Paper 2, Section II, C

Electromagnetism | Part IB, 2017

In special relativity, the electromagnetic fields can be derived from a 4-vector potential Aμ=(ϕ/c,A)A^{\mu}=(\phi / c, \mathbf{A}). Using the Minkowski metric tensor ημν\eta_{\mu \nu} and its inverse ημν\eta^{\mu \nu}, state how the electromagnetic tensor FμνF_{\mu \nu} is related to the 4-potential, and write out explicitly the components of both FμνF_{\mu \nu} and FμνF^{\mu \nu} in terms of those of E\mathbf{E} and B\mathbf{B}.

If xμ=Λνμxνx^{\prime \mu}=\Lambda_{\nu}^{\mu} x^{\nu} is a Lorentz transformation of the spacetime coordinates from one inertial frame S\mathcal{S} to another inertial frame S\mathcal{S}^{\prime}, state how FμνF^{\prime \mu \nu} is related to FμνF^{\mu \nu}.

Write down the Lorentz transformation matrix for a boost in standard configuration, such that frame S\mathcal{S}^{\prime} moves relative to frame S\mathcal{S} with speed vv in the +x+x direction. Deduce the transformation laws

Ex=ExEy=γ(EyvBz)Ez=γ(Ez+vBy)Bx=BxBy=γ(By+vc2Ez)Bz=γ(Bzvc2Ey)\begin{aligned} E_{x}^{\prime} &=E_{x} \\ E_{y}^{\prime} &=\gamma\left(E_{y}-v B_{z}\right) \\ E_{z}^{\prime} &=\gamma\left(E_{z}+v B_{y}\right) \\ B_{x}^{\prime} &=B_{x} \\ B_{y}^{\prime} &=\gamma\left(B_{y}+\frac{v}{c^{2}} E_{z}\right) \\ B_{z}^{\prime} &=\gamma\left(B_{z}-\frac{v}{c^{2}} E_{y}\right) \end{aligned}

where γ=(1v2c2)1/2\gamma=\left(1-\frac{v^{2}}{c^{2}}\right)^{-1 / 2}

In frame S\mathcal{S}, an infinitely long wire of negligible thickness lies along the xx axis. The wire carries nn positive charges +q+q per unit length, which travel at speed uu in the +x+x direction, and nn negative charges q-q per unit length, which travel at speed uu in the x-x direction. There are no other sources of the electromagnetic field. Write down the electric and magnetic fields in S\mathcal{S} in terms of Cartesian coordinates. Calculate the electric field in frame S\mathcal{S}^{\prime}, which is related to S\mathcal{S} by a boost by speed vv as described above. Give an explanation of the physical origin of your expression.

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