What causes blackbody radiation

In summary: If you were to paint the object black, however, then the atoms inside the object would still be jiggling, but they would no longer be able to reflect inwards. As a result, the object would be a good absorber and emitter of radiation in the visible spectrum.
  • #1
parsec
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I've seen the cavity resonator model but no one has been able to explain to me what the underlying physical mechanism is that generates the blackbody radiation continuum. Is there some intuitive explanation or is it deeply rooted in quantum mechanics?

An explanation I've heard is that it's doppler shifted collisional de-excitation, is this true?
 
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  • #2
Hot things glow. Objects with non-zero temperature have their constituent charged particles in continuous thermal, random motion. The acceleration of charged particles in this motion leads to electromagnetic radiation (acceleration of charged particles always leads to electromagnetic radiation). You don't need quantum theory to predict that hot things glow (with a continuous spectrum). You need it to avoid the http://en.wikipedia.org/wiki/Ultraviolet_catastrophe" .
 
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  • #3
So essentially it arises from Brem within atoms and molecules? What degree of freedom do the charges have? Is it typically a dipole vibration within a molecule, or some kind of orbital vibration in atoms?

A side question; Someone once mentioned to me that painting an object black increases the amount of radiation that object will emit. I didn't understand the blackbody process so I couldn't really comment, but now I'm having a hard time imagining how a pigment could alter the amount of brem radiation emitted by thermally agitating charges.

I figure the colour of a surface is either determined by absorption lines, or some kind of surface diffraction process, is this the case, and if so how can such surface properties mediate the amount of energy lost from that surface through radiation?
 
  • #4
Well, let's assume that the body is much hotter than its surroundings, so it's painted black specifically for the purpose of dissipating heat, and the amount of heat absorbed is negligible compared to the amount dissipated.

The context is that an air force pilot once told me that the SR-71 was painted black not only for stealth reasons, but also to facilitate heat removal from the skin (which is heated considerably at supersonic speeds). This shouldn't really have much to do with absorption.

Sorry, lazy shorthand. "brem" is "bremsstrahlung radiation"
 
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  • #5
That last post was a reply to a deleted post
 
  • #6
I don't believe the color of an object has that much to do with how much radiation it emits at a given temperature, if it has an effect at all.
 
  • #7
Drakkith said:
I don't believe the color of an object has that much to do with how much radiation it emits at a given temperature, if it has an effect at all.

Darker the object, wider wavelength spectrum is absorbed. This results in proportionally greater l.e.i. radiation.

If you were to desing a system without continuous spectrum, it (color) wouldn't matter then.
 
  • #8
PSz said:
Darker the object, wider wavelength spectrum is absorbed. This results in proportionally greater l.e.i. radiation.

If you were to desing a system without continuous spectrum, it (color) wouldn't matter then.

I wasn't talking about the radiation that is absorbed, but that is emitted. As I understood it the color of an object has no bearing on how radiation is emitted.
 
  • #9
Drakkith said:
I don't believe the color of an object has that much to do with how much radiation it emits at a given temperature, if it has an effect at all.
:) It has. I did an experiment about 1 year ago. I put a light bulb inside a metallic box with 1 face painted in white and another face painted in black (the 2 other faces had no paint but one was polished while the other no). I do remember that there was a difference of emission between the black and the white surfaces, the black emitted more; at a given temperature.
Edit: I don't have the data right now but I remember there was a small but noticeable difference, not as big as the polished vs non polished surfaces.
 
  • #11
Drakkith said:
I don't believe the color of an object has that much to do with how much radiation it emits at a given temperature, if it has an effect at all.

You can prove it with some clever thermodynamical arguments, but I always find those to be dissatisfying.

Here's my attempt at explaining it. Imagine an object that is painted white. That means if you shine light on it, the light will reflect off it. Now imagine all the atoms inside the object jiggling. If those atoms emit light, then when they encounter the surface, they will reflect inwards. So the object reflects light coming from both the outside and the inside, and hence is both a poor absorber and emitter of radiation in the visible spectrum (i.e., visible light). An object whose surface is painted white will not emit light because that light is trapped inside the object due to the surface.
 
  • #12
RedX said:
You can prove it with some clever thermodynamical arguments, but I always find those to be dissatisfying.

Here's my attempt at explaining it. Imagine an object that is painted white. That means if you shine light on it, the light will reflect off it. Now imagine all the atoms inside the object jiggling. If those atoms emit light, then when they encounter the surface, they will reflect inwards. So the object reflects light coming from both the outside and the inside, and hence is both a poor absorber and emitter of radiation in the visible spectrum (i.e., visible light). An object whose surface is painted white will not emit light because that light is trapped inside the object due to the surface.

See my post right above.
 
  • #13
Another simple example. Noticed this with a teflon frying pan. Warm the pan on the stove. Hold it up in front of your face(like a hand mirror) and you'll notice a dramatic difference in the heat you feel on your face as you flip the pan between dark inner coating and light metallic back.
 
  • #14
Drakkith said:
I wasn't talking about the radiation that is absorbed, but that is emitted. As I understood it the color of an object has no bearing on how radiation is emitted.

Again, the amount of radiation emitted is tied with spectrum of the wavelength absorbed. Cause -> effect, which is...

Drakkith said:
Ok, I did find this: http://en.wikipedia.org/wiki/Emissivity

So not all objects emit radiation equally.

... called Kirchhoff's law.

There are other factors as well, but they extent beyond the topic of this discussion.
 
  • #15
RedX said:
You can prove it with some clever thermodynamical arguments, but I always find those to be dissatisfying.

Here's my attempt at explaining it. Imagine an object that is painted white. That means if you shine light on it, the light will reflect off it. Now imagine all the atoms inside the object jiggling. If those atoms emit light, then when they encounter the surface, they will reflect inwards. So the object reflects light coming from both the outside and the inside, and hence is both a poor absorber and emitter of radiation in the visible spectrum (i.e., visible light). An object whose surface is painted white will not emit light because that light is trapped inside the object due to the surface.

I don't think reflection of photons occurs in a ballistic sense. Correct me if I'm wrong but from memory reflection of light must involve absorption and re-emission of photons. If this is indeed the case then I think non-unity emissitivity may be due to more than just reflection of photons from the inside surface of a pigment or coating.
 
  • #16
I think the differences in emission with respect to the objects color depends on the object not being a blackbody, which is what the topic is about. It seems like arguing that an ice rink is not really a friction-less surface, which is true but misses the point.
 
  • #17
Regardless, it would be nice to know why and how colour affects the emissitivity of a body.
 
  • #18
parsec said:
Regardless, it would be nice to know why and how colour affects the emissitivity of a body.

I find it best to think of the body as being one system coupled to the external electromagnetic field via its surface colors. A bright red object is weakly coupled to the red frequencies so red light scatters off instead of being absorbed. The object likewise is more strongly coupled to the non-red frequencies hence they are absorbed.

If you heat the red object up so that it has more energy internally than the surroundings then it will glow in all but the red colors since it couples weakly in red but strongly in other colors.

More generally you have a coupling strength (absorption/emission coefficient = emissivity) as a function of wavelength (or frequency). Call this function [itex]\epsilon(\lambda)[/itex]. Heat the body up to a certain temperature and it will glow with power spectrum equal to corresponding black-body temperature spectrum multiplied by the emissivity function.

Shine light on the object and it will scatter/reflect light at the source frequency times [itex]1-\epsilon(\lambda)[/itex].


Keep in mind that waves propagate because each point in space couples to neighboring points. (imagine a row of pendulums with weak springs linking them). You can have a boundary where the coupling changes and that boundary will partially reflect waves and partially transmit them. The boundary coupling may depend on frequency in which case you have more reflection of one frequency and less transmission across the boundary of that frequency... et vise versa. That is what surface color is for objects.
 
  • #19
Thanks, that's a good way to rationalise why absorptivity and emissivity are so intimately coupled.

Here's some insight into how what we perceive as surface colour affects the coupling of em radiation to a body:

http://en.wikipedia.org/wiki/Pigment#Physical_basis
 
  • #20
parsec said:
The context is that an air force pilot once told me that the SR-71 was painted black not only for stealth reasons, but also to facilitate heat removal from the skin (which is heated considerably at supersonic speeds). This shouldn't really have much to do with absorption.

The metal body of the aircraft has low emissivity, but high thermal conductivity. Perhaps heat moves from the metal body into the high-emissivity paint via conduction, and is then efficiently dumped into the atmosphere via a combination of conduction and radiation.
 
  • #21
parsec said:
I've seen the cavity resonator model but no one has been able to explain to me what the underlying physical mechanism is that generates the blackbody radiation continuum. Is there some intuitive explanation or is it deeply rooted in quantum mechanics?

An explanation I've heard is that it's doppler shifted collisional de-excitation, is this true?

Hot atoms wiggle around. Wiggling charges emit radiation.

The cavity model explains how the ultraviolet catastrophe is ultimately resolved by quantum mechanics, i.e. how quantisation of the EM field limits radiation in the short-wavelength region of the EM spectrum, rendering the total emitted energy finite as observed.

Claude.
 

What causes blackbody radiation?

Blackbody radiation is caused by the emission of electromagnetic radiation from a perfect blackbody, which is an object that absorbs all radiation incident upon it and emits radiation at all wavelengths.

How does temperature affect blackbody radiation?

According to Planck's law, the intensity and wavelength distribution of blackbody radiation is dependent on the temperature of the object. As an object's temperature increases, it emits more radiation at shorter wavelengths, resulting in a shift towards higher energy, visible light.

What is the relationship between blackbody radiation and the Stefan-Boltzmann law?

The Stefan-Boltzmann law states that the total amount of energy emitted by a blackbody is proportional to the fourth power of its absolute temperature. This means that as the temperature of a blackbody increases, the amount of radiation it emits also increases exponentially.

Why is blackbody radiation important in astronomy?

Blackbody radiation is important in astronomy because it allows us to determine the temperature of celestial objects. By analyzing the spectrum of radiation emitted by a star, for example, we can determine its temperature and gain insight into its chemical composition and other physical properties.

How does the color of an object relate to its blackbody radiation?

The color of an object is directly related to its blackbody radiation. As an object's temperature increases, it emits more radiation in the visible light range, resulting in a shift towards shorter wavelengths. This causes the object to appear "hotter" in color, ranging from red to orange to yellow to white as the temperature increases.

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