Question

Derive Planck’s law of black body radiation.

Answer 1

Answer:

Due to the linearity of the equations of the electromagnetic field, any of their solutions can be given in the form of a superposition of monochromatic waves, each with a certain frequency $\omega$ . The field energy can be represented as the sum of the energies of the corresponding field oscillators. As is known from quantum mechanics, the energy of the oscillator takes discrete values according to the following formula:

$Ep=h\omega(n-1/2).$

Since the equilibrium radiation is considered, using the canonical Gibbs distribution, we can determine the probability of the state of the oscillator with a given energy:

$W_p=1/Ze^{-E_p/kT};$

The statistical sum Z is equal to

$Z=\dfrac{e^{-1/2(h\omega/kT)}}{1-e^{-h\omega/kT}};$

Free energy is equal to

$\psi =\dfrac{h\omega}{2}+kTln(1-e^{-h\omega/kT});$

For the average (mathematical expectation) energy we use the Gibbs-Helmholtz equation

$\epsilon=\psi-kT\dfrac{\partial\psi}{\partial(kT)}=\dfrac{h\omega}{2}+\dfrac{h\omega}{exp(h\omega/kT)-1};$

The number of standing waves per unit volume in three-dimensional space in the range from $(\omega,\omega+d\omega)$ is equal to

$dn_\omega=\dfrac{\omega^2d\omega}{\pi^2c^3};$

Therefore, for the spectral power density of electromagnetic radiation we obtain:

$u(\omega,T)=\dfrac{h\omega^3}{2\pi^2c^3}+\dfrac{h\omega^3}{\pi^2c^3(exp(h\omega/kT)-1)}$ .