Energy, Photons, Blackbodies etc.

The energy of a photon of frequency $\nu$ is given by

$$E=h~\nu=\frac{h~c}{\lambda}~,$$

where

$$h\simeq6.626\cdot10^{-34}~\mathrm{J \: / \: s}~,$$

is Planck's constant and

$$c\simeq2.999\cdot10^8~\mathrm{m \: / \: s}~,$$

is the speed of light in a vacuum.
In spectroscopy, the energy of a photon is often expressed by the inverse of its wavelength, or Wave Number

$$\tilde{\nu}=\frac{1}{\lambda}=\frac{\nu}{c}~,$$

which is generally expressed in $\mathrm{cm}^{-1}$. When using an actual energy unit is appropriate, one generally uses the electronvolt

$$1~\mathrm{eV} = 1.602 \cdot 10^{-19} \mathrm{J}~,$$

i.e. the amount of energy equivalent to that gained by a single unbound electron when it is accelerated through an electrostatic potential difference of one volt (in a vacuum). In other words, an electronvolt is equal to one volt (1 volt = 1 joule per coulomb) multiplied by the (unsigned) charge of a single electron. The energy of a photon is related to its wavelength by the following

$$E_{[\mathrm{eV}]}=\frac{1.24}{\lambda_{ [ \mu \mathrm{m}]}}~.$$

Planck's law describing the radiation from a blackbody says that the frequency specific surface brightness, i.e. the energy emitted per unit time, unit area (of the emitter's surface), unit solid angle and unit frequency interval, of a blackbody at a temperature $T$ is

$$B_\nu(\nu,T)=\frac{2\,h\,\nu^3}{c^2}\left[\exp\left(\frac{h\nu}{kT}\right)-1\right]^{-1}~~~[\mathrm{J \: / \: s \: m^2 \: sr \: Hz}]~,$$

whereas the corresponding wavelength specific surface brightness, i.e. the energy emitted per unit time, unit area (of the emitter's surface), unit solid angle and unit wavelength interval (which is related to the previous quantity by Equation 25) is

$$B_\lambda(\lambda,T)=\frac{\nu^2}{c}\,B_\nu(\nu,T)= \frac{2\,h\,c... ...}{\lambda kT}\right)-1\right]^{-1}~~~[\mathrm{J \: / \: s \: m^2 \: sr \: m}]~,$$

where

$$k\simeq1.38 \cdot 10^{-23} \mathrm{J \: / \: K}~,$$

is Boltzmann's constant.
The corresponding specific brightness at the receiver, i.e. the energy received per unit time, per unit area (of the receiver's surface) per unit frequency/wavelength interval from the whole blackbody (assuming this has a spherical shape) is instead

$$\mathcal{SB}_{\nu\,(\lambda)}=\pi~B_{\nu\,(\lambda)}\,\left(\frac{R}{r}\right)^2~,$$

where $R$ is the radius of the blackbody, $r$ the distance between the blackbody and the observer, and the $\pi$ factor arises from integration.
Integration of one of the aforementioned formulae yields the energy emitted per unit time per unit area (of the emitter's surface) at all frequencies/wavelengths. This turns out to depend on the temperature of the blackbody only, a result also known as Stefan-Boltzmann's law

$$F_{bb}=\sigma\,T^4~~~[\mathrm{J \: / \: s \: m^2}]~,$$

where

$$\sigma = \frac{2\,\pi^5}{15}\,\frac{k^4}{c^2\,h^3} \simeq 5.67\cdot10^{-8}~\mathrm{J \: / \: s \: m^2 \: K^4}$$

is Stefan-Boltzmann's constant.
The corresponding brightness at the receiver, i.e. the energy received per unit time, per unit area (of the receiver's surface) at all frequencies/wavelengths from the whole blackbody (assuming this has a spherical shape) is instead

$$\mathcal{B}=\pi~F_{bb}\,\left(\frac{R}{r}\right)^2~,$$

where $R$ is the radius of the blackbody, $r$ the distance between the blackbody and the observer, and the $\pi$ factor arises from integration.
Similarly, the derivation of the expression given for $B_{\lambda}$ yields the relation between the wavelength of its maximum and the temperature, or Wien's displacement law

$$\lambda_{max}\,T \simeq 2.898 \cdot 10^{-3}\,\mathrm{m\:K}~.$$

The Emissivity $\varepsilon$ of a body is defined as the ratio

$$\varepsilon_{\lambda}=\frac{F_{true,\lambda}}{F_{bb,\lambda,T}}~,$$

between its wavelength specific brightness and the wavelength specific brightness of a blackbody at the same temperature $T$. The Absorptivity of a body is similarly defined as the ratio of the energy absorbed by a body and by a blackbody at the same temperature. Kirchoff's law states that at thermal equilibrium, the emissivity of a body equals its absorptivity. In general, the emissivity of a body is wavelength-dependent, but by definition a blackbody has got $\varepsilon_{\lambda}\equiv\varepsilon$. By extension, a graybody is a body for which $\varepsilon_{\lambda}\equiv\varepsilon$, and thus $F_{true,\lambda}=\varepsilon~F_{bb,\lambda,T}$.
In order to describe dust emission astronomers often use a modified graybody

$$S_\nu = K~\nu^\beta~B_{\nu,T_d}~,$$

where $K$ is a constant, and $\beta$ (whose value is between 1 and 2) is called the dust emissivity index. For such a modified graybody, in other words, $\varepsilon_\nu=K~\nu^\beta$.