Extinction

So far we have been assuming that the space between stars is a vacuum and does not do anything to the light. However in reality, the space between stars contains some interstellar medium. This causes some of the radiation to be absorbed (and perhaps re-emitted at a different wavelength), or scattered away from the line of sight. All of these radiation loses are called extinction.

Here we have a star radiating energy flux (i.e. light) \(L_0\) into a region (more formally, a solid angle \(\omega\)) in some wavelength range (Extinction rates are different for different wavelengths).

Since the medium absorbs and scatters radiation, the total light (energy flux) into the region \(L\) will now decrease with increasing distance \(r\).

From \(r\) to \(r+dr\), the amount of light/energy flux drops from \(L\) to \(L+dL\) (where \(dL\leq0\) is negative). This drop in light (\(dL\)) is proportional to the amount of light through the interstellar medium. (If you shine a greater amount of light through it, the additive value of the drop (\(dL\)) will increase)

\(dL\) is also proportional to the distance travelled in the interstellar medium (\(dr\)). (The longer light has to travel, the greater amount of light lost)

\[ \begin{align*} dL &\propto L \\ dL &\propto dr \\ dL &= -\alpha\cdot L \cdot dr \end{align*} \]

\(\alpha\) is the opacity, which is how good the interstellar medium is at blocking light. It is \(0\) for a pefect vacuum and \(\infty\) for a perfectly opaque substance.

Thus we can show that

\[ L = L_0e^{-\alpha r} \]

Derivation of \(L = L_0e^{-\alpha r}\)

\[\begin{align*} dL &= -\alpha L dr\\ \frac{dL}{L} &= -\alpha dr \end{align*} \]

As \(r\) goes from \(0\) to \(r\), light (energy flux) goes to \(L_0\) to \(L\):

\[\begin{align*} \int^L_{L_0} \frac{dL}{L} &= -\alpha \int^r_0 dr \\ \ln L - \ln L_0 &= -\alpha r \\ \ln\left(\frac{L}{L_0}\right) &= -\alpha r \\ \frac{L}{L_0} &= e^{-\alpha r} \\ L &= L_0e^{-\alpha r} \end{align*}\]

Optical Thickness

We define optical thickness \(\tau\) to be \(\tau = \alpha r\). Therefore

\[ L = L_0e^{-\tau} \]

Revisiting the distance modulus

We already know the old formula of

\[ m - M = 5\log\left(\frac{d}{10 \text{pc}}\right) \]

However this does not account for the decrease in flux desnity due to interstellar medium. The new formula becomes:

\[ \begin{align*} m - M &= 5\log\left(\frac{r}{10 \text{pc}}\right) + 2.5\tau\log e\\ &= 5\log\left(\frac{r}{10 \text{pc}}\right) + A \end{align*} \]

where \(A\) is the extinction (Difference in "expected" magnitude and actual magnitude).

If opacity \(\alpha\) is constant throughout the line of sight, we get

\[ \begin{align*} m - M &= 5\log\left(\frac{r}{10 \text{pc}}\right) + 2.5\tau\log e\\ &= 5\log\left(\frac{r}{10 \text{pc}}\right) + 2.5\alpha r\log e\\ &= 5\log\left(\frac{r}{10 \text{pc}}\right) + ar \end{align*} \]

where \(r\) is the distance to the star, and \(a=2.5\alpha\log e\) being the extinction in terms of magnitude per distance.

Derivation

In order to account fordecrease in flux desnity due to interstellar medium, we have to go back to

\[ m - M = -2.5\log\left(\frac{F_r}{F_{10 \text{pc}}}\right) \]

where \(F_r\) is the flux density at a distance \(r\) away and \(F_{10 \text{pc}\) is the flux density 10 parsecs away without extinction

Now we need to express \(F_r\) and \(F_{10 \text{pc}}\). We can show that the amount of light/energy flux at \(r\) is

\[L=\text{area}\times\text{flux density} = \omega r^2 F_r\]

Likewise we can define \(L_0\) to be \(\omega R^2 F_0\) where \(R\) is the radius of the star.

We can use \(L = e^{-\tau}L_0\) to express \(F_r\) in terms of \(F_0\)

\[\begin{align*} \omega r^2 F_r &= \omega R^2 F_0 e^{-\tau}\\ F_r &= \frac{R^2}{r^2} F_0 e^{-\tau}\\ \end{align*} \]

Now for the \(F_{10 \text{pc}}\) term:

\[F_{10\text{pc}} = \frac{L}{4\pi (10 \text{pc})^2 }\]

We can also express this in terms of \(F_0\). Since

\[ F_0 = \frac{L}{4\pi R^2}\]

we can find find \(F_{10\text{pc}}\):

\[\begin{align*} \frac{ F_{ 10\text{pc} } }{F_0} &= \frac{ \frac{L}{4\pi (10 \text{pc})^2 }} {\frac{L}{4\pi R^2}} \\ F_{10\text{pc}} &= F_0\frac{R^2}{(10\text{pc})^2} \end{align*}\]

Now we can find the distance modulus:

\[\begin{align*} m - M &= -2.5\log\left(\frac{F_r}{F_{10 \text{pc}}}\right)\\ &= -2.5\log\left(\frac{\frac{R^2}{r^2} F_0 e^{-\tau}}{F_0\frac{R^2}{r^2}}\right) \\ &= -2.5\log\left(e^{-\tau}\frac{(10\text{pc})^2}{r^2}\right) \\ &= -2.5\log(e^{-\tau}) -2.5\log\left(\frac{(10\text{pc})^2}{r^2}\right) \\ &= 2.5\tau\log(e) + 5\log\left(\frac{r}{10\text{pc}}\right) \end{align*}\]

Question

Fundamental Astronomy Example 4.6Solutions

The absolute magnitude of a star is M = −2 and the apparent magnitude m = 8. What is the distance to the star, given an interstellar extinction rate of 2 mag/kpc?

Where \(d\) is in parsecs:

\[\begin{align*} m-M &= 5\log\left(\frac{d}{10}\right)+A_Vd\\ 8-(-2) &= 5\log\left(\frac{d}{10}\right)+0.002d\\ 2-0.0004d &= \log\left(\frac{d}{10}\right)\\ 10^{2-0.0004d} &= \frac{d}{10}\\ d &= 10\times10^{2-0.0004d} \end{align*}\]

We unfortunately cannot solve this analytically. However, we can solve this numerically via iteration. This sounds intimidating and hard to do manually, but it is actually rather easy on a calculator with an "Ans" button.

We just get a random \(d\) by just randomly estimating or guessing, and get a slightly more accurate \(d_{new}\) using:

\[d_{new} = 10\times10^{2-0.0004\times d_{old}}\]

Now just plug the new \(d_{old}\) value and repeat until it approaches some value.

on a calculator you would want to input something like this, and repeatedly spam the equals button:

\[ 10\times10^{2-0.0004\times\text{Ans}} \]

you will get an answer of 584pc (actually 580pc because 2sf)

Question

Fundamental Astronomy Example 4.7Solution

What is the optical thickness of a layer of fog, if the Sun seen through the fog seems as bright as a full moon in a cloudless sky? (take \(m_\odot = -26.7\) and \(m_\text{moon} = -12.9\) for a full moon)

Extinction is really just the difference in magnitude, so A = \(m_{blocked} - m_{actual} = -12.9+26.7 = 13.8\)

Since

\[A = 2.5 \tau \log e\]

we get

\[\tau = \frac{A}{2.5 \log e} = \frac{13.8}{1.086} = 12.7\]

In reality, some of the observed light has been scattered multiple times such that they get scattered into the line of sight.

Therefore optical thickness must be slightly higher than calculated.

Intrinsic Color and Color Excess

Another effect caused by the interstellar medium is the reddening of light: blue light is scattered and absorbed more than red. Therefore the colour index \(B − V\) increases.

We can write the visual apparent magnitude \(V\) of a star as such:

\[ \begin{align*} m - M = V - M_V &= 5\log\left(\frac{d}{10\text{pc}}\right)+A_V\\ V &= M_V + 5\log\left(\frac{d}{10\text{pc}}\right)+A_V \end{align*} \]

Where \(A_V\) is the extinction in the V passband.

We can also get a similar expression for \(B\):

\[B = M_B + 5\log\left(\frac{d}{10\text{pc}}\right) + A_B\]

so the color index \(B-V\) is:

\[ \begin{align*} B-V &= M_B + 5\log\left(\frac{d}{10\text{pc}}\right) + A_B - \left(M_V + 5\log\left(\frac{d}{10\text{pc}}\right)+A_V\right)\\ &= M_B-M_V+A_B-A_V \end{align*} \]

since the \(5\log\left(\frac{d}{10\text{pc}}\right)\) cancel each other out.

we can rewrite this as:

\[ B-V = (B-V)_0+E_{B-V} \]

where \((B − V)_0 = M_B − M_V\) is the intrinsic color of the star and \(E_{B-V} = (B-V) - (B-V)_0\) is the color excess

One interesting fact is that the ratio between the extinction \(A_V\) and reddening \(E_{B-V}\) is almost constant for all stars:

\[ \frac{A_V}{E_{B-V}} \approx 3.0 \]

This means we can find \(A_V\) if \(E_{B-V}\) is known:

\[A_V \approx 3.0E_{B−V}\]