan \theta} + \\ + &= U \dv{f}{r} \cos\theta - \frac{U g}{r} \cos\theta + \frac{2 U f}{r} \cos\theta - \frac{U g}{r} \cos\theta + \\ + &= U \cos\theta \Big( \dv{f}{r} + \frac{2}{r} f - \frac{2}{r} g \Big) +\end{aligned}$$ + +The parenthesized expression must be zero for all $r$, +leading us to the following relation: + +$$\begin{aligned} + g(r) + = f + \frac{r}{2} \dv{f}{r} +\end{aligned}$$ + +Next, we take the divergence of the first Stokes equation, +and insert incompressibility: + +$$\begin{aligned} + \nabla^2 p + = \eta \nabla \cdot (\nabla^2 \va{v}) + = \eta \nabla^2 (\nabla \cdot \va{v}) + = 0 +\end{aligned}$$ + +This is simply the Laplace equation, +which is as follows for our ansatz $p(r, \theta)$: + +$$\begin{aligned} + 0 + = \nabla^2 p + &= \frac{1}{r^2} \pdv{r} \Big( r^2 \pdv{p}{r} \Big) + \frac{1}{r^2 \sin\theta} \pdv{\theta} \Big( \sin\theta \pdv{p}{\theta} \Big) + \\ + 0 + &= \frac{\eta U \cos\theta}{r^2} \dv{r} \Big( r^2 \dv{q}{r} \Big) + - \frac{\eta U q}{r^2 \sin\theta} \pdv{\theta} \Big( \sin^2\theta \Big) + \\ + &= \frac{\eta U \cos\theta}{r^2} \dv{r} \Big( r^2 \dv{q}{r} \Big) + - \frac{2 \eta U q}{r^2 \sin\theta} \sin\theta \cos\theta + \\ + &= \eta U \cos\theta \Big( \dv[2]{q}{r} + \frac{2}{r} \dv{q}{r} - \frac{2}{r^2} q \Big) +\end{aligned}$$ + +Again, the parenthesized expression must be zero for all $r$, +meaning it is an ODE for $q(r)$, +whose solution is straightforwardly found to be: + +$$\begin{aligned} + q(r) + = \frac{C_3}{r^2} + C_4 r +\end{aligned}$$ + +Where $C_3$ and $C_4$ are linearity constants ($C_1$ and $C_2$ appear later). +The pressure is therefore: + +$$\begin{aligned} + p + = \eta U \cos\theta \Big( \frac{C_3}{r^2} + C_4 r \Big) +\end{aligned}$$ + +Consequently, its gradient $\nabla p$ in spherical coordinates is as follows: + +$$\begin{aligned} + \nabla p + = \vu{e}_r \pdv{p}{r} + \vu{e}_\theta \frac{1}{r} \pdv{p}{\theta} + = \vu{e}_r \Big( \eta U \cos\theta \dv{q}{r} \Big) - \vu{e}_\theta \Big( \eta U \sin\theta \frac{q}{r} \Big) +\end{aligned}$$ + +According to the Stokes equation, this equals $\eta \nabla^2 \va{v}$. +Let us look at the $r$-component of $\nabla^2 \va{v}$: + +$$\begin{aligned} + (\nabla^2 \va{v})_r + &= \pdv[2]{v_r}{r} + \frac{1}{r^2} \pdv[2]{v_r}{\theta} + \frac{2}{r} \pdv{v_r}{r} + + \frac{\cot\theta}{r^2} \pdv{v_r}{\theta} - \frac{2}{r^2} \pdv{v_\theta}{\theta} - \frac{2}{r^2} v_r - \frac{2 \cot\theta}{r^2} v_\theta + \\ + &= U \cos\theta \Big( \dv[2]{f}{r} - \frac{1}{r^2} f + \frac{2}{r} \dv{f}{r} - \frac{1}{r^2} f + + \frac{2}{r^2} g - \frac{2}{r^2} f + \frac{2}{r^2} g \Big) + \\ + &= U \cos\theta \Big( \dv[2]{f}{r} + \frac{2}{r} \dv{f}{r} - \frac{4}{r^2} f + \frac{4}{r^2} g \Big) +\end{aligned}$$ + +Substituting $g$ for the expression we found from incompressibility lets us simplify this: + +$$\begin{aligned} + \eta (\nabla^2 \va{v})_r + &= \eta U \cos\theta \Big( \dv[2]{f}{r} + \frac{4}{r} \dv{f}{r} \Big) +\end{aligned}$$ + +The Stokes equation says that this must be equal to the $r$-component of $\nabla p$: + +$$\begin{aligned} + \eta U \cos\theta \Big( \dv[2]{f}{r} + \frac{4}{r} \dv{f}{r} \Big) + = \eta U \cos\theta \Big( \!-\! \frac{2 C_3}{r^3} + C_4 \Big) +\end{aligned}$$ + +Where we have inserted $\dv*{q}{r}$. +Dividing out $\eta U \cos\theta$ leaves an ODE for $f(r)$, +satisfied by: + +$$\begin{aligned} + f(r) + = C_1 + \frac{C_2}{r^3} + \frac{C_3}{r} + \frac{C_4 r^2}{10} +\end{aligned}$$ + +Then, thanks to our earlier relation again, +we know that $g(r)$ is as follows: + +$$\begin{aligned} + g(r) + = C_1 - \frac{C_2}{2 r^3} + \frac{C_3}{2 r} + \frac{C_4 r^2}{5} +\end{aligned}$$ + +So what about $C_1$, $C_2$, $C_3$ and $C_4$? +For $r\!\to\!\infty$, we expect that $\va{v}\!\to\!\va{U}$, +meaning that $f(r)\!\to\!1$ and $g(r)\!\to\!1$. +This implies that $C_4 = 0$ and $C_1 = 1$, leaving: + +$$\begin{aligned} + f(r) + = 1 + \frac{C_2}{r^3} + \frac{C_3}{r} + \qquad \quad + g(r) + = 1 - \frac{C_2}{2 r^3} + \frac{C_3}{2 r} +\end{aligned}$$ + +Furthermore, the viscous *no-slip* condition demands +that $\va{v} = 0$ at the sphere's surface $r = a$, so $f(a) = g(a) = 0$ there. +Inserting $a$ into $f$ and $g$, setting them to zero, +and solving the resulting system of equations +yields $C_2 = a^3 / 2$ and $C_3 = -3 a / 2$. +Therefore the full solution is: + +$$\begin{gathered} + \boxed{ + p + = - \frac{3 \eta U a}{2 r^2} \cos\theta + } + \\ + \boxed{ + v_r + = U \cos\theta \Big( 1 + \frac{a^3}{2 r^3} - \frac{3 a}{2 r} \Big) + \qquad + v_\theta + = - U \sin\theta \Big( 1 - \frac{a^3}{4 r^3} - \frac{3 a}{4 r} \Big) + } +\end{gathered}$$ + + +## Drag force + +From the definition of [viscosity](/know/concept/viscosity/), +we know that there must be shear stresses at the sphere surface, +described by the fluid's [Cauchy stress tensor](/know/concept/cauchy-stress-tensor/) $\hat{\sigma}$. +The drag force $\va{D}$ on the surface is: + +$$\begin{aligned} + \va{D} + = \oint \hat{\sigma} \cdot \dd{\va{S}} + = \int_0^{2\pi} \!\!\!\! \int_0^\pi \big( \hat{\sigma} \cdot \vu{e}_r \big) \:a^2 \sin\theta \dd{\theta} \dd{\phi} +\end{aligned}$$ + +Where $\vu{e}_r$ is the sphere's surface normal vector. +The integrand can be expanded as follows: + +$$\begin{aligned} + \hat{\sigma} \cdot \vu{e}_r + = \vu{e}_r \sigma_{rr} + \vu{e}_\theta \sigma_{\theta r} +\end{aligned}$$ + +To calculate this, we start by taking the gradient of the velocity field $\va{v}$: + +$$\begin{aligned} + \nabla\va{v} + &= \vu{e}_r \vu{e}_r \pdv{v_r}{r} + \vu{e}_r \vu{e}_\theta \pdv{v_\theta}{r} + + \vu{e}_\theta \vu{e}_r \Big( \frac{1}{r} \pdv{v_r}{\theta} - \frac{v_\theta}{r} \Big) + \\ + &\qquad + \vu{e}_\theta \vu{e}_\theta \Big( \frac{1}{r} \pdv{v_\theta}{\theta} - \frac{v_r}{r} \Big) + + \vu{e}_\phi \vu{e}_\phi \Big( \frac{v_\theta}{r \tan\theta} + \frac{v_r}{r} \Big) +\end{aligned}$$ + +Some of these terms are necessary to calculate the stress elements $\sigma_{rr}$ and $\sigma_{\theta r}$: + +$$\begin{aligned} + \sigma_{rr} + &= - p + 2 \eta (\nabla\va{v})_{rr} + = - p + 2 \eta \pdv{v_r}{r} + \\ + &= \frac{3 \eta U a}{2 r^2} \cos\theta + 2 \eta U \cos\theta \: \Big( \!-\! \frac{3 a^3}{2 r^4} + \frac{3 a}{2 r^2} \Big) + \\ + &= \frac{3 \eta U a}{2 r^2} \cos\theta \: \Big( 3 - 2 \frac{a^2}{r^2} \Big) +\end{aligned}$$ +$$\begin{aligned} + \sigma_{\theta r} + &= \eta \big( (\nabla\va{v})_{\theta r} + (\nabla\va{v})_{r \theta} \big) + = \eta \: \Big( \pdv{v_\theta}{r} + \frac{1}{r} \pdv{v_r}{\theta} - \frac{v_\theta}{r} \Big) + \\ + &= \eta U \sin\theta \: \Big( \!-\! \frac{3 a^3}{4 r^4} - \frac{3 a}{4 r^2} + - \frac{1}{r} - \frac{a^3}{2 r^4} + \frac{3 a}{2 r^2} + + \frac{1}{r} - \frac{a^3}{4 r^4} - \frac{3 a}{4 r^2} \Big) + \\ + &= - \frac{3 \eta U a^3}{2 r^4} \sin\theta +\end{aligned}$$ + +At the sphere's surface we set $r = a$, so these expressions reduce to the following: + +$$\begin{aligned} + \sigma_{rr} + = \frac{3 \eta U}{2 a} \cos\theta + \qquad \quad + \sigma_{\theta r} + = - \frac{3 \eta U}{2 a} \sin\theta +\end{aligned}$$ + +Now we can finally calculate the effective stress on the surface, +by converting the basis vectors $\vu{e}_r$ and $\vu{e}_\theta$ to Cartesian coordinates: + +$$\begin{aligned} + \hat{\sigma} \cdot \vu{e}_r + &= \vu{e}_r \frac{3 \eta U}{2 a} \cos\theta - \vu{e}_\theta \frac{3 \eta U}{2 a} \sin\theta + \\ + &= \Big( \vu{e}_x \sin\theta \cos\phi + \vu{e}_y \sin\theta \sin\phi + \vu{e}_z \cos\theta \Big) \frac{3 \eta U}{2 a} \cos\theta + \\ + &\qquad - \Big( \vu{e}_x \cos\theta \cos\phi + \vu{e}_y \cos\theta \sin\phi - \vu{e}_z \sin\theta \Big) \frac{3 \eta U}{2 a} \sin\theta + \\ + &= \Big( \vu{e}_z \cos^2\theta + \vu{e}_z \sin^2\theta \Big) \frac{3 \eta U}{2 a} + = \vu{e}_z \frac{3 \eta U}{2 a} +\end{aligned}$$ + +Remarkably, the stress at every point on the sphere is purely in the $z$-direction! +This is not entirely unexpected though: symmetry cancels out all other components. + +With this, we can do the integrals for $\va{D}$, +which reduce to a surface area factor $4 \pi a^2$: + +$$\begin{aligned} + \va{D} + %= \int_0^{2\pi} \!\!\! \int_0^\pi \big( \hat{\sigma} \cdot \vu{e}_r \big) \:r^2 \sin\theta \dd{\theta} \dd{\phi} + = \vu{e}_z \frac{3 \eta U}{2 a} \int_0^{2\pi} \!\!\!\! \int_0^\pi a^2 \sin\theta \dd{\theta} \dd{\phi} + = \vu{e}_z \frac{3 \eta U}{2 a} 2 \pi a^2 \int_0^\pi \sin\theta \dd{\theta} + = \vu{e}_z \: 6 \pi \eta U a +\end{aligned}$$ + +At last, we arrive at Stokes' law, +which simply expresses the magnitude of $\va{D}$: + +$$\begin{aligned} + \boxed{ + D + = 6 \pi \eta U a + } +\end{aligned}$$ + +To arrive at this result, +we assumed that the sphere was fixed, and the fluid was flowing past it. +We can equally well let the fluid be at rest, +with the sphere falling through it at $U$. +The force of gravity then exerts the following force $G$ on it, +subtracting [buoyancy](/know/concept/archimedes-principle/): + +$$\begin{aligned} + G + = \frac{4 \pi a^3}{3} (\rho_s - \rho_f) g_0 +\end{aligned}$$ + +Where $\rho_s$ and $\rho_f$ are the sphere's and fluid's densities, +and $g_0$ is the gravitational acceleration. +Since $D$ acts in the opposite sense of $G$, +after some time, they cancel out: + +$$\begin{aligned} + 6 \pi \eta U a + = \frac{4 \pi a^3}{3} (\rho_s - \rho_f) g_0 +\end{aligned}$$ + +This is an equation for the **terminal velocity** $U_t$, +which we find to be as follows: + +$$\begin{aligned} + U_t + = \frac{2 a^2 (\rho_s - \rho_f) g_0}{9 \eta} +\end{aligned}$$ + +The falling sphere will accelerate until $U_t$, +and then continue falling at constant speed. + + + +## References +1. B. Lautrup, + *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition, + CRC Press. diff --git a/sources/know/concept/repetition-code/circuit.tex b/sources/know/concept/repetition-code/circuit.tex new file mode 100644 index 0000000..8b30dde --- /dev/null +++ b/sources/know/concept/repetition-code/circuit.tex @@ -0,0 +1,186 @@ +\documentclass[11pt]{article} +\usepackage[utf8]{inputenc} +\usepackage{amsmath} +\usepackage{amsfonts} +\usepackage{physics} +\usepackage{yquant} + + +\begin{document} + +\yquantdefinebox{dots}[inner sep=0pt]{$\vdots$} + +\section{Spacer} + +\begin{tikzpicture} + \begin{yquant}[operator/separation=5mm, register/separation=0mm] + qubit {$\ket{\psi}$} q; + qubit {$\ket{0}$} z1; + qubit {$\ket{0}$} z2; + + cnot z1 | q; + cnot z2 | q; + + [value=$\ket{\overline{\psi}}$] + output (q, z1, z2); + \end{yquant} +\end{tikzpicture} + +\section{Spacer} + +\begin{tikzpicture} + \begin{yquant}[operator/separation=5mm, register/separation=0mm] + qubit {$\ket{\psi}$} q; + qubit {$\ket{0}$} z1; + qubit {$\ket{0}$} z2; + + cnot z1 | q; + cnot z2 | q; + + h q; + h z1; + h z2; + + [value=$\ket{\overline{\psi}}$] + output (q, z1, z2); + \end{yquant} +\end{tikzpicture} + +\section{Spacer} + +\begin{tikzpicture} + \begin{yquant}[operator/separation=2mm, register/separation=0mm] + qubit {} q1; + qubit {} q2; + qubit {} q3; + init {$\ket{\overline{\psi}}$} (q1, q2, q3); + qubit {$\ket{0}$} z; + + cnot z | q2; + cnot z | q3; + + measure z; + \end{yquant} +\end{tikzpicture} + +\section{Spacer} + +\begin{tikzpicture} + \begin{yquant}[operator/separation=2mm, register/separation=0mm] + qubit {} q1; + qubit {} q2; + qubit {} q3; + init {$\ket{\overline{\psi}}$} (q1, q2, q3); + qubit {$\ket{0}$} z1; + qubit {$\ket{0}$} z2; + + cnot z1 | q1; + cnot z1 | q2; + + cnot z2 | q2; + cnot z2 | q3; + + measure z1; + measure z2; + + box {$\mathrm{Correction}$} (q1, q2, q3) | z1, z2; + discard z1; + discard z2; + \end{yquant} +\end{tikzpicture} + +\section{Spacer} + +\begin{tikzpicture} + \begin{yquant}[operator/separation=2mm, register/separation=1mm] |
