# Calculate how far we could see through Earth's atmosphere if it had the opacity of the solar photosphere?

I've newly started studying astrophysics. There is a question in the book "An introduction to modern astrophysics" by W. Carroll: Calculate how far we could see through Earth's atmosphere if it had the opacity of the solar photosphere?

I was wondering if someone could guide me about it.

What I've tried: The value for the Sun's opacity at a wavelength of 500 nm is $$\kappa_{500} =0.03m^2kg^{-1}$$ and the density of Earth's atmosphere equals to $$1.2kgm^{-3}$$. The characteristic distance traveled by light before it can be absorbed by the photosphere is $$l=\frac{1}{k_{\lambda}\rho}=\frac{1}{0.03\times 1.2}=27m$$. Now we need to calculate the optical depth $$\tau_{\lambda}$$. But I don't have any ideal how to calculate it by $$\tau_{\lambda}=\int_0^s \kappa_{\lambda}\rho ds$$.

The writer says in the book that the optical depth may be thought of as the number of mean free path from the original position to the surface, as measured along the ray's path. But where do we know this number from?

• Issue: The usual modelling of absorption and scattering that results in opacity will use a negative exponential I = I₀exp(-k p x) That means that some light always gets through (even if not enough to be seen with our eyes) So this needs care on what "see through the atmosphere" means. We could see the sun through much more atmosphere than we could see Sirius. Oct 22, 2022 at 17:22
• – uhoh
Oct 23, 2022 at 1:58

If you want to be an astrophysicist, the first thing you'll have to do is to abandon SI units, and start using cgs units$$^\dagger$$. In these units the Sun's opacity at 5000 Å is $$\kappa_{5000} = 0.3\,\mathrm{cm}^2\,\mathrm{g}^{-1}$$ and the Earth's density is $$\rho = 1.2\times10^{-3}\,\mathrm{g}^2\,\mathrm{cm}^{-3}$$.

### Mean free path

You can calculate the answer in different ways, and you already did calculate it in terms of the mean free path $$\ell = \frac{1}{\kappa\rho} = \frac{1}{n\sigma}.$$ The second equality expresses the absorption as number density $$n$$ of particles, each of cross-sectional area $$\sigma$$. In other words, $$\kappa$$ and $$\sigma$$ express the same thing, namely a probability of being absorbed, but per unit mass and unit number, respectively).

The equation gives you the distance a beam of light can travel before the number of particles in the beam, times their cross section, equals the area $$A_\mathrm{beam}$$ of that beam. In other words, how far before you have encountered enough particles to completely cover your line of sight (LOS).

### Optical depth

Particles are not perfectly arranged to cover your LOS, but can "be behind each other"$$^\ddagger$$. If your beam travels a very short distance $$ds$$, so short that no particles "cover each other", it will cover a volume $$V = A_\mathrm{beam}\,ds$$, and the total number of particles inside this volume will be $$N = nV = n A_\mathrm{beam}\,ds.$$ The total area covered by particles is then $$A_\mathrm{par} = N\sigma.$$ The covered area fraction tells you the probability of being absorbed a particle: $$dP_\mathrm{abs} = \frac{A_\mathrm{par}}{A_\mathrm{beam}} = n\,\sigma\,ds.$$

The optical depth $$\tau$$ is defined as the integral of this probability over some distance $$s$$: $$\tau = \int_0^s\,n\sigma\,ds' = \int_0^s\,\kappa\rho\,ds'.$$ In general this can be a complex integral, but in the case of a constant density, the integral just becomes $$\int_0^s\,ds = s$$, so the optical depth is $$\tau = n\sigma s = \kappa\rho s.$$

If the intensity $$I$$ of the light in the beam decreases by $$dI$$ by traveling a distance $$ds$$, then $$dI = -In\sigma ds.$$ This differential equation has the solution $$I(s) = I_0 e^{-n\sigma s} = I_0 e^{-\tau},$$ where $$I_0$$ is the intensity of the beam before entering the medium.

Thus we see that the physical interpretation of $$\tau$$ is the distance that a beam of light can travel before its intensity is decreased by a factor $$e$$. Or we can say "before it's significantly decreased".

Very often, a medium will have either $$\tau\ll 1$$ or $$\tau\gg 1$$, which we call "optically thin" and "optically thick", respectively, or just "transparent" or "opaque". The intermediate distance corresponding to $$\tau=1$$ is the distance you can see in the medium before it becomes too opaque to see any further. This distance is $$\begin{array} {} 1 & = & n \sigma s = \kappa\rho s\\ & \Rightarrow & \\ s & = & \frac{1}{n \sigma} = \frac{1}{\kappa\rho}, \end{array}$$ and defines the mean free path (which we called $$\ell$$ above).

So now you have the origin of your result.

$$^\dagger$$Except for wavelengths in the UV/optical which should be in Ångström (Å), brightnesses which should be in negative $$\log_{2.51}$$ units, metallicities in log(O/H) + 12 units, and numerous other historical mishaps…

$$^\ddagger$$Though at the particle level, you shouldn't really think of each particle acting like a little disk that absorbs everything inside its are, and doesn't absorb anything outside its area. Rather each particle has "a probability of absorbing sufficiently nearby photons". This probability can then be expressed as a classical cross section with units of area.

• Thank you so much for your perfect and detailed explanation. I understood it properly. I appreciate for your nice comments too. Oct 24, 2022 at 13:07
• @M.Ramana You're welcome :)
– pela
Oct 24, 2022 at 13:31