Beer’s Law

One of the more pleasant aspects of laser physics is the existence of Beer’s law, a lovely little law with a fun name.

Up to now, when considering the interaction between light and matter, we’ve mostly focused on the perspective of the matter.  We’ve written rate equations for the density of atoms in excited states, for example.  The other perspective is that of the light.  Every time light is absorbed, an atom is promoted to an excited state, and a photon is destroyed, so that the optical energy is reduced by $latex h \nu$.  Similarly, when spontaneous or stimulated emission occurs, an atom drops out of an excited state, and a photon is created, so that the optical energy is increased by $latex h \nu$.

Let’s look at what happens to the intensity of a beam of light which is propagating in the $latex z$ direction as it travels some distance $latex \Delta z$ within a gas of atoms.  Let’s assume that the light’s photon energy matches the energy of an optically-active transition of the atoms, so that the beam interacts with the gas.  As described above, an optical transition that increases (decreases) the excited state density $latex N_2$ will decrease (increase) the density of photons $latex N_{ph}$ by the same amount.  So let’s try

$latex \displaystyle \frac{dN_{ph}}{dt} = – \frac{dN_2}{dt} \biggr \rvert_{radiative} $

However, not all optical processes are created equal from the beam’s point of view.  Absorption will remove a photon from the beam, and stimulated emission will add a photon to the beam with the same frequency, phase, and direction.  Spontaneous emission, on the other hand, creates a photon with random phase and direction – most likely, it won’t be traveling in the same direction as the beam.  Therefore, we’re going to neglect the spontaneous emission rate, and say

$latex \displaystyle \frac{dN_{ph}}{dt} = \frac{\sigma(\nu) I_{\nu}}{h \nu} \left[ N_2 – \frac{g_2}{g_1} N_1 \right] $

So now we have to relate the change in the density of photons in the beam $latex dN_{ph} / dt$ to the change in the intensity as it travels distance $latex \Delta z$.  If we multiply $latex dN_{ph} / dt$ by the energy per photon $latex h \nu$, we arrive at the change in energy density with time.  Assuming this change occurs uniformly within the region of the gas that we’re considering, we can say that the change in intensity is given by

$latex \displaystyle I_{\nu} (z + \Delta z) – I_{\nu} = h \nu \frac{dN_{ph}}{dt} \Delta z$


$latex \displaystyle \frac{I_{\nu} (z + \Delta z) – I_{\nu}}{\Delta z} = \sigma(\nu) I_{\nu} \left[ N_2 – \frac{g_2}{g_1} N_1 \right]$

Taking the limit as $latex \Delta z \rightarrow 0$, we have

$latex \displaystyle \frac{d I_{\nu}}{dz} = \sigma(\nu) I_{\nu} \left[ N_2 – \frac{g_2}{g_1} N_1 \right]$

If we define the optical gain per length as:

$latex \displaystyle \gamma(\nu) \equiv \sigma(\nu) [N_2 – (g_2 / g_1) N_1]$

we have

$latex \displaystyle \frac{d I_{\nu}}{dz} = \gamma(\nu) I_{\nu} $

$latex \gamma$ has units of (1 / length).  Assuming $latex \gamma(\nu)$ is constant with position, the solution to this equation is an exponential:

$latex \displaystyle I_{\nu}(z) = I_{\nu}(0) \exp{(\gamma z)} $

This is Beer’s law.  If $latex (g_2 / g_1) N_1 > N_2$, $latex \gamma$ will be negative.  We also know physically that absorption dominates over stimulated emission in this case.  Both the math and the physics tell us that the light intensity will decay exponentially as it propagates through the gas.  This is the ‘normal’ case: in thermal equilibrium, the density of atoms in the lower energy state $latex N_1 > N_2$.

On the other hand, if $latex N_2 > (g_2 / g_1) N_1$, stimulated emission dominates over absorption, and the light intensity will grow exponentially as it propagates through the gas.  This is the necessary condition for optical amplifiers and lasers.  But to achieve this, we need to inject energy somehow to promote a large number of atoms into the upper energy state (excited state); this is called ‘pumping’ the gas.  The condition where $latex N_2 > N_1$ is referred to as ‘inversion’.