Can Blackbody Radiation Explain Heat Absorption in Hollow Cavities?

[psholtz]

I'm reading from an introductory text on quantum physics, and came across this sentence:

According to classical electrodynamics, empty space containing electromagnetic radiation possesses energy. In fact, this radiant energy is responsible for the ability of a hollow cavity to absorb heat.
-- David Bohm, Quantum Mechanics

It's the second sentence that I don't understand: how can the energy in the EM field be responsible for the ability of a hollow cavity to absorb heat?

 

[Bill_K]

I agree this is pretty badly stated. What he means to say is that in thermal equilibrium a hollow cavity will contain black body radiation at a given temperature. If you supply heat to raise the temperature of the body, the temperature of the cavity will also increase and consequently the energy stored in it will increase. Thus the cavity has "absorbed heat."

 

[lightarrow]

I think the meaning could also be this: a hollow cavity can absorb heat because if you put it in front of a hot object, some radiation from the object goes in the cavity and cannot escape (and you notice that the cavity becomes hotter) even if there is void between object and cavity; so only a radiation can be responsible for this.

 

[_PJ_]

It's based on the origins of 'Blackbody radiation' which (I 'think' were german) the traslation of its (German if I'm correct) name was "Cavity radiation"

This came about as it was discovered and investigated by means of heating an insulated 'container' enclosing the 'cavity' with a tiny hole. The hole was smaller than the wavelength of thermal radiation.

The heat of the cavity, however, increased in the same, then unexplainable and "near-perfect" way, as a theoretical blackbody.

 

[psholtz]

So "black body" radiation really means "cavity" radiation?

That was my next question..

Why is the glowing red heat from an oven called "black" body radiation? :-)

 

[tom.stoer]

Using a cavity was the original way to derive the black body radiation, but based on statistical mechanics it is possible to derive the spectrum w/o using a specific shape of the radiating body.

It is called "black body" as the spectrum shall be determined entirely by the temperature, not by any surface properties like color, texture, reflection or something like that. That means that a ideal black body is really black in the sense of 100% absorption for all frequencies.

 

[psholtz]

So this is something that had always been a bit of a stumbling block for me, but I think I'm starting to grasp it .. at least partially.

What we really have going on is two distinct bodies: (a) the "radiator" itself, which might be the walls of an oven (i.e., the actual "solid" object that has been heated); and (b) we have an open "cavity" within this radiator. And what we are discussing is the exchange of energy that takes place between these two mediums.

So the "radiator" is going to emit radiation, at a whole range of frequencies and in a whole range of (more or less random) directions. This radiation will be emitted into the cavity. In turn, the radiator may absorb some of the energy that is present in the cavity. If we are modeling a "black body radiator", we are taking about a radiator that absorbs 100% of the energy that is incident on it from the cavity, correct?

Is this general line of reasoning correct?

So, when we talk about the black body radiator being in equilibrium, what we are taking about is a situation where (a) the black body is emitting radiation into the cavity; but (b) since the system is in equilibrium, the cavity "emits" this radiation back "into" the black body (or, better said, the black body absorbs this radiation from the cavity).. and since it is a "black body" radiator, it absorbs 100% of the radiation which is incident on it from the cavity... correct?

 

[tom.stoer]

I think this is much too complicated. E.g. the sun is to a rather good approximation a black body.

http://en.wikipedia.org/wiki/File:Solar_Spectrum.png

 

[psholtz]

I think this is much too complicated. E.g. the sun is to a rather good approximation a black body.

http://en.wikipedia.org/wiki/File:Solar_Spectrum.png

OK, but that's the part I don't get..

The Sun isn't black..

Nor is a red hot oven black.

Does the use of the term "black" here have nothing to do w/ the actual color of the object?

 

[lightarrow]

OK, but that's the part I don't get..

The Sun isn't black..

Nor is a red hot oven black.

Does the use of the term "black" here have nothing to do w/ the actual color of the object?

It's black at room temperature. The sun's surface is so poorly reflecting that it can be considered as a black body (= totally absorbing).

 

[haushofer]

OK, but that's the part I don't get..

The Sun isn't black..

Nor is a red hot oven black.

Does the use of the term "black" here have nothing to do w/ the actual color of the object?

A black body is an object which "absorbs all the radiation falling onto it", so radiation of all wavelengths. That's why people often come with this cavity with a small hole in it, to make this abundantly clear. At thermal equilibrium then, according to Kirchhoff's law, it will also emit radiation of all wavelengths. So this cavity looks a bit artificial; there are quite some more objects which have more or lesss this property, even though they are not cavities or black in colour.

The Sun approximates this property fairly well. :)

 

[lightarrow]

A black body is an object which "absorbs all the radiation falling onto it", so radiation of all wavelengths. That's why people often come with this cavity with a small hole in it, to make this abundantly clear. At thermal equilibrium then, according to Kirchhoff's law, it will also emit radiation of all wavelengths. So this cavity looks a bit artificial; there are quite some more objects which have more or lesss this property, even though they are not cavities or black in colour.

The Sun approximates this property fairly well. :)

Ok, but you should perhaps precise your statement "even though they are not cavities or black in colour". The "colour" black depends on temperature. A perfect cavity at 2000°C appears as bright white. Actually black it's a not a colour at all, is the fact it emits much less light than the surroundings, so we see it as black.

 

[tom.stoer]

As I already said:

... is called "black body" as the spectrum shall be determined entirely by the temperature, not by any surface properties like color, texture, reflection or something like that. That means that a ideal black body is ... "black" in the sense of 100% absorption for all frequencies.

In that sense the sun is nearly "black" as the absorbtion of incoming radiation is nearly 100%

 

[Penn.6-5000]

Hi.

Hot things glow. The hotter something is, the more light it emits. Also, hotter things emit light at higher frequencies. The burner on an electric stove glows red. The filament in your light bulb is hotter and so it glows yellowish-white. A deer's body temperature is much cooler so it glows infrared. Matter falling into black holes is extremely hot and glows at X-ray frequencies. The leftover glow from the big bang is very cold and glows at radio wave frequencies. This phenomenon is called "thermal radiation." Planck's blackbody equation tells exactly how much light (thermal radiation) you get at each wavelength for a given temperature for ideal objects.

If you look at a deer, it doesn't look infrared. It looks brown with tufts of white. Part of the problem is that humans can't see infrared. The other part of the problem is that the deer's fur is reflective, so when sunlight hits it the colors forming brown get reflected. Planck's equation doesn't predict this! But nobody ever expected it to. What we do is we say that there are two mechanisms to get light from a deer's fur: thermal radiation and reflected sunlight. Then we say that an "ideal deer" is one that doesn't reflect any sunlight; any light that hits it gets absorbed, so all the light you see from this ideal deer is thermal radiation. Now, "ideal deer" is a stupid name. It doesn't matter that it's a deer; what matters is that it doesn't reflect light. Black things don't reflect light, so we'll call it a "blackbody." So a blackbody is an ideal object that doesn't reflect any light, so when you look at it all the light you see can be described by Planck's "hot things glow" equation.

Imagine a hollow aluminum box with mass 500g. You're inside this box and you look around. What do you see? Nothing. The box is closed so it's dark in there. But if you have infrared goggles, you'll find that you're able to see anyway. Why? Is infrared light leaking in from the outside? No! The walls of the box are warm. Because they're warm, they glow. So there's all of this infrared thermal radiation inside the box with you.

Next question: What's the specific heat of the box? That is, how many joules of heat do you have to apply to the [outside of the] box in order to increase its temperature by 1 Kelvin? Well, everybody knows that the specific heat of aluminum is 900 Joules per kilogram-Kelvin, so if it's a half-kilogram box, you'd expect to need 450 joules of heat to make this box's temperature go up by one Kelvin. But when yo do this experiment, you find that 450 joules of heat cause the box's temperature to go up by only 0.998 Kelvins. (I made that number up, but it could be valid for a fantastically large box. Pay attention to the argument, not the actual numbers, from this point on.) So you really need 451 joules of heat to increase the box's temperature by a full Kelvin. Why? Where does that extra joule of energy go? It can't be in the aluminum, because aluminum heats up faster than that.

The extra energy is stored in those infrared photons that are bouncing around inside the box. It's true that every photon that jumps off one wall will soon be reabsorbed by another wall, but at any particular moment a certain number of photons are still "in flight" and they carry with them some energy. How much energy? For this particular box, all the blackbody photons in flight inside the box at any given time have energies that add up to the missing joule. Great, mystery solved! The empty space inside the box is like a heatsink. It can store thermal energy in the form of photons-in-flight, and those photons are thermal radiation photons.

Back to the original question: "How can the energy in the electromagnetic field be responsible for the ability of a hollow cavity to absorb heat?" Answer: Light is an electromagnetic wave. Replace "energy in the electromagnetic field" with "energy in all those thermal photons that are bouncing around" and you get my explanation.

[Edited in response to comments from lightarrow: Changed "blackbody radiation" to "thermal radiation" where appropriate. The thermal emission curves of most materials are very similar to blackbody radiation, up to a constant called `emissivity.' Lots of folks use the word 'blackbody' to cover all of these cases, even for approximations more appropriately called `greybody' and even some approximations that don't really fit it at all. But that isn't really correct, as lightarrow points out. Blackbody radiation is the spread of wavelengths for an ideal object, and "thermal radiation" is a better way to refer to similar light coming from non-ideal objects.]

 

[lightarrow]

When hot things glow, it's called blackbody radiation.

Not exactly. It's blackbody radiation if the body is in thermal equilibrium with the radiation and if the body is totally absorbing. A hot gold ingot doesn't generate blackbody radiation, for example.

 

[lightarrow]

Oh, one other thing: There seems to be a lot of argument about whether the sun deserves to be called black, and it's explained away by saying that the sun absorbs most of the light that hits it. That's true, I guess, but that's not the important part. Imagine a perfectly reflective ball in the middle of space. Now heat the ball to 6000 Kelvins. What light do you expect to see when you look at the ball? You'll see it glowing brightly because it's hot, and you'll also see it reflecting all the light that hits it. Does it deserve to be called a blackbody even though it's reflecting all the light that hits it? Sure. Because it's in space. Space is dark. There isn't much light to reflect. So there's a LOT of blackbody radiation coming from this ball but very little reflected light. When you look at the light coming from this ball, it will match the blackbody curve very closely, so you might as well call it a blackbody. But if you put this same ball in a very bright place, where there's a lot more light to reflect, then the light coming from it won't match Planck's equation so well and you might not want to call it a blackbody anymore.

No, it doesn't depend if there is or not light in the surrounding, it depends on the physical properties of the surface: if the body's surface is still reflecting at those temperatures (but at 6000 K it's quite difficult) then it doesn't emit BB radiation.

 

[Penn.6-5000]

In response to lightarrow: I have clarified the first post and outright deleted the second one.

[The point I was getting at in my second post was that the decision about whether to call an object a blackbody really has more to do with observations of the light coming off it rather than what happens to the light that happens to hit it. If, under the conditions of study, an object emits colors corresponding to Planck's blackbody curve, you might as well call it a blackbody regardless of what it's made of. What I forgot was that when an object is reflective, not only is it reflecting extra light that wasn't included in the curve, but also it's emitting less thermal radiation.]

To save face, here's a fun fact: How do vacuum flasks insulate so well? The idea is that you have an "inner bottle" and an "outer bottle," and between the inner and outer bottles there's a vacuum. Heat can't leak across a vacuum via conduction or convection, so that leaves radiation. How do you decrease the amount of thermal radiation across the vacuum between the inner and outer bottles in a vacuum flask? By making the surfaces facing the vacuum reflective. Reflective surfaces emit less thermal radiation. So the hot side stays hot and the cold side stays cold because there are fewer thermal photons jumping from the hot side to the cold side.

 

[psholtz]

Not exactly. It's blackbody radiation if the body is in thermal equilibrium with the radiation and if the body is totally absorbing. A hot gold ingot doesn't generate blackbody radiation, for example.

So what exactly is the difference between (a) a hot gold ingot; and (b) the Sun?

Why is one a blackbody, but the other isn't?

How can they "measure" or "determine" that the Sun absorbs 100% of the radiation incident upon it? What kind of experiments do they do to confirm this?

 

[Penn.6-5000]

It's easier to answer your questions in reverse order.

1. How can they "measure" or "determine" that the Sun absorbs 100% of the radiation incident upon it? What kind of experiments do they do to confirm this?
It's easiest to just look at the light coming from the Sun and see if it matches the ideal blackbody curve. If it does, then we wave our hands and say, "Good enough. If it glows like a blackbody, it's probably a blackbody." We don't actually try to reflect light off the sun and see how much comes back.

2. What exactly is the difference between (a) a hot gold ingot; and (b) the Sun?
Reflective objects emit less thermal radiation than their black counterparts. Find a shiny gold ingot and a perfectly black rock that have the same size and shape. Put them inside a box with ideal blackbody walls, so that the photons that are shooting past them follow the Planck distribution of wavelengths. Leave both objects there until they warm up (or cool down) to the same temperature as the box's walls. Now, because the gold and the black rock have the same surface area, we know that the same amount of light is hitting each one. But while the black rock absorbs all the light that hits it, the gold absorbs only 20% and reflects the rest because it's so shiny. But since the gold and the rock have both reached a steady-state temperature, they must be emitting energy at the same rate they're absorbing it. But we just established that gold is absorbing just 20% of what the ideal blackbody is absorbing... Therefore the gold must be emitting only 20% as much thermal radiation as the blackbody. This argument gives us "Kirchhoff's Law of Thermal Radiation."

It turns out that reflectivity and emissivity are equal on a per-wavelength basis. According to this http://en.wikipedia.org/wiki/File:Image-Metal-reflectance.png" , gold reflects 40% of the blue light that hits it but 90% of the yellow light hitting it. So a gold ingot heated to 300K will emit only 60% as much blue light as a 300K blackbody would, and only 10% as much yellow light. That means that gold's thermal radiation spectrum doesn't look very much like the blackbody spectrum, since each wavelength is adjusted by a different scale factor.

 

[lightarrow]

Ok. Penn.6-5000. Good post!
I would only add that probably even a star made of gold would be a black body, because in those conditions of temperature and high pressure, the plasma of gold atoms, ions and electrons, would have an absorbance near 100% and the emission spectrum would be almost totally continuous.