Revisiting Spectral Line Formation and Transitions (3)

 As discussed in Part 2 (12/11 post) we can express the equivalent width in two ways, based on frequency  (u)  or wavelength (l):

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 e.g. from the continuous spectrum outside of the spectral line where the quantity (I _c – I _u/ I _c ) is referred to as the “depth of the line”.  This is the analogous quantity to (F _c – F _l / F _c ) on the right side where we have radiant flux units.

     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 I_u / dτ _u  =  -  I_u – S _u

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

d I_u / ds   =  -k _u I_u  +   k _u S _u

Where:   k _u   =   d τ _u / ds

So: 

d I_u / ds   =  -( d τ _u / ds) I_u  +   (d τ _u / ds) S _u

d I_u =   - d τ _u  I_u  +   d τ _u  S _u  =   - d τ _u   (I_u – S _u)

whence:

d I_u / dτ _u  =  -  I_u – S _u

For which the solution is found (on integration):

I_u =  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 newly derived 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 such that:

exp (-x) »   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  »  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 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:  

                                                                      

Suggested Problem:  

How would the curve of growth appear if the broadening function:

 f _u  du  = 1/ Öp   [exp (u  - u _o)  /D u_D ]²  du / D u_D

Had:   (u)   »   (u_o)