Maxwell's equations in relativistic physicsby elivil Tags: equations, maxwell, physics, relativistic 

#1
May1508, 03:07 PM

P: 15

1. The problem statement, all variables and given/known data
It is often said that Maxwell's equations in differential form hold in special relativity while Maxwell's equations in integral form don't hold. Consider one of equations: [tex]\nabla \times \mathbf{E} = \frac{\partial \mathbf{B}} {\partial t}[/tex] [tex] \oint_{\partial S} \mathbf{E} \cdot \mathrm{d}\mathbf{l} =  \frac {1} {c} \frac {d} {dt} \int \mathbf {B} \cdot \mathrm{d} \mathbf {S}[/tex] Consider the integral form. It means that if one has an alternating magnetic field, then in any contour around this field circulation of electrical field immediately appears . If one takes a very long tube and at one end of the tube somehow generates an alternating magnetic field then in contour (radius R) around the tube circulation of electric field immediately appears. But in relativistic case it can't appear immediately because it will take time [itex]t=\frac {R} {c}[/itex] for news about alternating magnetic field to come to this contour. So here we have the violation of this equation in relativistic case. Still this 'paradox' can be solved in terms of classical electrodynamics. But how? 2. Relevant equations [tex]\nabla \times \mathbf{E} = \frac{\partial \mathbf{B}} {\partial t}[/tex] [tex] \oint_{\partial S} \mathbf{E} \cdot \mathrm{d}\mathbf{l} =  \frac {1} {c} \frac {d} {dt} \int \mathbf {B} \cdot \mathrm{d} \mathbf {S}[/tex] 3. The attempt at a solution Maybe it takes time for propagation of alternating magnetic field from one end of the tube to the point A where crosssection of the tube by the plane of the contour is. So in fact it takes equal time to propagate inside the tube and outside the tube so it appears as simultaneously arising magnetic field at point A in the tube and electric field in the contour around the tube. But I'm not quite sure. 


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