2008, Davisworld.org DSLs in Groovy Say what you mean ... - QCon

520Here ɛ- energy of electron, ω- energy of bremsstrahlung photon. The notations:Q = 2πZ2 α 2 ɛ(ɛ − ω)n a, Lm 4 1 =ln(a 2 s2ω/λ2 c ) = 2 ln(183Z−1/3 e −f ), (2)f =(Zα) 2 ∞ ∑k=11k(k 2 +(Zα) 2 )The determination of L c :L c = ln( a2 s2) (3)λ 2 cρ 2 cIf 4QL 1 ≤ 1 then ρ c =1.If 4QL 1 > 1 then ρ c is the solution of equation:4Qρ 4 c ln( a2 s2)=1λ 2 c ρ2 cLetν1 2 = ɛ ɛ − ωɛ 0 ω , ɛ m0 =8πZ 2 α 2 n a λ 3 cExtracting the imaginary part from (1) we getdW = α m 2 {[ ∫dω∞r 1 ( 1 2π ɛ ɛ − ω 0 z − 1sinhz ) sin( z√ ) exp(− √ z )dz+ν 1 2Lc ν 1 2Lc√∫ Lc ∞ν 12 r 2 ( 10 z − 12 (sinhz) ) exp(− z]zz√ )[sin( √ ) + cos( √ )]dz +2 ν 1 2Lc ν 1 2Lc ν 1 2Lc∫1 ∞exp(−z}zz√ )[A sin( √ ) − B cos( √ )]dz (5)2L c 0 ν 1 2Lc ν 1 2Lc ν 1 2LcHere:A = r 1Ref 1 (z)+2r 2 Imf 2 (z), B = r 1Imf 1 (z) − 2r 2 Ref 2 (z)(sinh z) 2 (sinh z) 2Ref 1 (z) = [ln(ρ 2 c ) + ln(ν 1Ref 2 (z) = ν 1sinh zImf 2 (z) = ν 1sinh z√L c ) − ln(sinh z) − C]g(z) − 2 cosh(z)G(z),Imf 1 (z) =− π 4 g(z)√Lc2 (Ref 1(z) − g(z)2 − Imf 1(z))√Lc2 (Ref 1(z) − g(z)2 + Imf 1(z))(4)