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I have read this post and from it was my understanding that the definition of a blackbody is:

Black body means a body which ABSORBS all wavelengths completely.

After reading this answer on another post

I gained the understanding that a $\color{green}{\text{black body doesn't have to be in thermal equilibrium.}}$

Wikipedia agrees with both the statements above.

So all is well, but then I read this (from the Imperial College London, Department of Physics):

Black Body Radiation

Stars have high densities -> frequent collisions that can lead to thermodynamic equilibrium where all particles (electrons, ions, photons) have a single temperature

Can often approximate stars as black bodies!

This has got to be a joke right? So does it seem that any object in thermal equilibrium with its surroundings will emit as a black body? I don't see how thermal equilibrium implies a black body (even if it is an approximation).

This also contradicts the line in green above.

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  • $\begingroup$ A black body can both absorb and emit radiation. $\endgroup$ – MaxW Jan 25 at 21:31
  • $\begingroup$ This isn't an implication, it's just two unrelated things. Stars can be approximated as being at a single temperature, and they can be approximated as blackbodies. $\endgroup$ – knzhou Jan 25 at 22:10
  • $\begingroup$ Individual particles do not have a temperature. The formulation of the anonymous slide is a bit awkward. $\endgroup$ – my2cts Jan 25 at 22:37
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    $\begingroup$ @BLAZE I edited the text into the question -- keeping it as an image isn't a good idea for accessibility reasons. Screen-readers won't be able to translate the image into something audible. In the future, please use native Markdown for text and equations, and only use images for things that cannot be expressed that way. Also, it would be good if you can update with a link/title/something for where you read this at the Imperial College. $\endgroup$ – tpg2114 Jan 27 at 6:09
  • $\begingroup$ Where emissivity approaches 1. $\endgroup$ – Manvendra Somvanshi Jan 27 at 7:51
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No, thermal equilibrium does not mean something is a blackbody. A red apple can be in thermal equilibrium with its surroundings. It is manifestly not a blackbody radiator.

The outer parts of a star, from whence the radiation we see, escapes, is approximately in thermal equilibrium at a certain temperature. A star absorbs nearly all radiation incident upon it. A star is approximately a blackbody radiator.

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  • $\begingroup$ Thanks for your answer, so you write "It is manifestly not a blackbody radiator." Why is it not? Or put in another way; under what conditions can an object be a black body emitter? So I guess the conditions are a) Body can be characterized by at a single finite temperature above absolute zero, b) Must absorb all incident radiation regardless of the angle of incidence and not reflect or transmit inbound radiation. So the red apple does not absorb all incident radiation and we know this because? $\endgroup$ – BLAZE Jan 27 at 5:56
  • $\begingroup$ @BLAZE because it is red. A blackbody at room temperature would appear black. $\endgroup$ – Rob Jeffries Jan 27 at 7:23
  • $\begingroup$ The sun is not black either, so why it fine to approximate the Sun as a black body emitter but not the red apple? $\endgroup$ – BLAZE Jan 27 at 7:39
  • $\begingroup$ @BLAZE The Sun is not at room temperature. The Sun emits a (sort-of) blackbody spectrum at around 5800K. This has a significant component (in fact peaks) in the visible band. A blackbody at room temperature should emit almost no visible radiation. $\endgroup$ – Rob Jeffries Jan 27 at 10:52
  • $\begingroup$ Thanks for your reply, I'm having a hard time understanding why "A blackbody at room temperature would appear black/A blackbody at room temperature should emit almost no visible radiation". So I looked into this and found there is an answer here that addresses this, as, actually, a blackbody at room temperature (in thermal equilibrium with its surroundings) will emit a finite amount of visible radiation, but the intensity of the visible radiation emitted will be very low. Is this the correct interpretation? $\endgroup$ – BLAZE Jan 28 at 5:03

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