# Non-ideal black body's thermal radiation?

Planck's law neatly describes the thermal radiation of an ideal black body in thermal equilibrium.

So the question is, what about things that are not black bodies? Like an object with a mirror finish? What about a perfect mirror?

This question came up in conversation, and I was despite my considerable qualitative knowledge of physics, stumped to come up with an answer.

• Objects that are not blackbodies can be among other things white bodies (total reflection), transparent bodies (total transmission) and opaque (no transmission, might reflect), if you want to look more into this. Oct 13 '15 at 14:25
• Usually, if a body might not be a perfect black-body, it is called a grey body - just to indicate the not-perfect black-body properties Oct 13 '15 at 14:35
• @Steeven a grey body is usually taken to be something with a frequency-independent emissivity. Oct 13 '15 at 15:30

For objects in thermal equilibrium, the radiance of the object is governed by Planck's law, which relates the temperature to the spectral radiance (that is, power per unit area per unit solid angle per unit spectrum, either wavelength or frequency).

Planck's law is typically given for a blackbody, i.e., an object that is a perfect absorber of all incident radiation. Because of this, the annotation $BB$ will often be subscripted on the radiance to clarify that the radiance computed is valid for a blackbody. $$L_{BB}(\lambda, T) = \frac{2 h c^2}{\lambda^5}\left( e^{\frac{h c}{\lambda k T}} - 1 \right)^{-1}$$ $$\text{ or }$$ $$L_{BB}(\nu, T) = \frac{2 h \nu^3}{c^2}\left( e^{\frac{h \nu}{k T}} - 1 \right)^{-1}$$ More generally, the radiance of an object is the product of the blackbody radiance and that objects emissivity - the ratio of the object's radiance at a given temperature and wavelength (frequency) to the radiance that a blackbody would emit at the same temperature and wavelength, i.e. $$\epsilon = \frac{L(\lambda, T)}{L_{BB}(\lambda, T)}$$ For a blackbody (a perfect emitter), $\epsilon=1$, while for a perfectly non-absorbing surface $\epsilon=0$. Most surfaces are neither perfectly emitting nor non-emitting, and so have values of emissivity $0 < \epsilon < 1$. (In general, $\epsilon$ may itself be a property of temperature and wavelength; if $\epsilon$ is independent of wavelength, the material is said to be a graybody emitter).

Kirchoff's law of thermal radiation is what gives rise to your question. These laws state, among other points, that the absorbance and emittance of a surface in thermal equillibrium must be equivalent, i.e., that $\alpha = \epsilon$. This means that an object which is an excellent absorber (e.g., soot) will also be excellent an excellent emitter, while an object that is an excellent reflector (a mirror) will be a poor emitter since it is a poor absorber.

More generally, Kirchoff's laws state that the change in total energy for a parcel of material in thermal equilibrium must be zero. Therefore, the incident energy must equal the exiting energy, otherwise energy would accumulate in the parcel.

Incident energy can do one of three things; it can

1. scatter (reflect) off of the parcel, leaving in a different direction than with which it arrived
2. transmit through the parcel, without interacting with the material
3. be absorbed by the parcel

Let's label these three parts by $R_{in}$, $T_{in}$, and $A_{in}$.

Exiting energy can come from one of three places; it can

1. Have been scattered off the parcel
2. Have been transmitted through the parcel
3. Have been emitted by the parcel

Here, we will label these parts by $R_{out}$, $T_{out}$, and $E_{out}$.

Since process 1 and 4 are the same process, $R_{out}$ = $R_{in}$. Likewise, process 2 and 5 are the same process, so $T_{out}$ = $T_{in}$. Setting the incident and exiting energy equal, we obtain $$R_{in} + T_{in} + A_{in} = R_{out} + T_{out} + E_{out}$$ which, removing the identical terms noted above, simplifies to $$A_{in} = E_{out}$$ i.e., absorbed energy must be equal to the emitted energy. What is interesting and worthy of particular note is that for a body in thermal equilibrium, this law is true for each and every wavelength, independently.

If we consider an amount of incident energy $I_{in}$, the relationship between each of the components (reflected, transmitted, and absorbed) is that $$R_{in} + T_{in} + A_{in} = I_{in}$$ We can therefore express these quantities in a way that is independent of the total amount of energy: $$R_{in} = \rho I_{in}$$ $$T_{in} = \tau I_{in}$$ $$A_{in} = \alpha I_{in}$$ subject to the constraint that $$\rho + \tau + \alpha = 1$$

Since $I_{out} = I_{in}$ and $E_{out} = A_{in}$, it therefore follows that if we write

$$E_{out} = \epsilon I_{out} = \epsilon I_{in}$$

the emissivity must be related to the absorbance, $$\epsilon = \alpha = 1 - (\rho + \tau)$$ This property is the reason that the self-emission of transparent or reflective materials is less than their opaque and non-reflective counterparts.

• Can you explain what you mean by "What is interesting and worthy of particular note is that for a body in thermal equilibrium, this law is true for each and every wavelength, independently." Oct 13 '15 at 16:53
• So the more mirrored a body is, the less radiation it emits? And a perfectly mirrored body emits none? Oct 13 '15 at 17:30
• @BowlOfRed What it means is that, for a body in thermodynamic equilibrium, $\epsilon(\lambda) = \alpha(\lambda)$ holds for all lambda. One might expect that it would be good enough for $\sum_\lambda \epsilon(\lambda) = \sum_\lambda \alpha(\lambda)$, but the equivalence is stronger than this; it holds for all wavelengths individually, not just for the collection of wavelengths.
– KDN
Oct 13 '15 at 18:11
• @BowlOfRed To elaborate, suppose I have a material that is a strong absorber at 5 microns and a weak absorber at 10 microns. I might suppose that energy absorbed at 5 microns might be strongly emitted at 10 microns; this is not the case. The strong 5 micron absorption corresponds to a strong 5 micron emission, while the weak 10 micron absorption corresponds to a weak 10 micron emission.
– KDN
Oct 13 '15 at 18:19
• @BowlOfRed That being said, if I shine 5 micron light on my object, the object will absorb that 5 micron light strongly and begin to warm up. As it warms, its output at 10 microns will increase. However, this scenario violates the premise that the object is in thermodynamic equilibrium; that assumption dictates that certain relative amounts of each wavelength are falling on the object. If the incident radiation field is out of thermodynamic equilibrium with the object, then the object will gradually change temperature until it has regained equilibrium with its environment.
– KDN
Oct 13 '15 at 18:23

Mirrors and other shinny things still emit black body radiation characterized by their temperature. What is different is the emissivity of the body. https://en.wikipedia.org/wiki/Emissivity

• I would agree that they emit thermal radiation. But some may have a spectrum that deviates considerably from that of an ideal black body. Do we still call it "black body radiation" even if the deviation is significant? Oct 13 '15 at 17:44
• @BowlOfRed, yeah at some point no. There was a device (recently) that would cool itself sitting on roof on a sunny day, because it radiated and absorbed strongly in the IR "window" in the sky. It could "see" the coolness of space. Thermodynamically, I think you can still assign a temperature to a light source even if it is far from a black body. (I don't have a good link for that.) What's the temperature of an LED for instance. Oct 13 '15 at 20:31

Planck's law neatly describes the thermal radiation of an ideal black body in thermal equilibrium.

Correct. The lynch pin here is "thermal equilibrium"

So the question is, what about things that are not black bodies? Like an object with a mirror finish? What about a perfect mirror?

As GeorgeHerold also says, everything in thermal equilibrium emits according to the Planck formula.

The Stefan Boltzmann law describes the energy radiated as a function of temperature:

The Stefan–Boltzmann law, also known as Stefan's law, describes the power radiated from a black body in terms of its temperature. Specifically, the Stefan–Boltzmann law states that the total energy radiated per unit surface area of a black body across all wavelengths per unit time (also known as the black-body radiant exitance or emissive power), j^{\star}, is directly proportional to the fourth power of the black body's thermodynamic temperature T: A body that does not absorb all incident radiation (sometimes known as a grey body) emits less total energy than a black body and is characterized by an emissivity, epsilon < 1 A thermos bottle utilizes the completely reflective body . The inside glass is made completely reflective, there is a vacuum so no convective energy exchanges with the atmosphere, and s whatever temperature is introduced is theoretically retained with no black body radiation as the radiation is trapped. In reality there will not be complete vacuum, there will be infrared radiation going through the vacuum,etc so the content will come into equillibrium at some time.