diff options
| -rw-r--r-- | content/know/concept/calculus-of-variations/index.pdc | 40 | ||||
| -rw-r--r-- | content/know/concept/heaviside-step-function/index.pdc | 91 | ||||
| -rw-r--r-- | content/know/concept/holomorphic-function/index.pdc | 232 | ||||
| -rw-r--r-- | content/know/concept/kramers-kronig-relations/index.pdc | 133 | ||||
| -rw-r--r-- | content/know/concept/schwartz-distribution/index.pdc | 119 |
5 files changed, 595 insertions, 20 deletions
diff --git a/content/know/concept/calculus-of-variations/index.pdc b/content/know/concept/calculus-of-variations/index.pdc index fb043e0..c5280e5 100644 --- a/content/know/concept/calculus-of-variations/index.pdc +++ b/content/know/concept/calculus-of-variations/index.pdc @@ -29,18 +29,18 @@ the path $f(x)$ taken by a physical system, the **principle of least action** states that $f$ will be a minimum of $J[f]$, so for example the expended energy will be minimized. -If $f(x, \alpha\!=\!0)$ is the optimal route, then a slightly +If $f(x, \varepsilon\!=\!0)$ is the optimal route, then a slightly different (and therefore worse) path between the same two points can be expressed -using the parameter $\alpha$: +using the parameter $\varepsilon$: $$\begin{aligned} - f(x, \alpha) = f(x, 0) + \alpha \eta(x) + f(x, \varepsilon) = f(x, 0) + \varepsilon \eta(x) \qquad \mathrm{or} \qquad - \delta f = \alpha \eta(x) + \delta f = \varepsilon \eta(x) \end{aligned}$$ Where $\eta(x)$ is an arbitrary differentiable deviation. -Since $f(x, \alpha)$ must start and end in the same points as $f(x,0)$, +Since $f(x, \varepsilon)$ must start and end in the same points as $f(x,0)$, we have the boundary conditions: $$\begin{aligned} @@ -50,16 +50,16 @@ $$\begin{aligned} Given $L$, the goal is to find an equation for the optimal path $f(x,0)$. Just like when finding the minimum of a real function, the minimum $f$ of a functional $J[f]$ is a stationary point -with respect to the deviation weight $\alpha$, +with respect to the deviation weight $\varepsilon$, a condition often written as $\delta J = 0$. In the following, the integration limits have been omitted: $$\begin{aligned} 0 &= \delta J - = \pdv{J}{\alpha} \Big|_{\alpha = 0} - = \int \pdv{L}{\alpha} \dd{x} - = \int \pdv{L}{f} \pdv{f}{\alpha} + \pdv{L}{f'} \pdv{f'}{\alpha} \dd{x} + = \pdv{J}{\varepsilon} \Big|_{\varepsilon = 0} + = \int \pdv{L}{\varepsilon} \dd{x} + = \int \pdv{L}{f} \pdv{f}{\varepsilon} + \pdv{L}{f'} \pdv{f'}{\varepsilon} \dd{x} \\ &= \int \pdv{L}{f} \eta + \pdv{L}{f'} \eta' \dd{x} = \Big[ \pdv{L}{f'} \eta \Big]_{x_0}^{x_1} + \int \pdv{L}{f} \eta - \frac{d}{dx} \Big( \pdv{L}{f'} \Big) \eta \dd{x} @@ -99,16 +99,16 @@ In this case, every $f_n(x)$ has its own deviation $\eta_n(x)$, satisfying $\eta_n(x_0) = \eta_n(x_1) = 0$: $$\begin{aligned} - f_n(x, \alpha) = f_n(x, 0) + \alpha \eta_n(x) + f_n(x, \varepsilon) = f_n(x, 0) + \varepsilon \eta_n(x) \end{aligned}$$ The derivation procedure is identical to the case $N = 1$ from earlier: $$\begin{aligned} 0 - &= \pdv{J}{\alpha} \Big|_{\alpha = 0} - = \int \pdv{L}{\alpha} \dd{x} - = \int \sum_{n} \Big( \pdv{L}{f_n} \pdv{f_n}{\alpha} + \pdv{L}{f_n'} \pdv{f_n'}{\alpha} \Big) \dd{x} + &= \pdv{J}{\varepsilon} \Big|_{\varepsilon = 0} + = \int \pdv{L}{\varepsilon} \dd{x} + = \int \sum_{n} \Big( \pdv{L}{f_n} \pdv{f_n}{\varepsilon} + \pdv{L}{f_n'} \pdv{f_n'}{\varepsilon} \Big) \dd{x} \\ &= \int \sum_{n} \Big( \pdv{L}{f_n} \eta_n + \pdv{L}{f_n'} \eta_n' \Big) \dd{x} \\ @@ -140,9 +140,9 @@ Once again, the derivation procedure is the same as before: $$\begin{aligned} 0 - &= \pdv{J}{\alpha} \Big|_{\alpha = 0} - = \int \pdv{L}{\alpha} \dd{x} - = \int \pdv{L}{f} \pdv{f}{\alpha} + \sum_{n} \pdv{L}{f^{(n)}} \pdv{f^{(n)}}{\alpha} \dd{x} + &= \pdv{J}{\varepsilon} \Big|_{\varepsilon = 0} + = \int \pdv{L}{\varepsilon} \dd{x} + = \int \pdv{L}{f} \pdv{f}{\varepsilon} + \sum_{n} \pdv{L}{f^{(n)}} \pdv{f^{(n)}}{\varepsilon} \dd{x} \\ &= \int \pdv{L}{f} \eta + \sum_{n} \pdv{L}{f^{(n)}} \eta^{(n)} \dd{x} \end{aligned}$$ @@ -187,17 +187,17 @@ $$\begin{aligned} The arbitrary deviation $\eta$ is then also a function of multiple variables: $$\begin{aligned} - f(x, y; \alpha) = f(x, y; 0) + \alpha \eta(x, y) + f(x, y; \varepsilon) = f(x, y; 0) + \varepsilon \eta(x, y) \end{aligned}$$ The derivation procedure starts in the exact same way as before: $$\begin{aligned} 0 - &= \pdv{J}{\alpha} \Big|_{\alpha = 0} - = \iint \pdv{L}{\alpha} \dd{x} \dd{y} + &= \pdv{J}{\varepsilon} \Big|_{\varepsilon = 0} + = \iint \pdv{L}{\varepsilon} \dd{x} \dd{y} \\ - &= \iint \pdv{L}{f} \pdv{f}{\alpha} + \pdv{L}{f_x} \pdv{f_x}{\alpha} + \pdv{L}{f_y} \pdv{f_y}{\alpha} \dd{x} \dd{y} + &= \iint \pdv{L}{f} \pdv{f}{\varepsilon} + \pdv{L}{f_x} \pdv{f_x}{\varepsilon} + \pdv{L}{f_y} \pdv{f_y}{\varepsilon} \dd{x} \dd{y} \\ &= \iint \pdv{L}{f} \eta + \pdv{L}{f_x} \eta_x + \pdv{L}{f_y} \eta_y \dd{x} \dd{y} \end{aligned}$$ diff --git a/content/know/concept/heaviside-step-function/index.pdc b/content/know/concept/heaviside-step-function/index.pdc new file mode 100644 index 0000000..0471acf --- /dev/null +++ b/content/know/concept/heaviside-step-function/index.pdc @@ -0,0 +1,91 @@ +--- +title: "Heaviside step function" +firstLetter: "H" +publishDate: 2021-02-25 +categories: +- Mathematics +- Physics + +date: 2021-02-25T11:28:02+01:00 +draft: false +markup: pandoc +--- + +# Heaviside step function + +The **Heaviside step function** $\Theta(t)$, +is a discontinuous function used for enforcing causality +or for representing a signal switched on at $t = 0$. +It is defined as: + +$$\begin{aligned} + \boxed{ + \Theta(t) = + \begin{cases} + 0 & \mathrm{if}\: t < 0 \\ + 1 & \mathrm{if}\: t > 1 + \end{cases} + } +\end{aligned}$$ + +The value of $\Theta(t \!=\! 0)$ varies between definitions; +common choices are $0$, $1$ and $1/2$. +In practice, this rarely matters, and some authors even +change their definition on the fly for convenience. +For physicists, $\Theta(0) = 1$ is generally best, such that: + +$$\begin{aligned} + \boxed{ + \forall n \in \mathbb{R}: \Theta^n(t) = \Theta(t) + } +\end{aligned}$$ + +Unsurprisingly, the first-order derivative of $\Theta(t)$ is +the [Dirac delta function](/know/concept/dirac-delta-function/): + +$$\begin{aligned} + \boxed{ + \Theta'(t) = \delta(t) + } +\end{aligned}$$ + +The [Fourier transform](/know/concept/fourier-transform/) +of $\Theta(t)$ is noteworthy. +In this case, it is easiest to use $\Theta(0) = 1/2$, +such that the Heaviside step function can be expressed +using the signum function $\mathrm{sgn}(t)$: + +$$\begin{aligned} + \Theta(t) = \frac{1}{2} + \frac{\mathrm{sgn}(t)}{2} +\end{aligned}$$ + +We then take the Fourier transform, +where $A$ and $s$ are constants from its definition: + +$$\begin{aligned} + \tilde{\Theta}(\omega) + = \hat{\mathcal{F}}\{\Theta(t)\} + = \frac{A}{2} \Big( \int_{-\infty}^\infty \exp(i s \omega t) \dd{t} + \int_{-\infty}^\infty \mathrm{sgn}(t) \exp(i s \omega t) \dd{t} \Big) +\end{aligned}$$ + +The first term is proportional to the Dirac delta function. +The second integral is problematic, so we take the Cauchy principal value $\pv{}$ +and look up the integral: + +$$\begin{aligned} + \tilde{\Theta}(\omega) + &= A \pi \delta(s \omega) + \frac{A}{2} \pv{\int_{-\infty}^\infty \mathrm{sgn}(t) \exp(i s \omega t) \dd{t}} + = \frac{A}{|s|} \pi \delta(\omega) + i \frac{A}{s} \pv{\frac{1}{\omega}} +\end{aligned}$$ + +The use of $\pv{}$ without an integral is an abuse of notation, +and means that this result only makes sense when wrapped in an integral. +Formally, $\pv{\{1 / \omega\}}$ is a [Schwartz distribution](/know/concept/schwartz-distribution/). +We thus have: + +$$\begin{aligned} + \boxed{ + \tilde{\Theta}(\omega) + = \frac{A}{|s|} \Big( \pi \delta(\omega) + i \: \mathrm{sgn}(s) \pv{\frac{1}{\omega}} \Big) + } +\end{aligned}$$ diff --git a/content/know/concept/holomorphic-function/index.pdc b/content/know/concept/holomorphic-function/index.pdc new file mode 100644 index 0000000..3e7a91e --- /dev/null +++ b/content/know/concept/holomorphic-function/index.pdc @@ -0,0 +1,232 @@ +--- +title: "Holomorphic function" +firstLetter: "H" +publishDate: 2021-02-25 +categories: +- Mathematics + +date: 2021-02-25T14:40:45+01:00 +draft: false +markup: pandoc +--- + +# Holomorphic function + +In complex analysis, a complex function $f(z)$ of a complex variable $z$ +is called **holomorphic** or **analytic** if it is complex differentiable in the +neighbourhood of every point of its domain. +This is a very strong condition. + +As a result, holomorphic functions are infinitely differentiable and +equal their Taylor expansion at every point. In physicists' terms, +they are extremely "well-behaved" throughout their domain. + +More formally, a given function $f(z)$ is holomorphic in a certain region +if the following limit exists for all $z$ in that region, +and for all directions of $\Delta z$: + +$$\begin{aligned} + \boxed{ + f'(z) = \lim_{\Delta z \to 0} \frac{f(z + \Delta z) - f(z)}{\Delta z} + } +\end{aligned}$$ + +We decompose $f$ into the real functions $u$ and $v$ of real variables $x$ and $y$: + +$$\begin{aligned} + f(z) = f(x + i y) = u(x, y) + i v(x, y) +\end{aligned}$$ + +Since we are free to choose the direction of $\Delta z$, we choose $\Delta x$ and $\Delta y$: + +$$\begin{aligned} + f'(z) + &= \lim_{\Delta x \to 0} \frac{f(z + \Delta x) - f(z)}{\Delta x} + = \pdv{u}{x} + i \pdv{v}{x} + \\ + &= \lim_{\Delta y \to 0} \frac{f(z + i \Delta y) - f(z)}{i \Delta y} + = \pdv{v}{y} - i \pdv{u}{y} +\end{aligned}$$ + +For $f(z)$ to be holomorphic, these two results must be equivalent. +Because $u$ and $v$ are real by definition, +we thus arrive at the **Cauchy-Riemann equations**: + +$$\begin{aligned} + \boxed{ + \pdv{u}{x} = \pdv{v}{y} + \qquad + \pdv{v}{x} = - \pdv{u}{y} + } +\end{aligned}$$ + +Therefore, a given function $f(z)$ is holomorphic if and only if its real +and imaginary parts satisfy these equations. This gives an idea of how +strict the criteria are to qualify as holomorphic. + + +## Integration formulas + +Holomorphic functions satisfy **Cauchy's integral theorem**, which states +that the integral of $f(z)$ over any closed curve $C$ in the complex plane is zero, +provided that $f(z)$ is holomorphic for all $z$ in the area enclosed by $C$: + +$$\begin{aligned} + \boxed{ + \oint_C f(z) \dd{z} = 0 + } +\end{aligned}$$ + +*__Proof__*. +*Just like before, we decompose $f(z)$ into its real and imaginary parts:* + +$$\begin{aligned} + \oint_C f(z) \:dz + &= \oint_C (u + i v) \dd{(x + i y)} + = \oint_C (u + i v) \:(\dd{x} + i \dd{y}) + \\ + &= \oint_C u \dd{x} - v \dd{y} + i \oint_C v \dd{x} + u \dd{y} +\end{aligned}$$ + +*Using Green's theorem, we integrate over the area $A$ enclosed by $C$:* + +$$\begin{aligned} + \oint_C f(z) \:dz + &= - \iint_A \pdv{v}{x} + \pdv{u}{y} \dd{x} \dd{y} + i \iint_A \pdv{u}{x} - \pdv{v}{y} \dd{x} \dd{y} +\end{aligned}$$ + +*Since $f(z)$ is holomorphic, $u$ and $v$ satisfy the Cauchy-Riemann +equations, such that the integrands disappear and the final result is zero.* +*__Q.E.D.__* + +An interesting consequence is **Cauchy's integral formula**, which +states that the value of $f(z)$ at an arbitrary point $z_0$ is +determined by its values on an arbitrary contour $C$ around $z_0$: + +$$\begin{aligned} + \boxed{ + f(z_0) = \frac{1}{2 \pi i} \oint_C \frac{f(z)}{z - z_0} \dd{z} + } +\end{aligned}$$ + +*__Proof__*. +*Thanks to the integral theorem, we know that the shape and size +of $C$ is irrelevant. Therefore we choose it to be a circle with radius $r$, +such that the integration variable becomes $z = z_0 + r e^{i \theta}$. Then +we integrate by substitution:* + +$$\begin{aligned} + \frac{1}{2 \pi i} \oint_C \frac{f(z)}{z - z_0} \dd{z} + &= \frac{1}{2 \pi i} \int_0^{2 \pi} f(z) \frac{i r e^{i \theta}}{r e^{i \theta}} \dd{\theta} + = \frac{1}{2 \pi} \int_0^{2 \pi} f(z_0 + r e^{i \theta}) \dd{\theta} +\end{aligned}$$ + +*We may choose an arbitrarily small radius $r$, such that the contour approaches $z_0$:* + +$$\begin{aligned} + \lim_{r \to 0}\:\: \frac{1}{2 \pi} \int_0^{2 \pi} f(z_0 + r e^{i \theta}) \dd{\theta} + &= \frac{f(z_0)}{2 \pi} \int_0^{2 \pi} \dd{\theta} + = f(z_0) +\end{aligned}$$ + +*__Q.E.D.__* + +Similarly, **Cauchy's differentiation formula**, +or **Cauchy's integral formula for derivatives** +gives all derivatives of a holomorphic function as follows, +and also guarantees their existence: + +$$\begin{aligned} + \boxed{ + f^{(n)}(z_0) + = \frac{n!}{2 \pi i} \oint_C \frac{f(z)}{(z - z_0)^{n + 1}} \dd{z} + } +\end{aligned}$$ + +*__Proof__*. +*By definition, the first derivative $f'(z)$ of a +holomorphic function $f(z)$ exists and is given by:* + +$$\begin{aligned} + f'(z_0) + = \lim_{z \to z_0} \frac{f(z) - f(z_0)}{z - z_0} +\end{aligned}$$ + +*We evaluate the numerator using Cauchy's integral theorem as follows:* + +$$\begin{aligned} + f'(z_0) + &= \lim_{z \to z_0} \frac{1}{z - z_0} + \bigg( \frac{1}{2 \pi i} \oint_C \frac{f(\zeta)}{\zeta - z} \dd{\zeta} - \frac{1}{2 \pi i} \oint_C \frac{f(\zeta)}{\zeta - z_0} \dd{\zeta} \bigg) + \\ + &= \frac{1}{2 \pi i} \lim_{z \to z_0} \frac{1}{z - z_0} + \oint_C \frac{f(\zeta)}{\zeta - z} - \frac{f(\zeta)}{\zeta - z_0} \dd{\zeta} + \\ + &= \frac{1}{2 \pi i} \lim_{z \to z_0} \frac{1}{z - z_0} + \oint_C \frac{f(\zeta) (z - z_0)}{(\zeta - z)(\zeta - z_0)} \dd{\zeta} +\end{aligned}$$ + +*This contour integral converges uniformly, so we may apply the limit on the inside:* + +$$\begin{aligned} + f'(z_0) + &= \frac{1}{2 \pi i} \oint_C \Big( \lim_{z \to z_0} \frac{f(\zeta)}{(\zeta - z)(\zeta - z_0)} \Big) \dd{\zeta} + = \frac{1}{2 \pi i} \oint_C \frac{f(\zeta)}{(\zeta - z_0)^2} \dd{\zeta} +\end{aligned}$$ + +*Since the second-order derivative $f''(z)$ is simply the derivative of $f'(z)$, +this proof works inductively for all higher orders $n$.* +*__Q.E.D.__* + + +## Residue theorem + +A function $f(z)$ is **meromorphic** if it is holomorphic except in +a finite number of **simple poles**, which are points $z_p$ where +$f(z_p)$ diverges, but where the product $(z - z_p) f(z)$ is non-zero and +still holomorphic close to $z_p$. + +The **residue** $R_p$ of a simple pole $z_p$ is defined as follows, and +represents the rate at which $f(z)$ diverges close to $z_p$: + +$$\begin{aligned} + \boxed{ + R_p = \lim_{z \to z_p} (z - z_p) f(z) + } +\end{aligned}$$ + +**Cauchy's residue theorem** generalizes Cauchy's integral theorem +to meromorphic functions, and states that the integral of a contour $C$, +depends on the simple poles $p$ it encloses: + +$$\begin{aligned} + \boxed{ + \oint_C f(z) \dd{z} = i 2 \pi \sum_{p} R_p + } +\end{aligned}$$ + +*__Proof__*. *From the definition of a meromorphic function, +we know that we can decompose $f(z)$ as follows, +where $h(z)$ is holomorphic and $p$ are all its poles:* + +$$\begin{aligned} + f(z) = h(z) + \sum_{p} \frac{R_p}{z - z_p} +\end{aligned}$$ + +*We integrate this over a contour $C$ which contains all poles, and apply +both Cauchy's integral theorem and Cauchy's integral formula to get:* + +$$\begin{aligned} + \oint_C f(z) \dd{z} + &= \oint_C h(z) \dd{z} + \sum_{p} R_p \oint_C \frac{1}{z - z_p} \dd{z} + = \sum_{p} R_p \: 2 \pi i +\end{aligned}$$ + +*__Q.E.D.__* + +This theorem might not seem very useful, +but in fact, thanks to some clever mathematical magic, +it allows us to evaluate many integrals along the real axis, +most notably [Fourier transforms](/know/concept/fourier-transform/). +It can also be used to derive the Kramers-Kronig relations. + diff --git a/content/know/concept/kramers-kronig-relations/index.pdc b/content/know/concept/kramers-kronig-relations/index.pdc new file mode 100644 index 0000000..1c2977e --- /dev/null +++ b/content/know/concept/kramers-kronig-relations/index.pdc @@ -0,0 +1,133 @@ +--- +title: "Kramers-Kronig relations" +firstLetter: "K" +publishDate: 2021-02-25 +categories: +- Mathematics +- Physics + +date: 2021-02-25T15:20:24+01:00 +draft: false +markup: pandoc +--- + +# Kramers-Kronig relations + +Let $\chi(t)$ be a complex function describing +the response of a system to an impulse $f(t)$ starting at $t = 0$. +The **Kramers-Kronig relations** connect the real and imaginary parts of $\chi(t)$, +such that one can be reconstructed from the other. +Suppose we can only measure $\chi_r(t)$ or $\chi_i(t)$: + +$$\begin{aligned} + \chi(t) = \chi_r(t) + i \chi_i(t) +\end{aligned}$$ + +Assuming that the system was at rest until $t = 0$, +the response $\chi(t)$ cannot depend on anything from $t < 0$, +since the known impulse $f(t)$ had not started yet, +This principle is called **causality**, and to enforce it, +we use the [Heaviside step function](/know/concept/heaviside-step-function/) +$\Theta(t)$ to create a **causality test** for $\chi(t)$: + +$$\begin{aligned} + \chi(t) = \chi(t) \: \Theta(t) +\end{aligned}$$ + +If we [Fourier transform](/know/concept/fourier-transform/) this equation, +then it will become a convolution in the frequency domain +thanks to the [convolution theorem](/know/concept/convolution-theorem/), +where $A$, $B$ and $s$ are constants from the FT definition: + +$$\begin{aligned} + \tilde{\chi}(\omega) + %= \hat{\mathcal{F}}\{\chi_c(t) \: \Theta(t)\} + = (\tilde{\chi} * \tilde{\Theta})(\omega) + = B \int_{-\infty}^\infty \tilde{\chi}(\omega') \: \tilde{\Theta}(\omega - \omega') \dd{\omega'} +\end{aligned}$$ + +We look up the FT of the step function $\tilde{\Theta}(\omega)$, +which involves the signum function $\mathrm{sgn}(t)$, +the [Dirac delta function](/know/concept/dirac-delta-function/) $\delta$, +and the Cauchy principal value $\pv{}$. +We arrive at: + +$$\begin{aligned} + \tilde{\chi}(\omega) + &= \frac{A B}{|s|} \: \pv{\int_{-\infty}^\infty \tilde{\chi}(\omega') + \Big( \pi \delta(\omega - \omega') + i \:\mathrm{sgn} \frac{1}{\omega - \omega'} \Big) \dd{\omega'}} + \\ + &= \Big( \frac{1}{2} \frac{2 \pi A B}{|s|} \Big) \tilde{\chi}(\omega) + + i \Big( \frac{\mathrm{sgn}(s)}{2 \pi} \frac{2 \pi A B}{|s|} \Big) + \: \pv{\int_{-\infty}^\infty \frac{\tilde{\chi}(\omega')}{\omega - \omega'} \dd{\omega'}} +\end{aligned}$$ + +From the definition of the Fourier transform we know that $2 \pi A B / |s| = 1$: + +$$\begin{aligned} + \tilde{\chi}(\omega) + &= \frac{1}{2} \tilde{\chi}(\omega) + + \mathrm{sgn}(s) \frac{i}{2 \pi} \: \pv{\int_{-\infty}^\infty \frac{\tilde{\chi}(\omega')}{\omega - \omega'} \dd{\omega'}} +\end{aligned}$$ + +We isolate this equation for $\tilde{\chi}(\omega)$ +to get the final version of the causality test: + +$$\begin{aligned} + \boxed{ + \tilde{\chi}(\omega) + = - \mathrm{sgn}(s) \frac{i}{\pi} \: \pv{\int_{-\infty}^\infty \frac{\tilde{\chi}(\omega')}{\omega - \omega'} \dd{\omega'}} + } +\end{aligned}$$ + +By inserting $\tilde{\chi}(\omega) = \tilde{\chi}_r(\omega) + i \tilde{\chi}_i(\omega)$ +and splitting the equation into real and imaginary parts, +we get the Kramers-Kronig relations: + +$$\begin{aligned} + \boxed{ + \begin{aligned} + \tilde{\chi}_r(\omega) + &= \mathrm{sgn}(s) \frac{1}{\pi} \: \pv{\int_{-\infty}^\infty \frac{\tilde{\chi}_i(\omega')}{\omega' - \omega} \dd{\omega'}} + \\ + \tilde{\chi}_i(\omega) + &= - \mathrm{sgn}(s) \frac{1}{\pi} \: \pv{\int_{-\infty}^\infty \frac{\tilde{\chi}_r(\omega')}{\omega' - \omega} \dd{\omega'}} + \end{aligned} + } +\end{aligned}$$ + +If the time-domain response function $\chi(t)$ is real +(so far we have assumed it to be complex), +then we can take advantage of the fact that +the FT of a real function satisfies +$\tilde{\chi}(-\omega) = \tilde{\chi}^*(\omega)$, i.e. $\tilde{\chi}_r(\omega)$ +is even and $\tilde{\chi}_i(\omega)$ is odd. We multiply the fractions by +$(\omega' + \omega)$ above and below: + +$$\begin{aligned} + \tilde{\chi}_r(\omega) + &= \mathrm{sgn}(s) \bigg( \frac{1}{\pi} \: \pv{\int_{-\infty}^\infty \frac{\omega' \tilde{\chi}_i(\omega')}{{\omega'}^2 - \omega^2} \dd{\omega'}} + + \frac{\omega}{\pi} \: \pv{\int_{-\infty}^\infty \frac{\tilde{\chi}_i(\omega')}{{\omega'}^2 - \omega^2} \dd{\omega'}} \bigg) + \\ + \tilde{\chi}_i(\omega) + &= - \mathrm{sgn}(s) \bigg( \frac{1}{\pi} \: \pv{\int_{-\infty}^\infty \frac{\omega' \tilde{\chi}_r(\omega')}{{\omega'}^2 - \omega^2} \dd{\omega'}} + + \frac{\omega}{\pi} \: \pv{\int_{-\infty}^\infty \frac{\tilde{\chi}_r(\omega')}{{\omega'}^2 - \omega^2} \dd{\omega'}} \bigg) +\end{aligned}$$ + +For $\tilde{\chi}_r(\omega)$, the second integrand is odd, so we can drop it. +Similarly, for $\tilde{\chi}_i(\omega)$, the first integrand is odd. +We therefore find the following variant of the Kramers-Kronig relations: + +$$\begin{aligned} + \boxed{ + \begin{aligned} + \tilde{\chi}_r(\omega) + &= \mathrm{sgn}(s) \frac{2}{\pi} \: \pv{\int_0^\infty \frac{\omega' \tilde{\chi}_i(\omega')}{{\omega'}^2 - \omega^2} \dd{\omega'}} + \\ + \tilde{\chi}_i(\omega) + &= - \mathrm{sgn}(s) \frac{2 \omega}{\pi} \: \pv{\int_0^\infty \frac{\tilde{\chi}_r(\omega')}{{\omega'}^2 - \omega^2} \dd{\omega'}} + \end{aligned} + } +\end{aligned}$$ + +To reiterate: this version is only valid if $\chi(t)$ is real in the time domain. diff --git a/content/know/concept/schwartz-distribution/index.pdc b/content/know/concept/schwartz-distribution/index.pdc new file mode 100644 index 0000000..2d9f9df --- /dev/null +++ b/content/know/concept/schwartz-distribution/index.pdc @@ -0,0 +1,119 @@ +--- +title: "Schwartz distribution" +firstLetter: "S" +publishDate: 2021-02-25 +categories: +- Mathematics + +date: 2021-02-25T13:47:16+01:00 +draft: false +markup: pandoc +--- + +# Schwartz distribution + +A **Schwartz distribution**, also known as a **generalized function**, +is a generalization of a function, +allowing us to work with otherwise pathological definitions. + +Notable examples of distributions are +the [Dirac delta function](/know/concept/dirac-delta-function/) +and the [Heaviside step function](/know/concept/heaviside-step-function/), +whose unusual properties are justified by this generalization. + +We define the **Schwartz space** $\mathcal{S}$ of functions, +whose members are often called **test functions**. +Every such $\phi(x) \in \mathcal{S}$ must satisfy +the following constraint for any $p, q \in \mathbb{N}$: + +$$\begin{aligned} + \mathrm{max} \big| x^p \phi^{(q)}(x) \big| < \infty +\end{aligned}$$ + +In other words, a test function and its derivatives +decay faster than any polynomial. +Furthermore, all test functions must be infinitely differentiable. +These are quite strict requirements. + +The **space of distributions** $\mathcal{S}'$ (note the prime) +is then said to consist of *functionals* $f[\phi]$ +which map a test function $\phi$ from $\mathcal{S}$, +to a number from $\mathbb{C}$, +which is often written as $\braket{f}{\phi}$. +This notation looks like the inner product of +a [Hilbert space](/know/concept/hilbert-space/), +for good reason: any well-behaved function $f(x)$ can be embedded +into $\mathcal{S}'$ by defining the corresponding functional $f[\phi]$ as follows: + +$$\begin{aligned} + f[\phi] + = \braket{f}{\phi} + = \int_{-\infty}^\infty f(x) \: \phi(x) \dd{x} +\end{aligned}$$ + +Not all functionals qualify for $\mathcal{S}'$: +they also need to be linear in $\phi$, and **continuous**, +which in this context means: if a series $\phi_n$ +converges to $\phi$, then $\braket{f}{\phi_n}$ +converges to $\braket{f}{\phi}$ for all $f$. + +The power of this generalization is that $f(x)$ does not need to be well-behaved: +for example, the Dirac delta function can also be used, +whose definition is nonsensical *outside* of an integral, +but perfectly reasonable *inside* one. +By treating it as a distribution, +we gain the ability to sanely define e.g. its derivatives. + +Using the example of embedding a well-behaved function $f(x)$ into $\mathcal{S}$, +we can work out what the derivative of a distribution is: + +$$\begin{aligned} + \braket{f'}{\phi} + = \int_{-\infty}^\infty f'(x) \: \phi(x) \dd{x} + = \Big[ f(x) \: \phi(x) \Big]_{-\infty}^\infty - \int_{-\infty}^\infty f(x) \: \phi'(x) \dd{x} +\end{aligned}$$ + +The test function removes the boundary term, yielding the result +$- \braket{f}{\phi'}$. Although this was an example for a specific $f(x)$, +we use it to define the derivative of any distribution: + +$$\begin{aligned} + \boxed{ + \braket{f'}{\phi} = - \braket{f}{\phi'} + } +\end{aligned}$$ + +Using the same trick, we can find the +[Fourier transform](/know/concept/fourier-transform/) (FT) +of a generalized function. +We define the FT as follows, +but be prepared for some switching of the names $k$ and $x$: + +$$\begin{aligned} + \tilde{\phi}(x) + = \int_{-\infty}^\infty \phi(k) \exp(- i k x) \dd{k} +\end{aligned}$$ + +The FT of a Schwartz distribution $f$ then turns out to be as follows: + +$$\begin{aligned} + \braket*{\tilde{f}}{\phi} + &= \int_{-\infty}^\infty \tilde{f}(k) \: \phi(k) \dd{k} + = \iint_{-\infty}^\infty f(x) \exp(- i k x) \: \phi(k) \dd{x} \dd{k} + \\ + &= \int_{-\infty}^\infty f(x) \: \tilde{\phi}(x) \dd{x} + = \braket*{f}{\tilde{\phi}} +\end{aligned}$$ + +Note that the ordinary FT $\tilde{f}(k) = \hat{\mathcal{F}}\{f(x)\}$ is +already a 1:1 mapping of test functions $\phi \leftrightarrow \tilde{\phi}$. +As it turns out, +in this generalization it is also a 1:1 mapping of distributions in $\mathcal{S}'$, +defined as: + +$$\begin{aligned} + \boxed{ + \braket*{\tilde{f}}{\phi} + = \braket*{f}{\tilde{\phi}} + } +\end{aligned}$$ |