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In a blackbody object the photons are reflected back and forward. If I heat up the the blackbody object with microwaves or other types of photons the electrons in the wall would vibrate. When the electrons absorbs the photons, does it take energy from the electron and becomes shorter in wavelength? And if you heat the blackbody even more, does the photons takes more energy from the electrons and gets even shorter in wavelength, leading to continuous spectrum ?

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2 Answers 2

In a blackbody object the photons are reflected back and forward

This is the wiki statement on the black body radiation

Black-body radiation is the type of electromagnetic radiation within or surrounding a body in thermodynamic equilibrium with its environment, or emitted by a black body (an opaque and non-reflective body) held at constant, uniform temperature. The radiation has a specific spectrum and intensity that depends only on the temperature of the body.

Considering that the number of molecules/atoms of a mole are ~10^23 the possibilities of interactions that would lead to photon radiation are enormous. If the object is solid, there are kinetic degrees of freedom within the solid vibrations and rotations and interactions due to this in the spill over fields of the atoms. The average temperature is connected statistically with the kinetic degrees of freedom

temperature

The spectrum of the emitted photons has to be continuous because of the overlap of the possible energy levels due to the large number of atoms. The same is true for liquids where also other degrees of kinetic freedom are involved. Gases also interact with collisions. In general various materials have modified formulas for black body radiation because the specific forms of kinetic energy of the atoms and molecules modifies the ideal curve.

If I heat up the the blackbody object with microwaves or other types of photons the electrons in the wall would vibrate.

Well, a kitchen stove blasts infrared radiation, and yes to start with the impinging photons will be absorbed by the external vibrational and rotational levels of the surface atoms. Then the kinetic energies will propagate to the bulk heating it to a higher temperature and thus the photons from the intrinsic electromagnetic interactions within the volume will on the average have higher energy.

When the electrons absorbs the photons,

The electron does not absorb photons. It interacts with them if free, or semi free as in metals, and will pick up some extra kinetic energy while the original photon is degraded, lower frequency and goes on to interact with further atoms and electrons. In non metals the photons interact with the whole molecule/atom transferring either the total energy, if an energy level is there, or part of it and then go on to interact with another .

does it take energy from the electron and becomes shorter in wavelength?

absolutely not. The incoming interacting photon can lose energy in interacting with the free electrons, ions, and the molecules/atoms as in compton scattering. It will transfer kinetic energy and become degraded lower frequency, less energy. It is the successive impacts of a lot of photons that raises the kinetic energies of the molecules/atoms to high temperatures and then the black body radiation can get harder than the incoming infrared by the internal interactions of the atoms/molecules of the object. They bounce against the fields of each other harder, and accelerating fields emit photons.

And if you heat the blackbody even more, does the photons takes more energy from the electrons and gets even shorter in wavelength, leading to continuous spectrum ?

This picture is wrong. The photons can be of low frequency and still the object get hot enough to emit black body with high frequencies ( energies, E=h*nu for photons). It is the additive effect of zillions of photons that heats up the object. Energy is conserved, the energy of the photons ends up as accumulated kinetic energy of the molecule/atoms, raising the temperature. ( see formula above)

Continuity of the spectrum comes from the high statistical number of interactions of the ~10^23 molecules per mole of the object. The vibrations and rotations in the field of each other are incoherent and the energies in total continuous.

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You never get continuous spectra -- Google "ultraviolet catastrophe" for details on the quantization of blackbody output.
Otherwise, your description is a bit naive but correct: a perfect black body absorbs every incoming photon and radiates photons at different wavelengths as described in the temperature vs. spectral density curves you probably have seen. There's no requirement that the photons be absorbed only via electron interaction. Phonon excitation, etc. is allowed as well.

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We never get mathematically continuous thermal spectrum only if it is derived based on statistics applied directly to EM modes in a box. This discreteness has nothing to do with the UV catastrophe or quantization - the spectra are discrete both in classical and quantum theory, if they are derived for box. If the box is expanded, the discreteness gets smaller. Alternatively, one can derive the Planck formula without any box, for example in the way Planck did it or as Boyer or Theimer did it, and the result is continuous spectrum. –  Ján Lalinský Feb 27 at 22:17

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