On the Kramers-Kronig relations

Abstract

We provide a new derivation of the Kramers-Kronig relations on the basis of the Sokhotski-Plemelj equation with detailed mathematical justifications. The relations hold for a causal function, whose Fourier transform is regular (holomorphic) and square-integrable. This implies analyticity in the lower complex plane and a Fourier transform that vanishes at the high-frequency limit. In viscoelasticity, we show that the complex and frequency-dependent modulus describing the stiffness does not satisfy the relation but the modulus minus its high-frequency value does it. This is due to the fact that despite its causality, the modulus is not square-integrable due to a non-null instantaneous response. The relations are obtained in addition for the wave velocity and attenuation factor. The Zener, Maxwell, and Kelvin-Voigt viscoelastic models illustrate these properties. We verify the Kramers-Kronig relations on experimental data of sound attenuation in seabottoms sediments.

Appendix: the Zener and Maxwell models verify the Kramers-Kronig relations

In this Appendix we derive, first, a very compact unified formulation of Kramers-Kronig relations, using Hilbert transform. Then we show, through detailed ad hoc computations, that the reduced complex modulus of the Zener model verifies this condition.

A.1 Compact formulation of Kramers-Kronig relations

As we have seen in the main text, Kramers-Kronig relations for a complex-valued function f of a real variable ω may be expressed as follows:

$$ \left\{ \begin{array}{llllllllllll} \text{Re} f &= \mathcal{H} (\text{Im} f ) \\ \text{Im} f &= - \mathcal{H} (\text{Re} f ) \end{array} \right. $$

(47)

where symbol $\mathcal {H }$ denotes Hilbert transform.

Summing (1)₁ and (1)₂ multiplied by the imaginary unit i, and rearranging, we obtain the following:

$$ f = - \mathrm{i} \mathcal{H } (f ) $$

(48)

where we have used the linearity of Hilbert transform.

Conversely, taking real and imaginary parts of Eq. 48, we obtain (47). Therefore, the single condition (48) is equivalent to Kramers-Kronig relations.

A.2 Zener model

We recall that the complex modulus of the Zener model may be expressed as follows:

$$ M(\omega) = M_{\infty} - \frac{ M_{\infty} - M_{0} }{ 1 + \mathrm{i} \tau \omega } $$

(49)

Now, we show that the reduced complex modulus

$$ \overline{M}(\omega) = M(\omega) - M_{\infty} = - \frac{ M_{\infty} - M_{0} }{ 1+ \mathrm{i} \tau \omega } $$

(50)

verifies formulation (48) of Kramers-Kronig relations, i.e.,

$$ \overline{M}(\omega) = (M_{\infty} - M_{0} ) \mathrm{i} \mathcal{H} \left( \frac{ 1 }{ 1 + \mathrm{i} \tau \omega } \right) $$

(51)

The Hilbert transform of a function f(ω) is given by the following:

$$\begin{array}{@{}rcl@{}} \mathcal{H} f(\omega ) &=& \frac{1}{\pi} \mathcal{P} \int_{-\infty}^{\infty} g(\omega, \omega^{\prime}) \mathrm{d} \omega^{\prime} = \frac{1}{\pi} \lim\limits_{L \rightarrow \infty} \mathcal{P} \int_{-L}^{L} g(\omega, \omega^{\prime}) \mathrm{d} \omega^{\prime} \\ &=& \frac{1}{\pi} \lim\limits_{L \rightarrow \infty} \lim\limits_{\epsilon \rightarrow 0^{+}} \left( \int_{-L}^{\omega - \epsilon} g(\omega , \omega^{\prime} ) \mathrm{d} \omega^{\prime} + \int_{\omega + \epsilon}^{L} g(\omega , \omega^{\prime} ) \mathrm{d} \omega^{\prime} \right) \\ &=& \frac{1}{\pi} \lim\limits_{L \rightarrow \infty} \lim\limits_{\epsilon \rightarrow 0^{+}} [ G(\omega, \omega - \epsilon ) - G(\omega, -L )\\&&+ G(\omega, L ) - G(\omega, \omega + \epsilon )] \end{array} $$

(52)

where

$$ g(\omega, \omega^{\prime}) = \frac{ f(\omega^{\prime}) }{ \omega - \omega^{\prime} } $$

(53)

and G is a primitive of the integrand in Eq. 52, i.e.,

$$ G(\omega , \omega^{\prime} ) = \int g(\omega, \omega^{\prime} ) \mathrm{d} \omega^{\prime} $$

(54)

Therefore, the Hilbert transform in Eq. 51 is given by the following:

$$ \mathcal{H} \left( \frac{ 1 }{ 1 + \mathrm{i} \tau \omega } \right) = \frac{1}{ {\tau \omega - \text i} } $$

(55)

as shown in the Lemma below.

Substituting (55) into (51) and rearranging, one obtains an identity, which proves that the reduced complex modulus $\overline {M}(\omega )$ verifies Kramers-Kronig relations.

Lemma

The Hilbert transform of functionf(ω) = 1/(1 + iτω) isgiven by the following:

$$ \mathcal{H} \left( \frac{ 1 }{ 1 + \mathrm{i} \tau \omega } \right) = \frac{1}{ {\tau \omega - \text i} } $$

(56)

Proof We rely on identity (52), where now

$$ g(\omega , \omega^{\prime} ) = \frac{ 1 }{ (\omega - \omega^{\prime}) (1+ \mathrm{i} \tau \omega^{\prime}) } $$

(57)

and hence its primitive is

$$ G(\omega , \omega^{\prime} ) = \frac{ 2 \tan^{-1}(\tau \omega^{\prime}) + 2 \mathrm{i} \ln(\omega^{\prime} - \omega) - \mathrm{i} \ln[1 + (\tau \omega^{\prime})^{2} ] }{ 2 (\tau \omega - \mathrm{i}) } $$

(58)

as can be checked by differentiating with respect to $\omega ^{\prime }$. Substituting (58) into (52), one obtains (56). Computations are cumbersome, but straightforward: indeed two divergent terms appear, $\ln (\epsilon )$ and $\ln (- \epsilon )$, but they are present only in the combination $\ln (\epsilon ) - \ln (- \epsilon ) = \ln (-1) = \mathrm {i} \pi$.

A.3 Maxwell model

The Maxwell model is obtained from Fig. 1 by removing the parallel spring. Its reduced complex modulus is as follows:

$$ \overline{M}(\omega) = M(\omega) - M_{\infty} = - \frac{ M_{\infty} }{ 1+ \mathrm{i} \tau \omega } $$

(59)

(e.g., Carcione 2014). From Eqs. 48 and 56, we have the following:

$$ \overline{M}(\omega) = - \mathrm{i} \mathcal{H } (\overline{M} ) = \mathrm{i} M_{\infty} \mathcal{H} \left( \frac{ 1 }{ 1 + \mathrm{i} \tau \omega } \right) = \frac{\mathrm{i} M_{\infty}}{\tau \omega - \text i} = \overline{M} (\omega) , $$

(60)

so we have shown that the Maxwell model satisfies the relations.

A.4 Kelvin-Voigt model

The Kelvin-Voigt model is obtained from Fig. 1 by removing the series spring. Its complex modulus is as follows:

$$ M(\omega) = M_{0} (1+ \mathrm{i} \tau \omega ) $$

(61)

(e.g., Carcione 2014). $M_{\infty } = \infty$ in this case. We may consider $\overline {M} = \mathrm {i} \tau \omega M_{0}$, since the Hilbert transform of a constant (M₀) is zero. This is the complex modulus of the dashpot.

The primitive function is as follows:

$$ G(\omega , \omega^{\prime} ) = - \mathrm{i} \tau M_{0} [\omega^{\prime}+\omega \ln (\omega^{\prime}-\omega)] . $$

(62)

It can be easily seen that the calculation (52) diverges, and therefore the Kelvin-Voigt solid does not satisfy the relations. Another explanation may be found by considering the relaxation rate $\dot \psi = M_{0} \delta + M_{0} \tau \delta$, which is the inverse Fourier transform of the complex modulus. One immeditely sees that it is causal, but both of its terms are singular distributions. Thus, there is no way, because of the lack of regularity, to introduce a “reduced complex modulus” as in the previous cases. Equivalently, one may notice that both terms in the complex modulus M = M₀ + iωτM₀ are not square-integrable.