3
$\begingroup$

I'm reading through a powerpoint presentation about Hawking Radiation (HR). They are explaining all of the reasons that built up to the postulate of HR, and one of the reasons is that if there is a change in the entropy of a black hole when it absorbs something, then the black hole must have a temperature. This kind of makes sense, but it's the next point that I'm a bit confused about: If an object has a temperature due to nonzero entropy, then it must radiate.

Why must it radiate?

$\endgroup$
9
  • $\begingroup$ from the Wikipedia article on Hawkins radiation "Hawking radiation is black-body radiation that is predicted to be released by black holes, due to quantum effects near the event horizon. ..." en.wikipedia.org/wiki/Hawking_radiation So the insight was to consider that a black hole would have a temperature, and the knowledge that all bodies with a temperature would radiate black body radiation. $\endgroup$
    – MaxW
    Commented Apr 19, 2016 at 2:43
  • $\begingroup$ and the knowledge that all bodies with a temperature would radiate black body radiation - why must this be true? $\endgroup$ Commented Apr 19, 2016 at 2:56
  • 2
    $\begingroup$ Every body with a temperature above 0K radiates $\endgroup$
    – hdhondt
    Commented Apr 19, 2016 at 3:39
  • 3
    $\begingroup$ Thermodynamics is valid for matter and radiation and both have to come to an equilibrium state. $\endgroup$
    – CuriousOne
    Commented Apr 19, 2016 at 3:44
  • 1
    $\begingroup$ @MaxW: "all bodies with a temperature would radiate black body radiation" - maybe I missed something, but this does not seem correct: not all bodies are black bodies, and their radiation may be different from that of a black body. $\endgroup$
    – akhmeteli
    Commented Apr 19, 2016 at 3:45

1 Answer 1

1
$\begingroup$

To understand the origin of black body radiation, which is what every material body with a temperature T has been measured to emit, one has to go to quantum statistical mechanics.

Atoms and molecules, to start with, in any ensemble, interact with the electromagnetic radiation. At that level, the processes are quantum mechanical. In quantum mechanics the exchanges of, for example, gas molecules as they bounce off each other, are changes in energy levels of the electrons that compose them . The kinetic energy of one molecule raises the energy level of an electron in the molecule it hit, (by exchanging a virtual photon). The now excited molecule releases the energy as a real photon in a random direction, the electron going back to the ground state. This photon has an arbitrary direction and there is a high probability it will escape the gas volume without interacting again, and be part of the black body radiation.

For fluids and solids, where the molecules are bonded and cannot freely move, there are the vibrational and rotational quantum mechanical degrees of freedom. The atoms and molecules still interact, transferring kinetic energy to electron pushing them to higher energy levels (by virtual photon exchanges), and electrons fall back to the quantum mechanical ground state and a real photon will be radiated, with a high probability of getting out of the material as black body radiation.

If an object has a temperature due to nonzero entropy, then it must radiate.

The internal motions of particles in an ensemble have a kinetic energy which is what builds up the thermodynamically defined temperature observed macroscopically . So any ensemble of molecules with a non zero temperature has kinetic energy which will result in radiation as described above.

In the statistical mechanics definition of entropy it is connected to the number of available microstates . "non zero entropy" is a confusing way to define the existence of temperature, the syllogism is: if there is entropy, microstates exist, therefore kinetic energy, therefor temperature. A simpler concept should be "due to nonzero kinetic energies of the ensemble particles". ( see this about zero temperature and entropy)

Why must it radiate?

Because of the inevitable interactions in large statistical ensembles due to kinetic vibrational and rotational motions/energy.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.