# Maurice Wilson's

## Dusty Disk Density Derivation

Prepare yourself for a lot of equations! I will just zoom through them since there are so many. I won't go into much detail explaining each equation like I usually do.

If you don't care about this analytical solution of the dust density distribution, then you should just wait for my next post to see the visualizations I created that portray a protoplanetary disk. Those beautiful visuals will be worth the wait, I assure you!

In my previous post, I stated that the gas and dust have very different characteristics. Because of this, deriving their distinct temperature profiles leads down two very different paths. For the sake of simplicity, I will guide you through only one of these paths: the simpler, albeit dusty, path. From hereon, everything I discuss will only be about the dust grains.

Also in my previous post, I briefly discussed the concept of radiative transfer. Its equation can be described as $\frac{dI_{\nu}}{d\tau_{\nu}}(\tau_{\nu}) = S_{\nu}(\tau_{\nu}) - I_{\nu}(\tau_{\nu}) ,$ where $$I_{\nu}$$ is the star light's intensity that fluctuates as it goes through the medium and $$S_{\nu}$$ is the ratio of the emission and absorption coefficients of the medium. Both vary with radiation frequency $$\nu$$ and optical depth $$\tau_{\nu}$$. The optical depth due to dust grains, $\tau_{\nu} = N_{gr} \sigma_{gr} Q_{ext},$ depends on their column density $$N_{gr}$$, the cross sectional area $$\sigma_{gr}$$, and extinction efficiency $$Q_{ext}$$. Now, as stated previously, you can see how the density of the dust greatly influences the radiative transfer.

With a few assumptions, we can derive an approximate dust density distribution function. One assumption being that the dust is a blackbody emitter. Considering this, the dust flux is $F_{cooling} = \sigma_{sb} T^4 ,$ where $$\sigma_{sb}$$ is the Stefan-Boltzmann constant and $$T$$ is the dust temperature. Because the grains are emitting radiation, this action makes them cool down. However, we know that the dust temperature barely fluctuates at all over time. Consequently, it is safe to assume that the stellar flux, which heats the dust, is equivalent to the dust's flux. In the star's case, $F_{\star} = \phi \frac{L_{\star}}{4 \pi r^2},$ where the angle between the surface of the star and the disk is $\phi \approx \frac{0.4 R_{\star}}{r}.$ $$L_{\star}$$ is the stellar luminosity, $$r$$ is the distance from the star, and $$R_{\star}$$ is the stellar radius. After equating dust flux to stellar flux, the radial temperature profile for a flat disk can be described as $T(r) = \bigg(\frac{0.4 R_{\star}L_{\star}}{4 \pi \sigma_{sb} r^3} \bigg)^{1/4}.$ Although this does not account for the disk having a third dimension, it is still a useful approximation for the midplane of a 3-dimensional disk. Using $$T(r)$$ assumes that the disk is vertically isothermal. (Note that we've estimated a 2D dust temperature profile in our derivation of the dust density distribution: a density distribution that will let us derive a 3D dust temperature profile of the disk.)

The equation of state $$P = nkT$$ can be equated to the pressure of the medium $$P = c_s^2 \rho$$ to find the medium's speed of sound: $c_s(r) = \sqrt{ \frac{kT(r)}{m} } ,$ where $$k$$ is the Boltzmann constant and $$m$$ is the mass of the grain.

If the disk is vertically in hydrostatic equilibrium, the vertical component of both the grains' pressure gradient and the stellar gravitational force must be equal. The gravitational force in spherical coordinates is $F_{grav} = -( sin(\phi)\hat{z} + cos(\phi)\hat{r} ) \frac{GmM_{\star}}{r^2} .$ Therefore, in cylindrical coordinates it is $F_{grav} \approx - \bigg(\frac{z}{r}\hat{z} +\hat{r} \bigg) \frac{GmM_{\star}}{r^2} ,$ showing us that $F_{grav,z} \approx - \frac{z}{r}\frac{GmM_{\star}}{r^2} .$ At hydrostatic equilibrium, $\frac{dP}{dz} = \frac{z}{r}\frac{GmM_{\star}}{r^2}$ where $$P$$, as previously stated, is proportional to $$\rho$$. Therefore, $\frac{d\rho}{dz} = \frac{z}{r}\frac{GmM_{\star}}{c_s^2 r^2} .$ The solution to this differential equation is $\rho(r,z) = \rho_0(r) exp\big[-\frac{1}{2} (z/H)^2\big]$ where $H = \bigg( \frac{r^3 c_s^2}{GM_{\star}} \bigg)^{1/2} .$ The scale height $$H$$ is usually equivalent to ~10% of the radial distance from the star.

With $$\rho(r,z)$$ we have derived our 3D dust density distribution function! However, we do not know $$\rho_0(r)$$, the density in the midplane, yet. This can be found by using the surface density profile, $\Sigma(r) = (2-\gamma)\frac{M_{disk}}{2\pi R_c^2} \bigg(\frac{r}{R_c} \bigg)^{-\gamma} exp\bigg[-\bigg(\frac{r}{R_c} \bigg)^{2-\gamma} \bigg] ,$ where $$R_c$$ is the outer boundary of the disk indicating the characteristic radius at which the surface density's power law behavior ($$\Sigma = \Sigma_0(r/R_c)^{-\gamma}$$) breaks down. Because $$\Sigma(r) = \int \rho(r,z)dz$$, the $$\rho_0(r)$$ can be expressed as $\rho_0(r) = \frac{\Sigma}{H\sqrt{2\pi}}.$

After all of this hard work, the disk's structure should look like a flared disk, as mentioned in my previous post. However, keep in mind that this analytical derivation required a few assumptions that, although gives us a decent estimate, were not realistic.

Next, I will show the structure that these equations conjure up. The temperature profile will be calculated and visualizations for both the density and temperature will serve as a nice treat for all of this hard work.

to be continued ...

Posted: May 31, 2017

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