# Emissivity

Blacksmiths work metal that is hot enough to emit plainly visible thermal radiation.

The emissivity of the surface of a material is its effectiveness in emitting thermal radiation compared to that of a "black body" at the same temperature; emissivity (usually written ε or e) is a simple number that is between 0 and 1. Thermal radiation is light, but for objects near room temperature (20 Celsius) this light is infrared and isn't visible. The thermal radiation from very hot objects (see photograph) is easily visible to the eye. There is a fundamental relationship (Kirchhoff's law of thermal radiation) that equates the absorption of incident light (the "absorptivity" of a surface) with the emissivity; Kirchoff's Law explains why emissivities cannot exceed 1, since the largest absorptivity - corresponding to complete absorption of all incident light by a truly black object - is also 1.[1] Mirror-like, metallic surfaces that reflect light well thus have low emissivities. A polished silver surface has an emissivity of about 0.01 near room temperature. Black soot has an emissivity as large as 0.97.[2][3]

## Explanation

Emissivity depends on factors such as temperature, emission angle, and wavelength. A typical physical assumption is that a surface's spectral emissivity and absorptivity do not depend on wavelength, so that the emissivity is a constant. This is known as the "gray body assumption".

Although it is common to discuss the "emissivity of a material" (such as the emissivity of highly polished silver), the emissivity of a material does in general depend on its thickness. The emissivities quoted for materials are for samples of infinite thickness (which, in practice, means samples which are optically thick) — thinner samples of material will have reduced emissivity.

When dealing with non-black surfaces, the deviations from ideal black body behavior are determined by both the geometrical structure and the chemical composition, and follow Kirchhoff's law of thermal radiation: emissivity equals absorptivity (for an object in thermal equilibrium), so that an object that does not absorb all incident light will also emit less radiation than an ideal black body.

Most emissivities found in handbooks and on websites of many infrared imaging and temperature sensor companies are the type discussed here, total emissivity. However, the distinction needs to be made that the wavelength-dependent or spectral emissivity is the more significant parameter to be used when one is seeking an emissivity correction for a temperature measurement device.

Thus, it is important to understand the distinction between total and spectral emissivity and where they apply.

## Emissivity of Earth's atmosphere

The emissivity of Earth's atmosphere varies according to cloud cover and the concentration of gases that absorb and emit energy in the thermal infrared (i.e., wavelengths around 8 to 14 micrometres). These gases are often called greenhouse gases, from their role in the greenhouse effect. The main naturally-occurring greenhouse gases are water vapor, carbon dioxide, methane, and ozone. The major constituents of the atmosphere, N2, O2, and Ar, do not absorb or emit in the thermal infrared.

## Astrophysical graybody

The monochromatic flux density radiated by a greybody at frequency $\nu$ through solid angle $\Omega$ is given by $F_{\nu} = B_{\nu}(T) Q_{\nu} \Omega$ where $B_{\nu}(T)$ is the Planck function for a blackbody at temperature T and emissivity $Q_{\nu}$.

For a uniform medium of optical depth $\tau_{\nu}$ radiative transfer means that the radiation will be reduced by a factor $e^{-\tau_{\nu}}$. The optical depth is often approximated by the ratio of the emitting frequency to the frequency where $\tau=1$ all raised to an exponent β. For cold dust clouds in the interstellar medium β is approximately two. Therefore Q becomes,

$Q_{\nu}=1-e^{-\tau_{\nu}}=1-e^{-\tau_0 (\nu / \nu_{0})^{\beta}}$. ($\tau_0=1$, $\nu_0$ is the frequency where $\tau_0=1$)

## Emissivity between two walls

Given two parallel walls whose facing surfaces have respective emissivities $\varepsilon_1$ and $\varepsilon_2$ at a given wavelength, a certain fraction of the radiation of that wavelength just inside one wall will leave that wall and enter the other. By Kirchhoff's law of thermal radiation for a given wavelength, whatever portion of the radiation incident on a surface, from either side, that does not pass through the surface as emission to the other side, is reflected. When this reflected radiation is neglected, the proportion of radiation emitted from the first wall is $\varepsilon_1$, and the proportion of that entering the second wall is therefore $\varepsilon_1\varepsilon_2$.

When reflection is taken into account, what does not enter the second wall is reflected back to the first wall, initially in an amount $\varepsilon_1(1-\varepsilon_2)$. A fraction $1-\varepsilon_1$ of this is then reflected back to the second wall, thereby augmenting the original emission from the first wall. These reflections bounce back and forth in diminishing quantity. Solving for the steady state then gives as the total proportion of radiation entering the second wall

${\varepsilon}_{1,2}=\frac{1}{{\frac{1}{\varepsilon_1}}+{\frac{1}{\varepsilon_2}}-1}=\frac{\varepsilon_1\varepsilon_2}{\varepsilon_1+\varepsilon_2-\varepsilon_1\varepsilon_2}$

This formula is symmetric, and the proportion of radiation just inside the second wall that enters the first wall is the same. This is true regardless of what reflections and absorptions take place inside the two walls away from their facing surfaces, since the formula only concerns the radiation leaving one wall for the other.

The quantities in these formulas are intensities rather than amplitudes, the appropriate choice when the walls are many wavelengths apart as the reflected and transmitted beams will then combine incoherently. When the walls are only a few wavelengths apart, as arises for example with the thin films used in the manufacture of optical coatings, the reflections tend to combine coherently, resulting in interference. In such a situation the above formula becomes invalid, and one must then add amplitudes instead of intensities, taking into account the phase shift as the gap is traversed and the phase reversal that occurs with reflection, concerns that did not arise in the incoherent large-gap or thick-film case.