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I read that good absorbers are good emitters - hence a blackbody, that absorbs all kinds of radiation, also emits all kinds of radiation? I'm not able to get my head around this.

What does it mean to absorb all kinds of radiation? Radiation of all frequencies? or are we saying that $100\%$ radiation incident on a blackbody is absorbed? None is reflected? Well then, how is it a good emitter?

Am I confusing good emitter and good reflector? I'd appreciate some clarifications. Thank you!

P.S. I'm coming back on Physics SE after years. I'm majoring in mathematics, but I decided to take a quantum physics course for fun anyway. Hence, I'm back here for the spring!

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    $\begingroup$ If an object absorbed well but emitted poorly it would be in a thermal runaway situation, forever heating up. It would passively heat itself on the heat radiating from colder objects. Not very intuitive. And in this house, we obey the laws of thermodynamics. So there is a balance that sums to exactly 1 between emissivity, reflectivity, transmissivity and absorbance. In must equal Out. Or else you have the runaway situation. (There are manufactured instances of this "balance" being broken though, read up on Meta Materials for instance) $\endgroup$ – Stian Yttervik Jan 18 at 11:17
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    $\begingroup$ @StianYttervik The short answer from first principles is (without knowing anything about matter or electromagnetism) "to keep the universe sane" ;-). $\endgroup$ – Peter - Reinstate Monica Jan 18 at 12:43
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    $\begingroup$ @StianYttervik that makes a degree of sense but it would be quite cool (see what I did there?) if you could have a material that would spontaneously have a higher or lower temperature than ambient. I'm not immediately certain that this would violate any thermodynamics laws. $\endgroup$ – Frog Jan 19 at 8:31
  • $\begingroup$ @Frog Well... Net heat spontaneously flowing from a cold object to a warm object would violate the Clausius statement (which is the exact opposite) - which is one of the better descriptions of the 2nd law. This just might be enough of a violation that would end up as conviction in the Court of Physics. $\endgroup$ – Stian Yttervik Jan 19 at 9:07
  • $\begingroup$ Time reversibility? $\endgroup$ – user253751 Jan 19 at 15:23
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Yeah, good emitter and good reflector are definitely not the same property. Maybe the best way to visualize it is just as the time inverse, under the mapping $t\mapsto-t$, of absorption is emission. This also gives something of a hint of why this relationship might hold, albeit I only know the derivation for the case of thermal radiation.

For thermal radiation, you bring two bodies of the same temperature into radiative contact, one of one material, one of the other. If either one emits more radiation than it absorbs, then it spontaneously cools and the other one spontaneously heats up, violating the second law of thermodynamics. Very simple argument. May require sealing the two in some idealized setup of mirrors or so, to make the proof work, but aside from that it appears to have immediate physical relevance.

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    $\begingroup$ time inversion really says it all. $\endgroup$ – Jeffrey Jan 18 at 22:00
  • $\begingroup$ @Jeffrey I wanted to say that for me it's the energy conservation, if a body would soak energy more than emit without ever reaching balance, it would be a thermal black hole, not a black body... but then from Noether's theorem time symmetry is equivalent to energy conservation, so it really does say it all. $\endgroup$ – luk32 Jan 19 at 10:32
  • $\begingroup$ @luk32 Technically energy conservation is equivalent to time translation invariance, not to time reversal. $\endgroup$ – J.G. Jan 19 at 22:05
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A blackbody must be in thermal equilibrium. Therefore if it absorbs everything incident upon it, it must emit just as much radiation. In principle this should be true at all frequencies. If it is not, then the object is not a blackbody.

At a microscopic level this boils down to the principle of detailed balance, that says that for systems in equilibrium, every process must be in balance with its reverse process.

When it comes to the absorption and emission of light, well these are each other's respective reverse processes. If we put an object into equilibrium (enclosed in a heat bath say) then it must be true that the absorption and emission processes are in balance (as in a blackbody). However, the relationship we could derive in this situation between the Einstein coefficients that govern emission and absorption at a microscopic level are properties of the material itself, and are not influenced by the fact that it is in equilibrium. Therefore it is generally true that an object that is a good absorber is also a good emitter, regardless of whether it is in equilibrium or not.

Reflection is an entirely different matter. If an object reflects light, then that light is obviously not absorbed. By definition a reflective object cannot be a blackbody and is both a poor absorber and emitter.

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According to Wikipedia:

Blackbody:
A black body or blackbody is an idealized physical body that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence. The name "black body" is given because it absorbs radiation in all frequencies, not because it only absorbs: a black body can emit black-body radiation.

So, you can see that a black body does emit radiation. An ideal blackbody must not only absorb radiations of all frequency but it must also not reflect and transmit them. An ideal blackbody must have two important properties:

  • The body must be an ideal emitter i.e., at every frequency it must emit radiation as much as any other body at that temperature.
  • It must also emit energy isotropically i.e., uniformly in all directions.

To add to these points a blackbody first absorbs all the electromagnetic radiations it comes in contact with. And then it emits thermal radiation. To be more clear, a perfect blackbody is one which is a perfect absorber and a perfect emitter of all kind of radiation. So, you can consider our sun(as well as other stars) as an almost perfect black body.
Wait a minute then does sun absorb light?
Believe me or not, it does!
And, that's one of the craziest things that I've heard so far. You can find more details about that over here in Astroquizzical

Now, coming to your question.
You're probably confusing with the meaning of reflection and emission.
According to Wikipedia,

Reflection: Reflection is the change in direction of a wavefront at an interface between two different media so that the wavefront returns into the medium from which it originated.

Emission:
In physics, emission is the process by which a higher energy quantum mechanical state of a particle becomes converted to a lower one through the emission of a photon, resulting in the production of light.

It is indeed clear from their definitions that reflection is the process due to which a light ray's direction is changed after hitting an interface. While, emission is the process by which an object gives off light on its own by absorbing energy.

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  • $\begingroup$ The sun absorbs light from sources of a temperature higher than its surface. If the source is of a lower temperature, the sun doesn't absorb light in a net sense. $\endgroup$ – Acccumulation Jan 18 at 6:35
  • $\begingroup$ Thermal radiation is electromagnetic radiation. Why are you making a distinction? $\endgroup$ – OrangeDog Jan 19 at 13:36
  • $\begingroup$ You're right @OrangeDog, thermal radiation is electromagnetic radiation. But all electromagnetic radiations are not thermal radiations. Thermal radiations are the ones which carry heat energy with them. $\endgroup$ – lee Jan 19 at 14:03
  • $\begingroup$ There are non thermal electromagnetic radiations too. So, I wanted to mean that a black body will absorb all kinds of electromagnetic radiation (which may or may not include thermal radiation) while it normally gives of thermal radiation(which contain heat energy with it). Anyway, thank you for asking that. Do you think that I should add this to the answer to avoid confusion? $\endgroup$ – lee Jan 19 at 14:13
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    $\begingroup$ @lee no, it adds confusion. If it absorbs all wavelengths then it must also emit all wavelengths. There is nothing special about "thermal" radiation or "heat" energy. $\endgroup$ – OrangeDog Jan 19 at 15:17
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All bodies radiate radiation at all temperatures. It's because it (the body)gives it's heat energy to electrons by changing their (of electrons)energy level ( lower energy level to higher energy level) after it as the electrons return to their initial energy level they emit electromagnetic radiation. And why do the body change heat energy into radiation ? It's because it want thermal equilibrium with the surrounding. This is how it radiates energy. And when the radiation emitted by surrounding bodies fall on the surface of this body it absorb some part of it and reflect the other.(because this radiation is em wave just like light wave). Now this increases the temperature of the body . Therefore in order to keep it's temperature equal to room temperature it emits more radiation. Therefore if a body absorb more radiation the it have emit more. That's why good absorbers are also good emitters

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Others have given good answers, so I'll just add flavor...

As weird as it sounds, absorption is not the opposite of emission, at least not the ordinary emission we see every day from light bulbs, fire, or LEDs. Those are all spontaneous emission. Absorption is instead the opposite of stimulated emission, which is what gives us lasers: if you have an excited electron, it can trigger the electron to emit another photon more or less identical to the first. But that photon could also be re-absorbed and excite electron back up from the ground state, getting you back where you started. Whether you have net absorption or net (spontaneous) emission depends on whether you start with >50% of your electrons in the excited state.

Even so, you can show that the strength of absorption, stimulated emission, and spontaneous emission are directly proportional to each other, as ProfRob pointed out. The derivation might have earned Einstein a Nobel prize (lasers!) if they hadn't already given him one for the photoelectric effect, of all things.

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Short answer: If "good absorbers are good emitters" wasn't  true, it would be possible to build  a PM machine. And since nature don't allow us to make PM machines "it" had to make the laws of QM so that it is true. :) :)

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  • $\begingroup$ Could you elaborate on the operation of this perpetual motion machine? $\endgroup$ – Gilbert Jan 18 at 15:50
  • $\begingroup$ Ok Assume you have 2 blocks of something sitting close , Block A ( good absorber ) Block T ( good transmitter ). both at the same temperature at start. Block T will emit radiation, but reflect any radiation from A. Block A will absorb radiation from T but not radiate anything.IE there is a net radiation from T to A. Temperature of A will increase and decrease in T. This temperature difference can be used to run a sterling engine or thermocouple. And voila you have energy for free :) $\endgroup$ – EnockNitti Jan 18 at 16:39
  • $\begingroup$ Check the other good answer for the details on how nature makes this PM impossible, you know: there is no free lunch. But I think my answer is easer to grok :) $\endgroup$ – EnockNitti Jan 18 at 16:53
  • $\begingroup$ So suppose you put an optical isolator between two identical blocks. Would this not also operate in the same way? Ultimately, it’s not energy for free, it’s just a scheme to harvest the thermal energy stored in the blocks. You could use that energy to drive a load, sure, but eventually the blocks would just go cold. Isn’t that why you can’t make this a perpetual motion machine? $\endgroup$ – Gilbert Jan 18 at 18:19
  • $\begingroup$ How would that optical isolator work ? If it is a one way isolator you would get the same effect as in my example, but there is no such animal . I just explained how my two imaginary blocks can be used in a PM. Not why could not. Btw: "it’s just a scheme to harvest the thermal energy stored in the blocks" Actually not. You can never extract thermal energy from something without a temperature gradient. It is the -<temperature difference<- that enables the energy extraction $\endgroup$ – EnockNitti Jan 18 at 19:14
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As others have said the key property is the time-reversibility of emission/absorption process. Lets say someone asks the analogous question "how can a good absorber of heat be a good absorber of cold?".

  1. A good heat absorber has a high thermal conductance, and good thermal contact with whatever its supposed to absorb heat from.

  2. A good cold absorber has exactly the same properties.

So good cold absorbers are also good heat absorbers, which one they act like is just a matter of whether they are hotter or colder than the other thing. You know this from everyday life, cold water makes you freeze faster than cold air and hot water makes you burn (boil) faster than hot air.

The emission and absorption of radiation is similar. Things that are white, or reflective (mirrors) have a very weak thermal coupling to the radiation field, its like they are separated from it by jacket. Things that are black have a very strong coupling to the radiation field, so they reach thermal equilibrium with it much faster.

A good experiment to try is this.

Sunlight has a temperature of about 6,000 degrees C, while even on a hot day the air will be much cooler than that. If you put a white and a black object out in the sun both will find a compromise temperature where the sun heating them and the air cooling them balances - the black one has better sun-contact so will end up hotter.

You can then repeat the same experiment at night (find somewhere very dark, no streetlights ideally). Now the light-field is (if its dark enough) really, really cold: about -270 degrees C (a bit above absolute zero, cosmic microwave background), so the air is now the heat source and the light the cold. Correspondingly the black object will become colder than the white one.

You can make the experiment better by trying to put a shield of some kind so that the white/black things are only interacting with light from space, ideally no infra red from the ground nearby.

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Every body loses or gains heat to be in thermal equilibrium with its surroundings. If the body is already at thermal equilibrium, it would maintain its temperature.

If radiation hits the body and gets absorbed, then it would raise the temperature of the body. Therefore, the body emits radiation to lose energy and return to thermal equilibrium.

Therefore if a body absorbs more energy, it must also lose more energy to maintain its temperature. This shows that good absorbers must also be good emitters.

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