# Detailed balance: What does one rigourously mean by the transition rate $dw$?

Physics Asked by catalogue_number on August 5, 2020

Consider (for example) non-relativistic electron-impact ionisation, in which an effectively stationary atom in state $$I$$ is struck by an electron of momentum $${p_1}$$, producing an atom in state $$J$$ with two additional electrons, momenta $${p_2}, {p’_2}$$. The electron-impact ionisation transition rate is then expressed as $$dw(final|initial) = dw(p_2, p_2′; B| p_1; A)$$

The total rate is then given, in the presence of an electron distribution $$f(p)$$, as
$$Gamma_{eii} = N_e n_A int d^3p_1 f(p_1) dw(p_2, p_2′; B| p_1; A)$$
where $$N_e$$ is electron population and $$n_A$$ the number density of atoms in state $$A$$. Here, the interpretation $$dw = frac{p_1}{m_e}sigma(p_1)$$ makes clear, intuitive sense – $$sigma(p)$$ is the effective area of an atom as seen by an electron of momentum $$p$$. However, I’m confused by the detailed-balance equation for this process, for which the relation reads

$$dw(p_2, p_2′; B| p_1; A) d^3p_1 g(p_1) = dw(p_1; A | p_2, p_2′; B) d^3p_2 g(p_2) d^3p’_2 g(p’_2)$$

I’ve seen the three-body rate expressed as

$$Gamma_{eii} = N^2_e n_A int d^3p’_2d^3p_2 f(p_2)f(p’_2) dw(p_1; A | p_2, p_2′; B)\ = N^2_e n_A int d^3p_1 f(p_2)f(p’_2) dw(p_2, p_2′; B | p_1; A ) frac{g(p_1)}{g(p_2)g(p’_2)}$$

The second line of this looks almost nonsensical – the integration in the non-$$p_1$$ variables must somehow be implicily carried out via $$dw$$, which conceals a number of Dirac deltas.

What exactly does one mean by $$dsigma$$ and $$dw$$ – are they differential forms on the manifold of all possible scattering parameters, and in this context, where do the density of states arise from in detailed balance? (i.e. why are they present in the detailed-balance equation, but absent from the total scattering rate integral?)

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