Physics 321 

Prof. Dale E. Gary
NJIT 
Stellar Interiors  I
Clusters and Main Sequence Fitting
A Globular Cluster, M3
If we look at the HR diagram of a star cluster, such as the one below, we see that it is quite different from the one we made from the nearby stars catalog. This difference is a very important clue about the way stars evolve.
The HR (colormagnitude) diagram for the globular cluster M3.
We can use the cluster HR diagram above to find the distance to the cluster by comparing it with a calibrated HR Diagram such as the one we constructed earlier from the Nearby Stars catalog. Such a comparison uses the technique of main sequence fitting as illustrated below. Here we have scaled the cluster diagram (shown in red) to the same scale as the calibrated HR diagram, and shifted it keeping the BV values aligned, until the main sequences overlap. We can then read the distance modulus directly off the graph. In this case it is about 15.4, which places the cluster at about 12 kpc.
We will come back to the use of cluster HR diagrams for another purpose, the determination of the age of the universe! But first we have to learn something about stellar evolution.
Stars are made of relatively simple stuff. By mass, our Sun is 73% hydrogen, 26% helium, and only 1% of higher Z (atomic number) atoms. We write these quantities in terms of the mass fraction, as
 X = m_{H}n_{H }/ r = density of hydrogen / total density
 Y = m_{He}n_{He }/ r = density of helium / total density
 Z = m_{Z}n_{Z }/ r = density of everything else / total density
where n is the number
density (particles cm^{3})
of particles of all types, i.e. n = n_{e}
+ n_{H} + n_{He}
+ n_{Z}. This is the number density
of electrons plus all of the kinds of atoms. A good approximation
for most star interiors is that the gas is fully ionized. For a fully
ionized gas, how many electrons will there be? Each hydrogen atom
will contribute 1 electron. Each helium atom will contribute 2, and
each higher element will contribute Z (that is, a number equal to the atomic
number of the element). In this case, we have the mean molecular
weight given by:
For the Sun, X = 0.73,
Y = 0.26, and Z = 0.01, so 1/m
~ 1.67. The other atoms besides hydrogen and helium are referred
to in astronomy as metals, and the
value of Z is called the metallicity.
Pressure is force/unit area, F / A. The atmospheric pressure at sea level is about 14 lbs/squareinch, or 10 Pascal, or 1000 millibars (mm of mercury). Pressure in the interior of a star arises from the weight of the atmosphere directly above the pointthat is, from the force of gravitybalanced by the outward pressure due to energy release inside the star. Nearly all stars are static, meaning that they are not expanding or contracting significantly. Such a state is called hydrostatic equilibrium, and like all equilibrium situations, it is characterized by force balance. Consider a column of gas of crosssectional area A. The forces on a section (parcel) of the gas is shown in the figure on the left:
where DP
is the pressure difference (P_{1}
 P_{2}) on the two ends of the column
section, and Dr
is the radius difference (r_{1}
 r_{2}). Taking the limit as
Dr
approaches zero, we obtain the equation of hydrostatic equilibrium:
There is another relationship
we can obtain between pressure and density (and temperature) called the
equation
of state.
The temperature, density, and pressure are not independent, but are joined by the equation of state, which describes the macroscopic manifestation of particle interactions. One wellknown example of an equation of state is the ideal gas law:Other Equations of Statewhere P = pressure, V = volume, N = number of particles, k = Boltzmann's constant, and T = temperature. We can write this more simply in terms of the number density n = N / V asPV = NkTAbove in equation (1), we saw that the number density is related to the mass density byP = nkT.so the equation of state becomesn = r/mm_{H}Inserting (3) into (2), and assuming for the moment that temperature T does not change with height (an isothermal atmosphere), we obtain an expression for the density:P = rkT / mm_{H }. (3)which is a firstorder differential equation with the solution:dr/dr =  (mm_{H}g / kT) r (4a)where H = kT / mm_{H}g is called the scale height. and r_{o} is called the base density since it is the value of the density at r = 0. Note that if we had solved for r in equation (3) and substituted it for r on the righthand side of equation (2), we would obtain a similar differential equation for the pressure:r = r_{o} exp(r / H)with the solution:dP/dr =  (mm_{H}g / kT) P (4b)Therefore, the pressure and density vary with the same scale height. What is the physical interpretation of the scale height? It means that the pressure and density drops with height by the same factor of 1/e = 0.368 for each increase in height of a distance H. Note that a hot atmosphere (large T) has a larger scale height than a cool atmosphere. Also, increasing gravity lowers the scale height. It is of interest to calculate the scale height for some atmospheres.P = P_{o} exp(r / H).Example 1: Scale height at Earth's surface. H = kT / mm_{H}g, where T ~ 300 K, g = 9.8 m/s^{2}, but what is m? The atmosphere is mostly (4/5) molecular nitrogen, N_{2}, which has a mean molecular weight of 28. Using these values, we obtain:
H = kT / mm_{H}g = (1.38x10^{23} J/K)(300 K) / [28(1.67x10^{27 }kg)(9.8 m/s^{2})]In fact, the Earth's atmosphere rapidly grows cooler with height, so the scale height is somewhat smaller.
= 9.0 kmExample 2: Scale height in the photosphere of the Sun. Here, T = 5770 K, m ~ 0.6, but what is g? Remember that mg is the force of gravity, which is also
so g = GM / R^{2}. At the surface of the Sun, we would use M = 1.989x10^{30} kg, R = 6.96x10^{8} m (from Appendix 7 in the text), soF = mg = GMm / R^{2}g = (6.67x10^{11} N m^{2} / kg^{2})(1.989x10^{30} kg)/(6.96x10^{8} m)^{2} = 296 m/s^{2}. ~ 30 gThen,H = kT / mm_{H}g = (1.38x10^{23} J/K)(5770 K) / [0.6(1.67x10^{27} kg)(296 m/s^{2})]
= 270 km
In the atmospheres of stars and planets, the ideal gas law is a pretty good approximation. However, in the interiors of stars some new effects can come into play. In any star, near the core, especially, the light itself can exert significant pressure. A photon has momentum p = hn/ c, and when it is absorbed or reflected it exerts a force F = dp/dt. Consequently, electromagnetic radiation has an associated pressure. It may come as no surprise that the pressure is related to the total radiant energy flux, which we saw from the StefanBoltzmann law was F = sT^{4}. One can show thatwhere a = 4s/c = 7.56591 x 10^{16} J m^{3} K^{4} is the radiation constant. In this case, the total pressure, equation (3), is replaced byP_{rad} = 1/3 aT ^{4}Finally, when relativistic and quantum effects are included to derive the equation of state, a very different pressure equation is found that applies to extremely dense states of matter such as found in white dwarfs and neutron stars. We will look into this later in the course.P = rkT/ mm_{H }+ 1/3 aT ^{4} (3')