## Friday, April 17, 2015

### Looking at Stellar Emission and Absorption (3)

The study of spectral lines is facilitated by using what is called the “equivalent width”. This provides the total strength of a line, yielding the same area of the line (for a rectangular facsimile) provided the depth is complete from the continuum to zero brightness.

We can express the equivalent width in two ways, based on frequency of wavelength, or wavelength: itself, e.g.

W =   ò ¥0  (I c – I u /  I c )du =  ò ¥0  (F c – F l / F c ) dl

The left side defines W in terms of the intensity from the continuous spectrum outside the spectral line where the quantity (I c – I u / I c ) is referred to as the “depth of the line”, the analogous quantity to (F c – F l / F c ) on the right side where we have radiant flux units. Technically, the integral should be taken only from one side of the line to the other but the limits can be as shown provided I c    (or  F c)  is kept constant in the neighborhood of the line.

The Curve of Growth:

Having obtained the equivalent width, the next logical step is to generate the “curve of growth”.  Recall from the previous section that W, the equivalent width, represents the line strength. Then the curve of growth describes how the latter increases as the optical depth, τ  increases.

To fix ideas, we use a model slab of  finite thickness ds and over which the optical depth increases by d τ.  The incident intensity is  I c  for frequencies in the neighborhood of the line,  and the intensity emerging from the opposite side is    I u  which we seek to find.

Starting with  the original basic transfer equation based on wavelength:

dI(l)/ds = -k(l)  I(l) + k(l)  S(l)

We can derive the frequency form:

d Iu / u  =  -  Iu – S u

We have (from the original equation) after adjusting for frequency:

d Iu / ds   =  -k u Iu  +   k u S u

Where:   k u   =   d τ u / ds

So:

d Iu / ds   =  -( d τ u / ds) Iu  +   (d τ u / ds) S u

d Iu =   - d τ u  Iu  +   d τ u  S u  =   - d τ u   (Iu – S u)

whence:

d Iu / u  =  -  Iu – S u

For which the solution is found (on integration):

Iu =  I c  exp (-τ u)

Where:  τ u   = N L a u  =  N L a o  f u

Where ‘N’ is the number of absorbing atoms per unit volume and we already saw that:

a o =  a u  /  f u = a u [1 - e -  h u o / kT] (p e2/ mc) f

In terms of this new information, the equivalent width of the line can then be written as:

W u =  ò¥0 [1  -   exp (- N L a o  f u) ] du

For small x we may use the approximation that:

exp (-x) much greater than   1 – x

So rewrite the equation for  W u:

W u »   ò¥0 [1  -  (1 - N L a o  f u)  du

»   ò¥0   N L a o  f u  du » N L a o  ò¥0  f u  du

So we find W u  depends only on the form for the broadening function.  For very weak lines, for example:

ò¥0  f u  du  much greater than   1 so that W u » N L a o

Or simply proportional to N, the number of absorbing atoms. For “strong” lines the absorption near the center is very large so we can expect: N L a o  f u    >> 1 and with sufficient working this can be shown to give the corresponding equivalent width: W u  = 2  D u D

And for the moderately strong lines:

W u  =  2  D u D {ln  (N L a o/Öp  D u D )} ½

While very strong lines yield:

W u  =   1/p  (N L a o  g)½

Where g is the damping constant.

Assembling all the diverse W u    and plotting log W u vs.  log (N L a o ) one gets the curve of growth shown below:

Here, each contribution can be separately considered, in terms of the “linear”, “saturated” and “damping” segments. The first then is for the weakest lines such that W u  is simply proportional to the number of absorbing atoms. But as more atoms are added this approximation becomes invalid and so we are in the saturated part. In effect,  the near zero slope shows the degree to which the equivalent width increases at an almost uniform rate. Finally, we have the damping segment wherein the line under consideration develops “damping wings” because the number of absorbing atoms is so large. (Bear in mind again that  a o  is proportional to the transition probability)

Problems:

1)Consider the ionization of helium, with three stages of ionization for which the partition functions are:    Z I =    Z 3    =   1,  and Z 2 =    2

With ionization energies:

c  1 =   24.58 eV    and c  2 =   13.6 eV

For a temperature of T = 5040 K write out two versions of the Saha equation for:

log( P e N 2 /N 1)   and log( P e N 3 /N 2)

And hence, compute:   (P e N 2 /N 1)   and (P e N 3 /N 2)

2)The diagram below was sketched by a first year astrophysics student. It represents part of the emission spectrum of atomic hydrogen. It contains a series of lines, and the wavelengths of some (in nm) are marked. '

There are no lines in the series less than 91.2 nm
(a)   In which region of the E-M spectrum would these lines occur?

(b) Obtain a relation between E and l   and thence find the photon energies equivalent for all the wavelengths marked.

(c)Use this information to map a partial energy level diagram for hydrogen. Show and label clearly the electronic transitions responsible for the emission lines labeled.

3)a) A stellar atmosphere is composed of pure hydrogen, at a temperature of 9600 K and with electron pressure, P e    =  200 dyne cm -2 .  Use the Saha equation to compute the fraction of atoms

that are ionized in relation to the total, i.e. find:   N II /( N I   + N II )
The partition functions are:  Z II =   1 and  Z I =   2

Hint: Check to see whether (E 2  -  E 1 ) >> kT with

k =  8.6174 x 10 -5 eV/K

So the Boltzman factor  exp (- E2 – E1) / kT << 1

b)Repeat the exercise again for a temperature of 5040 K and 20,000 K and thence account for why the Balmer lines attain maximum intensity at a temperature of 9520 K.

(4) Calculate how far you could see through the Earth’s atmosphere if it had the same opacity as the solar atmosphere, where:

k l   =   0.264 cm 2 /g.

Take the density of Earth’s atmosphere as:

r   =   0.0012 g cm -3

The density of the solar atmosphere is taken as :

r s   =  2.5 x 10  -7 g cm -3

And we know the intensity declines over a characteristic distance (length scale) such that:

=   1/ k l r