Interstellar Dust

When we look in different directions of the sky, we often see dark "holes" in the distributions of stars. These are not gaps where there are no stars, but instead are interstellar dust clouds.

Dust doesn't have to come just in thick clouds, it can also be spread diffusely throughout space.

What does dust do to star light? Several things:

it absorbs light
it reddens light
it scatters light

Question: What does this do to a star's observed position on the H-R diagram?

Interstellar absorption

Remember how absorption works (see this link for a refresher):

So we have

Rewrite as

If we define extinction as A=1.086 $\tau$ , we can then correct the observed magnitudes: m = m_obs - A. (Why is it minus A?)

If we didn't correct the magnitudes, what would this do to our derived distances?

Interstellar reddening

The extinction due to dust is not equally effective at all wavelengths. The shorter the wavelength, the higher the extinction -- blue light is extincted more strongly than red light. Therefore, stars behind a lot of dust look redder than they really are. This is called interstellar reddening.

So if we measure the B-V color of a star, we will be measuring a redder color than the true B-V color (usually called (B-V)₀). We define the reddening as

E(B-V) = (B-V) - (B-V)₀

Now it seems reasonable to expect that the more extinction there is, the more reddening there will be. In fact, studies have shown that the extinction and reddening are related by:

A_V=3.1xE(B-V)

Again, extinction is higher at shorter wavelengths, so for filters other than V:

A_B=4.1E(B-V) (blue)

A_R=2.7E(B-V) (red)

A_K=0.5E(B-V) (near infrared)

So if we know what the reddening is, we can calculate the extinction and correct our magnitude equation. How could we figure out the reddening?

Interstellar Scattering

The extinction is not due to absorption alone. Dust absorbs *and scatters* light.
So we see not only dark clouds, but reflection nebulae as well...