Frequency response of dielectric polarizability: when is the imaginary part of αe zero? In an alternating electric field of angular frequency ω, the complex electronic polarizability αe(ω) has a real and an imaginary part. For which frequency limits is the imaginary part equal to zero (i.e., no dielectric loss contribution from αe)?

Introduction / Context:Dielectric response to a time-varying field is commonly described by a complex polarizability αe(ω) or complex permittivity ε*(ω). The imaginary component represents energy dissipation (dielectric loss), while the real component represents stored energy. Recognizing where the loss vanishes helps interpret relaxation and resonance behavior.

Given Data / Assumptions:

• Linear, time-invariant response with a standard dispersion model (e.g., Debye-type for orientational or Lorentz-type for bound electrons/ions).
• We focus on qualitative limits: static (ω = 0) and very high frequency (ω → ∞).
• No extrinsic conduction or electrode polarization effects considered.

Concept / Approach:In canonical dispersion models, the imaginary part of the response is linked to phase lag between polarization and field. At DC (ω = 0), there is no phase lag for a relaxed system, so the loss term tends to zero. At extremely high frequencies (ω → ∞), massive or bound charges cannot follow the rapidly varying field, so the response amplitude tends to a limiting value with negligible phase lag, again driving the imaginary part to zero. Between these limits, around a characteristic time constant (Debye) or near a resonance (Lorentz), the imaginary part peaks.

Step-by-Step Solution:Recognize that αe(ω) = α′(ω) − j α″(ω).At ω = 0, polarization fully follows the field without lag ⇒ α″(0) = 0.At ω → ∞, charges cannot respond to rapid changes ⇒ lag goes to zero and α″(∞) → 0.Hence α″(ω) = 0 in both limiting cases: DC and infinite frequency.

Verification / Alternative check:For Debye relaxation, ε″(ω) = (εs − ε∞) * ωτ / (1 + ω^2 τ^2). This tends to zero as ω → 0 or ω → ∞, with a maximum at ωτ = 1. Analogous statements hold for αe(ω).

Why Other Options Are Wrong:Only ω = 0 or only ω → ∞ state one limit and miss the other; ω = ω0 corresponds to resonance where losses peak, not vanish; “any finite value” ignores the characteristic peak and decay behavior.

Common Pitfalls:Confusing conductivity loss (σ/ωε0 contribution) with dipolar loss; assuming constant nonzero loss across frequencies.

Final Answer:ω = 0 and ω → ∞