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Em field 6 degrees of freedoms & 4-potential

  1. Dec 11, 2013 #1


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    What does it mean that the electromagnetic field strength includes six degrees of freedom (three for the electric field strength, three for the magnetic field strength), whereas the four-potential includes only three degrees of freedom (two for the photon, one for the Higgs boson)? What is the four-potential? Why isn't it adjusted to be able to encompass 6 degrees of freedom too?
  2. jcsd
  3. Dec 11, 2013 #2
    You don't really have 6 degrees of freedom... You can't just arbitrarily pick all the electric and magnetic field components and expect them to satisfy Maxwell's equations. As a simple example, suppose [itex]E_x(x,y,z,t) = cos(\omega t + k z) [/itex] and set everything else to zero for all space and time ([itex]E_y = E_z = B_x = B_y = B_z = 0[/itex]). Those particular electric and magnetic fields are not allowed by Maxwell's equations. Specifically, if you plug them into Faraday's law
    [tex]\vec{\nabla}\times\vec{E} = -\frac{\partial \vec{B}}{\partial t}[/tex]
    you'll find that the left hand side is non-zero while the right hand side is zero, which is obviously a contradiction. There's no charge density/current density combination I can pick which will give me those particular fields.

    So you don't really have 6 degrees of freedom because not all combinations of [itex](E_x, E_y, E_z, B_x, B_y, B_z)[/itex] are allowed. They need to satisfy Maxwell's equations.
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