# What causes blackbody radiation

1. Aug 17, 2011

### parsec

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?

Last edited: Aug 17, 2011
2. Aug 17, 2011

### chrisbaird

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" [Broken].

Last edited by a moderator: May 5, 2017
3. Aug 17, 2011

### parsec

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. Aug 17, 2011

### parsec

Well, lets 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"

Last edited: Aug 17, 2011
5. Aug 17, 2011

### parsec

That last post was a reply to a deleted post

6. Aug 17, 2011

### Staff: Mentor

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. Aug 17, 2011

### PSz

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. Aug 17, 2011

### Staff: Mentor

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. Aug 17, 2011

### fluidistic

:) 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.

10. Aug 17, 2011

### Staff: Mentor

11. Aug 17, 2011

### RedX

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. Aug 17, 2011

### Staff: Mentor

See my post right above.

13. Aug 17, 2011

### jambaugh

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. Aug 18, 2011

### PSz

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

... called Kirchhoff's law.

There are other factors as well, but they extent beyond the topic of this discussion.

15. Aug 18, 2011

### parsec

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. Aug 18, 2011

### ModusPwnd

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. Aug 19, 2011

### parsec

Regardless, it would be nice to know why and how colour affects the emissitivity of a body.

18. Aug 19, 2011

### jambaugh

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 $\epsilon(\lambda)$. 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 $1-\epsilon(\lambda)$.

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. Aug 19, 2011

### parsec

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. Aug 20, 2011

### m.e.t.a.

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.