Dust Absorption and Scattering

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\usepackage{fullpage} \usepackage{amsmath} \usepackage{eufrak} \begin{document} \def\lya{Ly\alpha}

\section{Scattering in the Radiative Transfer Equation}

Scattering is naturally handled in the Radiative Transfer Equation as an emissivity that is proportional to the background radiation field: $$j_\nu=\sigma_\nu J_\nu,$$ where $J_\nu\equiv\frac1{4\pi}\int{I_\nu d\Omega}$ is the isotropic radiation field, and $\sigma_\nu$ is a frequency-dependent cross-section for scattering. We'll focus on this cross-section, which can depend on the size and composition of the scattering material.

\section{Geometric Interaction of Radiation with Grains: Babinet's Principle}

In general, we will be talking about a plane wave of wavelength $\lambda$ which is incident upon a particle (grain) of radius a. The cross-section for absorption, scattering, and emission will all be proportional to the physical cross-section of the grain, with some {\it absorption efficiency} Q: $$\sigma_i=Q_i\pi a^2$$ Kirchkoff's Law requires that $Q_{emit}=Q_{abs}$. If $a\gg\lambda$, then \def\qscat{Q_{scat}} \def\qabs{Q_{abs}} we have the geometric optics limit, and $\qscat+\qabs\sim1$. In fact, Babinet's Principle says: $$\boxed{\qscat+\qabs=2}$$ The proof of this goes as follows: suppose we have an infinite plane wave focused by an infinite lens onto a point on a wall. The power pattern should be a delta function at that point on the wall. Now suppose you place a screen with an aperture of diameter $a$ between the plane wave and the lens. We should now see an interference pattern on the wall. Call the power incident on a point on the wall $P_1$. Now put in a new aperture which is the exact compliment of our previous one (it is a scattering body of diameter $a$), we should see a new power at our point on the wall: $P_2$. However, the sum of waves incident on the wall under these two apertures should be the same as the wave from the sum of the apertures, which was a delta function. Thus: $$P_1=P_2$$ The first aperture (the slit) represented the scattering power from diffraction alone $(\qscat=1)$, and the second represented the absorbed power. The sum of these two powers, for everywhere but the true focus point, is actually twice the incident power, and since $P_1=P_2$, we get: $$\qscat+\qabs=2$$ Note that if $\qscat>1$, then the object is shiny: more light gets diffracted and less gets absorbed. Thus, although $\qscat$ changes, the sum remains the same.

\section{ Plane waves through a lens onto a backdrop}

We are considering a case of sending an infinite plane wave through an infinite lens with focal length $f$ onto a backdrop which is distance $f$ away. Next, we consider what happens if we have: (a) an occulting object at the center of the lens, (b) the exact opposite of that object--an aperture at that point. We'll get diffraction patterns on our backdrop in both of these cases. What is interesting is that the sum of the apertures (in our case, an infinite aperture minus a spot plus that spot) should also sum the diffraction patterns. The diffraction pattern for an infinite, unblocked aperture is just a delta function (everything focused at the focal point). Thus the fringe patterns for each of the partially blocked apertures must be the same, but $180^\circ$ out of phase. Of course, the (infinite aperture minus a spot) contains a delta function in addition to its diffraction pattern.\par

\section{ Small Grains}

Now let's consider the case that this spot (a small object obstructing a plane wave) is small, so that $a \ll \lambda$. This is called the Rayleigh Limit. In this case, $Q_{scat} \ll 1$ and $Q_{em} = Q_{abs} \ll 1$. If \def\qabs{Q_{abs}} \def\qscat{Q_{scat}} \def\qemis{Q_{emis}} we were to graph $\qabs$ vs. $\lambda$, we'd get something flat ($= 1$) out to $2\pi a$, after which, $\qabs$ would decrease as $(\inv{\lambda})^\beta$, where $\beta\eval{abs} = [1-3]$. On the other hand, graphing $\qscat$ vs. $\lambda$ , we get that after $2\pi a$, $\qscat$ drops as $(\inv{\lambda})^4$. Why does $Q$ decrease as $a\over \lambda$?\par

\begin{itemize} \item Let's consider the case of emitting with a dipole antenna. For a region close to the antenna (the near zone, of order $\lambda$), $P_{rad}\propto E^2$, and at the edge of the near zone, $E\propto{a\over\lambda}$. Since $P_{rad}\propto\qemis\propto\qabs$, $$\qabs\propto\left({a\over\lambda}\right)^2$$

\item Here's another crude way of looking at it: if you're in a boat that's small compared to the size of a wave coming at it, you aren't going to do anything to that wave (you won't scatter, everything transmits).

\item Another way of looking at it: the damped simple harmonic oscillator. In the context of dispersion relations, we wrote down the equation of motion of an electron in the presence of a plane wave: $$\ddot x+\gamma\dot x+\wz^2x={eE_0\eikrwt\over m_e}$$ where $\gamma$ is our dampening factor. This equation had a steady-state solution: $$x={-e\over m_e}{\left[(\wz^2-\omega^2)+i\omega\gamma\right]\over \left[(\wz^2-\omega^2)^2+(\omega\gamma)^2\right]}E_0\eikrwt$$ The imaginary component of this solution is the absorptive (dissipative) part. We can construct a cross-section from this: $$\sigma_{abs} = {4\pi e^2\over m_e c}{w\gamma \over \left[(w_0^2-w^2)^2 + (w\gamma)^2\right]}$$ If $w \ll w_0$, then: $$\sigma_{abs} \to {4\pi e^2\over m_e c}{w^2\gamma \over w_0^4} \propto w^2$$ Thus, $\qabs \propto \inv{\lambda^2}$. \end{itemize}

Finally, there is Mie Theory, which states that if $\eta = n +ik$, where $\eta$ is the complex index of refraction, then you can express $\qabs$ and $\qscat$ in terms of the size parameter $x \equiv {2\pi a\over \lambda}$. For this derivation that Mie did, $n, k$ are called optical constants. This is a bad name, because both of these are actually functions of $\lambda$.\par

Now we look at the handout Eugene gave us (Chiang et al, 2001, ApJ, 547, 1077). Notice on the $\qabs$ vs. $\lambda$ plots, in addition to the decay we described, there are bumps and wiggles. These are characteristic of the resonances of the systems (e.g. they leftrightarrowond to a rotational mode, etc.). The varying lines on each graph are for different sized particles. Notice the characteristic Rayleigh Limit for small particles. \par

\def\qext{Q_{ext}} There's another graph about how extinction $\qext = \qabs + \qscat$ goes as the size parameter is changed. There is a large hump where $\qext$ goes to about 4. It can do this (we'd said it couldn't go above 2) because we are not yet in the geometric optic limit. Notice that on this big hump, there is a region where increased wavelength causes increased extinction. This is the process which causes blue moons. If the size of particles in the atmosphere is just right, it will pass blue wavelengths while extinguishing red.\par

\section{ Some special results}

\def\sabs{\sigma_{abs}} \begin{itemize} \item Crystalline dielectrics: $\qabs \propto {a\over \lambda^2}$ (for $\lambda \gg a$. Then: $$\alpha_\nu \eval{abs} = n_d\sabs = n_d\pi a^2\qabs \propto {a^2\over \lambda^2}$$ Now what we measure in the field is $\tau$, the optical thickness. However, $\tau_\nu = \sum{K_\nu}$, where $K_\nu = {\sabs \over m_{particle}} \propto a^0$. This is how the mass of particles can measured. Then $\qscat \sim ({a\over \lambda})^4$, like Rayleigh scattering. \item In general: $j_\nu\eval{emission} = \alpha_\nu \cdot B_\nu(T)$. Then: $$j_\nu\eval{scattering}(\hat n) = n_d(\qscat\pi a^2)\int{I_\nu(\hat n) F(\hat n-\hat n^\prime)d\Omega^\prime)}$$ The term $F(\hat n - \hat n^\prime)$ is called the scattering phase function, and is used to add up all of the incoming paths of light, taking into account their relative phases. $n_d$ is the \# density of scattering grains. For small enough grains ($a\ll \lambda$), F is independent of the properties of the material the grains are made of: $$F = {1+\cos^2\theta\over 2}$$ where $\theta$ is the angle from the propagation direction of the incident beam. This gives us a characteristic dog bone scattering pattern. Small grains are (to within a factor of 2) isotropic scatterers. The factor of 2 is the dog bone pattern. Note that the missing half at right angles to the propagation direction is the portion of incident light which had no polarization component which aligns with the characteristic polarization which must exist for light detected at a right angle from a scattering body. \end{itemize}

\section{Increasing Grain Size}

We return to the model in which an infinite plane wave passes through an aperture, is focused by an infinite lens, and shines on a wall. In this model, as the slit aperture widens (a increases), then the diffraction pattern narrows. Thus, for larger grain sizes, there is more forward scattering than in other directions. The fitting formula for the power pattern for large grains is: $$F(\theta)={1-g^2\over1+g^2-2g\cos\theta}$$ where $g\equiv\mean{\cos\theta}={\int{F\cos\theta d\Omega}\over \int{Fd\Omega}}$. Note that this formula fails for $\theta=100^\circ$. If $g=1$, then we have isotropic scattering, and as $g\to1$, $F(\theta)$ peaks increasingly in the $\theta=0$ direction (it is increasingly forward throwing).

\section{ Forward Scattering}

Forward scattering can actually increase the intensity of light in some areas over if there were no scattering at all. If particles are smaller that the wavelength of light, there is isotropic scattering. For larger particles, the power is more concentrated in the perfectly forward and backward directions. This is why, where there is fog, you dim your headlights. The larger scattering particles in the fog actually increases the intensity of light reflected back at you and at oncoming cars.

\section{Absorption and Scattering by Astrophysical Dust Grains}

Here we will consider absorption and scattering due to dust grains in the UV-IR wavelength range (100 nm to 10 $\mu$ m). The dust grain sizes of concern for these wavelengths are of a size greater than 10 nm (containing about a million atoms). We write our complex refractive index as

$$m = n + ik$$

\subsection{Showing the refractive index $m$ is complex via Maxwell's equations}

Using the plane wave solutions

$$E = E_{0} e^{i(k y - \omega t)} \hat{z}, B = B_{0} e^{i(k y - \omega t)} \hat{x},$$

we then plug these in for our fields like so

$$\nabla \times B = - B_{0} i k e^{i(k y - \omega t)} \hat{z} = - i \frac{\omega}{c} \epsilon E_{0} e^{i(k y - \omega t)} \hat{z} + \frac{4 \pi \sigma}{c} E_{0} e^{i(k y - \omega t)} \hat{z}.$$

Now we use the fact that our waves have the same amplitude $E_{0} = B_{0}$, and we get a dispersion relation

$$-i k = -\frac{i \omega}{c} \epsilon + \frac{4 \pi \sigma}{c}$$

which we rewrite as

$$ck = \omega \epsilon + 4 \pi \sigma i$$

or

$$m = \frac{c k}{\omega} = \epsilon + \frac{4 \pi \sigma}{\omega}i.$$

So we see that a complex refractive index arises naturally from plane wave solutions in general matter. Connecting this to absorption and scattering:

$$e^{i(k y - \omega t)} = e^{i(\frac{m c}{\omega} y - \omega t)} = e^{i(\frac{(n + i k) c}{\omega} y - \omega t)}$$

We see that the imaginary part of the refractive index leads to an exponential attenuation - a.k.a. absorption in this material ($k$). The real part of this refractive index corresponds to reflection at the boundary of our matter - this is our scattering ($n$).

Let's now rewrite our index of refraction expression in terms of a new quantity that is simply the sum of our optical constants $n$ and $k$:

$$Q_{\rm{ext}} = Q_{\rm{abs}} + Q_{\rm{scat}}.$$

Our goal is now to determine $Q_{\rm{ext}} = Q_{\rm{abs}} + Q_{\rm{scat}}$ by finding $Q_{\rm{abs}}$ and $Q_{\rm{scat}}$.

\section{The three size regimes}

There are three particle size regimes relevant for our discussion of extinction ($Q_{{\rm{ext}}}$). We will discuss these regimes in terms of the quantity $x \equiv \frac{2 \pi a}{\lambda}$ - the ratio of the circumference of the dust particle to the wavelength of light we are considering. We will make different physical approximations in each regime to quantitatively describe the extinction processes of scattering and absorption.

\subsection{Rayleigh Limit}

In the Rayleigh limit, the size of the dust grain is much smaller than the wavelength of light we are considering (i.e. $x << 1$). We consider the dust grain as a dipole.

Here we treat the dust grain as a dipole oscillating at the frequency of the incident electromagnetic wave.

Writing the Larmor formula $P = \frac{e^{2} a^{2}}{c^{3}}$ where $x = \cos{\omega t}$, and $a = \omega^{2} \cos{\omega t}$, then we have $P \sim \lambda^{-4}$. Thus $Q_{\rm{scat}} \sim \lambda^{-4}$.

Returning to our complex index of refraction in our plane wave, we know the plane wave will have an absorption term $e^{-\frac{k \omega}{c} y} = e^{-\tau}$. So $\tau = \frac{k \omega}{c}y$, and factoring the optical depth in terms of number density, $n \sigma y = \frac{k \omega}{c}y$, we then have $n \sigma = \frac{2 \pi k}{\lambda}$ or $\sigma \sim \lambda^{-1}$. So $Q_{\rm{abs}} \sim \lambda^{-1}$.

Finally, we have $Q_{\rm{ext}} = Q_{\rm{abs}} + Q_{\rm{scat}} \sim \lambda^{-4} + \lambda^{-1}$ as our extinction expression in the Rayleigh limit.

\subsection{Geometric Limit}

In the geometric limit, the size of the dust grain is much larger than the wavelength of light we are considering (i.e. $x >> 1$). We consider the dust grain as a physical barrier.

By Babinet's principle, we know that $Q_{\rm{ext}} = 2$. As a result, we know that $Q_{\rm{abs}} \sim 1$ and $Q_{\rm{scat}} \sim 1$.

\subsection{Mie Regime}

The Mie regime is where things get complicated. Here the particles have a similar size to the wavelength of light we are considering (i.e. $x \sim 1$).

To appropriately describe the Mie regime, we need to solve Maxwell's equations inside the dust grain using an incident plane wave. This would then allow us to find the amount of our incident light that is scattered away. The spherical geometry of the boundary condition makes this complicated. If the grain is a perfect sphere there exists an analytic (and very messy) solution in terms of an infinite series of $x$. For any other geometry, this requires a numerical solution. We will ignore this for the moment.

To lowest order in the Mie regime, $Q_{\rm{scat}} \sim x^{4} + \mathcal{O}(x^{5})$ and $Q_{\rm{abs}} \sim x + \mathcal{O}(x^{2})$. So we have recovered the Rayleigh limit results in this way.

Having been unable to show the Mie solution, we will still connect all three regimes in the following plot:

%\begin{figure} %\end{figure}

Some overall properties of the extinction curve include the following. The "bump" in the curve produces increased extinction toward the IR and decreased extinction toward the UV (hence why astrophysicists often call extinction "reddening"). There is always a very high variety of dust grains in actual astrophysical observations, so it is worth remembering that this type of schematic curve will never be reproduced exactly.

\section{Observing Dust Extinction}

Here is how we would apply our equations to observations. Typically dust extinction is reported in terms of a ratio of magnitudes $\frac{A_{\lambda}}{A_{I}}$, where $A_{\lambda}$ is the amount of extinction measured in magnitudes at a given wavelength $\lambda$ and $A_{I}$ is the total extinction measured magnitudes within the I-band (for non-observers, this is approx. 150-800 nm / the near IR).

Our flux is then $F \propto e^{-\tau}$, or in magnitudes, $m \propto \tau$. This tells us that $A_{\lambda} \propto \tau$ and we know $\tau = n \sigma s$, where $\sigma \sim Q$ (from above). This allows us to compare the dust extinctions we expect from the regimes listed above to real data.

%\begin{figure} %\end{figure}

The above plot shows real observations with the wavelength power laws we expect from our theoretical extinction curves. (The 217.5 nm graphite transition bump does not fit into our theory, so ignore that for the purposes of this comparison.)

From above, in the Rayleigh regime, $Q_{\rm{ext}} = Q_{\rm{abs}} + Q_{\rm{scat}}$ and we know $Q_{\rm{abs}} \propto \lambda^{-1}$ and $Q_{\rm{scat}} \propto \lambda^{-4}$. In our plot, the absorption extinction curve is plotted in red, and we see it provides a pretty good overall fit to the data - implying our extinction is absorption dominated.

However, there is a deviation at low wavelengths between the data and the absorption extinction. The green line provides a better fit to the data, and has a scattering contribution incorporated (also in the Rayleigh regime).

At very high wavelengths, the data does not match our expectations very well at all, and we see it matches a power law in wavelength with an exponent of -1.6, and this reflects the fact that our optical "constants" are not truly constant (i.e. $k$ has some $\lambda$ dependence).

\end{document} <\latex>