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I Can we apply relativistic field transformations to electromagnetic waves?

  1. Jul 27, 2017 #1
    If we move towards a source of EM waves, in our reference frame the frequency appears (and it is) higher than what a stationary observer will see due to Doppler effect. The field transformations show that these two observers will se static fields differently so I would also expect that the peak amplitudes will rise in the moving reference frame. Also the wave momentum should increase and total energy of an EM pulse. This is what happens when we analyse objects. If instead of a source there is something that launches an object, in the stationary reference frame the object will have a certain momentum and energy, and in the moving reference frame the energy and momentum will be higher, with both frames being equivalent and valid.
     
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  3. Jul 27, 2017 #2

    Orodruin

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    The field is only stationary in one of the frames.
     
  4. Jul 27, 2017 #3

    Dale

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  5. Jul 27, 2017 #4
    Amplitude of EM waves is constant in coordinate transformation, I think.
     
  6. Jul 28, 2017 #5
    Momentum of a matter-antimatter bomb is ##\gamma##m0v

    So that's the momentum of the radiation that is produced when the bomb explodes turning completely into radiation.

    So velocity change ##\Delta##v causes a change of momentum ##\Delta##v * ##\Delta## ##\gamma##

    where ##\Delta##v is the change of the velocity of the bomb, at low speeds we can approximate
    ##\Delta##vobserver = ##\Delta##vbomb


    (A beam of photons can be though as the remnants of a very fast moving matter-antimatter bomb. An expanding cloud of photons can be thought as the remnants of a slow moving matter-antimatter bomb )

    (It was a fast moving, long rod shaped matter-antimatter bomb that produced the beam of photons)
     
    Last edited: Jul 28, 2017
  7. Jul 28, 2017 #6

    vanhees71

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    Let's first look at a scalar field since it has the most simple transformation properties. As the most simple example consider a free Klein-Gordon field fulfilling the Klein-Gordon equation
    $$(\Box+m^2)\phi(x)=0,$$
    where
    $$\Box=\partial_{\mu} \partial^{\mu} = \partial_t^2-\vec{\nabla}^2.$$
    I use the west-coast convention of the metric ("mostly minus") and set ##c=1##.

    The transformation property of the scalar field under a Lorentz transformation ##x'=\hat{\Lambda}x## is
    $$\phi'(x')=\phi(x)=\phi(\Lambda^{-1} x').$$
    Now consider a plane wave solution, i.e., we make the ansatz
    $$\phi(x)=\phi_0 \exp(-\mathrm{i} k \cdot x).$$
    Plugging this into the Klein-Gordon equation gives the dispersion relation
    $$k^2-m^2=0 \; \Rightarrow \; k^0=\omega=\sqrt{\vec{k}^2+m^2}.$$
    The ##k## is a Minkowski four-vector (since the dispersion relation is a covariant equation!). Thus you have
    $$\phi'(x')=\phi(\hat{\Lambda}^{-1} x')=\phi_0 \exp(-\mathrm{i} k \cdot \hat{\Lambda}^{-1} x').$$
    Now Minkowski products are scalar, i.e., you can write
    $$k \cdot \hat{\Lambda}^{-1} x' = (\hat{\Lambda} k) \cdot (\hat{\Lambda} \hat{\Lambda}^{-1} x')=(\hat{\Lambda} k') \cdot x',$$
    i.e., you get again a plain wave in the new frame with the wave four-vector
    $$k'=\hat{\Lambda} k$$
    Of course it also fulfills the dispersion relation, as it must be:
    $$(k')^2=(\hat{\Lambda} k)^2=k^2=m^2.$$
    Obviously the amplitude is a scalar, i.e., it doesn't change:
    $$\phi_0'=\phi_0.$$
    Now take a boost in ##x## direction
    $$t'=\gamma (t-v x), \quad x'=\gamma(-v t+x), \quad y'=y, \quad z'=z.$$
    As we've seen the wave vector transforms in the same way, i.e.,
    $$k^{\prime 0}=\omega'=\gamma(\omega-v k_x), \quad k_x'=\gamma(-v \omega+x), \quad k_y'=k_y, \quad k_z'=k_z.$$
    This formulae include the Doppler effect, i.e., the frequency changed as measured in the new frame compared to the old frame and the socalled "aberration", given by the change of the spatial components.

    For the electromagnetic field it's quite similar. You have only to consider that it is a vector field, given by the 2nd-rank Faraday tensor, i.e., the transformation properties are
    $$F^{\prime \mu \nu}(x')={\Lambda^{\mu}}_{\rho} {\Lambda^{\nu}}_{\sigma} F^{\rho \sigma}(x)={\Lambda^{\mu}}_{\rho} {\Lambda^{\nu}}_{\sigma} F^{\rho \sigma}(\hat{\Lambda}^{-1}x').$$
    If you work this transformation out in terms of the usual representation of the field in terms of electric and magnetic components ##\vec{E}## and ##\vec{B}## you get the transformation explained in

    https://en.wikipedia.org/wiki/Class...rmation_of_the_fields_between_inertial_frames

    If you put in plane waves you get of course massless dispersion relations, ##k^2=0 \; \Rightarrow \; \omega^2=\vec{k}^2##, and the Doppler shift and aberration formulae are the same as for scalar fields since ##k## is of course again a four vector. The polarization properties, of course also change according to the formulae given in the Wikipedia article.

    The em. field has two invariants (scalar and pseudoscalar). In conventional formulation they are given by ##\vec{E}^2-\vec{B}^2## and ##\vec{E} \cdot \vec{B}##. In covariant formulation these come from the scalar ##F_{\mu \nu} F^{\mu \nu}## and the pseudoscalar ##\epsilon_{\mu \nu \rho \sigma} F^{\mu \nu} F^{\rho \sigma}##. For plane waves both invariants are 0.
     
  8. Jul 28, 2017 #7

    Dale

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    Please show your math on this
     
  9. Jul 28, 2017 #8
    I was wrong. E^2 and H^2 are proportional to energy density W of EM field which is transformed as
    [tex]W=W'\frac{(1+\frac{V}{c}cos\alpha')^2}{1-\frac{V^2}{c^2}}[/tex] Thanks for your suggestion.
     
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