textbook: 6.8, 6.10 stress.pdf 6.8 recall results from 200: magnetic field from line-current I, via Amperian Loop: 2 pi r B = mu0 I electric filed from line-charge lambda, via Gaussian cylinder: 2 pi r L E = lambda L/epsilon0 J 4-vector in S (the cross sectional area A in invariant and cancels): J'=(i0,0,0,0) In[2]:= <0, e2->0, e3->0, b1->0, b2->0,b3->Bp} fuv=boost[-v,0,0].fpuv.Transpose[boost[-v,0,0]] Bp Bp -I Bp v Out[7]= {{0, ------------, 0, 0}, {-(------------), 0, 0, ------------}, {0, 0, 0, 0}, 2 2 2 Sqrt[1 - v ] Sqrt[1 - v ] Sqrt[1 - v ] I Bp v > {0, ------------, 0, 0}} 2 Sqrt[1 - v ] In[8]:= MatrixForm[%] Out[8]//MatrixForm= Bp ------------ 2 0 Sqrt[1 - v ] 0 0 Bp -I Bp v -(------------) ------------ 2 2 Sqrt[1 - v ] 0 0 Sqrt[1 - v ] 0 0 0 0 I Bp v ------------ 2 0 Sqrt[1 - v ] 0 0 SO Ey/c= Bp beta gamma Bz= gamma Bp consistent with above 6.10 fpuv={{0,b3,-b2,-I e1},{-b3,0,b1,-I e2},{b2,-b1,0,-I e3},{I e1,I e2,I e3,0}} /. {e1->0, e2->E0, e3->0, b1->0, b2->0,b3->0} fuv=boost[-v,0,0].fpuv.Transpose[boost[-v,0,0]] E0 v E0 v -I E0 Out[10]= {{0, ------------, 0, 0}, {-(------------), 0, 0, ------------}, {0, 0, 0, 0}, 2 2 2 Sqrt[1 - v ] Sqrt[1 - v ] Sqrt[1 - v ] I E0 > {0, ------------, 0, 0}} 2 Sqrt[1 - v ] In[11]:= MatrixForm[%] Out[11]//MatrixForm= E0 v ------------ 2 0 Sqrt[1 - v ] 0 0 E0 v -I E0 -(------------) ------------ 2 2 Sqrt[1 - v ] 0 0 Sqrt[1 - v ] 0 0 0 0 I E0 ------------ 2 0 Sqrt[1 - v ] 0 0 SO in the S frame Bz= E0/c beta gamma & Ey= E0 gamma Jp={j0,0,0,0} Out[3]= {j0, 0, 0, 0} In[4]:= J=Transpose[boost[v,0,0]].Jp j0 I j0 v Out[4]= {------------, 0, 0, ------------} 2 2 Sqrt[1 - v ] Sqrt[1 - v ] so in S: i=gamma j0; lambda=j0/c gamma beta fuv.J fuv.J Out[15]= {0, 0, 0, 0} In frame S there is a uniform electric field vector E=(0,a,0) and a uniform magnetic field vector B=(0,0,5a/3c). Find a frame S' in which the electric field is zero. Find B' in the S' frame. Check the values of the invariants: E.B and E^2-c^2B^2 are the results the same in S and S'? Note that in the given frame: E.B=0 and E^2-c^2B^2= a^2 (1-(5/3)^2)= -a^2 16/9 Note if we boost in the x direction both y & z are perpendicular directions and E'_perp = gamma ( E_perp + v cross B_perp ) if v is in the positive x direction, v cross z is in the minus y direction so we can cancel the positive Ey=a, thus require a=v(5 a/3c) or (3/5) c = v or beta = 3/5 in the postive x direction <0, e2->a/c, e3->0, b1->0, b2->0,b3->5/3 a/c} fpuv=boost[3/5,0,0].fuv.Transpose[boost[3/5,0,0]] MatrixForm[%] Out[4]//MatrixForm= 4 a --- 0 3 c 0 0 -4 a ---- 3 c 0 0 0 0 0 0 0 0 0 0 0 SO bz'=(4/3) (a/c) and the invariants are invariant stress.pdf see stress_answer.pdf