24. Optical depth II: mean free path

24.1. Introduction

Recall the definiton of the vertical optical depth defined by (22.2) and Stull Chapter 2 p. 43 or WH eq 4.17 p. 123:

\[ d\tau_\lambda = \rho \, r \, k_\lambda dz \]

where \(\rho\) is the air density (\(kg\,m^{-3}\)), \(r\) is the gas mixing ratio (kg gas/kg air) and \(k_\lambda\) is the mass absorption coefficient at wavelength \(\lambda\) (\(m^{2}\,kg^{-1}\)).

Stull eq. 2.32 on p. 43 also defines the volume absorption coefficient \(\gamma\) (\(m^2/m^3\)):

\[ \gamma_\lambda = \rho\,r\,k_\lambda \]

With that definiton we can go back to our (8.2) from week 1 and rewrite

\[ \frac{dI}{I} = -n b ds \]

where \(s\ (m)\) is the distance travelled (the path length), \(n\ (\#/m^3)\) is the number denstiy of absorbing particles and \(b\ (m^2)/molecule\) is the absorption cross section per molecule. Using \(\gamma_\lambda\) Beer’s law becomes:

(24.1)\[ \frac{dI}{I} = -\gamma ds \]

and if we’re pointing straight up, \(ds = dz\) so we can integrate (24.1) to get

(24.2)\[ I(z) = I_i \exp (-\gamma z) \]

with the big assumption that the mass absorption coefficient \(\gamma\) is independent of \(z\).

24.2. Photon lifetimes

So given (24.2), what is the probability that a photon is absorbed across a distance \(\Delta z\)?

Let’s call that probability \(p(z) dz\). Where \(p(z)\) is a probability distribution that satisfies the condition that:

\[ \int_0^\infty p(z) dz =1 \]

that is, 100% of all photons are absorbed over an infinite path. What about a shorter path? For a path of length \(\Delta z\), this is going to just be the difference between the fraction that reached \(z\) and the fraction that reaches \(z + \Delta z\).

\[ \int_z^{z+\Delta z} p(z) d z \approx p(z) \Delta z =\exp \left ( -\gamma z\ \right )-\exp \left ( -\gamma (z+\Delta z) \right ) \]

where we’ve assumed that \(p(z)\) is approximately constant over \(\Delta z\) and can be taken out of the integral.

Finally, divide by \(\Delta z\) and take the limit \(\Delta z \rightarrow 0\) and get

(24.3)\[ p(z) = \left [ \exp \left ( -\gamma z\ \right )-\exp \left ( -\gamma (z+\Delta z) \right ) \right ] / \Delta z = - \frac{d}{dz} \exp ( -\gamma z) = \gamma \exp ( -\gamma z ) \]

24.3. Mean free path

Finally, if we know the probability \(p(z)\) that a photon is going to be absorbed after travelling a distance \(z\), we can find the average distance that a photon travels before it is absorbed. The definition of the that average from statistics is just:

\[ \overline{z} = \int_0^\infty z\,p(z)\,dz \]

and using (24.3) that is just:

\[ \overline{z} = \int_0^\infty z \gamma \exp (- \gamma z ) \,dz = \frac{1}{\gamma} \]

24.4. Unit optical depth

We have shown that the volume absorption coefficient \(\gamma\) gives a measure for how far a photon can travel through the atmosphere before being absorbed. What is the optical depth that corresponds to that distance? Do the integration between \(z=0\) and \(z=\overline{z} = 1/\gamma\), again assuming that \(\gamma\) is independent of \(z\):

\[ \tau = \int_0^\overline{z} \gamma dz = \gamma \times \overline{z} = \gamma \times \frac{1}{\gamma} = 1 \]

So on average, the photons can travel through an optical depth of 1 before being absorbed.

24.5. Conclusion

How good is the assumption that \(\gamma\) is independent of height? In the atmosphere, not very good, because the gas density \(\rho(z)\) decreases exponentially with height. This is the subject of Walace and Hobbs problem 4.44 and the next lecture.