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Every object at a non-zero temperature radiates light, i.e. it glows. (Is that called blackbody radiation?)

What is the physical reason to this?

Is it because more heat implies that the atoms vibrate, and vibrating charges (the electrons or the nuclei) generate electromagnetic radiation?

SuperCiocia
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If the object has a temperature at absolute zero ( within the quantum uncertainties related to this statement) it means that all the atoms and molecules that compose it are at the lowest possible energy level.

Supplying energy to heat an object above absolute zero means increasing the kinetic energy of the component parts and raising them to higher energy levels.The degrees of freedom of atoms and molecules that compose a solid are vibrational and rotational . The energy supplied comes as electromagnetic radiation that is absorbed by kicking atoms/molecules to a higher energy level. Therefore yes, by heating they get to a higher energy level of vibrations and there is a probability that they will fall back to a lower energy and release a photon. (All first level interactions in an object are through the electromagnetic field).

This probability generates the black body radiation curve for a given temperature. The relaxation takes time, the electromagnetic interactions are fairly weak, but it is inexorable. A body will radiate away until its temperature reaches the ambient temperature and it is gaining as much energy from the environment as it is losing from the black body radiation.

anna v
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    And the mechanism behind the supply of energy is that this (radiation) energy is the sum of the action of photons. If one body is in energy balance, means has the same temperature as the objects around, and will be hidden by photons the body lose this energy again in the form of photons. And as you never will be in a situation of zero Kelvin for your body so the body in any time receive and radiate photons. – HolgerFiedler Jul 28 '14 at 05:13
  • Thanks. But certainly there are different materials with different emission spectra, which part of the Black body formula is distinctive and particular to the solid in question? Also, how do you know that the EM radiation shining on it is not increasing the solid's translational kinetic energy (eg via radiation pressure, unless we assume that the incident light isotropic?) – SuperCiocia Aug 04 '14 at 01:43
  • Ambient radiation is assumed isotropic, by definition of temperature. Directed radiation is another problem. The black body radiation curve is unique for a given temperature. http://en.wikipedia.org/wiki/Black-body_radiation . There exist modifications using emissivity and some assumptions to fit observed radiation curves for engineering purposes. So it is tables of emissivity that modify the black body spectrum for specific bodies http://en.wikipedia.org/wiki/Emissivity .see also http://en.wikipedia.org/wiki/Stephan-boltzmann_law – anna v Aug 04 '14 at 04:38
  • What about the part that makes the formula specific to a particular kind of solid/material? Or does it just take into account photons? But surely the energy levels must depend somehow on the object's constituents... – SuperCiocia Aug 05 '14 at 04:25
  • It is all covered with the emissivity number. The photons come from rotational and vibrational levels and these are practically continuous, the differences between them are very small. The emissivity takes care of these small differences in materials. – anna v Aug 05 '14 at 04:53
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Remember that when a crystal is cooled to absolute zero, the atoms don't stop moving, they still jiggle. Why? If they stopped moving, we would know where they were and that they had zero motion, and that is against the uncertainty principle. We can't know where they are and how fast they are moving, so they must be continually wiggling in there!$_1$

The atoms and molecules are composed of charged particles, i.e., protons and electrons, and kinetic interactions among matter particles result in charge-acceleration and dipole-oscillation. This results in the electrodynamic generation of coupled electric and magnetic fields, resulting in the emission of photons, radiating energy away from the body through its surface boundary.$_2$

So, every object at "any" temperature has certain probability of emitting photons.

Credits: $_1$*The Feynman Lectures on Physics-Vol.1Page-2-10* Page no's are subjected to change depending on editions.$_2$Wikipedia-Thermal raiation

Sensebe
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