Compton ScatteringInitial questions: How can we detect galaxy clusters at high redshift? Whatinformation could we get out of such detections?Last time we talked about scattering in the limit where the photon energy is muchsmaller than the mass-energy of an electron. However, when X-rays and gamma-rays areconsidered, this approximation is no longer so good. Therefore, today we’ll talk about themore general process of Compton scattering. We’ll start with the basic process, then discusshow it has an effect on spectra and the energetics of electrons.Think about a photon scattering off an electron that is initially at rest. Let’s treat boththe photon (of initial energy ¯hω0) and the electron as if they were particles. Suppose thatafter the photon scatters it moves off into a direction an angle θ from its original direction.Ask class: how would we figure out the new energy of the photon? We do this by energyand momentum conservation. Ask class: will the entire interaction take place in a singleplane? Yes, it will, so that simplifies things. Before we calculate, Ask class: will the finalphoton energy be less than, equal to, or greater than the initial photon energy? Less than,since electron recoil will take some energy. This is the difference between full Comptonscattering and our Thomson approximation, in which the photon energy was unchanged(thus producing coherent scattering).We can rephrase our conservation laws a la special relativity by saying that the total four-momentum is conserved. The initial four-momentum of the photon is~Pγi= (¯hω0/c)(1, ni),where niis the initial direction of the photon. Similarly, the final four-momentum of thephoton is~Pγf= (¯hω1/c)(1, nf). The initial four-momentum of the electron is~Pei= (mc, 0),since it isn’t moving, and the final is~Pef= (E/c, p). Setting~Pγi+~Pei=~Pγf+~Pefandmanipulating a bit gives¯hω1= ¯hω0/h1 + (¯hω0/mc2)(1 − cos θ)i. (1)In terms of wavelength it’s even easier:λ1− λ0= λc(1 − cos θ) (2)where λc≡ h/mc = 0.02426˚A(me/m) is the Compton wavelength.Does this make sense to us? Ask class: what limits can we check? Indeed thewavelength only increases, as we guessed beforehand. If θ = 0 there is no deflection and nochange, as is reasonable. The change in wavelength is independent of λ, meaning that forsmaller λ (larger photon energy) the fractional change is greater, which makes sense. Also,for a more massive particle (e.g., a proton instead of an electron) the wavelength change isless, meaning that the photon energy has to be larger for there to be a significant deviationfrom Thomson.Something not as obvious is that the cross section is also modified by this effect. Thedifferential cross section isdσ/dΩ = (r20/2)(ω1/ω0)2hω0/ω1+ ω1/ω0− sin2θi. (3)The main effect is to reduce the cross section from the Thomson value, but not very rapidly.Rybicki and Lightman give the expression for the total angle-integrated cross section, butit’s complicated and the most important parts are the limits at low and high energy:σ ≈ σT(1 − 2x + . . .), x ≪ 1,σ ≈38σTx−1(ln 2x + 1/2 + . . .), x ≫ 1 .(4)Here we define x ≡ ¯hω/mc2. As before, we’re usually thinking of photon-electron scattering,but it’s more general than that (e.g., a proton will do if we let m = mp). In any case, we seethat for low energies the cross section is close to (but less than) the Thomson cross section,whereas at high energies the cross section scales as ∼ (ln x)/x, ignoring smaller terms.Ultimately, this means that very high energy gamma rays have a substantially longer meanfree path to scattering (or absorption) than lower energy photons do.Fine, so we’ve done scattering off of a stationary electron. We found that in sucha scattering the photon always loses energy. Ask class: is this still true in a frame inwhich the electron was moving initially? No! Think of a photon incident from the left,and a rapidly-moving electron incident from the right. If the photon doesn’t have toooverwhelming an energy, the result will be that the electron moves more slowly afterthe collision (think of an elastic collision of two balls with different masses). Energyconservation then demands that the photon gain energy. This is still Compton scattering,but because the energy is transferred from the electron to the photon it is usually calledinverse Compton scattering. Note, by the way, that if the energy of the photon is largeenough, the electron will gain energy as before.As always in special relativity, the key to analyzing this is to pick a simple referenceframe for your analysis, then transform to the original frame. In this case, the simple frameis the one in which the electron is originally at rest, which we already did. If the electronsare relativistic (γ2− 1 ≫ ¯hω/mc2) then, if in the e−rest frame we have ¯hω ≪ mc2, theenergy of the photon before scattering, the energy of the photon in the electron rest frame,and the energy of the photon after scattering are in the rough ratio 1 : γ : γ2. That meansthat a gain in energy in a single scattering can be a factor of about γ2, which can beenormous. Of course, there are limits. Ask class: if the electron has energy γmec2, whatis the maximum gain of energy by the photon? It can only gain (γ − 1)mec2. Ask class:suppose that in the electron rest frame the condition ¯hω ≪ mc2does not hold. What arethe qualitative effects? One is that the electron will experience recoil, so the photon willeither lose energy or not gain as much energy. Another is that the cross section will drop,so interactions will be less frequent.Now let’s ask the following question: if (as measured in a particular reference frame)one has an isotropic distribution of electrons all of which have the same speed βc, andone has an isotropic distribution of photons, what will be the net power transferred fromelectrons to photons by inverse Compton scattering?Suppose that the number density of photons with energy in a range dǫ is (dnγ/dǫ)dǫ.Rybicki and Lightman use υ instead of dnγ/dǫ, but that looks too much like velocity to me:). If we ignore any change of energy of the photon in the rest frame, then Rybicki andLightman show that the scattered energy in a direction θ with respect to the motion of theelectron isdE1/dt = cσTγ2Z(1 − β cos θ)2ǫ(dnγ/dǫ)dǫ . (5)This makes sense: in a direction θ, the Doppler shift is γ(1− β cos θ), so by the 1 : γ : γ2ruleone