Power Absorption and the Lorentzian Lineshape
Statement
For a linear damped oscillator mẍ + bẋ + kx = F0cos(ωt) driven in steady state, the time-averaged power absorbed from the drive is P̄(ω) = (F02/2m) γω2 / [(ω02 − ω2)2 + γ2ω2], where ω02 = k/m and γ = b/m. Near resonance this reduces to a Lorentzian of full width at half maximum Δω = γ, peak P̄max = F02/2mγ, so the absorption bandwidth and the quality factor are locked together by Q = ω0/Δω.
Why it matters
The Lorentzian is the universal fingerprint of a weakly damped resonance responding linearly to a harmonic drive. It is what a spectrometer, a swept-frequency network analyser, or a nuclear-magnetic-resonance receiver actually measures: not the amplitude curve, but the power the system removes from the drive as a function of frequency. Its width is not a fitting nuisance — it is a direct read-out of the internal loss rate.
Because the width equals γ and the centre equals ω0, a single lineshape simultaneously yields the resonant frequency and the Q. This is the operational meaning of Q: the sharpness of a resonance in the frequency domain is the same physical quantity as the slowness of its energy decay in the time domain. Every downstream idea — cavity finesse, laser linewidth, filter selectivity, the energy–time reading of a decaying state — descends from this one calculation.
Assumptions
Derivation
Result
Reading. The power the drive pumps into the oscillator is a symmetric bell centred on ω0, whose full width between the half-power points equals the damping rate γ. A high-Q system (small γ) presents a tall, narrow line: it absorbs strongly but only within a sliver of frequency. The peak height F02/2mγ rises as damping falls, yet the area under the curve, ∫P̄ dω = πF02/4m, is independent of γ — sharpening the line trades width for height at fixed integrated absorption.
Units check. P̄max = F02/2mγ has units N2/(kg·s−1) = (kg m s−2)2/(kg s−1) = kg m2 s−3 = W. The Lorentzian factor is dimensionless (each term ~ s−2), and Δω = γ carries s−1 as an angular frequency should.
Limiting cases
- On resonance (ω = ω0): δ = π/2, velocity is exactly in phase with the force, absorption is maximal at F02/2mγ.
- Far off resonance (|ω − ω0| ≫ γ): P̄ → (F02γ/2m)/[4(ω − ω0)2], a Lorentzian tail falling as 1/(detuning)2.
- Zero damping (γ → 0): the line becomes a Dirac spike of zero width and infinite height but fixed area πF02/4m — a lossless oscillator absorbs only precisely at ω0.
- Half-power points (ω = ω0 ± γ/2): P̄ = ½P̄max, phase δ = π/4 and 3π/4; these are the −3 dB frequencies of the resonance.
Breaks when
- Strong damping (γ ≳ ω0): the factorisation ω02 − ω2 ≈ 2ω0(ω0 − ω) collapses, the exact curve is visibly asymmetric (heavier on the low-frequency side), and its peak shifts below ω0.
- Nonlinearity (large amplitude, Duffing-type stiffness): the resonance bends, becomes hysteretic and multivalued, and the notion of a single width γ loses meaning.
- Overlapping modes: when a second resonance lies within a few γ, the total power is a sum of Lorentzians and the individual widths cannot be read off directly.
- Non-viscous loss (dry friction, hysteretic/structural damping): the loss per cycle no longer scales as velocity squared, so the lineshape is not Lorentzian and its width becomes amplitude-dependent.
Failure modes
- Confusing the amplitude curve with the power curve. The amplitude A(ω) peaks slightly below ω0 (at √(ω02 − γ2/2)) and its FWHM is not γ; only the power peaks exactly at ω0 with width γ.
- Dropping the phase and writing P̄ = ½F0v0. Without the sin δ factor the reactive power is counted as absorbed; off resonance most of the velocity is 90° out of phase and does zero net work.
- Using half-width for full-width. The FWHM is γ, the half-width at half maximum is γ/2; halving this error doubles the inferred Q.
- Time-averaging F and v separately. ⟨Fv⟩ ≠ ⟨F⟩⟨v⟩ = 0; the product must be averaged, not the factors.
- Mixing conventions for γ. If one defines the equation with 2γẋ (amplitude decay rate γ), the width is 2γ, not γ. State whether γ is the energy or amplitude rate.
Discussion
The physical content of step 3 is that only the velocity component in phase with the driving force does net work over a cycle; the quadrature component merely sloshes energy in and out of the spring and mass without loss. On resonance the phase lag is exactly 90°, which places the velocity precisely in phase with the force (velocity leads displacement by 90°), maximising the overlap and hence the absorption. This is why the power peaks sharply at ω0 even though the amplitude peak is slightly detuned: absorption is governed by the velocity–force alignment, not by displacement.
The relation Δω = γ is the frequency-domain twin of the time-domain statement that stored energy decays as e−γt. A resonance that rings for a long time (small γ, long τ = 1/γ) must, by the reciprocity of Fourier conjugates, respond over a correspondingly narrow band. The product (lifetime)(bandwidth) ≈ 1 is the classical ancestor of the energy–time relation ΔEΔt ≈ ℏ for a decaying quantum state, whose spectral line is itself a Lorentzian of width ℏ/τ — the "natural linewidth." The mathematics is identical because both are a first-order linear system relaxing at a single rate.
The scale-invariance of the integrated absorption, ∫P̄ dω = πF02/4m, is a sum rule: it depends only on the drive strength and the inertia, never on the damping. It says that engineering a higher Q does not let a resonator absorb more total energy across the spectrum; it concentrates the same absorption into a narrower, taller line. This underlies why high-finesse cavities and sharp filters are prized for selectivity, not for raw throughput, and it is the classical form of the oscillator-strength sum rules of atomic spectroscopy.
The exact curve of step 6 is the modulus-squared of a complex susceptibility χ(ω) = 1/[m(ω02 − ω2 − iγω)], and the absorbed power is set by its imaginary part, P̄ = ½F02ω Im χ(ω). This is a fluctuation–dissipation statement: Im χ, the dissipative part of the response, controls both the energy the system takes from a coherent drive and the spectrum of its thermal fluctuations. Kramers–Kronig relations then force a matching dispersive (real) part — the phase lag δ and the amplitude curve — so the Lorentzian absorption line and the S-shaped dispersion curve are not independent facts but two projections of one analytic response function.
Common misconceptions. The width being γ does not mean the oscillator responds only within ±γ/2 and is dead elsewhere — the Lorentzian tails fall only as 1/(detuning)2 and carry non-negligible power far out. And "high Q" does not mean "absorbs more energy"; it means "absorbs over a narrower band," with the peak height rising to keep the integrated area fixed.
Worked examples
Reading. A low-Q woofer absorbs strongly but over a broad 12 Hz band — deliberately, so its response is smooth across the bass region rather than a peaky whistle.
Reading. A detuning of only 1 Hz — 30 parts per million — already cuts the absorbed power to 4% of peak. This razor-sharp line is exactly why such a crystal makes a precise clock reference.
Problems
- A driven oscillator has m = 0.10 kg, ω0 = 200 rad s−1, γ = 4.0 rad s−1, F0 = 5.0 N. Find P̄max, the FWHM in rad s−1, and Q.
Solution
P̄max = F02/2mγ = 25/[2(0.10)(4.0)] = 25/0.80 = 31.3 W. FWHM Δω = γ = 4.0 rad s−1. Q = ω0/γ = 200/4.0 = 50. - Show that the half-power points of the near-resonance Lorentzian lie at ω = ω0 ± γ/2, and confirm the full width is γ.
Solution
Set P̄ = ½P̄max: (γ/2)2/[(ω − ω0)2 + (γ/2)2] = ½. Cross-multiplying, 2(γ/2)2 = (ω − ω0)2 + (γ/2)2, so (ω − ω0)2 = (γ/2)2, giving ω = ω0 ± γ/2. The separation is γ/2 − (−γ/2) = γ. - An oscillator absorbs 8.0 W at resonance. By what factor does the absorbed power drop when the drive is detuned by exactly one full linewidth, |ω − ω0| = γ?
Solution
Ratio = (γ/2)2/[γ2 + (γ/2)2] = (1/4)/[1 + 1/4] = (1/4)/(5/4) = 1/5 = 0.20. So P̄ = 0.20(8.0) = 1.6 W, a factor-5 drop. - A tuning fork rings down: after being struck, its amplitude falls to 1/e in 3.0 s. Estimate its Q and absorption bandwidth if f0 = 440 Hz. (Amplitude decays as e−γt/2.)
Solution
Amplitude 1/e time: γ/2 = 1/3.0 s−1 ⟹ γ = 0.667 rad s−1. ω0 = 2π(440) = 2765 rad s−1. Q = ω0/γ = 2765/0.667 = 4.15×103. Bandwidth Δf = γ/2π = 0.106 Hz. (Equivalently Δf = f0/Q = 440/4150 = 0.106 Hz.) - Starting from the exact result P̄ = (F02/2m)γω2/[(ω02 − ω2)2 + γ2ω2], prove the sum rule ∫0∞ P̄(ω) dω = πF02/4m in the weak-damping limit, and comment on its independence from γ.
Solution
In the weak-damping limit the integrand is sharply peaked near ω0, so use the Lorentzian form P̄ ≈ (F02/2mγ)(γ/2)2/[(ω − ω0)2 + (γ/2)2] and extend the lower limit to −∞ (negligible error). The standard integral ∫−∞∞ dx/(x2 + a2) = π/a with a = γ/2 gives ∫ = (F02/2mγ)(γ/2)2(π/(γ/2)) = (F02/2mγ)(γ/2)(π) = πF02/4m. The γ cancels: peak height ∝ 1/γ and width ∝ γ compensate, so total integrated absorption depends only on F0 and m. Sharpening a resonance concentrates, but does not increase, its total absorption.