1 The Kompaneets equation

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1 Introduction to Particle Cosmology, Tutorial notes: The Sunyaev-Zeldovich effect 1 The Kompaneets equation Consider evolution of photon phase space density n(, t) in the presence of an electron gas. Assume that the electrons are thermal and hot: T e >> T γ. We re not going to force the photons to be strictly thermally distributed, but we will assume that they are not far from black-body, and so we define T γ as the corresponding temperature. The Boltzman equation (BE) describes the evolution of n(, t): n(, t) t = dω f e(p 1 )n( 1, t) (1 + n(, t)) f e (p)n(, t) (1 + n( 1, t))] (1) This equation describes energy transfer via scattering events p + p The first term describes population of states by incoming photons at 1, while the second term describes de-population similarly (the 1 + n factors are due to stimulated absorption). In the rest frame of the electron, 1 is related to through the scattering angle 1 (, Ω) = 1 + (1 cos θ) () and the cross section is given by Klein-Nishina formula dσ dω = 3σ T 16π ( 1 ) 1 + sin θ 1 ] (3) with σ T = 8πα 3m e =.66 barn. For soft photons, << me, and nonrelativistic electrons, elastic scattering dominates and we get dσ dω 3σ T 16π (1 + cos θ) (4) In a frame in which the initial electron velocity is β = p E e 1 = 1 β ˆn 1 β ˆn 1 + γ e (1 ˆn ˆn 1 ) 1 + p e (ˆn 1 ˆn) we have ( ) T + O e m e (5) for nonrelativistic electrons, β Te with p 1 = p + ˆn 1ˆn 1. < 1. Thus in (1) we should use (5) together 1

2 In our case typical energy transfer is small compared to electron kinetic energy: 1 T e << 1 (6) Find approximate solution to (BE) by expanding to second order in ( Fokker- Planck approximation ). Thermal nonrelativistic electrons: f e (E) = n e (πmt e ) 3/ e E/Te, E = p m we have E 1 = E T e and therefore: ) f e (E 1 ) (1 + + f e (E) (8) we also have: n( 1 ) n() + T e n() + 1 (T e ) n() (9) (7) Define: x = T e (1) so that T e n = x n n. With this (1) is approximated by (prove!) t n = n + n(n + 1)] dω f e 1 + n + (n + 1 ) ] n (1 + n) above we should use = x p e (ˆn 1 ˆn) ( ) Te + O dω f e (11) Note already at this point that the Bose-Einstein distribution n BE ( x) = 1 is e x+α 1 a steady state solution to (11) since: n BE + n 1 BE(n BE + 1) =, n BE + ( n BE + 1n BE) (1 + nbe ) =.] Use the cross section in the elastic limit to get the phase space integrals, to order ( O T m e ) (see appendix): dω f e = x n e σ T T e dω f e = n eσ T T e x(4 x) (1) (13)

3 Plug into (11) to get the Kompaneets equation: y n( x, y) = 1 x x x 4 (n + n + n ) ] (14) with time replaced by the Comptonization parameter : y = t dt neσ T T e, dy = dt. n eσ T T e The Sunyaev-Zeldovich effect Since Compton scattering conserves the total number of photons, the Kompaneets equation guarantees that dn γ d d x x n( x, y) = (15) dt dy however, the hot electron gas injects energy into the photon system through the (inverse) compton scattering. Therefore we know that photons are pumped to occupy higher energy states. In other words, the spectrum will no longer be Planckian: there is no way to increase the mean energy of a planckian distribution without changing the particle number 1. Lets assume that the photon spectrum was initially Planckian with temperature T γ << T e. This case is certainly relevant for CMBR photons of T γ 1 4 ev traversing the hot > 5KeV electron gas in clusters of galaxies. In this case we can neglect the term in (14) proportional to the radiation phase space density n(n + 1), obtaining: y n(x, y) = 1 x x x 4 x n(x, y) ], x = T γ (16) notice that we have replaced x = T e by x = T γ. We can derive quantitative estimations for our previous assertions already from the form of (16). The total photon number and energy density are obtained by direct integration: y n γ (y) n γ (y) = n γ () y ɛ γ (y) dxx y n(x, y) = dxx 3 y n(x, y) = ɛ γ (y) = e 4y ɛ γ () dxx 1 x x x 4 x n(x, y) ] = x 4 x n(x, y) ] = dxx 3 1 x x x 4 x n(x, y) ] = 4 (17) dxx 3 n(x, y) 1 In fact, if the hot electron gas is extended over a large enough region in space, than the photon distribution will have enough time to converge to the steady state solution n BE. Prove this point. How come we can neglect n(n + 1) compared to x n, when we just saw that for the steady state (Bose-Einstein) solution these terms cancel?.. 3

4 Equation (16) is solvable. The solution is: n(x, y) = 1 4πy ds s 1 e s 1 exp 1 ( 3y + ln x ) ] 4y s (18) where, again, we have assumed that n(x, ) was Planckian: n(x, ) = 1. When e x 1 the Compton parameter is small (optically thin gas), the solution (18) becomes n(x, y) = n(x) + δn(x, y) = 1 e x 1 {1 + yxex e x 1 ]} x tanh x/ 4 One way to estimate the effect is by comparing the low-frequency part of the distribution to that of a Planckian distribution. For a Planckian distribution, the Rayleigh-Jeans end x << 1 obeys n(x) = 1 e x 1 x 1 = T RJ. Since in our case there is a depletion of occupation numbers in the low energy end, the effective radiation temperature will seem lower!. Expanding (19) to lowest order in y we get: (19) n(x, y) x 1 (1 y) = T γ(1 y) T RJ = T γ (1 y) T RJ () therefore, large wavelength antenna detecting the CMBR will deduce that the radiation temperature in the direction of a massive galaxy cluster is reduced in comparison to that of the diffuse background, by the amount δt T = T RJ T γ T γ = dt T e σ T n e T e τ (1) The SZ effect has been measured in practice for various clusters. Since it is only sensitive to the Compton-y parameter, i.e. the integrated pressure of the hot electron gas, there is no direct way to extract the separate plasma properties from the measured effect. Experimentalists combine SZ results with measurements of the X-ray brightness in order to constrain model parameters for clusters. A Phase-space integrals The simplest way to compute the O( ) integral is to use the conservation of photon number in compton scattering, as follows: dn γ d x x t n =. () dt 4

5 This equation is guaranteed to hold provided that we can write t n = 1 x x x j( x)] (3) with the function j( x) vanishing at large x faster than any power. Comparing with (11) gives j = g( x)n + h(n, x)] (4) requiring j = for steady state (n = n BE ) gives h = n(n + 1). This leads to g = x n e σ T T dω f e = n eσ T T e x(4 x) (5) 5