1000 Solved Problems in Modern Physics

344 6 Special Theory of Relativity 6.49 T=(γ−1)m 0 c^2 =[1/(1−β^2 )^1 /^2 −1]m 0 c^2 =[(1−β^2 )−^1 /^2 −1]m 0 c^2 =[1+^12 β^2 +^ ...

6.3 Solutions 345 Fig. 6.6 (a)Scattering of a proton with a stationary electron(b)Momentum triangle 6.54 The energy released in ...

346 6 Special Theory of Relativity 6.56 TK=[(MD−MK)^2 −M^2 π]/ 2 MD=[(1865−494)^2 − 1402 ]/ 2 × 1865 = 498 .67 MeV PK=(TK^2 + 2 ...

6.3 Solutions 347 Fig. 6.7Symmetrical elastic collision between identical particles 6.61 (γ−1)mc^2 =mc^2 γ= 2 β=(1− 1 /γ^2 )^1 / ...

348 6 Special Theory of Relativity 6.64 (a)π+→μ++νμ T=Q(Q+ 2 mν)/2(mμ+mν+Q)=(mπ−mμ)^2 / 2 mπ (Q=mπ−mμ−mν=mπ−mμandmν=0) mμ= 206. ...

6.3 Solutions 349 Therefore,γ=[1+(p/m)^2 ]^1 /^2 =[1+(5/ 0 .938)^2 ]^1 /^2 = 5. 41 tanθtanφ=tan 82◦tan 2◦ 30 ′= 7. 115 × 0. 0436 ...

350 6 Special Theory of Relativity 6.73 Expressingθin terms ofθ∗ tanθ=sinθ∗/γc(cosθ∗+βc/β∗)(1) Differentiating with respect toθ∗ ...

6.3 Solutions 351 Because of (2) and (5) p′e=p=E/c (b)β=v/c=(1− 1 /γ^2 )^1 /^2 =(1−m^2 /Ee^2 )^1 /^2 =(1−m^2 /(pe^2 +m^2 ))^1 /^ ...

352 6 Special Theory of Relativity 6.78 1 2 (M^2 −m 12 −m 22 )=E 1 E 2 −p 1 p 2 cosθ θ= 900 P±=530 MeV/c E±=(P±^2 +mπ^2 )^1 /^2 ...

6.3 Solutions 353 Butγ= 10 / 0. 14 Therefore,φmin= 0 .028 rad or 1. 6 ◦ 6.83 Rest mass energy ofω^0 =Total available energy – (t ...

354 6 Special Theory of Relativity Fig. 6.11 (a)Decay ∧→P+π−in flight.(b) Momentum triangle (a) (b) 6.86 M 0 βγ=p 1 sinθ+p 2 cos ...

6.3 Solutions 355 M^2 =2(EpEπ−−PpPπ−cosθ)+mπ+^2 +mπ−^2 (6) Ep=(P+^2 +mp^2 )^1 /^2 (7) mp=^0 .938 GeV/c^2 (8) Using (3), (4), (5) ...

356 6 Special Theory of Relativity 6.91 (a)λ/λ′=[(1+β)/(1−β)]^1 /^2 = 656 / 434 = 1. 5115 v=βc= 1. 17 × 108 ms−^1 (b) the nebula ...

6.3 Solutions 357 From (1) we have E′^2 =(E 0 −E+m)^2 (4) Comparing (3) and (4) and simplifying cosφ= 1 −m(E 0 −E−m)/E 0 E≈ 1 −m ...

358 6 Special Theory of Relativity From (3) and (4),(Emax−Emin)/(Emax+Emin)=β (5a) From the measurement ofEmaxandEmin, the veloc ...

6.3 Solutions 359 Note that if theπ^0 s were to decay at rest (γ=1) then the rectangle would have reduced to a spike atE= 67 .5 ...

360 6 Special Theory of Relativity This shows that small emission angles of photons in the lab system are favored. 6.102 In Fig. ...

6.3 Solutions 361 6.103 The angle between the twoγ-rays in the LS can be found fom the formula m^2 c^4 =m 12 c^4 +m 22 c^4 +2(E ...

362 6 Special Theory of Relativity The minimum angle is found by setting dθ/dD= 0 This gives usD=1, that isE 1 =E 2 =E/2. θmin= ...

6.3 Solutions 363 6.107 (a), (b) In the CMS,m 2 will move with the velocityβcin a direction opposite to that ofm 1. By definitio ...