According to Kirchhoff's law, when two objects are in thermal equilibrium, both objects have the same temperature and each object emits as much energy as it absorbs. Also, the energy absorbed at each wavelength is equal to the energy emitted at the same wavelength.

In the case of thermal steady state (like the sun and planet earth), each object maintains its different temperature but they also emit as much as the absorb. However, the energy absorbed at each wavelength is not equal to the energy emitted at the same wavelength.In fact, the incident spectrum and the emitted spectrum are very different. Why?

Kirchoff's law, which state the emissivity =absorptivity at the same wavelength is only applicable for thermal equilibrium situations but is not true for steady state situations, correct?

  • 1
    $\begingroup$ Temperatures are different. Then Wien's law says that wavelengths will be different. $\endgroup$ – Pieter Jun 23 '18 at 19:48
  • $\begingroup$ Thanks. But why if earth absorbs the energy with the solar spectrum it does not end up emitting energy also with the solar spectrum? $\endgroup$ – Brett Cooper Jun 23 '18 at 21:05

Thermal equilibrium is elusive concept to this day. Currently the defining characteristic of thermal equilibrium is detailed balance. Kirchhoff's law, as you described, is manifestation of it. Practically, in many cases Einstein relation (or the generalization of it - the fluctuation dissipation theory) is used to characterize thermal states, and generalizations of it even non-thermal states, since it can describe systems in local (but not global) thermal equilibrium. Your second example seems like it, since it doesn't satisfy detailed balance, but seems to have local equilibrium across it.

Don't stick too much to the notion of temperature, which gain its importance by zeroth law of thermodynamics. The zeroth law most important statement is the equivalence of systems in thermal equilibrium (put two in contact and they will remain in thermal equilibrium). Temperature (also pressure, chemical potential and other intensive quantities) is just a quantitative measure of equilibrium, and different incompatible notions exist (Boltzmann and Gibbs temperatures for example)

Beyond fluctuation-diffipation, there are no generalizations of equilibrium (local or global) which are in scientific consensus.

For completnes I add two review articles about generalizations of temperature and fluctuation-dissipation theorems to system not in thermal state.


Since earth is receiving net input of energy from the sun, it is not in thermal equilibrium with the sun. Therefore the emission spectrum of earth and sun will differ. From the earth's perspective, the incident spectrum belongs to the sun and is unaffected by presence or absence of earth, while the emitted spectrum belongs to earth itself; as remarked before, since earth and sun are not in thermal equilibrium the two spectra will differ. Also steady-state does not imply thermal equilibrium (the converse is true, however).

  • $\begingroup$ thanks. So, the net input of energy and net output of energy are the same for earth (steady state even if not thermal equilibrium). $\endgroup$ – Brett Cooper Jun 24 '18 at 20:11
  • $\begingroup$ @BrettCooper Are you asking if it is true that the net input of energy and net output of energy are the same for earth? On average over a diurnal cycle I think it is true. $\endgroup$ – Deep Jun 25 '18 at 6:09
  • $\begingroup$ No, just saying that earth is, on average, at thermal steady-state (energy in = energy out). The earth temperature is at a steady ~300K. I am still wondering if our the condition emissivity(lambda)=absorptivity(lambda) is true or if it is only true at thermal equilibrium $\endgroup$ – Brett Cooper Jun 25 '18 at 13:49
  • $\begingroup$ @BrettCooper From my reading, it seems Kirchoff's law is strictly valid for a body in thermal equilibrium, while for a body in steady state (but not in thermal equlibrium) it remains a reasonably good assumption. See this paper (it's paywalled). $\endgroup$ – Deep Jun 26 '18 at 4:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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