Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Variance of the EM wave equation under Galilean transformation

  1. Jun 8, 2017 #1
    • Two very similar threads were merged. Some posts may seem redundant
    For using Galilean transformation, I have to assume that speed of light w.r.t. ether frame is c.
    W.r.t. ether frame,
    E = E0 eik(x-ct)

    W.r.t. S' frame which is moving with speed v along the direction of propagation of light,
    E' = E0 eik(x'-c't')
    Under Galilean transformation,
    x' = x-vt,
    t' = t,
    c' = c -v
    So, (x'-c't') = x -ct
    thus, E' = E
    i.e. under the Galilean transformation, em wave equation is invariant.

    Is this correct?
    I am asking it here instead of creating a new thread, because I think the topic is similar.
    If it is not allowed, sorry for it.
     
  2. jcsd
  3. Jun 8, 2017 #2

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    No, it isn't, because the speed of the wave changes. You go from a wave moving at speed ##c## to a wave moving at speed ##c - v##.

    (You are also assuming that the wavenumber ##k## does not change. I'm not sure that assumption is justified.)
     
  4. Jun 8, 2017 #3
    Then, x also changes from x to x - vt. And the two changes get canceled in the wave equation.
    Assuming that the frequency of an em wave depends on the source generating em wave.
    So, the frequency will remain same w.r.t. both reference frames.
    Then, ω=ω',
    k ≠k'.
    k = ω/c,
    k' = ω/(c -v)
    E = E0ei(kx-ωt) = E0eiω(x/c - t)
    E' = E'0eiω((x-vt)/(c -v) - t)
    E' ≠ E
    Is this correct?
    ,
     
  5. Jun 9, 2017 #4
    Any em wave could be written as a combination of plane waves.
    So, if I show the plane wave solution of em wave equation to be variant under Galilean transformation and then using superposition principle, I can say that any solution of em wave equation is variant under Galilean transformation.
    Will this be a proof of variance of em wave equation under Galilean transformation?


    For using Galilean transformation, I have to assume that speed of light w.r.t. ether frame is c.
    W.r.t. ether frame,
    E = E0 eik(x-ct)

    W.r.t. S' frame which is moving with speed v along the direction of propagation of light,
    E' = E0 eik'(x'-c't')

    Under Galilean transformation,
    x' = x-vt,
    t' = t,
    c' = c -v
    So, (x'-c't') = x -ct

    Assuming that the frequency of an em wave depends on the source generating em wave.
    So, the frequency will remain same w.r.t. both reference frames.
    Then, ω=ω',
    k ≠k'.
    k = ω/c,
    k' = ω/(c -v)
    E = E0 eiω(x/c - t)
    E' = E'0 eiω((x-vt)/(c -v) - t)
    E' ≠ E
    Is this correct?
     
  6. Jun 9, 2017 #5

    Dale

    Staff: Mentor

    The question isn't if E' is different from E. The question is if it satisfies Maxwell's equations in both frames.
     
  7. Jun 9, 2017 #6
    Do you mean that I have to check whether E satisfies Maxwell's equations(or Maxwell's wave equations ?) in ether frame and E' satisfies Maxwell's equations in S' frame?
    They have to (as I have taken them as solutions of Maxwell's wave equations in their respective frames).
     
  8. Jun 9, 2017 #7

    Dale

    Staff: Mentor

    Yes (and B)
     
  9. Jun 9, 2017 #8
    When you say Maxwell's equations, you mean Maxwell's equations of electrodynamics or Maxwell's wave equation?
     
  10. Jun 9, 2017 #9

    Dale

    Staff: Mentor

    This one. When I say "Maxwell's equations" that is always what I mean.
     
    Last edited: Jun 9, 2017
  11. Jun 9, 2017 #10

    Nugatory

    User Avatar

    Staff: Mentor

    Both. The wave equation is derived from the electrodynamic ones. That's why we say that a plane wave moving with speed c is a solution of Maxwell's equations and a plane wave moving at any other speed is not.

    The wave equation, from Maxwell's equations, is ##(c^2\frac{\partial^2}{\partial{x}^2}-\frac{\partial^2}{\partial{t}^2})E=0##.

    ##E=E_0e^{ik(x-ct)}## is a solution to that equation. ##E=E_0e^{ik(x-c't)}## where ##c'=c-v## is not.
     
  12. Jun 10, 2017 #11

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    The problem is to define (!) Galileo transformations on the quantities in Maxwell's equations such that these equations stay form-invariant. That turns out to be impossible, and that's how Einstein found the modern interpretation of Lorentz transformations, i.e., that observations clearly show that the Galilei-Newton spacetime is inaccurate and Einstein-Minkoski spacetime is a much better approximation, which is only invalidated by gravity, which needs General Relativity with its completely changed view of spacetime as a dynamical entity.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Variance of the EM wave equation under Galilean transformation
  1. EM wave equations (Replies: 19)

Loading...