# Blackbody Radiation and the Planck Function

A blackbody is an object which absorbs all the light which hits it: hence the name "blackbody". It also emits radiation, in a very particular manner.

#### Total energy emitted per second, at all wavelengths

The total amount of energy radiated per second by a perfect blackbody depends only on its temperature T and area A:

```                                    4
energy per second  =  sigma * T  * A
```
where
```
-8   Joules
sigma  is the Stefan-Boltzmann constant =  5.67 x 10    ---------
s*m^2*K^4

T      is the temperature of the object, in Kelvin

A      is the surface area of the object, in square meters
```

This relationship is called the Stefan-Boltzmann Law.

If we consider a patch of area exactly one square meter, then the energy radiated from it per second is

```                                    4
energy per second  =  sigma * T
per square meter
```

#### Energy emitted per second, as a function of wavelength

A blackbody doesn't emit equal amounts of radiation at all wavelengths; instead, most of the energy is radiated within a relatively narrow band of wavelengths. The location of that band varies with the body's temperature; for example,

```           object                        T (Kelvin)      radiates mostly
------------------------------------------------------------------------
very cold gas in insterstellar space        20             radio

a live human being                         310             infrared

the Sun                                  5,600             visible

interior of nuclear explosion        3,000,000             X-rays
```

The exact amount of energy emitted at a particular wavelength lambda is given by the Planck function:

```
(2*h*c^2) / lambda^5
B    (T)   =    ----------------------
lambda             h*nu/k*T
e           -  1
```

Or, in beautiful typeset format,

In this equation,

```           B     (T)        is the energy (Joules) emitted per second per unit
lambda                 wavelength per steradian from one square meter
of a perfect blackbody at temperature T

T                is the temperature of the blackbody

h                is Planck's constant      =  6.63 x 10^(-34) J*s

c                is the speed of light     =  3.00 x 10^(8)   m/s

lambda           is the wavelength

k                is Boltzmann's constant   =  1.38 x 10^(-23) J/K
```

The Planck function has a distinctive shape: it rises very sharply at short wavelengths (due to the exponential), reaches a peak at some wavelength, then falls gradually at longer wavelengths.

#### Integrating the Planck Function

In order to find the total energy emitted per second in some wavelength range

```              lambda 1  < lambda  < lambda 2
```
from one square meter of a perfect blackbody, one can integrate the Planck function
```                       lambda 1
/          (2*h*c^2) / lambda^5
energy emitted  =      |         ----------------------   d(lambda)
per second           /             h*nu/k*T
per square meter     lambda 2        e           -  1
```

and then multiply by (pi) steradians (we are interested in the energy emitted at all angles from one face of a flat plate, so we integrate over solid angle to cover half the sky; that ends up being the same as multiplying by a factor of (pi) = 6.28...)

Or, in nicely typeset form,

Obviously, if one integrates from the shortest possible wavelength (lambda = 0) to the longest possible wavelength (lambda = infinity), and multiplies by (pi), one ought to end up with the same total energy emitted per second as given by the Stefan-Boltzmann Law,

```                                    4
energy per second  =  sigma * T
per square meter
```