Cannot Concentrate Blackbody Radiation

In summary: I think I see what you're saying.Ok, so the increased radiation is caused by the absorption of the material, not the increased surface area?Correct.
  • #1
Warner
8
0
Hi, having a bit of trouble with a basic question on thermal radiation.

If you take a slab of material that has the characteristics of a blackbody and heat it to temperature where its only emitting in the IR, then its temperature will slowly decrease as the heat is rejected (assuming no incoming radiation).

If you take the same slab and drill a series of holes partially into one surface you are effectively increasing the surface area of the slab. However the rate of temperature decrease is the same because flux through the hole apertures cannot change otherwise it would violate the second law. I think I understand the mechanism of this as any increase in surface area is compensated for by the fact that all these surfaces are now absorbing radiation emitted by the opposing inner wall surfaces. Is this correct? I'm assuming the internal walls are perfectly smooth BTW.

My problem comes when I consider a slab made of some IR reflective material like gold. In this case the internal surfaces are highly, though not perfectly, reflective for IR. So the absorption of the inner walls should be reduced. What then is the mechanism that prevents this structure from breaking the second law? I'm sure I'm overlooking something obvious here.

Thanks!
 
Science news on Phys.org
  • #2
If you drill a cavity in the non-ideal body, it will emit more radiation than it did before. Is that not what you expected?
 
  • #3
I guess its what I would expect, but my understanding of the second law say's it should not happen. I've read about miniscule increases in radiation from surface roughening but nothing of any signficance. How would you calculate the increased radiation?
 
  • #4
Ever notice how a hole looks dark? A cavity behaves more like an ideal black body, hence the synonymous term "cavity radiation". You could get the same effect (of increased radiation) by painting over your gold surface with sufficiently black paint.

At a guess, you'd quantify the effect by altering the albedo in relation to the increase in surface area, noting that if there were an infinite surface area inside the cavity then the radiation from it would be a perfect black body spectrum (having infinite opportunity for the EM field to obtain thermodynamic equilibrium with the energy modes of the material).
 
Last edited:
  • #5
Yeah I understand the cavity radiator idea...looked at far too many of them lately. But doesn't that support the idea that there is no difference between the radiative flux from a hole with an aperture X (and some variable depth), and a body (black or non-ideal) with a side of area X?
 
  • #6
Could you clarify your question? There certainly is a difference in the radiation according to the depth or absence of a cavity and the blackness or otherwise of the material. If you're still thinking that the cavity surfaces "reflect IR" (as well as emitting more) that will mislead.
 
  • #7
I believe I am going down the misleading path. Will try to clarify. Let's take two slabs of gold with the same dimensions radiating into free space considering radiation from one side which has area 2L X 2L . We drill a square hole in one slab to a depth of say of L and sides L X L. So one slab has a surface area of 2L X 2L and the drilled slab now has surface area of 2L X 2L + 4L. I am seeing these surfaces reflecting IR so does the radiated flux go up by 4L?
 
  • #8
No, it goes up by less than that, because the cavity surfaces absorb some of the IR as well as reflecting and emitting. Moreover, no matter how many cavities you drill, the best you can do is approach the radiation that a perfect black body of area 2Lx2L would emit.
 
  • #9
Ok, I assumed that the black body was the limit, and I see that it is driven by the inherent absorption of the material, even if it is vanishingly small. That's where my fundamental issue lies. I presume that if your absorption were zero that would represent the blackbody limit.
 
  • #10
No.. absorption and emission of radiation are fundamentally the same, just call it the strength of interaction between the material and the field. So if your absorption were zero, that would be the white (?) body limit (zero radiation emission); the black body should obviously correspond with the 100% absorption limit. For a cavity in a poorly absorbent material, each surface emits only a fraction of the radiation necessary for equilibrium, but since it also absorbs only the same fraction of incident radiation (reflecting the rest), the net effect of each reflection is to increase the quantity of radiation and bring the material closer toward equilibrium with the field (i.e. ideal black body behaviour).
 
Last edited:
  • #11
Ok...I think it helps me to look at this in terms of the material achieving equilibrium with the field. So when you say our poorly absorbent material, due to mutiple reflections, increases the quantity of radiation to achieve equilibrium with the field I presume this is with respect to our white body material. The quantity of the radiation, increasing as the we make the material more blackbody like, approaches the blackbody limit. Can you decouple absorption from emission? Say by overcoating our low emissivity metal with a highly emissive oxide? If so, what happens to the fields then?
 
  • #12
Warner said:
Can you decouple absorption from emission?
I believe not.

At least, unless the two interactions are at different frequencies, like how a greenhouse can be opaque to the heat from the Earth and transparent to the heat from the sun.

There is actually a lab demonstration where you do what you are describing (produce a black and white pattern by painting over one material with another material of different emissivity). When you heat it (and turn out the lights), the image inverts.
 
Last edited:
  • #13
Hey thanks.
I think I'm settled for now. May have another question or two as I digest all this.
Really appreciate the responses!
 
  • #14
Another quick question...If you take the cavity and put an object in it at a certain temperature, then the body will cool via radiation and a certain flux will emerge from the cavity. If you put a second body, same temp same properties, in the cavity then the flux from the cavity remains the same but the objects achieve equilibrium with the field in such a way as to slow down the rate of cooling is that right?
 

Related to Cannot Concentrate Blackbody Radiation

What is blackbody radiation?

Blackbody radiation is the electromagnetic radiation emitted by a perfect absorber and emitter of energy, or a "perfect blackbody". It is also known as thermal radiation, and is emitted by all objects with a temperature above absolute zero.

Why is blackbody radiation important?

Blackbody radiation is important because it provides insight into the properties of objects that emit it, such as their temperature and composition. It is also used to study the behavior of light and its interaction with matter.

What is the relationship between blackbody radiation and temperature?

The intensity and spectrum of blackbody radiation are directly related to the temperature of the object emitting it. As the temperature increases, the intensity of the radiation also increases, and the peak wavelength shifts to shorter wavelengths.

What is the difference between a blackbody and a non-blackbody?

A blackbody is an idealized object that absorbs all radiation that falls on it, and emits radiation at all wavelengths. A non-blackbody, on the other hand, reflects or transmits some of the radiation that falls on it and does not emit radiation at all wavelengths.

How is blackbody radiation related to the color of an object?

The color of an object is determined by the wavelengths of light that it reflects or emits. Blackbody radiation has a continuous spectrum of wavelengths, so the color of an object emitting blackbody radiation will depend on its temperature. As the temperature increases, the color shifts from red to orange to yellow to white to blue, with shorter wavelengths corresponding to higher temperatures.

Similar threads

  • Thermodynamics
Replies
1
Views
1K
Replies
14
Views
1K
  • Thermodynamics
Replies
2
Views
1K
Replies
29
Views
2K
Replies
4
Views
1K
Replies
17
Views
3K
  • Thermodynamics
Replies
5
Views
1K
  • Quantum Physics
Replies
3
Views
1K
  • Thermodynamics
Replies
4
Views
2K
Replies
8
Views
1K
Back
Top