Copyright © Michael Richmond.
This work is licensed under a Creative Commons License.
We are interested in the details of light's passage through the outer layers of a star. One of the crucial factors in this process is the "temperature" of the gas. But -- what exactly do we mean by "temperature"? Is it possible to define a temperature in these exotic regions?
It will turn out that one can define a "temperature" in some very strict sense only under special conditions, called thermodynamic equilibrium, which do not actually apply in the atmospheres of real stars. However, for many purposes, one can define a sort of local thermodynamic equilibrium which is good enough to bring theory and observation into agreement.
Let's count the ways we could try to describe the temperature of the photosphere of a star.
Find the temperature of a blackbody which would produce a spectrum yielding the same difference in magnitudes when observed through those filters.
determine the temperature which would reproduce the observed ratio.
determine the temperature which would reproduce the observed ratio.
determine the temperature which would reproduce the observed distribution. This is often simplified to "measure some average speed of particles, and compare to the average speed of a Maxwell-Boltzmann distribution."
Q: Which of these is simplest to measure, in practice? What
observations would you need?
Q: Which of these is hardest to measure, in practice? What
observations would you need?
In some very simple situations, all these temperatures might be exactly the same. For example, if we build a box, fill it with some gas, heat the walls of the box to a uniform and steady temperature, and wait a bit, we will find that a single temperature T will fit into all of these equations. In such an ideal case, we say that the gas has achieved thermodynamic equilibrium (or TE for short).
Two of the hallmarks of thermodynamic equilibrium are isotropy -- properties are the same in all directions -- and homogeneity -- properties are the same in all places. If we consider an entire star as a whole, neither condition is true: the center of a star is much hotter than its outer regions, for example.
Rats! If we are trying to understand the detailed interaction of light and matter inside a star, it would be very, very convenient if we could use a single value, T, to predict the speed of gas molecules, the excitation of atoms, the spectrum of radiation, etc. But it appears that real life is more complicated ....
... unless, perhaps, we can find some way to limit ourselves to a small enough piece of the star that matter and radiation are at least approximately in thermodynamic equilibrium.
But how small does the piece have to be?
Well, one way to define such a piece is to choose a region that satisfies these two conditions:
In colloquial terms, we need to choose a region small enough that both particles and photons "see" a constant temperature in all directions. We can say that within this small region, there is an equilibrium between matter and energy which is "close enough" to thermal equilibrium. We might say that it satisfies a Local Thermodynamic Equilibrium, or LTE for short.
Just how far CAN a particle travel before it probably bumps into another particle?
Let's consider one particular particle, moving through a region filled with other little particles of diameter d, with a density of n particles per unit volume. How far will the blue particle travel, on average, before running into one of the other particles?
We can simplify the calculations by imagining an equivalent situation:
In this situation, collisions between the big blue particle and the teeny balls will occur at exactly the same rate as between the original blue particle and the other particles, but we avoid all the tricky math to detect collisions between spheres.
Now, as the blue particle moves through the sea of other particles, it will sweep out a cylindrical shape. When the volume inside that cylinder is equal to the typical volume occupied by a single particle, we can expect a collision.
Q: What is the volume of the cylinder swept out by the blue particle? Q: What is the typical volume occupied by a single particle? Q: At what distance L will the two volumes be equal?
We end up with an expression for the mean free path of a particle travelling through this region:
Sometimes the cross section area of the particles (shown as πd2 in the equation above) is denoted by the Greek letter σ, so that the mean free path can be written
Q: Ordinary air at room temperature has a density of about 1 kg
per cubic meter. Pretend that is made entirely of oxygen
molecules, each of which has a typical size of 10-10 m.
What is the mean free path of molecules in the air?
Let's repeat that calculation for particles in the solar photosphere.
Q: The outer regions of the solar atmosphere -- the photosphere --
are much less dense than the air on Earth. The density
there is roughly 2.5 x 10-4 kg/m3.
Pretend that is made entirely of hydrogen atoms, each of
which has a typical size of 10-10 m.
What is the mean free path of molecules in the solar photosphere?
How does it compare to that of molecules in an ordinary room on Earth?
We've seen how to calculate the mean free path of a particle of gas, but what about the mean free path of a photon? In order to do that, we need to look carefully at the way that light dims as it travels through a cloud of material.
Suppose that a beam of light with some original intensity I0 runs into a slab of material.
As the rays move through the material, some may be absorbed ...
... or scattered ...
... so that only a fraction I reaches the other side.
What is the ratio of the final intensity to the original intensity?
There are several factors which must affect the intensity in obvious ways:
So, if the original beam of intensity I passes through a tiny slab of length ds, it's reasonable to think that the change in intensity dI will depend linearly on these three factors:
Note that the change in intensity is NEGATIVE -- after the beam of light travels through the material, the resulting intensity must be smaller. Right?
Q: Are there any instances in which a beam of light travelling
through a slab of material can emerge with a
LARGER intensity?
So, the relationship between the initial intensity and the change in intensity can be written as
If we integrate this expression over some distance s, we find that the outgoing intensity I can be written in terms of the incoming intensity I0 like so:
How quickly does the intensity change? Well, the intensity will decrease to a factor of just 1/e of its original value after a distance s such that
This is a useful distance to keep in mind. We can call it the "characteristic distance" of light travelling through the material. We can even consider it to be the "mean free path" of a photon. Just how big might this be in some common situations?
In the photosphere of the Sun, the opacity to optical light is
very roughly
κ = 0.0264 m2 / kg
and the density
ρ = 2.5 x 10-4 kg / m3
Q: What is the mean free path of an optical photon in the photosphere?
Q: How does that compare to the mean free path of a hydrogen atom?
My answers.
Note that we have now seen two ways to compute the mean free path -- one more commonly used for photons, one more commonly used for particles.
But the two expressions can be exchanged for each other, in many cases. If one is given the cross-section area σ of photon-atom interactions, for example, and the density of atoms in the cloud, then one could compute the mean free path of a photon through the cloud.
Consider a beam of light entering some slab of material. The slab might be relatively narrow,
yet still block a considerable fraction of the light, if its atoms are densely packed and efficient at interacting with light.
On the other hand, a much wider slab of material
with a lower opacity and/or density might block exactly the same fraction of light,
If all we care about is the fraction of light which is blocked, or makes it through the slab, we might combine the three factors
We call this variable τ the optical depth of the slab. Look carefully at the definition of optical depth -- it is exactly the same as "the number of mean free paths through the slab."
The optical depth turns out have several meanings. Consider, for example, light which must makes its way through some scattering medium. Suppose that the mean free path of the light is small compared to the thickness of the material, so that the light bounces many times before finally escaping. Note the location of the last bounce.
If we send a second light ray into the medium, it will follow a different path ...
... and a third ray will again scatter randomly before finding its way out.
But note that the location of the final bounce tends to be roughly one mean free path, or one optical depth, from the outer edge of the medium. If we turn the situation around and ask, "How far can we see into the slab from the outside?"
the answer turns out to be "about one mean free path", or, equivalently, "about one optical depth."
Exercise:
- Estimate the optical depth of the Earth's atmosphere; consider the view across a landscape from the top of a mountain. Perhaps these pictures will help:
- Dawn from Lick Observatory, on Mount Hamilton with San Jose in the distance.
- The San Francisco Bay looking north from San Francisco; you can see Alcatraz and the hills of Marin in the background.
- The Blue Ridge Mountains from a photo gallery run by Richard Saylor.
- Compare the optical depth of the Earth's atmosphere to that of the Sun's photosphere you calculated earlier.
The overall optical depth of a cloud of gas is an important number. It tells us right away if the cloud falls into one of two useful regimes:
Consider this example: suppose that you look at a portion of the Earth's atmosphere on a clear, sunny day, over a distance L = 1 km; it has an optical depth of τ = 0.05.
Q: Send a beam of intensity I
through the gas. What fraction of
the light is blocked, and what fraction
escapes?
Q: If you double the length of the sample,
what happens to the fraction which
is blocked? What happens to the fraction
which escapes?
In the optically thin regime, the amount of extinction (absorption plus scattering) is simply related to the amount of material: double the amount of stuff, double the extinction. There is hope, then, that if we can measure the amount of light absorbed (or emitted) by the gas, we can calculate exactly how much gas there is.
Consider this example: suppose that you look at a typical portion of the atmosphere in smoggy downtown LA on a bad day. A length of L = 1 km corresponds to an optical depth of τ = 3.
Q: Send a beam of intensity I
through the gas. What fraction of
the light is blocked, and what fraction
escapes?
Q: If you double the length of the sample,
what happens to the fraction which
is blocked? What happens to the fraction
which escapes?
You can't see through an optically thick medium; you can only see light emitted by the very outermost layers. Bummer.
But there is one convenient feature of optically thick materials: the spectrum of the light they emit is a blackbody spectrum, or very close to it.
The opacity κ depends on the detailed interaction of light with the matter inside some sample of material. It varies with a large number of factors, such as temperature, density, chemical composition, ionization state, and so forth. There are many different physical mechanisms by which matter can absorb or scatter light. Let's try to put them into some categories.
In these interactions, only photons with particular energies may participate -- all other photons will pass through untouched. Moreover, there is no simple, general formula describing this sort of opacity as a function of temperature, etc.
In these interactions, photons of any energy (above or below some threshold) can be emitted or absorbed; in other words, these interactions produce a continuous spectrum. In some cases, one can use a simple formula to describe the bound-free opacity.
Note that in order for a photon to be absorbed in this manner by an atom, the photon must have at least enough energy to lift an electron from its current energy level to n = ∞ -- that is, to ionize the atom.
Q: Consider a hydrogen atom in the n = 2 state.
How much energy must a photon have to ionize
this atom?
Q: What is the wavelength corresponding to this critical
energy?
Below is a spectrum of an A0 main sequence star. Do you see any evidence for bound-free absorption in this spectrum?
This process is sometimes called "brehmsstrahlung", which is German for "braking radiation," because a fast electron will slow down and radiate away its energy as it passes through some material.
The opacity due to electron scattering can be written down in a general formula, as the cross-section for interaction has a constant value, independent of the photon's energy or wavelength: σT = 6.65 x 10-29 m2
The total opacity in some material is just the sum of all these components. It will, in general, depend on the wavelength of the light, and on the properties of the material as well: density, temperature, chemical composition, etc. There are various formulae which can provide some rough approximations to the opacity averaged over all wavelengths for a particular choice of material and temperature; you can see two written in section 9.2 of your textbook. If you prefer tables with opacities for various combinations of parameters, one of the references in "for more information" provides hundreds for your perusal. The figure below shows the overall opacity for a solar composition over the range of 104 to 108 Kelvin.
Figure 2a taken from
Rogers, F. J., and Iglesias, C. A., ApJS 79, 507 (1992)
and slightly modified
Copyright © Michael Richmond.
This work is licensed under a Creative Commons License.