Damping Regimes Slides Damping Regimes
Characterizing the Mathematical Landscape of Energy Dissipation in Harmonic Systems
Physics Graduate Sequence
The Hook
"Why is critical damping essential for the suspension systems of precision laboratory equipment and luxury vehicles alike?"
The Problem
Uncontrolled oscillations lead to instability, wear, and data noise. Energy must be removed efficiently.
The Solution
Finding the "Sweet Spot" where the system returns to equilibrium as fast as possible without overshooting.
The Mathematical Foundation
The second-order linear homogeneous ODE for a damped harmonic oscillator:
\[ \ddot{x} + 2\beta\dot{x} + \omega_0^2 x = 0 \]
Parameters
\(\omega_0 = \sqrt{k/m}\) : Natural frequency
\(\beta = b/2m\) : Damping coefficient
Trial Solution
Assume \(x(t) = e^{rt}\). This leads to the characteristic equation: \(r^2 + 2\beta r + \omega_0^2 = 0\)
Three Regimes of Motion
Underdamped
Condition: \(\beta < \omega_0\)
Oscillatory behavior with exponentially decaying amplitude.
\(x(t) = A e^{-\beta t} \cos(\omega_1 t - \delta)\)
Overdamped
Condition: \(\beta > \omega_0\)
No oscillation. The system slowly creeps back to equilibrium.
\(x(t) = e^{-\beta t}(A e^{\gamma t} + B e^{-\gamma t})\)
Critically Damped
Condition: \(\beta = \omega_0\)
Fastest return to equilibrium without oscillation.
\(x(t) = (A + Bt)e^{-\beta t}\)
The Quality Factor (Q)
Q measures how "good" an oscillator is—specifically, how many cycles it performs before losing its energy.
\[ Q = \frac{\omega_0}{2\beta} \]
Dimensionless parameter representing the damping strength.
High Q (> 1/2)
Underdamped. Many oscillations. Examples: Atomic clocks, musical strings.
Q = 1/2
Critically damped.
Low Q (< 1/2)
Overdamped. Sluggish response. Examples: Door closers, car shocks.
Damping Regimes Worksheet Damping Regimes Worksheet
Response Theory Mastery | Lesson 1
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Part 1: Parameter Identification
Consider a mass-spring-damper system with \(m = 2.0 \, \text{kg}\), \(k = 50 \, \text{N/m}\), and a damping coefficient \(b\). The governing equation is given by \(m\ddot{x} + b\dot{x} + kx = 0\).
1.1 Calculate the natural frequency \(\omega_0\) and define the parameter \(\beta\) in terms of \(b\).
1.2 Determine the critical damping coefficient \(b_{crit}\) for this system. At what value of \(b\) will the system transition to the underdamped regime?
Part 2: Analytical Solutions
Assume the system is underdamped with \(b = 4.0 \, \text{kg/s}\). At \(t=0\), the mass is displaced to \(x_0 = 0.1 \, \text{m}\) and released from rest (\(v_0 = 0\)).
2.1 Calculate the damped frequency \(\omega_1\) and the Quality Factor \(Q\).
2.2 Derive the specific equation of motion \(x(t)\) for these initial conditions. Express your answer in the form \(x(t) = A e^{-\beta t} \cos(\omega_1 t - \delta)\).
Part 3: Comparative Dynamics
3.1 On the axes below, sketch the displacement \(x(t)\) for three cases: underdamped (high Q), overdamped (low Q), and critically damped. Label each curve clearly.
Time (t) Pos (x)
3.2 Engineering Application
A high-precision optical table is being designed. Why might an engineer choose to tune the isolation system to slightly overdamped rather than underdamped ? What is the trade-off compared to critical damping ?
Damping Regimes Answer Key Answer Key: Damping Regimes
Instructor Resource | Lesson 1
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Part 1: Parameter Identification
1.1 Calculate \(\omega_0\) and define \(\beta\).
Natural Frequency: \(\omega_0 = \sqrt{k/m} = \sqrt{50/2} = \sqrt{25} = 5.0 \, \text{rad/s}\)
Damping Parameter: \(\beta = b/(2m) = b/4\)
1.2 Critical damping coefficient \(b_{crit}\).
Critical condition: \(\beta = \omega_0 \implies b/(2m) = \omega_0 \implies b = 2m\omega_0\)
\(b_{crit} = 2(2.0)(5.0) = 20.0 \, \text{kg/s}\)
Transition: The system is underdamped when \(b < 20.0 \, \text{kg/s}\).
Part 2: Analytical Solutions
2.1 Calculate \(\omega_1\) and \(Q\).
\(\beta = b/(2m) = 4/4 = 1.0 \, \text{s}^{-1}\)
\(\omega_1 = \sqrt{\omega_0^2 - \beta^2} = \sqrt{25 - 1} = \sqrt{24} \approx 4.90 \, \text{rad/s}\)
\(Q = \omega_0 / (2\beta) = 5.0 / (2.0) = 2.5\)
2.2 Derive equation of motion \(x(t)\).
General solution: \(x(t) = A e^{-\beta t} \cos(\omega_1 t - \delta)\)
1. \(x(0) = 0.1 \implies A \cos(-\delta) = 0.1\)
2. \(v(0) = 0 \implies \dot{x}(0) = -\beta A \cos(-\delta) + \omega_1 A \sin(-\delta) = 0\)
\(\tan \delta = -\beta / \omega_1 = -1.0 / 4.90 \implies \delta \approx -0.201 \, \text{rad}\)
\(A = 0.1 / \cos(-0.201) \approx 0.1 / 0.98 \approx 0.102 \, \text{m}\)
\(x(t) = 0.102 e^{-t} \cos(4.90t + 0.201) \, \text{m}\)
Part 3: Conceptual Analysis
3.2 Engineering Application Discussion
Overdamped Choice: Engineers may choose slight overdamping to ensure that the system never overshoots the zero position, even if there is uncertainty in the load mass (which changes \(\omega_0\) and thus the critical damping point). Overdamping is "safer" to avoid oscillations that might disrupt sensitive laser paths.
Trade-off: The recovery time is slower than critical damping. The system takes longer to return to the "ready" state after a disturbance compared to the optimal (critically damped) case.
Resonance Dynamics Slides Resonance Dynamics
Steady-State Solutions and the Physics of Forced Oscillations
Lesson 2 Driven Systems
The Catastrophe Principle
"How can a small, periodic wind force destroy a massive bridge?"
The Tacoma Narrows disaster wasn't just about force; it was about matching frequencies and the accumulation of energy.
Aeroelastic Flutter
A feedback loop between aerodynamic forces and the structure's natural modes.
The Inhomogeneous ODE
We add a sinusoidal driving force \(F(t) = F_0 \cos(\omega t)\):
\[ \ddot{x} + 2\beta\dot{x} + \omega_0^2 x = \frac{F_0}{m} \cos(\omega t) \]
Transient Solution
The solution to the homogeneous equation (\(x_h\)). Decays over time due to damping.
Steady-State Solution
The particular solution (\(x_p\)). Persists as long as the driving force is active.
The Steady-State Response
Assuming a solution of the form: \(x(t) = A(\omega) \cos(\omega t - \delta)\)
Amplitude \(A(\omega)\)
\[ A(\omega) = \frac{F_0/m}{\sqrt{(\omega_0^2 - \omega^2)^2 + (2\beta\omega)^2}} \]
Phase Lag \(\delta\)
\[ \tan \delta = \frac{2\beta\omega}{\omega_0^2 - \omega^2} \]
Key Phases of \(\delta\)
Low freq (\(\omega \ll \omega_0\)) \(\delta \to 0\)
At resonance (\(\omega = \omega_0\)) \(\delta = \pi/2\)
High freq (\(\omega \gg \omega_0\)) \(\delta \to \pi\)
Phase shift is the critical indicator of energy transfer efficiency.
Visualizing Resonance
Lorentzian Profiles
The amplitude curve peaks near \(\omega_0\). As damping \(\beta\) decreases (Higher Q):
The peak height increases significantly.
The width (FWHM) of the resonance narrows.
Full Width at Half Maximum
\[ \Delta \omega \approx 2\beta = \frac{\omega_0}{Q} \]
Measuring the peak width is a standard experimental method to determine the Quality Factor of a physical system.
Resonance Dynamics Worksheet Resonance Analysis Sheet
Lesson 2 | Steady-State Response
Researcher:
System ID:
1
The Lorentzian Response
A system has a natural frequency \(\omega_0 = 100 \, \text{rad/s}\) and a quality factor \(Q = 50\). It is driven by a force \(F(t) = F_0 \cos(\omega t)\).
1.1 Calculate the damping parameter \(\beta\) and the resonance width \(\Delta\omega\).
1.2 At what driving frequency \(\omega_r\) is the amplitude exactly maximized? (Note: It is slightly less than \(\omega_0\)).
2
Phase Lag Interpretation
The phase lag \(\delta\) is given by \(\tan \delta = \frac{2\beta\omega}{\omega_0^2 - \omega^2}\).
2.1 Prove that at resonance (\(\omega = \omega_0\)), the displacement lags the force by exactly \(\pi/2\). What does this imply about the direction of the velocity relative to the driving force?
2.2 Consider the "In-Phase" (\(\omega \to 0\)) and "Out-of-Phase" (\(\omega \to \infty\)) limits. Briefly explain why the system is unable to follow the force at very high frequencies.
3
Power Absorption
The average power \(\langle P \rangle\) absorbed by the oscillator is given by \(\langle P \rangle = \frac{F_0^2 \omega^2 \beta}{m [(\omega_0^2 - \omega^2)^2 + 4\beta^2 \omega^2]}\).
Analysis Task:
Sketch the power absorption curve for a low-damping system (\(Q=100\)) versus a high-damping system (\(Q=5\)). Label the peak height and the Full Width at Half Maximum (FWHM) on your sketch.
Driving Frequency (\(\omega\)) Avg Power (\(\langle P \rangle\))
Resonance Dynamics Answer Key Answer Key: Resonance Dynamics
Instructor Resource | Lesson 2
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Part 1: The Lorentzian Response
1.1 Calculate \(\beta\) and \(\Delta\omega\).
Given \(\omega_0 = 100\) and \(Q = 50\).
\(Q = \omega_0 / (2\beta) \implies 50 = 100 / (2\beta) \implies 2\beta = 2 \implies \beta = 1.0 \, \text{s}^{-1}\).
Resonance width (FWHM) \(\Delta\omega \approx 2\beta = 2.0 \, \text{rad/s}\).
1.2 Amplitude Maximization Frequency \(\omega_r\).
The resonant frequency of the amplitude is \(\omega_r = \sqrt{\omega_0^2 - 2\beta^2}\).
\(\omega_r = \sqrt{100^2 - 2(1^2)} = \sqrt{10000 - 2} \approx 99.99 \, \text{rad/s}\).
Note: For high Q, \(\omega_r\) is extremely close to \(\omega_0\).
Part 2: Phase Dynamics
2.1 Phase Proof at \(\omega = \omega_0\).
\(\tan \delta = \frac{2\beta\omega}{\omega_0^2 - \omega^2}\). If \(\omega = \omega_0\), the denominator is zero.
\(\tan \delta \to \infty \implies \delta = \pi/2\).
Implication: Since displacement lags by \(\pi/2\), the velocity (\(\dot{x}\))—which is \(\pi/2\) ahead of displacement—is exactly in phase with the driving force. This is why power transfer is maximized at resonance: the force always pushes in the direction of motion.
2.2 High Frequency Limit (\(\omega \to \infty\)).
At high frequencies, \(\delta \to \pi\), meaning the oscillator moves in the exact opposite direction of the force. This occurs because the inertia of the mass prevents it from accelerating quickly enough to keep up with the rapid changes in the driving force, leading to a complete phase reversal.
Part 3: Power Absorption
Sketching Guidelines:
Low Damping (Q=100): Should show a very tall, very sharp (narrow) spike centered at \(\omega_0\). FWHM is small (\(2\beta = 1\)).High Damping (Q=5): Should show a broad, low hump. FWHM is large (\(2\beta = 20\)).The area under the power curve is related to the total energy dissipated per cycle.
Impulse Foundations Slides Impulse Foundations
The Dirac Delta Function and the Birth of Green's Functions
Lesson 3 Impulse Response
Predicting the Chaos
"If we kick a harmonic oscillator once, can we use that information to predict its motion under a complex, time-varying earthquake?"
Linear systems obey the Superposition Principle . If we know the response to a "unit kick" (impulse), we can build the response to any force.
The Goal
Find a universal kernel \(G(t, t')\) that describes the system's inherent physics.
The Method
Decompose a continuous force into an infinite sequence of infinitesimal impulses.
The Dirac Delta Function \(\delta(t)\)
The idealization of an instantaneous, unit-area impulse:
\(\delta(t - t') = 0\) for \(t \neq t'\)
\(\int_{-\infty}^{\infty} \delta(t - t') dt = 1\)
\(\int f(t) \delta(t - t') dt = f(t')\)
\(t'\)
Amplitude \(\to \infty\)
The Green's Function Definition
For a linear operator \(\mathcal{L}\), the Green's function \(G(t, t')\) is the solution to:
\[ \mathcal{L} G(t, t') = \delta(t - t') \]
Physical Causality
In physical systems, an effect cannot precede its cause. Therefore: \(G(t, t') = 0\) for \(t < t'\)
Boundary Conditions
For the oscillator, we typically look for the response starting from equilibrium: \(G(0, t') = \dot{G}(0, t') = 0\)
The Impulse Response of SHM
The impulse \(\delta(t)\) at \(t=0\) transfers exactly 1 unit of momentum to the mass \(m\).
Initial Conditions for \(t > 0\):
\(G(0^+) = 0\)
\(\dot{G}(0^+) = 1/m\)
Final Result (Underdamped)
\[ G(t) = \frac{1}{m\omega_1} e^{-\beta t} \sin(\omega_1 t) \]
This single function contains everything we need to know about how the system reacts to any input.
Impulse Foundations Worksheet Impulse Response Seminar
Lesson 3 | Mathematical Foundations of G(t, t')
Student ID:
1
The Dirac Delta Tool
1.1 Evaluate the following integral using the sifting property:
\[ \int_{-\infty}^{\infty} (t^2 + 2t + 5) \delta(t - 3) dt \]
1.2 Show that \(\delta(at) = \frac{1}{|a|} \delta(t)\). (Hint: Use a change of variables in an integral test).
2
Finding the Oscillator's Green's Function
Consider the undamped oscillator equation: \(\ddot{x} + \omega_0^2 x = \frac{1}{m} \delta(t)\).
2.1 Integration over the impulse:
Integrate the ODE from \(t = -\epsilon\) to \(t = +\epsilon\) as \(\epsilon \to 0\). Show that this implies \(\dot{x}(0^+) = 1/m\), assuming \(x\) is continuous at \(t=0\).
2.2 Solve the homogeneous equation for \(t > 0\) with the initial conditions \(x(0^+) = 0\) and \(\dot{x}(0^+) = 1/m\). This is your Green's function \(G(t)\).
3
Causality & Physical Reality
3.1 Short Answer Reflection
A mathematical solution for \(G(t)\) might technically exist for \(t < 0\). Why do we enforce \(G(t) = 0\) for all \(t < 0\) by multiplying the solution by the Heaviside step function \(\Theta(t)\)? What physical principle would be violated if \(G(t)\) were non-zero before the impulse occurred?
Impulse Foundations Answer Key Answer Key: Impulse Foundations
Instructor Resource | Lesson 3
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Part 1: The Dirac Delta Tool
1.1 Evaluate the integral.
Using \(\int f(t) \delta(t-a) dt = f(a)\):
\(f(3) = 3^2 + 2(3) + 5 = 9 + 6 + 5 = 20\).
1.2 Show that \(\delta(at) = \frac{1}{|a|} \delta(t)\).
Consider \(I = \int_{-\infty}^{\infty} f(t) \delta(at) dt\).
Let \(u = at \implies du = a dt\). Case \(a > 0\): Limits remain the same.
\(I = \int_{-\infty}^{\infty} f(u/a) \delta(u) \frac{du}{a} = \frac{1}{a} f(0)\).
Case \(a < 0\): Limits swap, providing a negative sign that cancels with the swap, resulting in \(1/|a| f(0)\). Thus, \(\delta(at) = \frac{1}{|a|} \delta(t)\).
Part 2: Green's Function Derivation
2.1 Integration over the impulse.
\[ \int_{-\epsilon}^{\epsilon} (\ddot{x} + \omega_0^2 x) dt = \int_{-\epsilon}^{\epsilon} \frac{1}{m} \delta(t) dt \]
\[ [\dot{x}]_{ -\epsilon}^{ \epsilon} + \omega_0^2 \int_{-\epsilon}^{\epsilon} x dt = \frac{1}{m} \]
As \(\epsilon \to 0\), the integral of the continuous function \(x\) vanishes. Thus, \(\dot{x}(0^+) - \dot{x}(0^-) = 1/m\). Since the system starts from rest (\(\dot{x}(0^-)=0\)), then \(\dot{x}(0^+) = 1/m\).
2.2 Solve for \(G(t)\).
Homogeneous soln: \(x(t) = A \cos(\omega_0 t) + B \sin(\omega_0 t)\).
\(x(0) = 0 \implies A = 0\).
\(\dot{x}(t) = B \omega_0 \cos(\omega_0 t) \implies \dot{x}(0) = B \omega_0 = 1/m \implies B = \frac{1}{m \omega_0}\).
\(G(t) = \frac{1}{m \omega_0} \sin(\omega_0 t) \Theta(t)\).
Part 3: Causality Reflection
The Causality Principle: Physical causality states that the effect (response) cannot occur before the cause (force). If \(G(t) \neq 0\) for \(t < 0\), it would imply the mass started moving before the impulse hit it—a "precognitive" response.
By enforcing \(G(t)=0\) for \(t < 0\), we ensure the math reflects the linear, causal nature of physical reality. This is mathematically handled by the Heaviside step function \(\Theta(t)\).
Convolution Mastery Slides Convolution Mastery
Building Reality from Impulses: The Power of the Convolution Integral
Lesson 4 System Synthesis
Predicting the Present
"How does the mathematical operation of 'sliding and multiplying' allow us to synthesize the past to predict the present?"
The present state of an oscillator is the sum of all past impulses, weighted by how much the system has "remembered" (or decayed) since each impulse occurred.
Memory
The Green's function \(G(t - t')\) is the "memory kernel" of the system.
Integration
Integration is the bridge from discrete "kicks" to continuous "forces".
The Convolution Solution
For any driving force \(F(t)\), the response \(x(t)\) is:
\[ x(t) = \int_{0}^{t} F(t') G(t - t') dt' \]
Input
\(F(t')\)
The force applied at time \(t'\)
Physics
\(G(t - t')\)
Response at time \(t\) due to a kick at time \(t'\)
Output
\(x(t)\)
The total resulting displacement
Canonical Forcing Examples
The Step Response
\(F(t) = F_0\) for \(t > 0\)
The oscillator settles to a new equilibrium. The response reveals the Steady State offset: \(F_0/k\).
The Ramp Response
\(F(t) = at\) for \(t > 0\)
The oscillator tracks the increasing force but with a constant Time Lag proportional to damping.
Computational Strategy
Convolution can be difficult to compute analytically. In the lab, we use:
Numerical Integration (Simpson's, Trapezoid)
FFT-based Convolution (The Convolution Theorem)
The Anatomy of a System Response
Past History weighs Present Outcome
For highly damped systems, the memory is short. For low damped (High Q) systems, the oscillator "remembers" forces from much earlier in time.
Convolution Mastery Worksheet Convolution Workshop
Lesson 4 | Synthesizing Forced Motion
Engineer:
1
The Step Response
Consider an undamped oscillator with natural frequency \(\omega_0\) and mass \(m\). The Green's function is \(G(t) = \frac{1}{m\omega_0} \sin(\omega_0 t)\) for \(t > 0\). At \(t=0\), a constant force \(F_0\) is switched on and remains constant for all \(t > 0\).
1.1 Set up and evaluate the convolution integral \(x(t) = \int_0^t F(t') G(t - t') dt'\) to find the displacement of the mass.
1.2 Does the mass settle to its new equilibrium \(x = F_0/k\)? Explain why or why not based on your result from 1.1. (Recall \(k = m\omega_0^2\)).
2
The Finite Impulse (Pulse)
A system is subjected to a "top-hat" force pulse: \(F(t) = F_0\) for \(0 < t < T\), and \(F(t) = 0\) elsewhere.
2.1 Calculate the response \(x(t)\) for the interval \(t > T\). Use the linearity property of convolution: treat the pulse as the sum of a step function at \(t=0\) and a negative step function at \(t=T\).
2.2 Under what condition (relating \(T\) and \(\omega_0\)) will the oscillator be left perfectly at rest after the pulse ends? This is a core concept in "Input Shaping" for robotics.
3
Numerical Synthesis
Suppose you have a recorded earthquake acceleration profile \(a(t)\) and the impulse response \(G(t)\) of a skyscraper model.
Task 3.1
Describe, in 3-4 steps, how you would numerically approximate the convolution integral if you only had discrete samples of \(a(t)\) at intervals of \(\Delta t\).
Write steps here...
Convolution Mastery Answer Key Answer Key: Convolution Mastery
Instructor Resource | Lesson 4
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Part 1: The Step Response
1.1 Evaluate the convolution integral.
\[ x(t) = \int_0^t F_0 \frac{1}{m\omega_0} \sin(\omega_0(t-t')) dt' \]
Let \(u = \omega_0(t-t') \implies du = -\omega_0 dt'\).
\[ x(t) = \frac{F_0}{m\omega_0} \int_{0}^{\omega_0 t} \sin(u) \frac{du}{\omega_0} = \frac{F_0}{m\omega_0^2} [-\cos u]_0^{\omega_0 t} \]
\(x(t) = \frac{F_0}{k} (1 - \cos \omega_0 t)\).
1.2 Analysis of Equilibrium.
No, the mass does not settle. In an undamped system, energy is conserved. The mass oscillates forever between \(x=0\) and \(x=2F_0/k\), centered at the new equilibrium \(F_0/k\). To settle, we would need to include the damping term \(e^{-\beta t}\) in the Green's function.
Part 2: The Finite Pulse
2.1 Response for \(t > T\).
By superposition: \(x(t) = x_{step}(t) - x_{step}(t-T)\).
\(x(t) = \frac{F_0}{k} (1 - \cos \omega_0 t) - \frac{F_0}{k} (1 - \cos \omega_0(t-T))\)
\(x(t) = \frac{F_0}{k} [\cos(\omega_0(t-T)) - \cos(\omega_0 t)]\).
2.2 Perfect Rest Condition.
The oscillator is at rest if \(x(t) = 0\) for all \(t > T\).
This occurs if \(\cos(\omega_0(t-T)) = \cos(\omega_0 t)\).
This happens if \(\omega_0 T\) is an integer multiple of \(2\pi\), i.e., \(T = n \tau\) where \(\tau\) is the period of oscillation. If the pulse lasts exactly one (or more) full periods, the system returns to zero velocity and zero displacement simultaneously.
Part 3: Numerical Synthesis
Numerical Steps:
Discretize time axis with step \(\Delta t\).
For each target time \(t_i\), compute the sum \(\sum_{j=0}^{i} F(t_j) G(t_i - t_j) \Delta t\).
Use the Convolution Theorem : Multiply the FFT of the input by the FFT of the Green's function.
Perform an Inverse FFT to return to the time domain (most efficient for long signals).
Spectral Causality Slides Spectral Causality
The Frequency Domain and the Kramers-Kronig Constraints
Lesson 5 Final Synthesis
The Price of Causality
"Why does the fact that 'an effect cannot precede its cause' impose strict mathematical limits on how materials absorb and refract waves?"
Causality in the Time Domain implies analyticity in the Frequency Domain . This forces a deep connection between how a system stores energy (refraction) and how it loses energy (absorption).
The Core Insight
You cannot have a refractive index that changes with frequency without also having absorption. They are two sides of the same coin.
From \(G(t)\) to \(\chi(\omega)\)
The frequency-domain response (Susceptibility) is the Fourier Transform of the Green's function:
\[ \chi(\omega) = \int_{0}^{\infty} G(t) e^{i\omega t} dt \]
Real Part \(\chi'(\omega)\)
Represents Dispersion (Refraction). Energy stored in the system's "springs".
Imaginary Part \(\chi''(\omega)\)
Represents Absorption (Dissipation). Energy converted to heat.
Analyticity in the Complex Plane
Because \(G(t) = 0\) for \(t < 0\), the integral for \(\chi(\omega)\) converges for all complex frequencies \(\omega\) with a positive imaginary part (\(\text{Im}(\omega) > 0\)).
Mathematical Fact
\(\chi(\omega)\) is analytic in the Upper Half Plane of complex \(\omega\).
Analytic Region Re(\(\omega\))
Upper Half Plane (UHP) stability
The Kramers-Kronig Relations
If you measure the absorption spectrum of a material at all frequencies, you can calculate the refractive index at any frequency without additional measurements.
\[ \chi'(\omega) = \frac{2}{\pi} \mathcal{P} \int_{0}^{\infty} \frac{\Omega \chi''(\Omega)}{\Omega^2 - \omega^2} d\Omega \]
Real from Imaginary
\[ \chi''(\omega) = -\frac{2\omega}{\pi} \mathcal{P} \int_{0}^{\infty} \frac{\chi'(\Omega)}{\Omega^2 - \omega^2} d\Omega \]
Imaginary from Real
"Physics is constrained by the geometry of time itself."
Spectral Causality Worksheet Spectral Causality Analysis
Lesson 5 | Dispersion & Absorption Case Study
Researcher Name:
1
The Spectral Susceptibility
For a damped harmonic oscillator, the Green's function is \(G(t) = \frac{1}{m\omega_1} e^{-\beta t} \sin(\omega_1 t) \Theta(t)\).
1.1 By taking the Fourier Transform of \(G(t)\), derive the complex susceptibility \(\chi(\omega)\). Show that it takes the form:
\[ \chi(\omega) = \frac{1/m}{\omega_0^2 - \omega^2 - 2i\beta\omega} \]
1.2 Decompose your result into real and imaginary parts: \(\chi(\omega) = \chi'(\omega) + i\chi''(\omega)\). Identify which part describes the phase-lagged energy dissipation.
2
Poles and Stability
The poles of \(\chi(\omega)\) represent the system's "natural modes" in the complex frequency plane.
Analysis Task 2.1
Find the locations of the two poles of \(\chi(\omega)\) for the underdamped case (\(\beta < \omega_0\)).
Task 2.2: Stability Criterion
Causality requires that all poles of \(\chi(\omega)\) must lie in the Lower Half Plane (\(\text{Im}(\omega) < 0\)). Briefly explain why a pole in the Upper Half Plane would correspond to an unphysical, exponentially growing response in the time domain.
3
Kramers-Kronig Significance
A researcher measures the absorption of a new polymer (\(\chi''\)) and finds a very sharp peak at \(\omega_{res}\).
3.1 Qualitative Prediction
Based on the Kramers-Kronig relations, describe how the refractive index (\(\chi'\)) must change as the frequency sweeps across the absorption peak. Does it increase or decrease? Does it show an "anomalous" region?
Student Response Space...
Spectral Causality Answer Key Answer Key: Spectral Causality
Instructor Resource | Lesson 5
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Part 1: The Spectral Susceptibility
1.1 Derivation of \(\chi(\omega)\).
Method 1 (Fourier Transform): \(\chi(\omega) = \int_0^\infty \frac{1}{m\omega_1} e^{-\beta t} \sin(\omega_1 t) e^{i\omega t} dt\).
Method 2 (Transforming the ODE): \(\mathcal{F}[\ddot{x} + 2\beta\dot{x} + \omega_0^2 x] = \mathcal{F}[\frac{F(t)}{m}]\)
\((- \omega^2 - 2i\beta\omega + \omega_0^2) \chi(\omega) = 1/m\)
\[ \chi(\omega) = \frac{1/m}{\omega_0^2 - \omega^2 - 2i\beta\omega} \]
1.2 Real and Imaginary Parts.
\[ \chi'(\omega) = \frac{1}{m} \frac{\omega_0^2 - \omega^2}{(\omega_0^2 - \omega^2)^2 + (2\beta\omega)^2} \]
\[ \chi''(\omega) = \frac{1}{m} \frac{2\beta\omega}{(\omega_0^2 - \omega^2)^2 + (2\beta\omega)^2} \]
Dissipation: The imaginary part \(\chi''(\omega)\) describes dissipation (energy loss), as it is associated with the velocity term and the work done by the damping force.
Part 2: Poles and Stability
2.1 Pole Locations.
Roots of denominator: \(\omega^2 + 2i\beta\omega - \omega_0^2 = 0\).
\[ \omega = \frac{-2i\beta \pm \sqrt{-4\beta^2 + 4\omega_0^2}}{2} = -i\beta \pm \omega_1 \]
The poles are at \(\pm \omega_1 - i\beta\). Both poles are in the Lower Half Plane (LHP) because \(\beta > 0\).
2.2 Stability Logic.
A pole at \(\omega = \omega_r + i\alpha\) with \(\alpha > 0\) (Upper Half Plane) would result in a time-domain term \(e^{-i(\omega_r + i\alpha)t} = e^{\alpha t} e^{-i\omega_r t}\). This is an exponentially growing oscillation, which violates conservation of energy (unless the system is active/unstable). For a passive linear system like a damped oscillator, causality and energy conservation force the poles into the LHP.
Part 3: Kramers-Kronig Application
Qualitative Prediction: As the frequency approaches the resonance \(\omega_{res}\) from below, \(\chi'\) (refractive index) increases. Right at resonance, it passes through the center point. Just above resonance, \(\chi'\) drops sharply. This region where the refractive index decreases as frequency increases is called the "Anomalous Dispersion" region. The sharp absorption peak requires this specific S-curve shape in the dispersion via the Kramers-Kronig integral.
