advanced35 min studyLesson 275 in Physics

Kramers–Kronig dispersion relations

Connect causal memory to linked real and imaginary response, with explicit convergence assumptions.

Before you begin

On this page

The question for today

What you’ll understand

Explain why a causal response is analytic in the upper frequency half-plane.
State a convergent dispersion relation and its subtraction requirement.
Use a causal relaxation model to check real and imaginary parts.

A material cannot begin responding before an applied perturbation arrives. In a linear stationary medium that statement constrains the entire frequency response, rather than only its behavior at one frequency. The Kramers–Kronig relations express this constraint as integral links between dispersive and absorptive components. They are valuable consistency tests and reconstruction tools, but their use requires attention to high-frequency behavior, singularities and the unmeasured spectral range.

Write causality as memory with one direction

Let a real response kernel vanish for negative delays. Polarization at a given time is then a weighted sum of earlier electric fields. Stationarity makes the kernel depend on elapsed delay rather than absolute clock time; linearity makes the weighting independent of field amplitude. These assumptions exclude pumped systems whose properties change during the measurement and strong fields that alter the material response. They should be checked before interpreting a fitted susceptibility as a stationary causal function.

With the negative time-exponential convention, the Fourier transform of the kernel uses a positive frequency exponential. When frequency has positive imaginary part, that exponential decreases with delay. Under suitable stability and integrability conditions the transform is analytic in the upper half-plane. A real time-domain kernel also imposes conjugate symmetry between positive and negative frequencies. These two properties provide the contour argument linking real and imaginary parts; passivity adds a sign constraint but is not identical to causality.

P(t)=\epsilon_0\int_0^\infty g(s)E(t-s)\,ds,\qquad \chi(z)=\int_0^\infty g(s)e^{izs}\,ds,\qquad \chi(-\omega)=\chi(\omega)^*

The causal scalar kernel g has units s⁻¹ when χ is dimensionless; s is a delay in seconds and z is complex angular frequency. Analyticity needs stable response and an appropriate convergent transform.↔ Scroll wide equations horizontally.

State the integral with its assumptions

Suppose the susceptibility after removing any instantaneous constant decays sufficiently at large complex frequency, has no upper-half-plane poles, and has no untreated real-axis singularities. A contour closed in the upper half-plane then gives a principal-value relation. Combining positive and negative frequencies yields an integral over positive frequencies only. The principal value handles the singular kernel at the observation frequency by symmetric exclusion; it does not mean simply omitting one arbitrary data point from a measured spectrum.

A nonzero high-frequency constant must be subtracted before applying a vanishing-at-infinity formula. More general growth can require additional subtracted dispersion relations with reference values supplied separately. A conducting permittivity also has a zero-frequency pole that needs explicit treatment; applying a regular dielectric formula blindly to that pole can give a false inconsistency. In practice finite spectral measurements need extrapolated tails, and those assumptions contribute uncertainty to the reconstructed real response even when the measured imaginary data are excellent.

\chi^{\prime}(\omega)=\frac{2}{\pi}\,\mathcal P\int_0^\infty\frac{\Omega\chi^{\prime\prime}(\Omega)}{\Omega^2-\omega^2}\,d\Omega,\qquad \chi^{\prime}(0)=\frac{2}{\pi}\int_0^\infty\frac{\chi^{\prime\prime}(\Omega)}{\Omega}\,d\Omega

This displayed form assumes χ tends to zero sufficiently rapidly and has no unhandled dc pole. For εr tending to ε∞, apply it to εr − ε∞. The static integral also requires convergence at both endpoints.↔ Scroll wide equations horizontally.

Check a complete causal example

A simple relaxation kernel decreases exponentially after the perturbation. Its transform has a single pole below the real frequency axis, where a stable causal response permits it. The imaginary part is positive at positive frequency, peaks at the inverse relaxation time, and falls again. Its real part decreases smoothly from the static response to zero. These curves cannot be adjusted independently without changing the kernel or violating the assumptions that linked them.

The static dispersion relation provides an especially transparent check because its denominator contains no interior principal-value singularity. Inserting the relaxation loss curve reduces the integral to an arctangent and exactly recovers the static susceptibility. A narrow transparent window can still have appreciable dispersion because absorption elsewhere contributes to the integral. Conversely, an isolated measurement showing little absorption cannot establish frequency-independent permittivity throughout that window or beyond it.

g(s)=\frac{\chi_s}{\tau}e^{-s/\tau}\ (s\ge0),\qquad \chi(\omega)=\frac{\chi_s}{1-i\omega\tau},\qquad \chi^{\prime\prime}=\frac{\chi_s\omega\tau}{1+\omega^2\tau^2}

χs is a dimensionless static susceptibility and τ > 0 is in seconds. This is a causal scalar Debye relaxation model, not a resonant Lorentz oscillator or a general reconstruction algorithm.↔ Scroll wide equations horizontally.

Work it through

A relaxation curve at one frequency

For $\chi_s=6$ and $\tau=2\,\mathrm s$ , evaluate susceptibility at $\omega=0.5\,\mathrm{s^{-1}}$ , and verify the location of its pole.

The dimensionless product is $\omega\tau=1$ . The response is therefore $6/(1-i)$ ; rationalizing gives $3+3i$ . Its real and imaginary parts are equal at this frequency.
The kernel is $g(s)=3e^{-s/2}\,\mathrm{s^{-1}}$ for nonnegative delay and zero otherwise. Integrating it over delay gives six, the static susceptibility rather than its finite-frequency value.
Set $1-iz\tau=0$ for complex frequency z. The pole is $z=-i/\tau=-0.5i\,\mathrm{s^{-1}}$ , below the real axis. A sign-reversed pole would imply exponential growth or a noncausal transform prescription.

The result

The response is 3 + 3i at the chosen frequency, and its pole lies at −0.5i s⁻¹ in the lower half-plane.

The living notebook

Explore the model

Interactive lab

Interactive lab · oscillator

Prepared relaxation is not a spectral reconstruction

Reset and set BOTH Damping ratio ζ = 0.50 and Natural frequency ω₀ = 1; observe a decaying oscillatory release. Reset and set ζ = 1 and ω₀ = 1; predict decay without zero crossings. Step each fresh run from x = 1 and zero initial velocity. These are prepared mechanical relaxations; the lab does not Fourier-transform them, implement the Debye kernel, drive a frequency sweep or compute a Kramers–Kronig integral.

Displacement x(t)Current time

Displacement x(t): 1
Velocity v(t): 0
Regime: underdamped

Damping ratio ζ0.2Natural frequency ω₀1.4

t = 0 s

Unit mass, x(0) = 1 and v(0) = 0. The analytic solution includes underdamped, critically damped and overdamped cases; time is measured in seconds. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA time response contains memory information, but connecting it to spectra requires the specified response experiment.

Pause & reason

Check your understanding

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Read the checkpoint explanation

The constant represents an instantaneous contribution and must be handled separately. The contour argument for the displayed formula requires decay of the transformed remainder.

Your turn

Try it yourself

Use $\chi^{\prime\prime}(\Omega)=\chi_s\Omega\tau/(1+\Omega^2\tau^2)$ to evaluate the static Kramers–Kronig integral exactly. Also find the fraction supplied by frequencies from zero through $1/\tau$ .

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Cancel Ω, substitute u = Ωτ, and use the antiderivative arctan u.

Compare with the solution

The integral becomes $(2\chi_s/\pi)\int_0^\infty du/(1+u^2)=(2\chi_s/\pi)(\pi/2)=\chi_s$ . Restricting its upper limit to one gives $(2\chi_s/\pi)(\pi/4)=\chi_s/2$ . Thus half the static response comes from frequencies above the loss-peak frequency in this model. Neglecting that entire tail would underestimate the static susceptibility by a factor of two, even with perfect low-frequency data. This exact result relies on the full relaxation law, including its convergent high-frequency tail.

Keep the idea

Take it with you

Causality constrains analytic structure and links dispersive and absorptive response.
Constants, dc poles and insufficient decay require explicit treatment or subtractions.
Finite-band reconstruction depends on assumptions about unmeasured frequencies.

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