Interactions of High Energy Particles with Nuclei |
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Page 2
... m ) db exp ( ia.b ) ( 1 - eixi ( b ) ) , is shifted to the position of the jth nucleon : ik fi ( 8 ) = * S ) b dụb exp ( ia.b ) { 1- exp [ ix ; ( b - s ; ) ] } , where k is the momentum of the incident particle in laboratory frame A ...
... m ) db exp ( ia.b ) ( 1 - eixi ( b ) ) , is shifted to the position of the jth nucleon : ik fi ( 8 ) = * S ) b dụb exp ( ia.b ) { 1- exp [ ix ; ( b - s ; ) ] } , where k is the momentum of the incident particle in laboratory frame A ...
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and assuming that the particle goes through the target so fast that all the nucleons are ' frozen ' at certain positions , we get for the amplitude ik Msi = db ) 2 . l dru ... dor \ , * ( no ... ra ) sa ra ) | < ob exp ( ia.b ) x { 1- ...
and assuming that the particle goes through the target so fast that all the nucleons are ' frozen ' at certain positions , we get for the amplitude ik Msi = db ) 2 . l dru ... dor \ , * ( no ... ra ) sa ra ) | < ob exp ( ia.b ) x { 1- ...
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... k ) der exp ( -ik ' • r ) V ( r ) ** ( r ) 21 +0 m s :) a2b exp ( ia . b ) S ** xp ( -__'_ ) 2 ) S ab exp ( ia - b ) [ 1– exp ( -_ L * d V ( b , - ) ) ] :) dzeik : V ( b , z ) exp L. dzV ( b , z ' ) 27 00 ik + 了 db ) dz'V ( b z ...
... k ) der exp ( -ik ' • r ) V ( r ) ** ( r ) 21 +0 m s :) a2b exp ( ia . b ) S ** xp ( -__'_ ) 2 ) S ab exp ( ia - b ) [ 1– exp ( -_ L * d V ( b , - ) ) ] :) dzeik : V ( b , z ) exp L. dzV ( b , z ' ) 27 00 ik + 了 db ) dz'V ( b z ...
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... ia • E ) + eV ] y = erite = exp ( izvE - m ? ) ceilin , ( E ) . Inserting this into the Dirac equation and noting a -i qiExp = EeiExp + eie : eizp дz ( -1 ) ♡ we get Əz E ( 1 - a3 ) o = [ - ia : v + Bm - KB ( D.B - ia E ) + eV ] .
... ia • E ) + eV ] y = erite = exp ( izvE - m ? ) ceilin , ( E ) . Inserting this into the Dirac equation and noting a -i qiExp = EeiExp + eie : eizp дz ( -1 ) ♡ we get Əz E ( 1 - a3 ) o = [ - ia : v + Bm - KB ( D.B - ia E ) + eV ] .
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Without going into any details of the calculation let us quote the results . In the case when only one Coulomb potential is present ( hence B = 0 , but E + 0 ) , we have + M ( A ) ixt { D ) db exp ( ia.b ) [ 1 exp dzV ( b 2 ) ) Xin ...
Without going into any details of the calculation let us quote the results . In the case when only one Coulomb potential is present ( hence B = 0 , but E + 0 ) , we have + M ( A ) ixt { D ) db exp ( ia.b ) [ 1 exp dzV ( b 2 ) ) Xin ...
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