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Electromagnetic field has two invariant quatities. Usingthe transformation formulas (3.6 i)demonstrate that thesequantiesare (a) EB, (b) E^(2) - c^(2) B^(2). |
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Answer» Solution :(a) We see that, `vec(E').vec(B') = vec(E')_(|\|).vec(B')_(|\|) + vec(E')_(_|_).vec(B')_(_|_)` `= vec(E)_(|\|).vec(B)_(|\|) + ((vec(B)_(_|_) + vec(v) xx vec(B)).(vec(B)_(_|_)- (vec(v) xx vec(E))/(C^(2))))/(1 - (v^(2))/(c^(2)))` `= vec(E)_(|\|).vec(B)_(|\|) + (vec(E)_(_|_).vec(B)_(_|_) - (vec(v) xx vec(B)).(vec(v) xx vec(E))//c^(2))/(1 - v^(2)//c^(2))` `= vec(E)_(|\|).vec(B)_(|\|) + (vec(E)_(_|_).vec(B)_(_|_) - (vec(v) xx vec(B)_(_|_)).(vec(v) xx vec(E_(_|_)))//c^(2))/(1 - (v^(2))/(C^(2)))` But, `vec(A) xx vec(B).vec(C) xx vec(D) = A. C B. D - A.D B. C`, so, `vec(E').vec(B') = vec(E)_(|\|).vec(B)_(|\|) + vec(E)_(_|_).vec(B)_(_|_) ((1 - (v^(2))/(c^(2))))/(1 - (v^(2))/(c^(2))) = vec(E).vec(B)` (b) `E^(2) - c^(2) B'^(2) = E'_(|\|)^(2) - c^(2) B'_(|\|)^(2) + E_(_|_)^(2) - c^(2) B'_(_|_)^(2)` `=E_(|\|)^(2)c^(2) B_(|\|)^(2) + (1)/(1 - (v^(2))/(c^(2))) [(vec(E)_(_|_) + vec(v) xx vec(B))^(2) - c^(2) (B_(_|_)^(2) - (vec(v) xx vec(E))/(c^(2)))^(2)]` `=E_(|\|)^(2)c^(2) B_(|\|)^(2) + (1)/(1 - (v^(2))/(c^(2))) [E_(_|_)^(2) -c^(2) B_(_|_)^(2) + (vec(v) xx B_(_|_)^(2) - 91)/(c^(2)) (vec(v) xx E_(_|_))^(2)]` `=E_(|\|)^(2)c^(2) B_(|\|)^(2) + (1)/(1 - (v^(2))/(c^(2))) [E_(_|_)^(2) - c^(2) B_(_|_)^(2)] (1 - (v^(2))/(c^(2))) = E^(2) - c^(2) B^(2)`. since, `(vec(v) xx vec(A)_(_|_))^(2) = v^(2) A_(_|_)^(2)` |
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