Interactions of High Energy Particles with Nuclei |
From inside the book
Results 1-5 of 8
Page 4
In fact it is amazing that ( 2.1 ) works so well . Even in the conceptually simplest cases of rela- tivistic potential scattering one can give examples in which it breaks down . Examples Example 1. Dirac particle with anomalous magnetic ...
In fact it is amazing that ( 2.1 ) works so well . Even in the conceptually simplest cases of rela- tivistic potential scattering one can give examples in which it breaks down . Examples Example 1. Dirac particle with anomalous magnetic ...
Page 5
Əz ( 2.4 ) So , if the anamalous magnetic moment K = 0 , we end up with an expression which is virtually the same as in the case of the Schrödinger equation : a ( -i + ev ) p = 0 dz whose solution ❤ = u ...
Əz ( 2.4 ) So , if the anamalous magnetic moment K = 0 , we end up with an expression which is virtually the same as in the case of the Schrödinger equation : a ( -i + ev ) p = 0 dz whose solution ❤ = u ...
Page 7
For instance , the presence of an anomalous magnetic moment can be looked upon as a mark of " compositeness . " Suppose Anm ( F , F1 , ... ... ... , r ^ ) = Ĉ q☺ ( r − r ; ) . j1 m Diagonalization should produce a diagonal matrix of ...
For instance , the presence of an anomalous magnetic moment can be looked upon as a mark of " compositeness . " Suppose Anm ( F , F1 , ... ... ... , r ^ ) = Ĉ q☺ ( r − r ; ) . j1 m Diagonalization should produce a diagonal matrix of ...
Page 8
... than one scattering center if the Dirac particle has an anomalous magnetic moment , K # 0 . Without going into any details of the calculation let us quote the results . In the case when only one Coulomb potential is present ( hence ...
... than one scattering center if the Dirac particle has an anomalous magnetic moment , K # 0 . Without going into any details of the calculation let us quote the results . In the case when only one Coulomb potential is present ( hence ...
Page 9
Let us allow for our vector particle to have an arbitrary magnetic moment and define the magnetic moment operator e M = ( 1 + x ) - s , 2m ( 2.6 ) ( S - spin operator ) , where i determines the value of the magnetic moment .
Let us allow for our vector particle to have an arbitrary magnetic moment and define the magnetic moment operator e M = ( 1 + x ) - s , 2m ( 2.6 ) ( S - spin operator ) , where i determines the value of the magnetic moment .
What people are saying - Write a review
We haven't found any reviews in the usual places.
Common terms and phrases
absorption additivity analysis approximately assume attenuation beam coherent collision complete components compute consider contribution corrections Coulomb Coulomb interactions coupling cross section db exp db exp i▲·b depend describe deuteron diffractive production processes discussed effects elastic scattering elastic scattering amplitude equation example excited existence experimental experiments expression fact factor field final formula forward given gives Glauber ground hadrons Hence high energy limit important incident particle inelastic initial Institute interactions introduce magnetic mass measurement momentum transfer multiple scattering Note nuclear nuclear targets nuclei nucleon numbers objects obtained parameters phase shifts photon photoproduction physical position possible problem profiles regeneration shadowing single Standards step strong structure technical vector meson wave function weak