Abstract

Relying on a simple analytic two-atom model in which the anisotropy of the interaction dipole polarizability obeys an inverse power law as a function of separation, we offer mathematical and numerical evidence that, in a monoatomic gas, the free-free Raman spectrum for a collisional pair of two different isotopes, a-a', may vastly differ from that for a-a. This result is obtained even if a and a' are assumed to have the same mass and zero nuclear spin and even if a-a and a-a' are subject to the same interaction polarizability and potential. The mechanism responsible for this effect is inherent in the parity of the partial-wave rotational quantum number J: given that the contribution of each partial wave to the Raman cross section is controlled by a polarizability-transition matrix-element and that each of those matrix-elements has a radial component with a magnitude slightly smaller than that of the preceding partial wave, a deficit which disfavors the odd-numbered waves is accumulated upon summing over J. In the far high-frequency wing, this deficit tends to generate spectral intensities for a-a' about half as great as the a-a ones, a tendency which becomes all the more effective as temperature is decreased. We show for instance that, for the spectral branch ΔJ = 2, the fractional difference between the free-free differential cross sections for a-a and a-a' is 12(1-x(2))(3)1+3x(4), with x=√[E/E(')] (E (E') being the initial (final) state energy of the pair and E' - E = hcν (ν > 0)). Remarkably, this quantity is zero at ν ≈ 0 but goes to 12 for ν ≫ 0. For ΔJ = 0, analogous conclusions may be drawn from the expression (1+ln(1+x1-x)2arctanx)(-1).

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