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I Applicability of E and B field Transformations

  1. Oct 31, 2016 #1
    I have a question regarding these transformation formulas:

    ##\begin{align}
    & E'_x = E_x & \qquad & B'_x = B_x \\
    & E'_y = \gamma \left( E_y - v B_z \right) & & B'_y = \gamma \left( B_y + \frac{v}{c^2} E_z \right) \\
    & E'_z = \gamma \left( E_z + v B_y \right) & & B'_z = \gamma \left( B_z - \frac{v}{c^2} E_y \right). \\
    \end{align}##

    Suppose we have E and B vector fields in one frame, and we wish to transform them in another frame.

    To what extent are the above formulae applicable? Can they be applied to E and B fields of any shape or form, or are they applicable only under certain circumstances?
     
  2. jcsd
  3. Oct 31, 2016 #2

    dextercioby

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    They are applicable in the same manner that Special Relativity is.
     
  4. Oct 31, 2016 #3
    Nooo, you have to stop talking like that.
     
  5. Oct 31, 2016 #4

    dextercioby

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    OK, let me elaborate, then :) Those formulas are particular cases of the very general one:

    [tex] F'^{\mu\nu}(x) = \Lambda^{\mu}_{~\alpha}\Lambda^{\nu}_{~\beta} F^{\alpha\beta}(x)[/tex]

    if you plug in a particular form of the Lorentz transformation. You can also consider the GR transformation formulas for the double contravariant components of the em tensor if you want. Just replace ##\Lambda## by the partial derivatives.
     
  6. Oct 31, 2016 #5
    So you were just trolling? ?:) Damn.

    What if we consider flat spacetime only?
     
  7. Oct 31, 2016 #6

    robphy

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    This assumes a boost along the x-direction.
     
  8. Oct 31, 2016 #7
    Yeah. But can they be applied to E and B fields of any shape or form, say spherical wavefronts? Or only to plane waves?
     
  9. Oct 31, 2016 #8

    dextercioby

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    No trolling, just wanting you to do some thinking/textbook research by yourself. The Lorentz transformations are applicable to flat space-time. There's a theorem in differential geometry (proved in an archaic version by Riemann himself) which states that if a space-time manifold is globally flat, then at each point of it you can set the components of the metric tensor to +1 or -1. With this choice of metric, you necessarily have SR with its Lorentz transformations between inertial frames.

    The shape of the wavefronts is not important. What is important is how the system of coordinates is chosen.
     
  10. Oct 31, 2016 #9

    robphy

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    Yes... apply it at each point in space [in an inertial frame] to an arbitrary E and B field.
    For a plane wave, it is simple because of the symmetry of that configuration.
    In other configurations, it is more complicated.

    If one is trying to highlight a general feature, it might be good to choose a highly symmetrical situation first... or else your message might be clouded by the complications of a less symmetric choice. Certainly, you can [with more work] show that the feature persists in a less symmetrical situation.
     
  11. Oct 31, 2016 #10

    PeterDonis

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    To the extent that you are applying a Lorentz boost in the ##x## direction. More general Lorentz transformations required more complicated formulas.
     
  12. Oct 31, 2016 #11

    dextercioby

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  13. Oct 31, 2016 #12
    In that case might I direct you to an earlier thread of mine: (post #31 onwards, full convo also posted below)
    https://www.physicsforums.com/threa...vs-momentum-models.890533/page-2#post-5607457

    Note that the image posted in that thread has the fields as a function of (x) and (x'). That wasn't my intention, but was in the image I plucked from the net. I'm still thinking about the question I posed in #31.


     
  14. Oct 31, 2016 #13

    PeterDonis

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    That conversation is about whether the fields, in a given frame, are a function of ##x## only or of all the spatial coordinates.

    What I posted earlier in this thread is about what kind of Lorentz transformation you have to make to transform from one frame to another. That's a separate question.
     
  15. Oct 31, 2016 #14
    Sorry, I guess my question wasn't stated specifically enough. :oops:

    So are the fields a function of ##x## only or of all the spatial coordinates?
     
  16. Oct 31, 2016 #15

    PeterDonis

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    It depends on the situation.
     
  17. Oct 31, 2016 #16

    robphy

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    In physics, a vector field is an assignment of a vector to every point in space.
    It may turn out that because of certain symmetries, there is no variation in the vectors when one moves along (say) the y-axis.
    So, we can say, in that case, that "it doesn't depend on [the specific value of] the y-coordinate"... but it's still technically a function of all of the spatial coordinates.
     
  18. Oct 31, 2016 #17
    do you want to move back to the older thread?
     
  19. Oct 31, 2016 #18

    PeterDonis

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    Regarding the spherical wavefront situation you were asking about in the older thread, your question was already answered by Dale.
     
  20. Oct 31, 2016 #19
    Are these two sets of transformations considered identical, or is there a slight difference?

    ##\begin{align}
    & E'_x = E_x & \qquad & B'_x = B_x \\
    & E'_y = \gamma \left( E_y - v B_z \right) & & B'_y = \gamma \left( B_y + \frac{v}{c^2} E_z \right) \\
    & E'_z = \gamma \left( E_z + v B_y \right) & & B'_z = \gamma \left( B_z - \frac{v}{c^2} E_y \right). \\
    \end{align}##

    proxy.php?image=http%3A%2F%2Fwww7b.biglobe.ne.jp%2F~kcy05t%2Fsiki%2Felein%2Feeb3.gif
     
  21. Oct 31, 2016 #20
    I do apologize for being vexing, I am easily confused, so I spend a lot of effort trying to get to the bottom of things.
     
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