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I do understand how oscillating (accelerating) charges emits Electromagnetic waves through the propagation of kinks(back and forth) in the electric field of the oscillating charge which propagates at the speed of light.

But how does linearly accelerating charges emits "waves" as I can't see how a linearly accelerating charge emits waves which are periodic in nature since they move in a straight path. Also I thought of asking this question, since it its been said that accelerating charges emit EM waves. So does this apply in the case of linear acceleration too?

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    $\begingroup$ The linearly accelerating particle will not emit a pure sine wave, true. But neither does almost any antenna. $\endgroup$
    – Jon Custer
    Commented Jan 3 at 13:21
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    $\begingroup$ @Jon Custer How does the EM wave emitted from an antenna look like, I mean the shape? $\endgroup$
    – Jeffy James
    Commented Jan 3 at 13:25
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    $\begingroup$ see en.wikipedia.org/wiki/Li%C3%A9nard%E2%80%93Wiechert_potential $\endgroup$
    – anna v
    Commented Jan 3 at 13:29
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    $\begingroup$ In everyday language "wave" does almost always imply "waving" (i.e. oscillatory) but in physics technical language a "wave" can be a more general time-dependent form, not necessarily oscillating. A rough equivalent from everyday language is the "tidal wave". $\endgroup$
    – Andrew Steane
    Commented Jan 3 at 13:34
  • $\begingroup$ Yes, it does. For a non-relativistic case, see Larmor_formula for accelerated charge at low speed total power radiated in EM waves spectrum. $\endgroup$
    – Agnius Vasiliauskas
    Commented Jan 3 at 14:56

2 Answers 2

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Accelerating charge produces propagating EM field pattern in space whose intensity falls off with distance as function $1/r$. This field behaves as a wave in the sense it obeys the wave equation (any rigid moving pattern in space obeys it). It is not a wave in the colloquial sense of having peaks and troughs and nodes, it does not have a wavelength, and it does oscillate like the sine function.

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  • $\begingroup$ "it does not have a wavelength, and it does oscillate like the sine function" could you further elaborate this, and if possible could you give the wave equation for this particular field $\endgroup$
    – Jeffy James
    Commented Jan 3 at 20:24
  • $\begingroup$ @JeffyJames sorry I think you better read this in a book on EM theory, Griffiths or Feynman's lectures are nice. The wave equation here is the standard 3D wave equation of EM theory. $\endgroup$
    – Ján Lalinský
    Commented Jan 3 at 22:16
  • $\begingroup$ I'm referring to the standard retarded solution to the 3D wave equation with point source, this is formally a wave solution everywhere except at the source particle, because it obeys the 3D wave equation, but it is a dipole radiation field pattern with field strength monotonically decaying with distance, no oscillating pattern, because the particle does not oscillate. To get "usual" waves with peaks and troughs, you need the particle to perform oscillatory motion. $\endgroup$
    – Ján Lalinský
    Commented Jan 3 at 22:33
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For example, according to Larmor formula, electron freely falling in Earth gravitational field (assuming vacuum environment, i.e. no scattering), should radiate away about $10^{-52}~W$ in EM waves. This is of course a very tiny number, but technically is not zero. And because power adds from multiple sources if coherent,- if we assume that us reaches about $10^{23}$ electrons from solar winds each second, then this total power can increase to $10^{−29}~W$.

Now this would be just a radiation for electron acceleration in Earth gravity, which still is too small and probably obscured by electron scattering in atmosphere. However, we know that at first electrons from Solar winds enters Earth magnetosphere, where it experiences Lorentz force and due to which it's trajectory becomes curved. That's why we can see aurora borealis, because curved path of electron means stopping acceleration, which radiates EM waves due to the same Larmor (or relativistic) effect. Alhought it's technically not a linear path requested by post author, but Larmor principle doesn't care about acceleration roots.

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  • $\begingroup$ This does not answer the question - the Larmor formula is about energy per unit time radiated, but the question is not about energy at all. $\endgroup$
    – Ján Lalinský
    Commented Jan 3 at 16:10
  • $\begingroup$ Larmor relation explains that for charge to radiate EM waves,- it's enough for it to move in accelerated or decelerated fashion (linear or not). Which was the basic question "Do linear acceleration produces EM waves ?", no ? $\endgroup$
    – Agnius Vasiliauskas
    Commented Jan 3 at 17:08
  • $\begingroup$ No - the Larmor formula quantifies energy radiated, not that it is carried away by waves in the colloquial sense. Understanding the Larmor formula is further down the road, first one should understand what radiating waves means. Also, one can have radiating waves with zero energy radiated, there are cases where the Larmor formula does not apply, such as radiation of point particles. $\endgroup$
    – Ján Lalinský
    Commented Jan 3 at 19:04
  • $\begingroup$ waves with zero energy Can you expand on it more ?,- I though all waves has some energy at least. there are cases where the Larmor formula does not apply, such as radiation of point particles. Wiki says : "In electrodynamics, the Larmor formula is used to calculate the total power radiated by a nonrelativistic point charge as it accelerates". So you are against wiki in this case ? $\endgroup$
    – Agnius Vasiliauskas
    Commented Jan 3 at 19:23
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    $\begingroup$ The main problem with using the Larmor formula for point particles is that it gave generations of physicists the wrong idea that Poynting energy expressions apply generally, implying accelerating point particle too has to lose energy to radiation, and thus there has to be a self-force acting on the particle, saving local conservation of energy. No consistent model of this self-force was ever found (not for lack of trying), but people moved on to quantum theory and only few realized the problem was with the very energy interpretation of the Poynting expressions and the Larmor formula. $\endgroup$
    – Ján Lalinský
    Commented Jan 3 at 22:13

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