0 1 2 3 4 5 6 7 8
x 10−6010002000300040005000600070008000900010000wavelength / minternal energy density u / J/m3T=2000 K
T=2200 KFigure 5: Planck radiation - internal energy as a function of wavelength atT= 2000KOut of these resultsPlanck’s radiation lawcan be derived. From the internal energyu=
∫∞
0 u(ω)dω
the energy density can be found with substitutingx=k~BωT:
u(ω) =~ω
ω^2
c^3 π^21
e
k~ω
BT− 1dωThis is Planck’s radiation law in terms ofω. The first term is the energy of one mode, the second
term is the density of modes/states, the third term is theBose-Einstein factor, which comes from
the geometric series. Planck’s law can also be written in terms of the wavelengthλby substituting
x=λkhcBT, which is more familiar (see also fig. 5):
u(λ) =
8 πhc
λ^51
ehc
λkBT − 1dλThe position of the peak of the curve just depends on the temperature - that isWien’s displacement
law. It is used to determine the temperature of stars.
λmax=
2897 , 8 · 10 −^6 mK
T4.1 Thermodynamic Quantities
After an integration over all frequencies (all wavelengths) we get the thermodynamic quantities. The
free energyF has the form
F=−kBTln(Z) =
− 4 σV T^4
3 c