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E&M and relativity

  1. Aug 11, 2013 #1
    I remember when I took E&M my teacher said that special relativity was built into Maxwells equations.
    Lets look at a line charge moving at non-relativistic speeds first.
    We use amperes Law to find the B field.
    frame 1: [itex] B=\frac{\mu_0 v \lambda }{2 \pi r } [/itex]

    frame 2: relativistic speed [itex] B'=\frac{\mu_0 \lambda v' \gamma }{2 \pi r } [/itex]
    in frame 2 the line charge becomes length contracted so the gamma factor takes care of that.
    v and v' are the speeds relative to the moving charge.
    Now if relativity was built into maxwells equations why did I need to correct for frame 2.
    Or if I knew the charge per length in frame 2 at speed v' then would I calculate it just like frame 1.
    If I know the correct charge density at any speed do I just calculate it normally with amperes law and there is no need for a relativistic correction.
    I know there is an E field in each frame but Im just looking at the B field.
     
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  3. Aug 11, 2013 #2

    tiny-tim

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    hi cragar! :smile:
    ah, that's where you're missing the point …

    E&M isn't just E and M

    the E&M field is one field, and it transforms (as a 2-form) according to the lorentz transformation, mixing up the E and M bits

    that mixing by the lorentz transformation is what is built into maxwell's equations :wink:
    you measure velocity differently in each frame, so what is strange about measuring distance differently in each frame? :smile:
     
  4. Aug 11, 2013 #3
    ok ya I should pay attention to both E and B cause in some frames processes are Electric and magnetic.
    Ok I understand how if v changes in different frames then so will the length contraction.
    I still don't see why relativity is built into E&M yet. I do know that you can derive c from the wave equations of E and B.
     
  5. Aug 11, 2013 #4

    tiny-tim

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    from the pf library on Maxwell's equations

    Changing to units in which [itex]\varepsilon_0[/itex] [itex]\mu_0[/itex] and [itex]c[/itex] are 1, we may combine the two 3-vectors [itex]\mathbf{E}[/itex] and [itex]\mathbf{B}[/itex] into the 6-component Faraday 2-form [itex](\mathbf{E};\mathbf{B})[/itex], or its dual, the Maxwell 2-form [itex](\mathbf{E};\mathbf{B})^*[/itex].

    And we may define the current 4-vector J as [itex](Q_f,\mathbf{j}_f)[/itex].

    Then the differential versions of Gauss' Law and the Ampère-Maxwell Law can be combined as:

    [tex]\nabla \times (\mathbf{E};\mathbf{B})^*\,=\,(\nabla \cdot \mathbf{E}\ ,\ \frac{\partial\mathbf{E}}{\partial t}\,+\,\nabla\times\mathbf{B})^*\,=\,J^*[/tex]

    and those of Gauss' Law for Magnetism and Faraday's Law can be combined as:

    [tex]\nabla \times (\mathbf{E};\mathbf{B}) = (\nabla \cdot \mathbf{B}\ ,\ \frac{\partial\mathbf{B}}{\partial t}\,+\,\nabla\times\mathbf{E})^*\,=\,0[/tex]​

    the last two equations are clearly lorentz covariant! :smile:

    (to see how to transform (E;B), write it as …

    ##E_x\mathbf{x}\wedge\mathbf{t} + E_y\mathbf{y}\wedge\mathbf{t} + E_z\mathbf{z}\wedge\mathbf{t} + B_x\mathbf{y}\wedge\mathbf{z} + B_y\mathbf{z}\wedge\mathbf{x} + B_z\mathbf{x}\wedge\mathbf{y}##​

    and then use the lorentz transformation on the individual x y z and t , together of course with ##\mathbf{x}\wedge\mathbf{x} = 0, \mathbf{x}\wedge\mathbf{y} = -\mathbf{y}\wedge\mathbf{x}##)
     
  6. Aug 11, 2013 #5

    WannabeNewton

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    It is a manifestly Lorentz covariant field theory. All you have to do is take the standard Maxwell equations written using vector calculus, define the electromagnetic field strength tensor, and rewrite the equations using the space-time derivative operator to put it into a form that is manifestly Lorentz covariant.
     
  7. Aug 11, 2013 #6
    ok thanks, wouldn't a lot of Newtonian mechanics be Lorentz covariant. like momentum.
     
  8. Aug 12, 2013 #7

    BruceW

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