How things radiate electromagnetic radiation? I don't ask why they radiate (higher temperature than 0K) but how they radiate this electromagnetic waves?
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$\begingroup$ Any electric charge create electric field. Hint: what happens when the charge is moving? $\endgroup$– user5402Jul 12, 2013 at 18:07
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$\begingroup$ Are you asking about blackbody radiation? $\endgroup$– NeutronStarJul 12, 2013 at 18:19
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$\begingroup$ @metacompactness: But even the "non-charged" particles do radiate electromagnetic radiation. $\endgroup$– HakimJul 12, 2013 at 18:24
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$\begingroup$ The radiation is duo to the charged particles (like electrons) in matter. $\endgroup$– MostafaJul 12, 2013 at 18:27
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2$\begingroup$ The question is vague. I don't understand what's being asked. If the dialog in comments has helped the OP to refine the question, then the OP should edit the question appropriately. Voting to close as unclear. $\endgroup$– user4552Jul 12, 2013 at 19:29
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3 Answers
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There are two ways of getting electromagnetic radiation from matter.
Matter is usually neutral, the electrons and protons are equal in number to each other and any fields are spill over giving rise to Van der Waals forces which bind neutral atoms into molecules etc.
At this micro level nature is quantum mechanical. That means that all electrons are in energy levels some of which energy levels are practically a continuum, i.e. the difference between them is very small. This means that vibrations of the atoms and molecules in their solid structure, as an example, will excite by kinematics these levels and fall back by the emission of a photon ( de-excitation); the ensemble of these photons gives rise to black body radiation. When the temperature is high the corresponding energy levels have larger gaps, and the photons are of higher energy.
A filament lamp has high enough temperature to emit visible light . Liquids have similar behavior, gases only have molecular energy levels and vibrations but the process is the same. Kinetic energy from temperature is transformed into photons from de-excitations .
The bulk of light we see comes from this mechanisms, even the light from the sun.
There are the LED lights, again a quantum mechanical effect, but of different origin:
"when electrons cross the junction from the n- to the p-type material, the electron-hole recombination process produces some photons in the IR or visible in a process called electroluminescence."
The second way of getting light is how the other answers state, by accelerating charges, ions and electrons, as in sparks and lightning, plasma etc.
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Electromagnetic Radiation is a time-changing component of an electric (and magnetic) field in the direction transverse to the direction of propagation. Generally, this radiation is emitted by accelerating charges.
Consider placing a positive point charge in a vacuum; electric field lines spread out radially from its position at the speed of light, c. There is only a radial component to the field.
If the particle starts moving moving with a constant velocity, then the field lines will shift and this shift will propagate away from the particle at a speed c. There is now a transverse component to the field as well as a radial one, but it's constant in time (ignoring the initial kink, of course). The moving charge generates a current, so we also have a magnetic field, but it is constant in time.
If the particle is accelerating, then the transverse component becomes time-changing. we also have a time-changing current, so the magnetic field changes in time too.
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$\begingroup$ Everything with a temperature more than 0 Kelvin radiate electromagnetic radiation but if something is 0 Kelvin then it will not radiate EM radiation. But the electron move even at 0 Kelvin so it will radiate following the logic EM but it actually doesn't. $\endgroup$– HakimJul 12, 2013 at 19:16
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Electromagnetic radiation is a result of the movement of charges.
If you take a look at this lecture by Dr Lewin, he has a nice visualization on how waves are formed.
It can also be a result of blackbody radiation
answered Jul 12, 2013 at 18:30
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$\begingroup$ Isn't the latter a special case of the former? $\endgroup$– user5402Jul 12, 2013 at 18:53