every time an object accelerates it accelerates along with itself its constituent electrons and protons both of which being charged particles must generate their own set of EM waves because they are accelerated. However in the real world we don't see visible (As in detectable) EM waves. One way this can be explained is that the EM waves produced by the positive nucleus and the negative electron cancel out each other. This explanation seems (Frankly) lame and kind of unbelievable. Is there a better reason for this. or the waves are too tiny to be ever detected
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$\begingroup$ A neutral object do not emit any electromagnetic radiation. Charged particles do. EM waves produced by electrons and nucleus do not cancel each other. EM waves are just energy. How they get cancelled? $\endgroup$– UKHJan 19, 2017 at 14:12
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$\begingroup$ @Unnikrishnan I am talking about the individual charged particles (protons and electrons) inside the real world object $\endgroup$– Suhrid MulayJan 19, 2017 at 14:14
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$\begingroup$ I understand that. Every material objects are made of charged particles. Your question is "Why doesn't an accelerating object in real life always generate EM waves?". the "object" if neutral, will not emit radiation. There should be a net charge (and hence an electromagnetic field) to create EM waves. I think you haven't understood what "net" charge means $\endgroup$– UKHJan 19, 2017 at 14:17
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$\begingroup$ See the Poynting's theorem: web.mit.edu/6.013_book/www/chapter11/11.2.html $\endgroup$– UKHJan 19, 2017 at 14:18
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$\begingroup$ No but shouldn't the individual protons and electrons accelerate with the object because they simply are a part of it. $\endgroup$– Suhrid MulayJan 19, 2017 at 14:19
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2 Answers
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Look at J. J. Thomson's and Edward Purcell's beautiful explanation of radiation by accelerating charge (summarized below). The acceleration causes a "kink" between two electrostatic field distributions that travels outwards at $c$.
Now, if you have a cancelling charge almost on top of one of opposite sign, the dipole field resulting drops of much faster than $1/r^2$, i.e. as $1/r^3$. The electrostatic fields of opposing, neighboring charges really do cancel out in this way. Work out the static distribution from a dipole and see for yourself.
But if there is a much lower static field, then it follows, through the Thomson/ Purcell's argument, that the radiation field - the kink - has to be much smaller too if the opposite charges are accelerated together, so that their separations do not change.
Thomson / Purcell Reasoning Summary
J. J. Thomson's gave us this wonderfully elegant description: we imagine a stationary charge, whose equilibrium electric field line distribution is the radial lines outside the circles in this image.
Now the charge begins to move uniformly suddenly, making its steady state field line distribution look like the field inside the circles in the image above. But the changes to the field can only propagate outwards at a maximum speed of $c$ (special relativity), and because the field lines cannot break (since there is no charge in the diagram aside from the accelerated one, and Gauss's law tells us that lines can only terminate on a charge), we must have a configuration like that in the diagram where there is a transition region between the two circles where the E-field is bent nonradially to link the two steady state configurations. This outwardly running kink is the radiation.
Edward Purcell uses this visualization to derive the Larmor radiation formula.
See:
Hat tip to user Ján Lalinský for pointing out an error in my history - I had always thought it was Purcell who came up with the travelling kink in the field visualization, but in fact it was Thomson. Purcell used the visualization in Larmor formula derivation.
answered Jan 19, 2017 at 14:24
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$\begingroup$ The visual explanation of radiation as kinks in electrostatic field was introduced by J. J. Thomson, the discoverer of electron. See for example physics.buffalo.edu/phy514/w02 $\endgroup$ Apr 20, 2017 at 9:33
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$\begingroup$ Thanks @JánLalinský - MTW indeed state pretty clearly that it was J. J. Thomson - I'd trust them on matters of history. See updates. I'm kind of surprised I'd never noticed my error before, because I must have read that bit of MTW dozens of times ... $\endgroup$ Apr 20, 2017 at 10:14
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It is too bad that explanation seems lame or unbelievable. It is correct for objects with equal numbers of positive and negative charges.
For charged objects, another reason is that the accelerations are too small and too slow. The frequency of visible light is 430 - 770 THz (https://en.wikipedia.org/wiki/Visible_spectrum) You would have to vibrate the object that fast to generate visible light.
When Maxwell first realized that light was an electromagnetic wave back in the 1800's, he was puzzled by this. This was before the atomic structure of matter was known. Hot objects glow, which means something vibrates that fast when heated. But he could not imagine what it was.
answered Jan 19, 2017 at 14:19