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Hall effect and relativity

  1. Aug 1, 2013 #1
    In the regular hall effect we calculate like this
    F=qE=qvB then E=vB , now we assume our conductor has width d so we multiply both sides d.
    then Ed=vBd=V so now our Hall voltage is vBd v is the speed of the charge carriers and B is the magnetic field. But what if our charge carriers were moving at relativistic speeds? How would we correct our derivation. Could I just write my initial velocity in terms of momentum p=mv and then use
    relativistic momentum.
  2. jcsd
  3. Aug 1, 2013 #2
    I am not certain that I can answer your question but in the 'regular Hall effect' as you call it, the velocity of charge carriers is always going to be very much less than c.
    In a metal carrying a current v is of the order of mm/s so there is no real problem in the 'regular hall effect'
  4. Aug 2, 2013 #3


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    Relativity applies at ALL speeds. The effects are not always obvious, at first glance but, for example, you can explain Magnetism in terms of the relativistic effects on the moving charges in any conductor and reduce it to a simple Electric force.
    Extending your question: you could replace the charge carriers, moving in a solid, with high speed electrons in a vacuum and you would then be dealing with charge carriers that are, in fact, moving at high enough speeds to be regarded as 'relativistic', in the conventional sense. (Ha - relativistic and conventional in the same sentence.) You would be dealing with what's effectively, a high energy cyclotron situation - which is dealt with in many text books.
  5. Aug 3, 2013 #4


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    Maxwell's equations are fully relativistic. (Maxwell's equations are invariant under the Lorentz transform).

    So the answer is probably yes, though I don't know a lot about the quantum end of things.
  6. Aug 3, 2013 #5


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    You might want to take a look at http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html#Length_Contraction, from the FAQ on experimental effects of special relativity linked at the top of this forum.
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