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I'm trying to start understanding quantum mechanics, and the first thing I've come across that needs to be understood are black bodies. But I've hit a roadblock at the very first paragraphs. :( According to Wikipedia:

A black body (also, blackbody) is an idealized physical body that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence.

OK, that's nice. It's an object that absorbs (takes in itself and stores/annihilates forever) any electromagnetic radiation that happens to hit it. An object that always looks totally black, no matter under what light you view it. Good. But then it follows with:

A black body in thermal equilibrium (that is, at a constant temperature) emits electromagnetic radiation called black-body radiation.

Say what? Which part of "absorbs" does this go with? How can it absorb anything if it just spits it right back out, even if modified? That's not a black body, that's a pretty white body if you ask me. Or a colored one, depending on how it transforms the incoming waves.

What am I missing here?

Vilx-
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    That it absorbs everything means that every radiation coming from it is truly emitted rather than reflected. – ACuriousMind Feb 23 '15 at 21:58
  • @ACuriousMind - and if it always emitted the radiation in a way that mimics reflection (preserving the direction, wavelengths, etc) - would that also count as a "blackbody"? – Vilx- Feb 23 '15 at 22:11
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    How did "absorbs all incident radiation" translate as "never lets any radiation out and looks totally black"? "Absorbs" means "takes in", it doesn't mean "doesn't let out". A perfect absorber doesn't reflect or transmit any light. That is all. – Jim Feb 23 '15 at 22:12
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    @Vilx- There's no physical way to emit absorbed radiation in a way that mimics reflection without it actually being reflection. So no, that wouldn't count as a blackbody – Jim Feb 23 '15 at 22:16
  • @JimdalftheGrey - Hmm... I guess I had a different idea of what "absorbs" means. And also... well, if you allow it to absorb and then re-emit, then how can you tell that that was what happened and not just some reflection/filtering/whatever thing? – Vilx- Feb 23 '15 at 22:17
  • Blackbodies emit along a blackbody spectrum that corresponds to their temperature. Reflection preserves information about wavelength (at least), transmission preserves incident angle and wavelength. Absorption and re-emission preserves none of this – Jim Feb 23 '15 at 22:19
  • @JimdalftheGrey - Hmm... this is going in circles. Blackbodies are emitting blackbody radiation because they are blackbodies that emit blackbody radiation. I'm kinda trying to approach this like maths - start with some definitions (and possibly axioms), then layer theorems on top of them. But I find it difficult to find the bottom of it all (the definitions). :( – Vilx- Feb 23 '15 at 22:23
  • Vilx, an object with non-zero temperature will emit energy until it no longer can (until it has reached a ground state). It seems to me, from your comments, that you're imagining a perfect absorber as an object that absorbs all incident energy while emitting none, i.e, while remaining in the ground state. How could that be? – Alfred Centauri Feb 23 '15 at 23:02
  • @AlfredCentauri - In real life it couldn't, but we're talking about idealized stuff here anyway. Or, well, that was my reasoning. I see now that the word "absorbs" has a different meaning than I thought. – Vilx- Feb 23 '15 at 23:26
  • Dont use the math approach in physics, you have to build up understanding from concrete experiments/examples/objects and then move up the abstraction ladder. – lalala Dec 25 '17 at 08:27

5 Answers5

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Say what? Which part of "absorbs" does this go with?

The key to understanding this is to carefully note the phrase "in thermal equilibrium".

This means that the rates of absorption and emission are the same.

If a body were at a lower temperature than the environment, the rate of absorption would be higher and the body would then heat up.

If a body were at a higher temperature then the environment, the rate of emission would be higher and the body would then cool down.

But, in thermal equilibrium, the temperature is constant and, thus, the rates of absorption and emission must be equal.

Now, put this all together:

  • A black body is an ideal absorber, i.e., a black body does not reflect or transmit any incident electromagnetic radiation.
  • An object in thermal equilibrium with the environment emits energy at the same rate that it absorbs energy.

Then, it follows that, a black body in thermal equilibrium emits more energy than any other object (non-black body) in the same thermal equilibrium since it absorbs more energy.

Imagine several various objects, including one black body, in an oven and in thermal equilibrium. The black body will 'glow' brighter than the other bodies.

Jim
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Alfred Centauri
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    OK, I'll accept this. I guess I understand it now, but really it raises more questions in other directions. I'll have to think about it more before I ask my next question. – Vilx- Feb 23 '15 at 22:24
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"OK, that's nice. It's an object that absorbs (takes in itself and stores/annihilates forever) any electromagnetic radiation that happens to hit it."

Right! But that frontslash is important, and you want the first option. It can absorb the energy as heat and it never reflects anything. What black-body theory then goes on to detail is what the black-body emits. This isn't ambient light--a black body would emit radiation even if it never absorbed any, as long as its temperature were nonzero.

As you note, the name is kind of a misnomer. Black bodies (interesting ones) hardly ever appear black. To look at, the sun is basically a black body with a temperature of 5500K. As you may have noticed, the sun is not black$^\mathrm{[citation \ needed]}$.

zeldredge
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The name "black body" is maybe a bit misleading.

The "Idealizations" section of the wikipedia entry illustrates the idea in the way that always stuck with me. Put a pinprick in a totally "black" cavity. Here "black" means the sides of the cavity do not allow radiation, and the cavity is large enough that the photons that happen to wonder in are unlikely to find their way back out. As @SirElderberry said, the phenomenon to study, that eventually landed Planck on quantization, are those few photons that do happen to make their way out of the blackbody.

jjstankowicz
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Any body above 0K emits radiation. Same is the case with black body, it absorbs radiation and it also emits, now the rate at which it absorbs /emits depends upon the surrounding.

Sordan
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In simple words

Black body radiation means a body - independent of its color - that absorbs all the wavelength falling on it in the form of energy and does not reflect any of that wavelength, but instead it radiates what it absorbed with different wavelength.

That is why a star is considered a black body, since it just radiates and doesn't reflect.

sherif Elhosainy
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