Why do spontaneous emission and absorption alone violate thermodynamic laws? Einstein suggested that stimulated emissions must also occur along with spontaneous emission and absorption, because the latter two alone violate thermodynamic laws.
How exactly do spontaneous emission and absorption alone violate thermodynamic laws and which ones? And how does introducing stimulated emission solve this?
 A: Consider an ensemble of two level system with energies $E_2>E_1$ such that $E_2 - E_1 = \hbar\omega$. Let their populations be $n_1$ and $n_2$. The process of an $E_2$ atom emitting a photon is called spontaneous emission and let its associated probability be $A$.
With the benefit of hindsight (it will become clear in the end) let us also consider a process where a photon of frequency $\omega$ induces an $E_2$ atom to emit a photon and reach $E_1$. This is the stimulated emission with the associated probability $B$ times the intensity of light $I(\omega)$.
The final process under consideration is the absorption which increases the $n_2$ population, with its associated probability $B'$ times $I(\omega)$.
With these processes, the rate of change of population $n_2$ in the ensemble will be given by the following:
$$dn_2 = -\left[A + BI(\omega)\right]n_2+ B'I(\omega)n_1$$
Under the equilibrium condition we must have $dn_1 + dn_2 = 0$, furthermore the level populations for an ensemble of $N$ atoms will be given by the Boltzmann distributions: $n_i = Ne^{-\beta E_i}$. Plugging this into our rate equation and rearranging gives us:
$$I(\omega) = \frac{A}{B'e^{\beta\hbar\omega}-B}$$
But from Planck has shown that for this system the intensity goes as :
$$I(\omega) = \frac{\hbar}{\pi^2}\frac{\omega^3}{e^{\beta\hbar\omega}-1}$$
As you can see, without the stimulated emission ($B=0$), we cannot get back the Plank distribution for a blackbody.
A: Without stimulated emission, the only transition process that would be taking molecules down to lower energy states would be spontaneous emission, which does not depend on EM radiation intensity, and so would not be strong enough to keep molecules predominantly in the lower energy states they should be in if they are to be in thermodynamic equilibrium at some temperature $T$ (Boltzmann's distribution).
Let's look at a two-level system with populations $N_1$ and $N_2$. If there was no stimulated emission, rate of increase of population of the excited state would be
$$
\frac{dN_2}{dt} = N_1 B\rho -N_2A .
$$
Dynamical equilibrium happens when this rate is zero, which is when
$$
N_1 B\rho = N_2 A.
$$
From statistical physics, we expect that in thermodynamic equilibrium, whatever the temperature, $N_1 > N_2$ (Boltzmann's distribution). But our system without stimulated emission behaves differently;if EM radiation intensity $\rho$ is high enough (higher than $\frac{A}{B}$), equilibrium is possible only when $N_2$ becomes bigger than $N_1$. That is, if radiation is stronger than some limit, our system population will invert, higher energy states will have bigger population, which is against the standard idea of thermodynamic equilibrium.
When stimulated emission is kept in the model, the rate equation is
$$
\frac{dN_2}{dt} = N_1 B\rho -N_2A - N_2B\rho
$$
and the inversion in dynamical equiliubrium does not happen, no matter how high $\rho$ is.
