# What causes emissivity to be less than 1?

1. Oct 11, 2011

### SDEric

Everything I have read about emissivity simply says that in thermal equilibrium, emitted radiation has to be equal to absorbed radiation.

That's a glib explanation. What about an object that is not in thermal equilibrium with its surroundings? What about an object that is in a vacuum that extends to infinity in all directions? Say it starts at some temperature; it will radiate away energy until it reaches absolute zero.

Why would it radiate energy faster if it is a reflective material versus, say, carbon black? What is going on at the atomic level that makes emissivity equal to absorption? What is it about carbon, for example, that makes it very good both at absorbing and emitting radiation? What is it about a reflective material that makes it not emit as much radiation for a given temperature? Is it that the photons emitted by the thermally vibrating atoms are reflected back or otherwise not allowed to leave the atom?

2. Oct 12, 2011

### Drakkith

Staff Emeritus
If the object is not in equilibrium, then it will either absorb or emit radiation until it reaches it. This doesn't mean that it quits absorbing or emitting, but that the two equal out. A perfect vacuum around an object that causes zero radiation to be given to an object would mean that all of the thermal energy of the object gets converted to radiation and radiated away over a large amount of time.

The properties of absorbtion and emission vary widely, but I believe that in general the two are usually about equal in most objects. I'm not sure on the details of why and how though.

3. Oct 12, 2011

### SDEric

Thanks. I can see why emissivity has to be equal to reflectance, from a thermodynamic perspective: If they weren't, you could have an object that is good at absorbing and bad at emitting, and vice versa, so if you put those two objects in a room together, heat would flow from one to the other and entropy would decrease. Maybe I am wrong (I did lousy in thermodynamics) but it seems to make sense to me.

That still does not explain why, at an atomic level, emissivity has to equal absorption. Why would the energy of emitted photons necessarily equal the energy of absorbed photons?

Someone suggested that if you heat a teflon frying pan, you will feel more heat from the teflon side than the aluminum side. Why does the teflon side emit more photons than the aluminum? Certainly both sides are emitting more photons, more photon energy, than they are absorbing. Both sides are the same temperature; why does the teflon side emit more energy?

Last edited: Oct 12, 2011
4. Oct 13, 2011

### lifestil

Fundamentally, emissivity and absorption are time symmetric--the process looks identical forward and backwards--because the mechanism in blackbody radiation is time reversal invariant.

Blackbody radiation is derived using an EM field in thermal equilibrium with quantum oscillators. The process of absorption and emission are due to the same physical laws which are the same forwards and backwards. This property is also called Kirchoff's Law, and can be explained by a simple yet subtle fact: if the blackbody surface absorbed more photons of a certain wavelength than it emitted, there would be a net loss of them over time and the spectrum would no longer be that of a blackbody!

A true blackbody is a hollow cavity with a small hole in it: photons that enter are trapped, and photons that leave obey the blackbody spectrum. By definition, it has emissivity of 1. Everything real object only approximates a blackbody, and so will have emissivity less than 1.

You are confused on two other points, I think:

1) When a photon is absorbed, and identical photon is not necessarily emitted. The process is statistical in nature, so that on average the energy flux in and out will be the same for a blackbody in equilibrium with its surroundings. For example, a high energy photon could be absorbed, and that energy emitted sometime later as two lower energy photons. All that matters is an energy balance on a reasonable time scale.

2) In the pan example, the hot pan is radiating photons. Remember that even if it were in thermal equilibrium with the surroundings, it would STILL emit photons. In this case, though, it emits more than it receives because it's not in thermal equilibrium. Emissivity is higher for the teflon side because (presumably) it's darker, and so better approximates a blackbody and has higher emissivity, and so more photons.

5. Oct 13, 2011

### lifestil

Oops, I should have said for point 1)

All that matters is an energy balance and maintaining the blackbody spectrum on a reasonable time scale.

6. Oct 13, 2011

### SDEric

Thanks! I think I understand Kirchoff's Law pretty well; it's a necessary result from the conditions of thermal equilibrium. I also understand both (1) and (2). My question is regarding what happens when not at thermal equilibrium; I think you hinted at the answer, and to understand this I need to understand the mechanism of emission and absorption. The only way emissivity and absorptance are automatically equal is if, as you say, they are fundamentally the same process. I don't know what that process is.

What makes carbon atoms a good emitter and silver atoms a poor emitter?

Is it because the free electrons in the surface of the silver reflect the photons emitted by the thermally jiggling silver ions - that is, they reflect the photons back into the silver?

Do the photons emitted by the carbon atoms either strike another carbon atom or wander free?

Presumably if you got a neutron hot it would not radiate photons.

Again, I can see why this has to be true from a thermodynamic perspective, but it seems like it also has to be true from a mechanical perspective. Thermodynamics does not tell the individual atoms and photons how to behave.

Thanks for your help.

7. Oct 13, 2011

### Drakkith

Staff Emeritus
It boils down to moving charged particles. A photon enters the material and interacts with the EM fields of the particles in the atoms. Once this photon is absorbed the energy is transferred to whatever absorbed it, whether thats the entire atom or molecule or just one electron.

On the flip side, thermal energy (heat) causes the particles to move and vibrate and other things. Since they are moving and they are charged they will send out EM radiation.

As for not being in thermal equilibrium, I don't understand what your question is. It takes time to absorb enough energy for the object to heat up enough to reach equilibrium and it also takes time for a hot object to radiate and cool down to equilibrium.

I think that answer is fairly complicated. Not only is it the actual atoms that cause it to vary, but also the way they are put together. For example, diamond has very different properties than graphite, yet both are made up of carbon atoms.

I think it would. Neutrons are composed of electrically charged Quarks so they should radiate if they get excited.

8. Oct 13, 2011

### fluidistic

I'd like someone to confirm this. If this is true (actually that makes sense to me!), then I wonder why it took so much time to realize that neutrons weren't elementary particles.

Edit: About emissivity and absorption coefficients, maybe see https://www.physicsforums.com/showthread.php?t=537456, especially the last post. This isn't a quantum description though.

9. Oct 13, 2011

### Astronuc

Staff Emeritus
neutron = (udd)
http://hyperphysics.phy-astr.gsu.edu/Hbase/particles/proton.html#c3

A free neutron decays into a proton, electron and anti-neutrino.
http://hyperphysics.phy-astr.gsu.edu/Hbase/particles/proton.html#c4

10. Oct 13, 2011

### fluidistic

11. Oct 13, 2011

### Drakkith

Staff Emeritus
Yeah I'm not sure it would radiate simply from being accelerated since as far as I know neutral matter such as atoms don't radiate like that.

12. Oct 14, 2011

### vela

Staff Emeritus
A neutron in the first excited state is what's called the Δ0 baryon. Both consist of one up quark and two down quark, but the Δ0 is in a higher energy state. This would be analogous to an atom in an excited state. The mass of the Δ0 is about 300 MeV higher than the mass of the neutron, so you'd have to hit a neutron with a very high energy photon before you saw any sort of interaction. Going in the other direction, when the Δ0 decays, it usually does it through a strong interaction, but to get a photon out, you need an electromagnetic interaction. The branching fraction for that type of decay is pretty small, on the order of 0.5%.

The magnetic moment of the neutron, however, was an indication that the neutron had substructure.

Last edited: Oct 14, 2011