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Energy dissipation in a travelling EM wave

  1. Jul 15, 2015 #1
    How is energy dissipated in a travelling electromagnetic wave ? Will there be any dissipation if it were to travel through vaccum ?
     
  2. jcsd
  3. Jul 15, 2015 #2
    There may be a slight redshift if it is travelling away from a large mass. But in a true vacuum with constant gravitational potential, there would be no energy dissipation.
     
  4. Jul 15, 2015 #3
    why a large mass is needed? How are electromagnetic waves affected by configuration of nearby masses or the gravitational potential at a point ?
     
  5. Jul 15, 2015 #4
    General Relativity explains it, in fact It was predicted by Albert Einstein before being observed, Gravity doesn't only change the wavelength of the wave, it does even bend it, the shift is due the constancy of c !!
     
  6. Jul 15, 2015 #5
    that is good, but how is the energy dissipated ?
     
  7. Jul 15, 2015 #6
    I wouldn't say that energy is dissipated, gravitationnal red/blue shifts are basically doppler shifts, they depend on the frame of reference (you can google equivalence principle), if you put a moving light source btw two people and ask them about what they record, one of them would say blueshift whereas the other will go for a redshift, but if you look from a far (the dumb way) you'll say that the energy lost from one side has transferred (magically) to the other, the "same" thing happens near massive planets, but it's just energy is a frame dependant .,
     
  8. Jul 15, 2015 #7

    davenn

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    That is incorrect

    the dissipation aka dropping off in strength is due to the inverse square law ... the dispersion of the wave
    It applies to any EM wave ... vacuum or no vacuum is irrelevent

    examples ....

    108834-004-B7B1A692.gif

    inverse-square-law.png



    Dave
     
  9. Jul 15, 2015 #8
    Don't confuse intensity with energy.
     
  10. Jul 15, 2015 #9

    davenn

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    Im not
     
  11. Jul 15, 2015 #10
    https://en.wikipedia.org/wiki/Dissipation

    Dissipation is the result of an irreversible process that takes place in inhomogeneous thermodynamic systems. A dissipative process is a process in which energy (internal, bulk flow kinetic, or system potential) is transformed from some initial form to some final form; the capacity of the final form to do mechanical work is less than that of the initial form. For example, heat transfer is dissipative because it is a transfer of internal energy from a hotter body to a colder one. Following the second law of thermodynamics, the entropy varies with temperature (reduces the capacity of the combination of the two bodies to do mechanical work), but never decreases in an isolated system.

    Losing intensity with distance because the area over which light spreads has increased is not dissipation.
     
  12. Jul 23, 2015 #11

    vanhees71

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    The gravitational redshift is not the same as a Doppler shift, which is due to the relative motion between light source and observer. The gravitational redshift is due to the interaction between the electromagnetic field and the gravitational field. The latter is described, within standard GR, as the curvature of space-time. It is not possible to completely disentangle the two effects, because this "disentanglement" is a frame-dependent issue. The math is the only way to make this clear. So here it is:

    In eikonal approximation you can calcuate the redshift as follows. You solve the eikonal equation
    $$g^{\mu \nu} \partial_{\mu} \psi \partial_{\nu} \psi.$$
    The characteristics of this partial differential equation give the direction of wave propagation, and thus
    $$k_{\mu}=\partial_{\mu} \psi$$
    is the frequency-wavenumber fourvector of the light wave. The eikonal equation implies that this is a light-like vector
    $$k_{\mu} k^{\mu}=0,$$
    and the "light rays" in the sense of geometrical optics given as the null-geodesics of space-time.

    A "pointlike observer" is uniquely and covariantly specified by giving his four-velocity, ##u^{\mu}##, which is a time-like vector. In natural units, where ##c=1##, you have ##u_{\mu} u^{\nu}=1##. Then the frequency of the light measured by this observer at his location in space-time is
    $$\omega_{\text{obs}}=k_{\mu} u^{\mu}.$$
    That's it, and this formula includes both the gravitational redshift for light emitted from a heavy object (like the sun) and the Doppler effect due to the relative motion between the observer and the light source. If the observer is at rest relative to the light source, you have a pure gravitational redshift; otherwise gravitational redshift and Dopplereffect are intermingled.

    Another special gravitational redshift is the Hubble redshift. Here you can define this specific redshift, because you can define it as the redshift of light emitted from a distant source as measured by a comoving observer who is defined as being at rest relative to the large-scale coarsegrained "cosmic substrate". Note that for such an observer the cosmic substrate and thus the light source are uniquely defined to be at rest. This is because of the symmetry of the underlying space-time model, the Friedmann-Lemaitre-Robertson-Walker pseudometric, which is the realization of the cosmic principle, i.e., the assumption that the local laws on the large-scale coarse-grained level are everywhere and always the same and that spacetime is thus maximally symmetric.
     
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