DIFFERENTIAL CROSS SECTION FOR COMPTON SCATTERING. 1. introduction. 2. Differential cross section with respect to the square of momentum transfer

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1 DIFFERENTIAL CROSS SECTION FOR COMPTON SCATTERING E. LIZARAZO Abstract. The i erential cross section for Compton scattering (e e ) in Feynman gauge has been calculate. Results are shown with respect to two i erent variables: square momentum transfer an cosine of the scattere angle. For completeness the i erential cross section is given both in the centre of mass an the fixe target frames.. introuction The i erential cross section for Compton scattering (e e ) -c.f Fig. - is investigate. Section gives the basic steps to calculate the formula for the i erential cross section with respect to the square of momentum transfer, leaving the square matrix element as a variable to be calculate. In section 3 the square matrix element is worke out in Feynman gauge by means of the algebraic properties of the gamma matrices, section shows how to work out the i erential cross section as a function of the cosine of the scattere angle by two i erent methos: by evaluating the phase space integral in the fixe target frame, an irect substitution. Finally, in section 5 the i erential cross section is calculate for three energy regimes: non relativistic, relativistic an ultra relativistic.. Differential cross section with respect to the square of momentum transfer The relation between the S-matrix elements an cross section for Compton scattering is given by Eq. () = LIPS X p M () k CM s LIPS = ( ) k 0 0 k 0 0 (p s k 0 k) 0 3 k 0 3 k 0 ()

2 E. LIZARAZO where the photon an electron are represente by the subinices an respectively, unprime variables are given to the incoming particles, prime variables to the outgoing particles, k represents four momentum, k inicates three momentum, s =(k +k ),anlips is the -boy Lorentz-invariant phase-space measure which in the centre of mass frame (CM) is given by (3) LIPS = = ( ) EE 0 0 ( ) EE 0 0 (E 0 + E 0 (E 0 + E 0 p s) 3 (k 0 + k 0 ) 3 k 0 3 k 0 (3) p s) 3 k 0 () Integrating Eq. (3) over 3 k 0 results in Eq. (). It shoul be highlighte that this integration transforms Eq. (5), that is, the original E 0 of Eq. (3) into Eq. (6). q E 0 = m 0 + k 0 q E 0 = m 0 + k 0 (5) (6) Now, in terms of the soli angle CM it is possible to rewrite 3 k 0 as Eq. (7). Substituting this into Eq. () an integrating the resulting equation over k 0 the -boy Lorentz-invariant phase-space measure Eq. (8) is obtaine. 3 k 0 = k 0 k 0 CM (7) LIPS = k0 6 p s CM (8) Substituting Eq. (8) in Eq. () the i erential cross section Eq. (9) is obtaine. = k0 CM 6s k X M (9) Now, by i erentiating the square momentum transfer t Eq. (0) with respect to the cosine of the angle between k an k 0 Eq. () is obtaine. This can be rewritten as aerivativeoft in terms of soli angle Eq. (), which can then be use to rewrite Eq. (9) in terms of t Eq. (3)

3 DIFFERENTIAL CROSS SECTION FOR COMPTON SCATTERING 3, k, k 0, k e, k 0 e, k e, k 0 e, k, k 0 Figure. Feynman iagrams for Compton scattering t =(k k 0 ) = E E 0 + k k 0 cos (0) t cos = k k 0 () t = CM k k 0 () t = X M (3) 6 s k 3. The square matrix element The matrix element M Eq. () for Compton scattering is obtaine by applying Feynman rules to the two iagrams -c.f Fig. -, apple µ M = e µ(k) 0 (k )ū(k) 0 /k + µ k + k k /k 0 µ + k µ k k 0 u(k ) () where (k ), µ(k 0 ), u(k )anū(k 0 ) are the initial an final polarisation vectors for photons an electrons respectively. X u(k )ū(k )= /k + m (5) X polarizations µ g µ (6)

4 E. LIZARAZO By using the ientity Eq. (5) an the replacement Eq. (6), the average square matrix element Eq. (7) is obtaine X M = e g µ g Tr apple (/k + m) (/k 0 + m) /k + k k k + apple µ /k + µ k k k + Evaluating the traces of Eq. (7) results in Eq. (8) /k 0 + k k k 0 /k 0 µ + k µ k k 0 (7) " X # M =e k k 0 +m + m k k k k k 0 k k k k 0 k (8). Differential cross section with respect to scattering angle.. From phase-integral in lab-frame. Eq. (9) follows from conservation of - momentum, an the Minkowski norm k 0 = m.solvingthisequationfortheenergy of the final photon 0 gives Eq. (0) m = m +m( 0 ) 0 ( cos ) (9) 0 m = (0) m + ( cos ) where m is the mass of the electron an is the energy of the incoming photon. Di erentiating Eq. (0) with respect to cos Eq. () is obtaine by rewriting 3 k 0 as Eq. () 0 = 0 cos m () 3 k 0 = 0 0 FT () substituting this in Eq. (), an integrating with respect to 3 k 0,thefixetarget frame -boy Lorentz-invariant phase space measure Eq. (3) is obtaine,

5 DIFFERENTIAL CROSS SECTION FOR COMPTON SCATTERING 5 LIPS = 0 FT ( 0 ) ( 0 + E 0 ( ) 3 0 E 0 m ) (3) = cos 8 ( 0 ) m () where E 0 = p m + +( 0 ) 0 cos, Eq. () follows from integration of Eq. (3) with respect to 0,anusingEq.(0)tosimplify. ReplacingEq.(3)an Eq. (5) into Eq. () results in Eq. (6). The Klein-Nishima formula Eq. (7) follows from Eq. (6) by evaluating Eq. (8) with k k = m an k 0 k = 0 m. m k FT = p s k CM (5) cos = (0 ) X M (6) 3 m apple cos = (0 ) 0 3 m + sin (7) 0.. Direct conversion of t into cos. The Manelstam variables satisfy Eq. (8), which i erentiate with respect to 0 gives Eq. (9) t =m s u =m( 0 ) (8) t =m 0 (9) From Eq. (9) an Eq. (5) it follows that Eq. (3) can be rewritten as Eq. (30) cos = (0 ) X M 3 m (30) which turns into the Klein Nishima formula Eq. (7) when the average square matrix is evaluate. Using the ientity R f(x) (g(x))x = P ((x i )/ g 0 (x i ) ). This is the relation between the magnitue of the 3-momentum of the photon in the fixe target an the centre of mass frames.

6 6 E. LIZARAZO 0.5 σ / cos(θ) [b] s =.m e s = m e s = 5 m e cos(θ) Figure. / cos as a function of cos for i erent centre of mass energy square s 5. Differential cross section for ifferent energies The i erential cross section was evaluate for three i erent energies, s<<m, s m an s>>m,theresultsfor / cos against cos an /t against t are shown in Fig. an Fig. 3 respectively. In Fig. it can be seen that for low energies, that is s<<m,thescatteringprobabilityisnearlysymmetricaboutcos =0, which implies that it is almost equally likely for a photon to be scattere in the forwar or backwar irection. However as the energy increases, the scattering becomes more likely in the backwars irection, with an almost constant scattering probability with respect to cos in the forwar irection. Fig. 3 shows also that for low energies there is symmetry of the scattering probability for the square momentum transfer roughly about the mipoint of the t interval, however this symmetry breaks for higher energies. In particular, for the ultra relativistic regime -i.e s>>m e- the scattering probability is nearly constant for the right half of the interval, an increases rapily on its left half.

7 DIFFERENTIAL CROSS SECTION FOR COMPTON SCATTERING 7.5 x 08 σ / t [b/gev ] t [GeV ] x 0 9 (a) s =.m e 5.5 x 06 5 σ / t [b/gev ] t [GeV ] x 0 7 (b) s =m e 7 x 05 6 σ / t [b/gev ] t [GeV ] x 0 7 (c) s =5m e Figure 3. /t as a function of the square momentum transfer t for i erent centre of mass energy square s

8 8 E. LIZARAZO 6. Conclusion The square matrix element for Compton scattering has been calculate in Feynman gauge, the resulting element was use to fin an expression for the i erential cross section with respect to the square momentum transfer in the centre of mass frame. The i erential cross section was also foun with respect to the cosine of the scattering angle in two i erent ways: Firstly, by performing the phase space integral in the fixe target frame, an seconly, by irect conversion of the infinitesimal element t into cos, bothmethosyielethesameresult. References [] M. E. Peskin an D. V. Schroeer, An Introuction To Quantum Fiel Theory (Frontiers in Physics). Westview Press, 995. [] M. Srenicki, Quantum Fiel Theory. Cambrige University Press, e., February 007.