If radiation with a blackbody spectrum with temperature $T$ is incident on a blackbody, the blackbody heats up to the temperature $T$. What happens if the spectrum of incoming radiation is not that of a blackbody? What temperature will the blackbody reach?

  • $\begingroup$ A blackbody in thermal equilibrium at temperature $T$ emits radiation that resembles a blackbody spectrum. I do not understand what do you mean by blackbody spectrum being incident upon a blackbody. $\endgroup$ – SRS Oct 3 '16 at 5:48
  • $\begingroup$ @SRS for example, one can shine monochromatic light on a blackbody $\endgroup$ – Brian Bi Oct 3 '16 at 6:04
  • 4
    $\begingroup$ Your assumption about blackbody absorption is wrong. To quote wikipedia (where you should have looked before asking) "A black body is an idealized physical body that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence" $\endgroup$ – Alexander Oct 3 '16 at 6:23
  • $\begingroup$ Sorry, that should have been obvious. Silly me $\endgroup$ – Brian Bi Oct 3 '16 at 6:48

An object absorbs some fraction of the incoming radiation. If the object is indeed an ideal black body, all the power is absorbed, and none is reflected.

The body will heat up, and emit a black body spectrum corresponding to its temperature. The emitted power is given by $P=A\sigma T^4$ (this is known as the Stefan–Boltzmann law).

When the absorbed power is equal to the emitted power, the temperature of the blackbody stops changing. By equating the two expressions, you can solve for temperature. So, if the incoming power is $P_\text{in}$, then the steady-state temperature of the blackbody is

$$T_\text{steady state} = \left( \frac{P_\text{in}}{A \sigma} \right)^{1/4} \, .$$

Note that a real body (not an ideal black body) has a finite, wavelength-dependent reflectivity. In other words, real objects don't absorb all of the incoming radiation.


I think you're confusing it with the equilibrium state where the incident radiation is exactly the same as the emitted radiation. An ideal blackbody absorbs all radiation and re-emits it with the spectrum given by Planck's Law. You consider this equilibrium state only to calculate the conversion factor between the energy density and the radiated power (Wikipedia and related question).

In your case, the spectrum is irrelevant. The temperature of the blackbody will reach an equilibrium at $T = (I/\sigma)^{1/4}$, where $I$ is the total power absorbed per unit surface area of the blackbody (for all frequencies) and $\sigma$ is the Stefan–Boltzmann constant. This is known as the Stefan–Boltzmann law.


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.