Neutrinos, nonzero rest mass particles, and production of high energyphotonsParticle interactionsPreviously we considered interactions from the standpoint of photons: a photon travelsalong, what happens to it? Now, we’ll think about interactions of particles: an electron,proton, or nucleus zips along, what happens to it?Ask class: generically, what could, say, an electron interact with? Photons, protons ornuclei, magnetic fields, neutrinos. Let’s first consider interactions of electrons with photons.Ask class: for a low-energy electron interacting with low-energy photons, what is the crosssection (not a trick question)? Just the Thomson cross section. In fact, since this is exactlythe same process as we considered before, the cross section for general energies is again theKlein-Nishina value.Compton scatteringIn the absence of a magnetic field, th e cross section for the interaction of a photon withan electron is just the Thomson cross section (σT= 6.65 × 10−25cm2) for low energies, butbecomes more complicated at higher energies. The general Klein-Nishina cross section is inRybicki and Lightman and other standard references. Let us define x ≡¯hωmc2, where ¯hω isthe energy of the photon in the rest frame of the electron. For x ≪ 1 this reduces to theThomson value, whereas for x ≫ 1σ ≈38σTx−1(ln 2x + 0.5) . (1)Radiation can exert a force on matter, via scattering or other interactions. Radiationforce is often referred to as radiation pressure in the literature. However, let’s give somethought to this. Suppose that an electron is in an isotropic bath of radiation. The radiationpressure is nonzero; Pr= aT4, in fact. Ask class: is there any net radiation force on theelectron? No, because the bath is isotropic. In th is situation it would be more accurateto say that the force is due to a pressure gradient. This is the same reason why we’re notcurrently being cru shed by the atmosphere, d espite the pressure of about 1 kg per squarecentimeter (many tons over the whole body). Even more accurately, it’s the net radiationflux that matters.By balancing radiation force with gravitational force we can define the Eddingtonluminosity, which is very important for high-energy astrophysics. For a flux F on a particleof mass m and scattering cross section σ around a star of mass M, the balance impliesGMmr2=σcF =σcL4πr2, (2)where L is the luminosity. The r−2factors cancel, leaving us with the Eddington luminosityLE:LE=4πGcMmσ, (3)For fully ionized hydrogen, we assume that the electrons and protons are electricallycoupled (otherwise a huge electric field would be generated), so the light scattersoff the electrons with cross section σTand the protons provide the mass mp. ThenLE= 4πGcMmp/σT= 1.3 × 1038(M/M⊙) erg s−1. If the luminosity of the star is greaterthan this, radiation will drive matter away. This is also the maximum luminosity for steadyspherical accretion.Ask class: does the dependence on m and σ make sense, that is, should m be in thenumerator and σ in the denominator? Yes, because if gravity is stronger (m is higher) thenmore luminosity is needed; if σ is greater, radiation couples m ore effectively and the criticalluminosity is less. This means that for a fixed density, large things are less affected byradiation than small things (because for a size a, the mass goes like a3whereas the areagoes like a2. So, asteroids are not affected significantly by radiation on times comparableto their orbital time! Over millions of years, however, the effects can add up to produceinteresting spin effects (e.g., look up YORP).Curvature radiationWe know that any accelerated charge radiates. If there is a magnetic field around, thereare two ways the charge can be acclerated. One is if it moves perp endicular to the field(synchrotron radiation). The other is if it moves along the field, but the field is curved(curvature radiation). We’ll just state a couple of results here.The power emitted by cu rvature radiation is (see, e.g., Jackson 1975)P ≈23e2cR2γ4(4)where γ = (1 − β2)−1/2and β = |v|/c. Ask class: is the dependence on R r easonable? Yes,because for smaller radius of curvature and a fixed energy, the acceleration is greater andhence so is the radiation.Synchrotron radiationIf a particle of charge e and energy E is moving perpendicular to a static magnetic fieldof strength B, the frequency of its orbit around the field isωc=eBcE. (5)If the particle has velocity v, this means that its orbital radius is d = v/ωc, so for highlyrelativistic particles with v ≈ c, d = E/eB. The power emitted by synchrotron radiationfor an isotropic distribution of velocities isP =43σTcβ2γ2(B2/8π) (6)where σT= 8πr20/3 and r0is the classical radius of the charged particle: r0= e2/(mc2). Foran electron r0≈ 2.8 × 10−13cm, whereas for a proton r0≈ 1.5 × 10−16cm.A particle may acquire a nonnegligible motion perpendicular to the magnetic fieldif, e.g., it is created from a photon which was moving with some angle to the magneticfield. If γBBc≪ 1, then a classical treatment of synchrotron radiation is approximatelyvalid. Ask class: recalling that synchrotron radiation is due to acceleration of a charge,suppose you have an electron and a proton with the same Lorentz factor. Which would youexpect to lose energy on a faster time scale? The electron, because the proton is not aseasily accelerated, hence the proton does not lose energy as rapidly. That’s why circularaccelerators accelerate protons or ions instead of electrons, and why electron acceleratorsare straight: the radiation losses are too significant otherwise. However, when the magneticfields are weak, relativistic electrons have a long synchrotron cooling time. In fact, radioemission from many AGN is dominated by synchrotron radiation.Pair annihilation, e−e+→ γγIn the extreme relativistic limit, the cross section for two-photon annihilation isσan≈38σTln 2γγ. (7)We will consider one-photon annihilation in neutron star section.BremsstrahlungThe energy lost per unit length to bremsstrahlung radiation by an electron or positrontraversing a region of number density n fixed charges per unit volume is approximatelyd(ln E)dx≈ n1603 e2¯hc! e2mec2!2. (8)This expression contains two combinations of symbols that are useful to remember.e2/¯hc = 1/137 is the fine structure constant, and is dimensionless. e2/mec2= 2.8× 10−13cmis the classical radius of the electron. Evaluating this, we