Extinction and Optical Thickness

Equation (4.11) shows how the apparent magnitude increases (and brightness decreases!) with increasing distance. If the space between the radiation source and the observer is not completely empty, but contains some interstellar medium, (4.11) no longer holds, because part of the radiation is absorbed by the medium (and usually re-emitted at a different wavelength, which may be outside the band defining the magnitude), or scattered away from the line of sight. All these radiation losses are called the extinction.

Now we want to find out how the extinction depends on the distance. Assume we have a star radiating a flux L0 into a solid angle w in some wavelength range. Since the medium absorbs and scatters radiation, the r flux L will now decrease with increasing distance r (Fig. 4.8). In a short distance interval [r, r + dr], the extinction dL is proportional to the flux L and the distance travelled in the medium:

The factor a tells how effectively the medium can obscure radiation. It is called the opacity. From (4.14) we see that its dimension is [a] = m-1. The opacity is zero for a perfect vacuum and approaches infinity when the substance becomes really murky. We can now define a dimensionless quantity, the optical thickness r by dr = a dr . (4.15)

Next we integrate this from the source (where L = L 0 and r = 0) to the observer:

Here, r is the optical thickness of the material between the source and the observer and L, the observed flux. Now, the flux L falls off exponentially with increasing m — M =—2.51g r

10 pc

10 pc

10 pc

optical thickness. Empty space is perfectly transparent, i. e. its opacity is a = 0; thus the optical thickness does not increase in empty space, and the flux remains constant.

Let F0 be the flux density on the surface of a star and F(r), the flux density at a distance r. We can express the fluxes as

L = rnr2 F(r) , L0 = rnR2 F0 , where R is the radius of the star. Substitution into (4.16)


R2 r

For the absolute magnitude we need the flux density at a distance of 10 parsecs, F(10), which is still evaluated without extinction: R2

The distance modulus m - M is now

Fig. 4.8. The interste11ar medium absorbs and scatters radiation; this usua11y reduces the energy flux L in the so1id ang1e m (dL < 0)

where A > 0 is the extinction in magnitudes due to the entire medium between the star and the observer. If the opacity is constant along the line of sight, we have r t = a j dr = ar , and (4.17) becomes m - M = 5 lg

10 pc

where the constant a = 2.5a lg e gives the extinction in magnitudes per unit distance.

Colour 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. The visual magnitude of a star is, from (4.17), r

10 pc where My is the absolute visual magnitude and Ay is the extinction in the y passband. Similarly, we get for the blue magnitudes r

The observed colour index is now

This makes it possible to find the visual extinction if the colour excess is known:

When AV is obtained, the distance can be solved directly from (4.19), when V and MV are known.

We shall study interstellar extinction in more detail in Sect. 15.1 ("Interstellar Dust").

Atmospheric Extinction. As we mentioned in Sect. 3.1, the Earth's atmosphere also causes extinction. The observed magnitude m depends on the location of the observer and the zenith distance of the object, since these factors determine the distance the light has to travel in the atmosphere. To compare different observations, we must first reduce them, i. e. remove the atmospheric effects somehow. The magnitude m0 thus obtained can then be compared with other observations.

If the zenith distance z is not too large, we can approximate the atmosphere by a plane layer of constant thickness (Fig. 4.9). If the thickness of the atmosphere is used as a unit, the light must travel a distance

in the atmosphere. The quantity X is the air mass. According to (4.18), the magnitude increases linearly with the distance X:

where k is the extinction coefficient.

The extinction coefficient can be determined by observing the same source several times during a night with as wide a zenith distance range as possible. The observed magnitudes are plotted in a diagram as a function of the air mass X. The points lie on a straight line the slope of which gives the extinction coefficient k. When where (B — V)0 = MB — My is the intrinsic colour of the star and EB—y = (B — V) — (B — V)0 is the colour excess. Studies of the interstellar medium show that the ratio of the visual extinction Ay to the colour excess EB—y is almost constant for all stars:

Fig. 4.9. If the zenith distance of a star is z, the light of the star travels a distance H/ cos z in the atmosphere; H is the height of the atmosphere

this line is extrapolated to X = 0, we get the magnitude m0, which is the apparent magnitude outside the atmosphere.

In practice, observations with zenith distances higher than 70° (or altitudes less than 20°) are not used to determine k and m 0, since at low altitudes the curvature of the atmosphere begins to complicate matters. The value of the extinction coefficient k depends on the observation site and time and also on the wavelength, since extinction increases strongly towards short wavelengths.

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