Calculating the optical depth of an inhomogeous gas

[colorofeternity]

TL;DR

Trying to find the required integral to calculate the optical depth for a cloud with a mass density that follows the inverse square law with a finite density at the center.

My question emerges from my desire to calculate the optical depth, which should be unitless, for an inhomgeneous cloud of radius $r$. For a homogeneous medium, the optical depth can be defined in terms of the density of a cloud relative to the density of the condensed medium:

$$\tau = \alpha \frac{\rho}{\rho_c}l$$

Where $\alpha$ is the linear attenuation coefficient of light at a given wavelength for the material in question, $\rho$ is the density of the cloud and $\rho_c$ is the mass density of the condensed phase (assuming the cloud is all made of one material, lets say Iron). #l# is the path length of light through the cloud.
This form is used when the linear attenuation coefficient of the gaseous form of the material is unknown, as it is for many metallic gases. What I wish to do is calculate the optical depth for light passing through the spherical cloud of radius, $r_0$ given that the mass density of the cloud follows the inverse-square law with a finite mass density at $r = 0$, $\rho_0$ ($kg/m^3$).
My attempt at the required integral is as follows:

$$\tau = \int_{0}^{r_0} \frac{\rho_0}{r^2} dr$$

However, I am unsure about this as this expression would give a diverging density at $r = 0$, which isn't what is happening physically. If I add in an extra $r^2$ term as would be the case for spherical coordinates, then I would end up with a path length of $r_0$, which seems odd. In addition, it seems that the units don't fully add up as I would expect $\tau$ to be unitless.

How should I approach this problem?

 

[member38644]

To calculate the optical depth for an inhomogeneous gas, you can use a similar approach as for a homogeneous medium, but instead of using the density of the cloud relative to the density of the condensed medium, you can use the density of the gas at a given point relative to the average gas density. This will give you a unitless optical depth that takes into account the varying density of the gas throughout the cloud.

The integral you have attempted is not correct as it does not take into account the varying density of the gas. Instead, you can use the following expression:

$$\tau = \int_{0}^{r_0} \frac{\rho(r)}{\bar{\rho}} \alpha(r) dr$$

Where $\rho(r)$ is the density of the gas at a given point, $\bar{\rho}$ is the average gas density, and $\alpha(r)$ is the linear attenuation coefficient at that point. This integral will give you the optical depth for light passing through the spherical cloud.

To address the issue of the diverging density at $r = 0$, you can use a small cutoff radius, $r_{min}$, and integrate from that radius instead of from $r = 0$. This will prevent the density from becoming infinite and will still give you a good approximation of the optical depth.

In terms of units, the integral will give you a unitless optical depth, as desired. However, the linear attenuation coefficient, $\alpha$, will have units of $m^{-1}$. You can convert this to a unitless coefficient by dividing by the path length, $l$, giving you a unitless attenuation coefficient, $\alpha/l$. This can then be used in the integral to calculate the optical depth.

Overall, the key is to take into account the varying density of the gas throughout the cloud and to use a cutoff radius to prevent any diverging values. This approach will give you a more accurate and physically meaningful calculation of the optical depth for an inhomogeneous gas cloud.