We start with a reminder of a few basic concepts in probability. Let x be a discrete random variable with some probability function p(x).

Transcription

1 1 Probability We start with a reminder of a few basic concepts in probability. 1.1 discrete random variables Let x be a discrete random variable with some probability function p(x). 1. The Expectation value of x is the average value of x that we will measure by doing many repetitions of the experiment described by the probability function p(x): µ E(x) x i p(x i ) (1) i where the sum goes over all (perhaps infinite) possible values for x. This is not necessarily the peak of p(x), nor does it have to be the median value of the distribution (the value of x for which half the possible values are above and half are below). 2. The Variance of x measures the spread of the distribution function: V ar(x) E[(x µ) 2 ] i (x i µ) 2 p(x i ) i x 2 i p(x i ) 2µ i x i p(x i ) + µ 2 i p(x i ) E(x 2 ) 2E(x) 2 + E(x) 2 E(x 2 ) E(x) 2. The variance is of course non-negative. A related term is the standard deviation which is defined as: σ(x) V ar(x). 3. The Cumulative distribution function describes the probability that x will be found to have a value less than or equal to X: F (X) P (x X) p(x i ). (3) x i X This is a monotonically rising function, ranging between and Continuous random variables For continuous random variables, the probability function is replaced with a probability density function (PDF). A PDF, is a function that describes the relative likelihood for a random continuous variable to take on a given value. Let f(x) be such a PDF. The chance of measuring x to lie in the range a x b is given by integration over the PDF in the said range: P (a x b) b a (2) f(x)dx. (4) 1

2 Notice that the chance of measuring any specific value of x is zero (even if this x is at the peak of the PDF), since the integral goes to as a b. In the case of a continuous random variable, the sums in the expressions for the expectation value, variance, cumulative distribution functions are simply replaced by integrals: E(x) xf(x)dx V ar(x) (x µ) 2 f(x)dx (5) F (X) X f(x)dx 2 Black body radiation A black body is a physical body that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence. A black body in thermal equilibrium emits electromagnetic radiation called black-body radiation. The radiation is emitted according to Planck s law: R ν 23 1 c 2, (6) e 1 where R ν is the power emitted per unit area of the body, per unit solid angle that the radiation is measured over, per unit frequency. The spectrum is determined by the temperature alone, not by the body s shape or composition. One can also write the spectrum per unit wavelength instead of per unit frequency. To do so, we must ensure that the emitted energy in a given wavelength band is the same as that in the corresponding frequency band: R ν dν R λ dλ R λ dν dλ R ν ν2 c R ν (7) Dividing R ν by the energy carried by each photon () we can obtain the number of photons emitted per unit area of the body, per unit solid angle that the radiation is measured over, per unit frequency: N ν 2ν2 1 c 2. (8) e 1 N ν can be thought of as a PDF describing the chance of the Black Body emitting a photon with frequency ν (per unit area of the body, per unit solid angle that the radiation is measured over). Question: what is the average energy of photons emitted from a black body with temperature T? To answer this question we must find the expectation value of N ν according to Eq. 5. Since the integral of N ν presented above is not normalized to 1, we 2

3 need to divide by this integral to get the expectation value: < E > N ν dν N ν dν 3 e 1 ν 2 e 1 dν dν K 4 B T 4 h 3 KB 3 T 3 h 3 y 3 e y 1 dy y 2 e y 1 dy K BT y 3 e y 1 dy y 2 e y 1 dy 2.7K BT (9) where in the third equality we have switched variables in the integral to y /( ). We see that the average energy is proportional to the temperature of the black body. Notice that this energy is different than the peak of R λ given by Wien s law, which is: E W ien An interesting mistake - why should we not expect astrophysical X-ray sources? Notice that R ν (T ) is always rising as a function of temperature, since: dr ν (T ) dt 23 c 2 e 2 (e 1) 2 > (1) The implication is that black bodies with higher temperatures emit more then those with lower temperatures in any frequency. Since in early astrophysical observations, no such excess radiation was observed in optical telescopes, the prevalent thought was that there could not be astrophysical objects which spectrum peaks at larger frequencies (hotter). For this reason it was argued that there is no justification in building X-ray telescopes. This argument turned out to be wrong since astrophysical objects are often not in thermal equilibrium and do not emit a Planck spectrum. Indeed, today we know that the sky is full of objects that peak in the X-rays and even γ-rays. 3 Compton Scattering Compton scattering is the inelastic scattering of a photon by a free charged particle, usually an electron. It results in a decrease in energy (and hence frequency) of the photon, called the Compton effect. The frequency of the scattered photon is given by: ν ν ν(1 cos θ)λ c /c + 1 (11) where λ c h mc is the Compton length scale. We see that the frequency of the outgoing photon changes depending on the scattering angle. Since for the electron λ c Å you need x-ray photons to see the effect. Example - The maximum energy loss of a photon due to Compton A photon of energy E 5MeV strikes a proton at rest. What is the maximal energy gain of the proton? 3

4 The maximum energy transfer to the photon occurs for a head-on collision. In this case: (12) m pc 2 The energy gain of the proton is the energy loss of the photon: E ( ) 2E 2 2 m pc 2 m pc 2 m pc E. (13) 2 m pc 2 With E 5MeV and m p c 2 94MeV we obtain E 26MeV. 3.1 Inverse Compton If the electron is moving relativistically it can increase rather than decrease the energy of the photon. In many cases in astrophysics you see two similar spectra but one is shifted to a higher frequency than the other. This is due to some photons being up-scattered by relativistic electrons. Assume that we have an electron moving at a velocity βc and a Lorentz factor γ which hits a photon. A simple Lorentz transformation to the electron rest frame will give us the same formula for the scattering as above except that the original frequency will now be blueshifted by the factor γ(1 + β) since the transformation is ( Ẽ/c p ) ( γ βγ βγ γ ) ( E/c p In the electron s rest frame the frequency after scattering will be ) (14) ν νγ(1 + β) ν(1 cos θ)γ(1 + β)λ c /c + 1 (15) For simplicity let us assume θ π so that the photon is involved in a head-on collision with the electron and changes its direction by 18 degrees. Transforming back into the lab frame will result in an additional factor of γ(1 + β) so the final frequency will be ν νγ 2 (1 + β) 2 (16) 2νγ(1 + β)λ c /c + 1 In the relativistic limit (β 1) and for γ mc 2 the electron s frequency is boosted by a factor of 4γ 2. We also get that there is a maximum boost which occurs for γ mc 2 that gives us a maximum frequency of γm e c 2 /h. 4 Pair production The Compton effect is not the only way in which photons and massive particles can interact. Another result achieved by combining the particle nature of light with special relativity, is that a collision between two photons in a vacuum may 4

5 lead to the production of a pair of particle and anti-particle (an anti-particle has the same mass as the particle but opposite charge). The opposite effect, annihilation of matter and anti-matter, resulting in two or more photons, can also occur. In what follows, we aim to find the condition for production of an electron-positron pair. Denote by P 1, P 2 the four-momenta of the incoming photons, and by P 3, P 4 the four-momenta of the outgoing electron and positron. P 1, P 2 can be written using the directions of motion of the incident photons, î 1, î 2 : P 1 [E 1 /c, (E 1 /c)î 1 ]; P 2 [E 2 /c, (E 2 /c)î 2 ]; (17) Conservation of four-momenta requires: P 1 + P 2 P 3 + P 4 (18) To find the threshold for creation we assume the pair is created without kinetic energy (and momentum), such that the only energy resides in the mass of the newly created particles: P 3 P 4 [m e c,,, ]. (19) Squaring both sides of Eq. 18 and remembering that P 1 P 1 P 2 P 2 (because photons are mass-less particles) and that P 3 P 3 P 3 P 4 P 4 P 4 m 2 ec 2 we obtain: 2P 1 P 2 4m 2 ec 2 2( E 1E 2 c 2 E 1E 2 c 2 cosθ) 4m 2 ec 2 E 2 2m 2 ec 4 E 1 (1 cosθ) (2) where θ is the incident angle between the direction of the photons. Thus, if electron positron pairs are created, the threshold for the process occurs for head-on collisions, θ π, and hence: E 2 m2 ec ev (21) E 1 E 1 where E 1 is in electron Volts. This process thus provides not only a means for creating electron positron pairs but also results in an important source of opacity for high-energy γ-rays. For instance, the energy of optical light (such as produced by the sun, for example) is of order ev. Plugging this number to E 1 shows that photons with energies above 3 GeV can be absorbed by starlight. 5