I know that the eletromagnetic field tensor Fμν, can be transfomed to another reference frame by Fαβ=ΛαμΛβνFμν
Since these tensors can be represented by matrices so I thought that I could represent the electromagnetic field tensor in another inertial reference frame by doing matrix multiplications, but I ended up with:
Fαβ=[γ−γβ00−γβγ0000100001][γ−γβ00−γβγ0000100001][0−1cEx−1cEy−1cEz1cEx0−BzBy1cEyBz0−Bx1cEz−ByBx0]= =[−2γ2βcEx−γ2cEx(1+β2)−γ2cEy(1+β2)+2γ2β2Bz−γ2cEz(1+β2)−2γ2βBy1cEx2γ2βcEx2γ2βcEy−γ2Bz(β2+1)2γ2βcEz+γ2By(β2+1)1cEyBz0−Bx1cEz−ByBx0] But as it can easily be seen, this matix is not anti-symmetric as an eletromagnetic field tensor should be, by definition, so does this mean that I cannot use matrix multiplication on tensors or does it mean that I made a mistake somewhere in the calculations of the matrix products?
Answer
Fαβ=[−γ−γβ−0−0−12−γβ−γ−0−0−12−0−0−1−0−12−0−0−0−1−12][0−Ex−Ey−Ez12Ex−0−cBz−cBy12Ey−cBz−0−cBx12Ez−cBy−cBx−012][−γ−γβ−0−0−12−γβ−γ−0−0−12−0−0−1−0−12−0−0−0−1−12]=[−γβEx−γEx−γ(Ey−βcBz)−γ(Ez+βcBy)12−−γExγβEx−γ(βEy−cBz)−γ(βEz+cBy)12−γβEy−cBz0−cBx12−γβEz−cBycBx−012−][−γ−γβ−0−0−12−γβ−γ−0−0−12−0−0−1−0−12−0−0−0−1−12]=[−0−Ex−γ(Ey−βcBz)−γ(Ez+βcBy)−12−Ex0−γ(βEy−cBz)−γ(βEz+cBy)−12−γ(Ey−βcBz)−γ(βEy−cBz)−0−cBx−12−γ(Ez+βcBy)−γ(βEz+cBy)−cBx−0−12]=[−0−E′x−E′y−E′z−12−E′x−0−cB′z−cB′y−12−E′y−cB′z−0−cB′x−12−E′z−cB′y−cB′x−0−12] Since β=υ/c E′x=−Ex12E′y=γ(Ey−υBz)12E′z=γ(Ez+υBy)12B′x=−Bx12B′y=γ(By+υc2Ez)B′z=γ(Bz−υc2Ey)
For Your Information :
The equations of a more general Lorentz Transformation between two systems S(x,t) and S′(x′,t′), the latter translating with constant velocity v=υn,‖n‖=1,υ∈(−c,+c), with respect to the former, are : x′=x+(γ−1)(n⋅x)n−γvtt′=γ(t−v⋅xc2)γ=(1−υ2c2)−1/2 see Figure.(1)
Under (ft-01) the vectors E,B of the electromagnetic field in empty space are transformed as follows :
E′=γE−(γ−1)(n⋅E)n+γ(v×B)B′=γB−(γ−1)(n⋅B)n−γc2(v×E) Equations (02),(03) are a special case of (ft-02) for n=(1,0,0).
(1) See a 3D version of this Figure here : Figure 3D version
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