Copyright © Michael Richmond.
This work is licensed under a Creative Commons License.
Today, we introduce one more tool used by astronomers to study the atmospheres of stars: the curve of growth, which converts the equivalent widths of absorption lines into a measure of the number of absorbing atoms in the atmosphere. After that, we will put many of the ideas from the past few weeks into practice, as we go through the steps required to use the spectrum of a star to figure out the abundance of some element in its photosphere.
For a recent cautionary tale on the difficulties of deriving chemical abundances from spectral data, see Observational constraints on the origin of the elements, Magg et al., A&A 661, A140 (2022)
When we look at the spectrum of a star, we see a mixture of strong and weak lines. In the spectrum below, for example, the line centered at 6563 Angstroms is very strong, while the line at 6580 Angstroms is weak.
Last time, we defined the equivalent width of a line as the integrated area "above the curve" of an absorption line. We can measure the equivalent widths of these two lines -- let's say that the strong line has an equivalent width 100 times larger than that of the weak line.
Q: Does that mean that there are 100 times
as many atoms absorbing the 6563 line as the 6580 line?
No, it doesn't, in general. The only time that the equivalent width of a line is linearly proportional to the number of absorbing atoms is when the line is optically thin (τ << 1); in other words, when the absorbing layer of gas blocks only a small fraction of all the light passing through it. The 6580 line in the spectrum above qualifies as such a line -- but not the 6580 one. As a simple rule, if a line has a small depth, less than, say, 50 percent, then it may be optically thin.
The graph below shows the curve of growth:
Curve based on a digitized version of Fig 9.22 of Carroll and Ostlie (1996),
which is in turn based on data from Aller, "Atoms, Stars
and Nebulae" (1971).
As you can see, the relationship is linear only for small equivalent widths.
When the absorbing layer blocks most of the light at the center of a line, that simple relationship no longer holds. Instead, as the gas becomes optically thick (τ ≈ 1), the equivalent width grows very slowly as the number of absorbing atoms increases. A line with a depth close to 90 or 95 percent has reached this stage. We might characterize such lines as "saturated".
Finally, when the number of absorbing atoms is VERY large, pressure/collisional broadening starts to create strong wings. Those wings cause the equivalent width to start growing again as the number of atoms increase, albeit at a slower rate.
Now, let's look more closely at the label of the horizontal axis of this diagram. The label indicates that the graph shows the logarithm-base-10 of some quantity involving three terms:
What is the "oscillator strength" of the transition? It is a measure of the probability that an atom starting in some lower energy state will absorb a photon and jump up to some particular higher energy state. Consider the neutral sodium (Na I) atom, for example. If it starts in its ground state, labelled "3s", it may absorb photons of many different energies. The oscillator strength indicates the relative probability of these possible transitions.
Figure 1 taken from
Erman, Brzozowski, and Smith, ApJ 192, 59 (1974)
For example, the oscillator strengths for the two labelled transitions are
It is much more likely for an atom to jump to the 3p level.
In the solar photosphere, atoms of neutral sodium in the ground state might absorb light of many different wavelengths. The equivalent width of the 3302 line tells us about the number of atoms originally in the ground state which jumped to the 4p state; the equivalent width of the 5589 line tells us about the number of atoms originally in the ground state which jumped to the 3p state. It would be much more useful if we could find out the number of neutral sodium atoms in the ground state, period, not just the number which made some particular transition. By including this oscillator strength in the curve of growth, we can (with a little extra work) use the observations to recover the number of atoms in the ground state, period.
A common way to use the curve of growth follows this recipe:
Let's try it! We'll start with a spectrum of the Sun, and end up with the abundance of sodium relative to hydrogen in the solar photosphere.
Our target is the neutral sodium atom (Na I). From its ground energy state (3s), it may absorb light of wavelengths 3302.38 and 5589.97 Angstroms -- and many other wavelengths, too, but let's concentrate on these. We'll focus most of our attention on the 5589 line.
First, the 3302.38 line, based on measurements made by the Kitt Peak Solar Fourier Transform Spectrometer.
Q: Estimate the equivalent width of this line.
The 5589 line is much, much stronger.
Since this line has curvy wings, a triangle is not a very good approximation. I'll just do the math for you here; my numerical integration yields an equivalent width of 0.875 Angstroms.
The next step is to use the measured equivalent widths to compute the quantity plotted on the vertical scale of the curve of growth. I'll call it "y" for short.
Q: The equivalent widths of these two lines are
3302.38: EW = 0.13 Angstroms
5589.97: EW = 0.875 Angstroms
What is the "y" value for each line?
For each line, we locate its vertical position on the curve of growth, and then look down at the horizontal axis to read the quantity there -- which I'll call "x" for short.
Now, these two lines lie on somewhat different portions of the curve of growth.
One can't read values accurately from that large-scale graph, so let me zoom in on each one. First, the 3302 line:
Q: Estimate the value of "x" for the 3302 line.
I get a value of about 13.77.
Next, the 5589 line:
Q: Estimate the value of "x" for the 5589 line.
I get a value of about 14.91.
We can now use our value of this "x" coordinate on the curve of growth to solve for N, the number of atoms in the lower state of the transition. The common convention for this quantity, which was used in constructing the graph we've examined, is that this N is the column density of atoms in the lower state:
All we need to do is to invert this equation
to solve for "x". We'll need to plug in the appropriate oscillation strength and wavelength, of course. From Caroll and Ostlie, section 9.4,
Q: What is the value of N using the 3302 line? Q: What is the value of N using the 5589 line?
Let's choose the value from the 5589 line, which is much more well defined than the 3302 line. I worry that the 3302 line might be a blend of Na I and some other absorption line.
So, have we finished at this point? The column density of hydrogen, for comparison, is 6.56 x 1023 atoms per sq.cm. -- so are we ready to compute the abundance of sodium relative to hydrogen?
NO!
All we have done so far is to determine the number of neutral sodium atoms in the ground state. What we want to do is to find the total number of sodium atoms, in any energy level, neutral or ionized.
Q: How can we make these corrections?
If we know the number of neutral sodium atoms in the ground (3s) state, can we figure out the number in ALL the energy states? Hmmm ... if only we had some way to calculate the ratio of atomic populations in different energy levels ...
Of course! We'll use the Boltzmann equation!
It can tell us the ratio of populations in different energy states; only two states at a time, but perhaps we can see a pattern and quickly come to a general conclusion.
Let's try to use this equation to compute the ratio of neutral atoms in the ground state (3s) and the excited state (3p), the two levels involved in the 5589 Å absorption line. One can look up the energies and statistical weights in any number of places, such as the NIST Atomic Spectra Database Levels Form. Let's adopt these values, with the "a" denoting the ground state and "b" the excited state.
We'll continue to use a temperature of T = 5800 K, and the Boltzmann constant is, as usual, k = 1.38 x 10-23 J/K.
Q: What is the ratio of the number of neutral atoms in the
(3p) excited state to those in the (3s) ground state?
Hmmm. That's a pretty small number. If one repeats this exercise with the values for the (4p) excited state, which is responsible for the 3302 Å line, one finds a much smaller fraction. There are many other energy levels with higher energies, but for each of them, the negative exponential term becomes more and more dominant.
To a rough approximation, then, we might just say that all the neutral sodium atoms are in the ground state. But let's make at least a small correction: the number of neutral sodium atoms in all states should be
Well, that wasn't much of a correction, but it can't hurt.
Q: Have we completed our analysis, or is there some other
correction that must be made to the abundance of sodium?
There is another correction we might need to make. All our calculations so far have assumed that NEUTRAL sodium atoms are the only ones found in the solar photosphere. But what about IONIZED sodium atoms? If a significant fraction of the sodium atoms are ionized, then our count so far will be missing a lot of sodium.
So, how can we compute the relative numbers of ionized and neutral atoms? By using the Saha equation, of course!
Here, let's choose state i to be neutral, and state i+1 to be singly ionized. Once again, we can look up some of the relevant atomic properties of sodium, and, once again, the NIST websites are a good place to look. Let's adopt the following values:
There, that's all the ... wait a minute. One item is missing: ne, the density of free electrons in the solar photosphere. This number is crucial to the ionization fraction, because the more free electrons there are, the more likely an ion will grab one and turn into a neutral atom.
Could we figure this out ourselves? We know (from other sources) that the density of hydrogen atoms in the photosphere is roughly nH = 1.5 x 1023 atom cm-3. If -- and this is a big if -- all that hydrogen were ionized, then the density of free electrons would be the same: ne = 1.5 x 1023 atom cm-3. But is all the hydrogen ionized?
Q: Is all the hydrogen ionized at a temperature of 5800 K?
Well, as you may recall, we used the Saha equation in a previous class to calculate the fraction of hydrogen atoms which would be ionized at different temperatures. This was our result:
That pretty clearly shows that the fraction of hydrogen which is ionized at 5800 K is very small; thus, we can't assume that the density of free electrons is dominated by those contributed by hydrogen.
Instead, at this relatively low temperature, the great majority of free electrons are contributed by the "heavy" elements: carbon, oxygen, iron, calcium, etc. They are much less abundant than hydrogen, but their electrons are (in general) much easier to remove. Fortunately for us, astronomers have measured this electron density in the solar photosphere, so let's just adopt this representative value: ne = 1.25 x 1019 atom m-3. Note that the units here are "per cubic meter", not "per cubic cm", so that we can use it together with all the other MKS values I've provided.
Q: What is the ratio of ionized to neutral sodium atoms? Q: Does this have a large effect on the overall abundance of sodium?
Holy cow! The ionized sodium atoms GREATLY outnumber the neutral ones, so if one doesn't include them in the abundance calculations, one will derive an abundance which is much, much too small.
What about doubly-ionized sodium atoms, or those with even more electrons removed. Should we include them in our calculations, too? Fortunately, it turns out that all ionization states above Na II are vanishingly uncommon, and so we may safely ignore them.
If we now compute the total abundance of all sodium atoms, ionized and neutral, in all energy states, we find
How does our result compare to that of other studies? Well, in order to make an easy comparison, we need to express our value in the standard way.
One might think that astronomers would simply divide the number of sodium atoms by that of hydrogen atoms to derive a relative abundance:
But astronomers (and some physicists and geologists) have adopted a system which is designed, I guess, to avoid terms with exponential notation. They define the elemental abundance on a logarithmic scale, on which hydrogen tops the list at a value of 12:
Q: What is our elemental abundance of sodium on this scale?
Let's compare that value to the numbers in the table below, one based on solar spectra (left-hand side), and one based on measurements of ancient meteorites (right-hand side).
Table 4 taken from
Lodders, Palme, and Gall, Landolt-Bornstein, New Series, Astronomy
Astrophysics, p. 560 (2009)
I think we did quite well, actually. The close agreement is in large part coincidental, I would guess, but it's nice to end up so close to the truth.
Note that sodium is actually relatively common, as elements go.
Figure 6 taken from
Lodders, Palme, and Gall, Landolt-Bornstein, New Series, Astronomy
Astrophysics, p. 560 (2009)
Copyright © Michael Richmond.
This work is licensed under a Creative Commons License.