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Absorption, scattering and reddening

What is the effect of dust on the light we detect from distant stars? Suppose the wavelength of the light is much smaller than the size of the dust grains (bricks, say) - then you could compute the dimming just from the geometric cross-section of the bricks along the line of sight. Consider a ray of light with intensity $ I$. The amount of light of this ray absorbed per unit distance, will satisfy

$\displaystyle {dI\over dr}=-A\,I\,,$ (2.3)

which expresses the fact that each distance $ dr$ will absorb a constant fraction $ dI/I=-A\,dr$ of the light. This is what we want since, if the light source were twice as bright, then the bricks will stop twice as many photons. $ A$ is some constant, which will depend on the size of the bricks, and their number density. This equation is easy to integrate, and the solution is $ I=I_0\exp(-Ar)$. Taking the $ \log$ of both sides, we find that the effect on the magnitude will be $ \Delta m=A\,r$. So the relation between apparent ($ m$) and absolute magnitude ($ M$) will change from $ m-M=5\log(r)-5$ to

$\displaystyle m-M=5\log(r)-5+A\,r\,,$ (2.4)

due to the effect of absorption. So the unit of $ A$ is magnitudes per kpc. (Careful readers will notice that I have slightly changed the definition of $ A$ along the way, but you get the idea.)

If the size of the particles is of order of the wavelength $ \lambda$ of the light, then it is not a good approximation to use the geometric cross-section of the dust grains to estimate absorption. The amount of absorption (the value of $ A$) will then depend on $ \lambda$, and so $ A_{\hbox{\sc V}}\ne A_\B$, say: absorption in the B band will differ from that in the V band, and so absorption will also lead to a colour change (reddening, since the longer wavelengths get less absorbed than the shorter ones).

Applying Eq.(2.4) to two bands, we find

$\displaystyle (m-M)_\B$ $\displaystyle =$ $\displaystyle 5\log(r)-5+A_\B\,r\,$  
$\displaystyle (m-M)_{\hbox{\sc V}}$ $\displaystyle =$ $\displaystyle 5\log(r)-5+A_{\hbox{\sc V}}\,r\,$  
$\displaystyle E_{\B -{\hbox{\sc V}}}$ $\displaystyle \equiv$ $\displaystyle (m_\B -m_{\hbox{\sc V}})-(M_\B -M_{\hbox{\sc V}})=(A_\B -A_{\hbox{\sc V}})\,r\,.$ (2.5)

The quantity $ E_{\B -{\hbox{\sc V}}}$ is called the colour excess, note that it is the difference between the observed and intrinsic colour of the star,

$\displaystyle E_{\B -{\hbox{\sc V}}}=(\B -{\hbox{\sc V}})_{\rm observed}-(\B -{\hbox{\sc V}})_{\rm intrinsic}\,.$ (2.6)

Trumpler's measurements, and also laboratory measurements, show that for interstellar dust grains

$\displaystyle E_{\B -{\hbox{\sc V}}}\approx {1\over 3}A_{\hbox{\sc V}}\,r\,.$ (2.7)

This is a crucial result. Reddening and hence $ E_{\B -{\hbox{\sc V}}}$, is easy to measure, and so if we do this for stars of known distance, we find2.3 $ A_{\hbox{\sc V}}\approx
1{\rm mag}\,{\hbox{\rm kpc}}^{-1}$. If we now measure $ E_{\B -{\hbox{\sc V}}}$ for another star of known colour (from stellar evolutionary models say) we can estimate $ r$.


next up previous contents
Next: Summary Up: The discovery of the Previous: Time-line
Tom Theuns
平成19年2月7日