# Thermal radiation, Kirchhoff's law, and black bodies (1/4)

By 0x7df, Sun 17 May 2015, modified Fri 18 September 2020, in category Physics All matter continuously emits electromagnetic radiation as a consequence of its temperature. This radiation is called thermal radiation or heat radiation (although of course it isn't intrinsically different from electromagnetic radiation generated by any other means). Thermal radiation is what makes thermal imaging possible, and why hot embers glow, etc. From our everyday experience and from experimentation we can see that both the wavelength and intensity of radiation emitted depend in some way on the temperature of the matter.

We can understand a lot about the properties of thermal radiation from thought experiments, in particular by considering a hollow enclosure of any shape, whose walls are opaque to radiation and which are held everywhere at a constant temperature. The inner surface emits thermal radiation, and therefore the interior space is filled with a radiation field. The walls also absorb radiation. If at a particular time a small volume $$dv$$ of space around a point $$P$$ contains an amount of radiation $$Q$$, then the quantity $$Q/dv$$ is called the energy density at point $$P$$. We can show that the energy density in a constant-temperature enclosure is independent of the nature of the walls of the enclosure, and depends only on the temperature of the walls.

First, imagine two such containers, $$A$$ and $$B$$, with the same wall temperatures, but in which, for some reason (e.g. the material of the walls perhaps) the energy density is different. It is higher in $$B$$. Then imagine we can bring the two enclosures together to form a single enclosure - perhaps they each have at least one flat face of a given shape and size, which we can match up, and then instantaneously remove these walls so the two cavities are joined into one. If $$B$$ had the higher initial energy density, then the energy density in $$A$$ will begin to increase, and the energy density in $$B$$ will decrease. Correspondingly, the walls of $$A$$ will increase in temperature by absorbing the excess radiation, and the walls of $$B$$ will cool. The result is that we are causing heat to flow from one body to another at higher temperature without doing any work, which contravenes the second law of thermodynamics. From this we can conclude that, if the walls are at the same temperature, then the energy densities must be the same, no matter what. Hence, the energy density in each enclosure is dependent on only the wall temperature.

The result also applies to the energy density in any restricted range of wavelengths,  between $$\lambda$$ and $$\lambda + d\lambda$$. If, when we first conjoin the two enclosures, instead of simply removing the interior walls we replace them with a screen that is transparent to radiation only in the wavelength range of interest, then the situation is the same. Therefore not only must the total energy density be the same in the two enclosures (if their wall temperatures are the same), but the energy density in any given range of wavelengths must also be the same; i.e. the energy in both enclosures must have the same spectrum. This spectrum is called the Planck spectrum or the black-body spectrum, and is evidently a function of temperature only.

## Kirchhoff's Law (1859)

For radiation of wavelengths between $$\lambda$$ and $$\lambda + d\lambda$$, the absorptive power (or absorptivity), $$a_\lambda$$, of a surface is defined as the fraction of the energy incident on the surface that is absorbed. The emissive power, $$e_\lambda$$, is the energy emitted per unit area per unit time (per unit wavelength); such that $$e_\lambda d\lambda$$ is the energy emitted per unit area per unit time.

Knowing this, we can determine that, if a new body is inserted into a constant-temperature enclosure of the kind discussed earlier, then some amount of radiation, $$dQ$$, will be incident on each unit area in each unit time, and an amount $$a_\lambda dQ$$ will be absorbed. Since the nature of the walls of the outer container cannot have any effect on the density or spectrum of the radiation inside the enclosure, then the body cannot either; it must be in equilibrium and the emission per unit area per unit time - $$e_\lambda d\lambda$$ - must equate to the absorption. Hence:

$$a_\lambda dQ = e_\lambda d\lambda$$

and:

$$\frac{dQ}{d\lambda} = \frac{e_\lambda}{a_\lambda}$$

Because $$dQ$$ depends only on the temperature, then for a given temperature both sides of the above equation are equal to a constant, whose value depends on the temperature and the wavelengths in question, but not on the composition of the body. This is Kirchhoff's law:

The ratio of the emissive to absorptive power for radiation of a given wavelength is the same for all bodies at the same temperature.

A way of clarifying this is to compare two different bodies, placed separately in the interior. The body discussed above, has equilibrium state:

$$a_\lambda dQ = e_\lambda d\lambda$$

but if this is replaced with a different body having different surface properties $$a_\lambda'$$ and $$e_\lambda'$$, then the new equilibrium state is:

$$a_\lambda' dQ = e_\lambda' d\lambda$$

The ratios $$a_\lambda/e_\lambda$$ and $$a_\lambda' / e_\lambda'$$ must clearly be equal. Alternatively, you can think of a body inside and therefore in equilibrium with one enclosure, being instantaneously transferred to a different enclosure at the same temperature, and arrive at the same conclusion again.

## Black bodies and perfect radiators

One interesting consequence of the fact that:

$$a_\lambda dQ = e_\lambda d\lambda$$

is that a stronger absorber (larger $$a_\lambda$$) is also a stronger emitter (larger $$e_\lambda$$), at a given temperature and for a given wavelength of radiation. In fact a perfect absorber, for which $$a_\lambda = 1$$, and which we therefore refer to as a black body, also radiates as intensely as it's possible to do so under the given conditions; so it is also sometimes called a full or perfect radiator. An enclosure with a small hole acts nearly as a block body, absorbing radiation incident on it with a very small probability of escape (as long as the walls absorb a non-zero fraction of the radiation incident on them).

A black body is a theoretical construct - a substance like lamp black is an excellent but not perfect absorber - but one way to approach a black body very nearly, is by constructing a sealed enclosure with only a small hole through the walls. Radiation incident on this hole has a very tiny probability of escaping again (i.e. by being reflected off the interior walls) as long as the hole is small in comparison with the dimensions of the enclosure; therefore this system acts as a perfect absorber or black body.

Interestingly, such an enclosure doesn't have the characteristics of a black body only in terms of its ability to absorb the radiation incident on it; radiation emitted from the hole also has identical characteristics to radiation emitted from a black body. We can see this from the fact that, if we were to place a black body of the same temperature into the interior of the cavity, then for it to remain at this original temperature (which it must), the radiation emitted from the cavity walls must be incident on the black body inside it with just the same rate as the rate at which the black body is emitting radiation itself. Hence, the radiation in the enclosure - or any enclosure with constant wall temperature - is black-body radiation.

Since the radiation in the interior of any constant-temperature enclosure has the intensity of black-body radiation, then another way of expressing Kirchhoff's law is that:

The ratio of the emissive power of a body to the emissive power of a black body at the same temperature is equal to the absorptive power of the body.

(Recall that the absorptive power is just a number between 0 and 1). This ratio is also called the emissivity, or absorptivity.