When/How does thermal radiation stop (if it stops) and conduction starts?

Question

Consider two solid objects in the vacuum (of different materials, if you will) at different temperatures approaching each other until they make "perfect contact" through flat surfaces (no gaps or defects, so that thermal contact conductance effects are absent, even though interfacial thermal resistance may still be present). Of course, before making contact, they exchange heat via electromagnetic radiation. My question is: is this heat exchange via thermal radiation still present once the bodies are making contact with each other and exchanging heat via thermal conduction? (Please explain what's going on in terms of the constituent particles of the bodies.)

Answer 1

Background: Assume a volume of particles in a medium with a thermal distribution of kinetic energies. In addition to the transfer of kinetic energy between particle 1 and 2 during collisions, thermal radiation is also produced by the acceleration and deceleration of atoms and molecules during each collision.

Emphasis: Thermal radiation is not due to the velocity of the thermal kinetic energy of atoms/molecules. Rather, thermal radiation is a type of Bremsstrahlung, which is a photonic/electromagnetic emission produced by the acceleration and deceleration of charged particles during collisions.

Particles in a thermal system continuously both absorb and emit photons. If a particle system has a mode of emission at a particular frequency (emitted during deceleration), it also has that same mode of absorption available to accelerate and regain the energy lost by another particle’s radiation of energy.

The power emitted/absorbed vs. frequency is unique for each substance (e.g., the absorption/emission spectrum will probably not be the same as blackbody radiation). But, as long as the absorption/emission frequencies of two substances overlap, radiation can emit and absorb between the particles in two masses.

In summary, "Will two materials continue to exchange radiation energy once they are in intimate thermal conductive contact."

The answer is yes. Body A and B will continue to radiate according to their own power absorption/emission vs frequency curve. There is no fundamental change in the configuration of the system before and after conductive contact. The particles are separated by space before conductive contact as well as after contact, so there is a space between particles for generation, transit, and absorption of radiation.

Note: the depth of penetration of the thermal radiation depends upon the materials, but the depth of penetration and mutual absorption/emission of photons are unaffected by whether the two materials are in conductive contact or separated by a vacuum gap.
- The absorption of a photon occurs when a photon intersects within the absorptive cross-section of the target particle. Both masses have thermal energy before and after conductive contact, so both will radiate photons before contact, and both will radiate photons after contact. The distance between the photon's source and target does not affect whether a photon is generated or absorbed.
- Even when in conductive contact, the emitting particles and absorbing particles are never superimposed. There is still a distance between particle A and B over which a photon can transit between its point of emission and absorption.
- Thus, when two bulk masses are isolated from thermal conduction by a vacuum gap, they will exchange energy by mutual emission/absorption of radiation. And, when those two masses make conductive contact, the atoms internal to the two masses are still separated, so they continue to mutually emit and absorb photons.

Example 1: Net Radiation Loss/Gain by Body A:
- Consider Body A and B, with unequal temperatures, and separated by a distance. Both Body A and B have thermal energy, so both will radiate and can absorb photons. Because of the temperature difference between the two, the hotter body will experience a net energy loss. But, both bodies will absorb and emit radiative energy, regardless of their relative temperatures.
- The instantaneous rate of heat gain from Body B to A and heat loss from A to Body B due to thermal radiation is:

$$dQ/dt_{net A} = dQ/dt_{gain:AfromB}-dQ/dt_{loss:AtoB}$$

Example 2: Total heat loss/gain due to conduction and radiation:
The total rate of heat gain or loss between A and B will include a sum of both the net rate of energy flow due to thermal radiation and the net rate of energy flow due to conductive heat transfer.

$$dQ/dt_{net A all} = dQ/dt_{Gain: Rad + Cond} - dQ/dt_{loss: Rad + Cond}$$