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Last Latexed: April 22, 2010 at 14:08 1 504: Lecture 23 Last Latexed: April 22, 2010 at 14:08 2

Physics 504, Lecture 23 at the high frequency end because it seems to say that integrating over all April 22, 2010 frequencies, the total power into any solid angle is infinite. This is not a serious problem, however, because for high frequency our approximation that Copyright c 2009 by Joel A. Shapiro the acceleration occurs in a time interval small compared to 1/ω is unphysical, and we expect there to be modifications at that end. At the other end, 1 Bremsstrahlung however, our approximation is valid. There is no divergence in the energy radiated, but the quantum-mechanical interpretation is a problem. Quantum We have seen that whenever charged particles are accelerated, will mechanically, radiation consists of , each of energyhω ¯ ,andifweget be produced. For elementary particles this effect was first observed when a classical prediction that we should have an energy deposit of ∆E in a small high-energy entered material and quickly lost their energy, which range of frequencies near ω, we need to interpret that quantum-mechanically was therefore called braking radiation (in german). For relativistic particles that we have an expected number of photons ∆E/hω¯ in that range. But we this can be an appreciable deposition of the energy when a particle enters are predicting energy/unit frequency to be constant and nonzero down to zero material, in addition to the energy transferred to the particles on which the frequency, so the expected number of photons per energy range diverges as ω 0. Usually in quantum physics we ask for the probability that, starting scattering takes place. Experimentalists use this to measure the energy of → electrons, as the radiation produced (as photons) often pair-produce, so the from some initial state, we wind up in a given final state, and here we are measurement of all the energy of charged particles produced measures that finding that the probability that an scatters from a nucleus, for of the original electron. example, without any additional particles in the final state, is zero. This Another place where understanding bremsstrahlung is important is at the is known as infrared divergence, and in quantum field theory we need to opposite end of the energy spectrum, where we consider low frequency radi- take into account that any experimental arrangement has a lowest energy ation. The process of scattering occurs on an atomic or subatomic scale, and cutoff δE, below which a could not have been detected, so when we measure the cross section for elastic scattering β~ β~ , we are actually therefore over a very short time interval, so the details of the scattering should i → f be irrelevant when discussing frequencies corresponding to larger including processes for which an arbitrary number of very soft photons, with than an angstrom. Thus in the equation we derived last time, total energy totaling less than δE. It turns out the cross section for this is finite and calculable. 2 ~ q iωR/c ∞ iω(t nˆ ~r(t)/c) d nˆ (ˆn β) Beyond those difficulties, the most noticeable feature of bremsstrahlung A(ω)= e e − · × × dt. (1) s8π2c dt  1 nˆ β~  radiation for ultrarelativistic particles is the strong peaking in the directions Z−∞ − ·   of the final particle and the initial particle. We can think of these as parts of we can assume that the d[]/dt factor contributes over such a small range the Coulomb field that have travelled on without noticing that the particle of t values that the phase iω(t nˆ ~r(t)/c can be taken to be an irrelevant got bumped. In quantum field theory, what is relevant is that the state of the constant phase, and therefore we− have· an integral of a total derivative, given charged particle, of mass m with momentum pα, together with the photon by the difference of its final and initial values. Projecting the contribution of momentum kα, has a mass of a single polarization ~, which is perpendicular ton ˆ,wehave 2 1 α α 2 2 2 α 2 m k M = (p + k ) = m + p kα = m +2 | |γ(1 nˆ β~) 2 2 2 2 2 c c c − · d I q β~f β~i = ~ ∗ . 2 m k 2 2 2  ~ ~  m + | |(1 + γ θ ), dωdΩ 4π c · 1 nˆ βf − 1 nˆ βi − · − · ≈ γc   2 This formula has a disturbing feature with difficulties at both ends of the which goes to m (goes on-shell in QFT language) as β 1 and the angle frequency spectrum, as it is independent of frequency. This is problematic θ between the photon and the charged particle goes to zero.→ Thus in some 504: Lecture 23 Last Latexed: April 22, 2010 at 14:08 3 504: Lecture 23 Last Latexed: April 22, 2010 at 14:08 4

ways it acts as if this combination were the single charged particle. dI e2 1 β2(1 u2) e2 1+β β2 1+2x x2 = 2π − du = − − dx 2 Bremsstrahlung in dω 4π2c 1 (1 βu)2 2πβc 1 β x2 Z− − Z − e2 1 1+β Beta decay is a weak interaction process in which a or proton in a = ln 2 πc "β 1 β − # nucleus emits an electron or and an antineutrino or , turning − into a proton or neutron respectively, and thereby changing the charge of the The predicted intensity distribution has no ω dependence, but we have nucleus by one unit already discussed that there is a cutoff ωmax. The energy per frequency Z (Z 1) + e∓ + ν. value is not large. For small β the term in brackets is 2β2/3, and we have → ± 2 2 While the electron or positron was created, and not previously residing in dI/dω =2e β /3πc. Let’s compare the energy radiated to the lost rest mass the nucleus, its charge was there, so as far as the electromagnetic field is energy ∆E which gives the upper limit ωmax =∆E/h¯. Multiplying dI/dω by concerned, we have a charge at rest until the sudden decay, and then a the cutoff frequency we see that the fraction of energy lost going into radiation is less than 2e2β2/3πhc¯ . As the fine structure constant α = e2/hc¯ 1/137, pretty-much instantaneous acceleration to a large final velocity. Assuming ≈ the decay takes place at t =0,wehave we see this is well less than 1%. As for bremsstrahlung in scattering, the low frequency limit predicts an infinity at zero frequency in the number spectrum 2 2 2 for photons emitted, though the total energy emitted is small. d I e ~ ∗ β~ = · , There is some interest in this lost energy, though it is often ignored, 2 dωdΩ 4π c 1 nˆ β~ because one way to try to measure neutrino masses is to look for a cutoff − · in the energy spectrum of the electrons. The energy difference between the ~ where cβ is the final velocity of the e±, and we are assuming ω is small parent and daughter nucleus is shared between the electron, the neutrino, and enough. Small enough might mean compared to the scale of the nucleus or any photons which accompany them. If the neutrino has a mass, there is a size, as we might imagine the acceleration occurs on this distance lower limit on the energy it can have, and thus an upper limit on the electron scale, but more obviously, if we take into account the fact that any radiation 2 energy distribution which is slightly less than (mZ mZ 1)c . The expected 2 ± will be quantized in units ofhω ¯ ,wemusthave¯hω < (mZ mZ 1 me)c . − − ± − energy distribution from , ignoring photon production, The radiation in then ˆ direction is polar- might be misleading, however, if inner bremsstrahlung photons are taking off ized, as the component perpendicular to some of the energy. the plane containingn ˆ and β~ is not ex- α cited (by ~ ∗ β~). Then, as we see in the · 3 Orbital Electron Capture diagram, for the direction that is excited, α = π − θ ~ ∗ β~ = β sin θ,and 2 · β From an elementary particle point of view another process which is basically d2I e2 sin2 θ the same, or the inverse of, beta decay, is electron capture. Here a nucleus = β2 . ε 2 2 of charge Ze grabs one of the atomic electrons, converting a proton into a dωdΩ 4π c 1 nˆ β~ − · θ neutron, and emits a neutrino, Z + e− (Z 1) + ν.   Let us first give a classical model for→ electron− capture, and then discuss Integrating over all angles, dΩ= 1 n how to interpret this quantum mechanically. Our classical model has a non- sin θdθdφ =2π 1 du,wehaveR − relativistic electron moving in a circle of radius a about the z axis with R R 504: Lecture 23 Last Latexed: April 22, 2010 at 14:08 5 504: Lecture 23 Last Latexed: April 22, 2010 at 14:08 6

angular frequency ω0,upuntiltimet =0,atwhichpointitdisappears.So The quantum mechanical interpretation of this classical model is not com- for negative t the position and velocity are pletely clear to me. Electron capture requires the electron to be within the nucleus, which is very small compared to atomic orbitals, even in heavy atoms ~r(t)=a cos(ω0t + α)ˆex + a sin(ω0t + α)ˆey, for which the radius is the Bohr radius a0 divided by Z.Onlys wave func- ~v(t)= ω a sin(ω t + α)ˆe + ω a cos(ω t + α)ˆe . − 0 0 x 0 0 y tions have a nonzero amplitude to be at r = 0, so it is an s orbital, usually Let’s look at this from a point in the xz plane at an angle θ from the z axis, in the K shell (n =1,`= 0) that is captured, and the circular orbit is not a good description of the spherically symmetric s orbital wave function. On the son ˆ =sinθeˆx +cosθeˆz. The frequency spectrum is best evaluated by doing the integration by parts in Eq. 1, other hand, very soon after the nucleus gobbles up the innermost electron, an electron in the 2p orbital will make a transition into the newly emptied 2 ~ q iωR/c ∞ nˆ (ˆn β) d iω(t nˆ ~r(t)/c) 1s state, so probably that is what we are describing. The energy emission of ( )= − · A ω 2 e × × e dt 2 2 −s 8π c 1 nˆ β~ dt a2p 1s transition is approximately 3Z e /8a0 which is the best thing to Z−∞ h i − · use for→ ¯ in our classical approximation. Here =¯2 2 =53pmisthe q2 0 hω0 a0 h /me = iω eiωR/c nˆ (ˆn β~)eiωtdt, Bohr radius. The 2p radius is roughly a = a /Z. Thus the expected number − s8π2c × × 0 Z−∞ of photons emitted per energy is wherewehaveassumedtheωnˆ ~r(t)