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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. |