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Reference frames in Magnetism

  1. Jul 23, 2009 #1
    What if i was travelling in a car with an e- with me. W.r.t to a positive charge at rest , it would seem accelerating to me i.e. it would have a magnetic field but acc. to a person on the ground it would just have an electric field. If there were a car moving relative to me , the person would know that my electron has an electric field but if it were for the person on the ground , he would say it has a magnetic field too.
    How to get around this dilemma?
  2. jcsd
  3. Jul 23, 2009 #2
  4. Jul 23, 2009 #3
    But , that seems a bit confusing.
    I dont get that level of mathematics.
  5. Jul 23, 2009 #4


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    Staff: Mentor

    In relativity, the electric and magnetic fields are two aspects of a single entity, the electromagnetic field. It's described by a 16-component tensor whose components are the components of the electric and magnetic fields. Observers who move relative to each other see the components as sort of "rotated" among each other. They see different electric and magnetic fields, but the net physical effect is the same for all observers.
  6. Jul 23, 2009 #5
    You've hit the nail right on the head. They're the same thing. If you actually did the calculation from either reference frame you'd get exactly the same predicted result (say the person in the car tossed the e- in the air or some such) but it's simply semantics whether person A says it is an electric phenomena or person B says it is an induced magnetic phenomena. One of the lucky conveniences for Einstein when developing the special theory of relativity was that maxwell's equatiosn (the heart of classical E&M) we ALREADY compliant with the idea of refernces frames. Do the calculations in any reference frame and they'll be consistent with any other reference frame. You might, in later E&M endeavours, consider the perspective of things if say you were sitting ON an electron traversing an electrical circuit or somesuch. In terms of results you'd make identical predictions but you'd see the reason for these results very differently.
  7. Jul 24, 2009 #6
    So Maxwell could see this through his equations?
  8. Jul 24, 2009 #7
    Sorry for not getting back to you earlier. You are correct, the observer sees both an electric field and a magnetic field. See the 4 equations at bottom of page 1 in
    The unprimed E and B are electric and magnetic fields as seen by you in the car with the charge. The charge has only an electric field so B = 0. v= your velocity relative to the observer in the primed reference system: E' and B'.. The formulas separate the E and B fields into two components: the longitudinal fields EL and BL, and the transverse fields ET and BT.
    The most important relevant equation for you is

    B'T = γ[BT - (1/c2) v x E]

    γ is the relativistic factor 1/sqrt[1-(v/c)2] which equals1 in your case, and v x B is the vector cross product of the velocity v and the electric field E. The result of the v x E is a vector perpendicular to both v and E, and in this case has to be completely azimuthal. So the observer sees both an electric field (equations 1 and 2) and the azimuthal B field given by equation (4).

    α β γ δ ε ζ η θ ι κ λ μ ν ξ ο π ρ ς σ τ υ φ χ ψ ω .
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