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How can be the exponential decay of the intensity $I$ via the optical depth $\tau$ be derived?

$$I(\tau)=I(0)e^{-\tau} $$

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Fundamentally, there are three processes that can affect a beam of light passing through a medium: absorption, emission, and stimulated emission. Emission is independent of the beam of preexisting photons, so we neglect it here. On the other hand, both absorption and stimulated emission (the latter by virtue of quantum mechanical photon statistics) are proportional to beam strength, so we can group them together.

Define the attenuation coefficient to be the difference between how much absorption there is per unit length and how much stimulated emission there is: $$ \kappa = n_\mathrm{abs} \sigma_\mathrm{abs} - n_\mathrm{s.e.} \sigma_\mathrm{s.e.}. $$ Here $n$ is a number density and $\sigma$ is the relevant (frequency-dependent) cross section. $\kappa$ is not be confused with opacity (despite using the same letter), the latter of which can be obtained by neglecting stimulated emission and multiplying by mass density.

By construction, $\kappa$ is the probability per unit distance any given photon will be lost from the beam via absorption not compensated by stimulated emission. If $N$ is the number of photons in the beam,1 $$ \frac{\mathrm{d}N}{\mathrm{d}x} = -\kappa N. $$ Since intensity is proportional to photon number, $$ \frac{\mathrm{d}I}{\mathrm{d}x} = -\kappa I, $$ or $$ \frac{\mathrm{d}I}{I} = -\kappa\,\mathrm{d}x. $$ This differential equation is easily solved given $I = I_0$ at $x = 0$: $$ I = I_0 \exp\left(-\int \kappa\ \mathrm{d}x\right). $$

All the physics is done now. Optical depth is simply introduced to simplify that integral inside the exponential. Define $\tau$ according to $$ \mathrm{d}\tau = -\kappa\,\mathrm{d}x, $$ with $\tau = 0$ at $x = 0$. Then clearly $$ I = I_0 \mathrm{e}^{-\tau}. $$

Optical depth has the added advantage of rescaling "depth" from being measured in length to being measured in photon attenuation, which in many problems is the more relevant measure. Ultimately, though, all we did was define a quantity $\tau$ such that the OP's original equation holds.


1 To see how this relation holds, suppose we have a single photon propagating a small distance $\Delta x$. We can picture the photon as having $0$ width and as going through a medium with particles with number density $n$ and cross sectional area $\sigma$. The photon will be lost if and only if it intersects a particle. Equivalently, suppose the photon has cross sectional area $\sigma$ and the absorbers are point-like, with complete absorption happening if and only if the extended photon's path intersects a particle.

Over a distance $\Delta x$, the extended photon will trace out a volume $\sigma \Delta x$. What are the chances it will be absorbed? More importantly, how many times on average will it be absorbed if it were instantly respawned each time? (We ask this because our $1$ photon is a stand-in for a large number $N$ of them, so $2$ expected absorptions over $\Delta x$ really is twice as effective as $1$ absorption.) The answer is simply the expected number of particles in the volume, $n \sigma \Delta x$.

Calling $n \sigma$ by $\kappa$ now, we see that $1$ photon becomes effectively $1 - \kappa \Delta x$ photons over this distance, or $N$ photons become $N - N \kappa \Delta x$. Then $$ \frac{\mathrm{d}N}{\mathrm{d}x} = \lim_{\Delta x\to0} \frac{(N-N\kappa\Delta x)-N}{(x+\Delta x)-x} = -\kappa N, $$ as desired.

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Hi Christ, thanks. I'm still stuck seeing why your first ODE holds. The following steps are cristal clear. –  Jorge May 16 '13 at 19:27
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@Nivalth Take a look at the footnote now. Hopefully that explains things better. (btw, attenuation, mean free paths, scattering, and all sorts of things use very similar reasoning, so it's good to study this derivation and make sure it makes sense.) –  Chris White May 16 '13 at 20:37
    
Understood, thank you very much! –  Jorge May 17 '13 at 6:15

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