PC Chapter 02

PC Chapter 02 - 2 Relativity II 2-1 p= mv 1 ( v 2 c 2 ) p=...

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17 2 Relativity II 2-1 ( 29 =  -  12 22 1 mv p vc (a) ( 29 (29 ( - - × = = ×⋅ - 27 21 2 1.67 10 kg 0.01 5.01 10 kg m s 1 0.01 c p cc (b) ( ( - - × = = - 27 19 2 1.67 10 kg 0.5 2.89 10 kg m s 1 0.5 c p (c) ( ( - - × = = - 27 18 2 1.67 10 kg 0.9 1.03 10 kg m s 1 0.9 c p (d) - - × = = × 13 22 8 1.00 MeV 1.602 10 J 5.34 10 kg m s 2.998 10 m s c so for (a) ( ( - - == 21 22 5.01 10 kg m s 100 MeV 9.38 MeV 5.34 10 kg m s c pc Similarly, for (b) = 540 MeV and for (c) = 1 930 MeV . 2-2 (a) Scalar equations can be considered in this case because relativistic and classical velocities are in the same direction. ( 29 ( 29 ( 29 γ = = = = ⇒=-  -- = 2 11 1.90 1.90 1 1.90 0.85 mv p m v m v c (b) No change, because the masses cancel each other.
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18 CHAPTER 2 RELATIVITY II 2-3 As F is parallel to v , scalar equations are used. Relativistic momentum is given by ( 29 γ ==  -  12 2 1 mv p mv vc , and relativistic force is given by    -     - 2 32 22 1 1 dp d mv F dt dt dp m dv F dt dt 2-4 (a) Using the results of Problem 2-3, - =- 2 2 1 v dv q Em dt c , or = 2 2 1 qE d vv a d tm c . Here v is a function of t and q , E , M , and c are parameters. (b) As expected; as , 0 a because in general no speed can exceed c , the speed of light. (c) Separating variables, ( 29 = - 1 qE dv dt m , or ( = - ∫∫ 00 1 vt qE dv dt m , ( ( ( ( (29 ( 29 = - -   +=      = + = + 0 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1 v qEt v m qEt qEt v mm qEt qEt v v c qEt qEt v qEt mc v qEt mc qEct v mc qEt Note that the limiting behavior of v as 0 t and →∞ t is reasonable. As + 2 2 , qEct dx v dt mc qEt
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MODERN PHYSICS 19 ( 29 ( 29 ( 29 ( 29 ( 29 { }  = + = +-   12 22 2 0 1 t c x qEc mc qEt mc qEt mc qE qE . As 0 t , 0 x , and →∞ t , x ct ; reasonable results. 2-5 This is the case where we use the relativistic form of Newton’s second law, but unlike Problem 2-3 in which F is parallel to v , here F is perpendicular to v and = d dt p F so that ( 29   = × ==  -  2 1 d d mv q dt dt vc p F vB . Assuming that B is perpendicular to the plane of the orbit of q , the force is radially inward, and we find - radial 2 1 d mv q dt F vB . As the force is perpendicular to v , it does no work on the charge and the magnitude (but not the direction) of v remains constant in time. Thus, = -- 11 d mv m dv dt dt . Identifying    dv dt as the centripetal acceleration where the scalar equation = 2 radial d vv d tr gives = - 2 radial radial 1 mv qvB r or =- 2 2 1 qBr v v m c . Finally, the period T is π 2 r v and ( 29 ( ( ππ rm T q B q B . As = 1 f T , π 2 2 1 2 qB v f m c . 2-6 Using Equation 2.4 ( 29 - = = × = ×⋅ 19 19 1.60 10 C kg m C s 1.60 10 kg m s p e BR BR BR . To convert kgms to MeV c , use ( ( ( - - × = = × 6 19 22 8 10 1.60 10 C 1 J C 1 MeV 5.34 10 kg m s 3.00 10 m s c , so that ( 29 ( - - 19 22 1.60 10 kg m s 1 MeV 300 MeV 5.34 10 kg m s B Rc p B .
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20 CHAPTER 2 RELATIVITY II 2-7 γ = 2 E mc , γ = p mu ; ( 29 γ = 2 22 E mc ; ( 29 γ = 2 2 p mu ; ( 29( 29 ( 29 (29 { } γ γγ - - = - =-  = --=   =+ 2 2 2 2 2 2 2 2 1 2 2 2 1 1 Q.E.D.
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PC Chapter 02 - 2 Relativity II 2-1 p= mv 1 ( v 2 c 2 ) p=...

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