Gray body

A gray body in terms of radiation physics is a body whose surface does not completely absorb incident radiation and accordingly does not emit maximum radiation ( black body radiation ) at a given temperature (see Planck's law of radiation ). However, it has a wavelength- independent degree of emission or absorption - it appears “gray”, whereby the missing “color” does not refer to the visible, but to the range of the spectrum that is relevant for the measurement .

Due to Wien's law of displacement , a wavelength-dependent spectral emissivity leads to a temperature-dependent total emissivity. With many materials and in large temperature ranges, however, the temperature dependence of is so small that it can be neglected. ${\displaystyle \varepsilon _{T}}$

But there are exceptions: on metal surfaces , the change in the spectral distribution at low temperatures has such an effect that it is almost proportional to the temperature . As a result, the radiation is not only proportional to , but almost proportional to . ${\displaystyle \varepsilon _{T}}$ ${\displaystyle T^{4}}$ ${\displaystyle T^{5}}$

The numerical value of how "gray" the surface is is expressed by the absorption coefficient - in the corresponding context also referred to as the emission coefficient : ${\displaystyle \varepsilon }$

${\displaystyle 0<\varepsilon <1}$

without the ideal values ​​being achieved:

• ${\displaystyle \varepsilon =0}$ would be an ideal white body
• ${\displaystyle \varepsilon =1}$would be the ideal black body .

As a rule, it depends on the wavelength or the frequency of the radiation: ${\displaystyle \varepsilon }$ ${\displaystyle \lambda }$ ${\displaystyle \nu }$

${\displaystyle \varepsilon _{\lambda }:=\varepsilon (\lambda )}$

${\displaystyle \varepsilon _{\nu }:=\varepsilon (\nu )}$

This becomes the law of the black body ${\displaystyle T^{4}}$

${\displaystyle M^{o}=\sigma \cdot T^{4}}$ because of ${\displaystyle I(\nu )\cdot \mathrm {d} \nu \cdot \mathrm {d} \Omega ={\frac {2\cdot h\cdot \nu ^{3}}{c^{2}}}{\frac {1}{e^{\left({\frac {h\cdot \nu }{k\cdot T}}\right)}-1}}\mathrm {d} \nu \cdot \mathrm {d} \Omega }$

for the gray body (real surfaces):

${\displaystyle M_{\varepsilon _{T}}^{o}=\varepsilon _{T}\cdot \sigma \cdot T^{4}}$ because of ${\displaystyle I_{\varepsilon _{\nu }}(\nu )\cdot \mathrm {d} \nu \cdot \mathrm {d} \Omega =\varepsilon _{\nu }{\frac {2\cdot h\cdot \nu ^{3}}{c^{2}}}{\frac {1}{e^{\left({\frac {h\cdot \nu }{k\cdot T}}\right)}-1}}\mathrm {d} \nu \cdot \mathrm {d} \Omega }$

The weighted means of or , which are equal, correspond to : ${\displaystyle \varepsilon _{T}}$ ${\displaystyle \varepsilon _{\nu }}$ ${\displaystyle \varepsilon _{\lambda }}$

${\displaystyle \varepsilon _{T}={\frac {\int \int \limits _{0}^{\infty }\varepsilon _{\nu }\cdot I(\nu )\cdot \mathrm {d} \nu \cdot \mathrm {d} \Omega }{\int \int \limits _{0}^{\infty }I(\nu )\cdot \mathrm {d} \nu \cdot \mathrm {d} \Omega }}={\frac {\int \int \limits _{0}^{\infty }\varepsilon _{\lambda }\cdot I(\lambda )\cdot \mathrm {d} \lambda \cdot \mathrm {d} \Omega }{\int \int \limits _{0}^{\infty }I(\lambda )\cdot \mathrm {d} \lambda \cdot \mathrm {d} \Omega }}}$