Thursday, 13 October 2016

quantum mechanics - Is it a typo in David Tong's derivation of spin-orbit interaction?


A few lines below equation 7.8 D. Tong writes



The final fact is the Lorentz transformation of the electric field: as electron moving with velocity v in an electric field E will experience a magnetic field B=γc2(v×E).



The note says that it was derived in another note but I couldn't find this expression.



Is this coefficient γ/c2 correct? Griffiths derives this to be 1/c2 and I did not find anything wrong there. See Griffiths electrodynamics, third edition, equation 12.109.


Then I looked at this book which uses Griffiths' expression in Sec. 20.5, but uses p=mv instead to p=γmv to derive the same result. Which one is correct and why?



Answer



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In above Figure-01 an inertial system S is translated with respect to the inertial system S with constant velocity
υ=(υ1,υ2,υ3)υ=υ=υ21+υ22+υ23(0,c)


The Lorentz transformation is x=x+γ2c2(γ+1)(υx)υγυcctct=γ(ctυxc)γ=(1υ2c2)12


For the Lorentz transformation (03a)-(03b), the vectors E and B of the electromagnetic field are transformed as follows E=γEγ2c2(γ+1)(Eυ)υ+γ(υ×B)B=γBγ2c2(γ+1)(Bυ)υγc2(υ×E) Now, if in system S we have B=0, then from (04a)-(04b) E=γEγ2c2(γ+1)(Eυ)υB=γc2(υ×E) Equation (05b) corresponds to Tong's equation (it remains to explain the minus sign).


From equations (05a)-(05b) we have B=γc2(υ×E)=1c2(υ×γE)=1c2(υ×[γEγ2c2(γ+1)(Eυ)υ])=1c2(υ×E) that is B=1c2(υ×E) Equation (06) corresponds to Griffiths' equation.


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ADDENDUM


If in system S we have E=0, then from (04a)-(04b) E=γ(υ×B)B=γBγ2c2(γ+1)(Bυ)υ so that E=γ(υ×B)=(υ×γB)=υ×[γBγ2c2(γ+1)(Bυ)υ]=υ×B that is E=υ×B Equations (06) and (08) are the following equations (12.109) and (12.110) respectively B=1c2(v×E).abab


E=v×B.abab as shown in ''Introduction to Electrodynamics'' by David J.Griffiths, 3rd Edition 1999.


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