Physics 321 

Prof. Dale E. Gary
NJIT 
Stellar Interiors  II
Stellar Energy Sources
Stellar Energy Sources
Of course, coal is much
less efficient than the extremes that we assumed, and the answer was only
about 2000 years in that case.
and is negative because
we define the zero of potential energy at infinity. Starting
with an initial configuration, or arrangement of mass, as we move
mass toward the center of gravity of the arrangement the potential energy
decreases (becomes more negative) indicating that the kinetic energy must
have increased. Considering now the case of a spherical distribution
inside a star, we can add up all of the contributions to potential energy
by integrating over the mass of the star:
where dm
is the element of mass at radius r
from the center of the star, and M_{r}
is the interior massthat is, the
mass contained within the sphere of radius r
interior to the point in question. We have made use of two facts
in going from (1) to (2):
Both of these can be
proved using simple calculus, and were first shown by Isaac Newton.
This applies only to spherical shells of uniform density, which happily
is a condition well met by stars and many other objects. Note that
each shell can have a different density from all of the others, rather
like an onion.
Given this onionlike structure, we can consider the shell of uniform density r, at radius r and thickness dr. Such a shell will have a mass dm = 4pr^{2}r dr. Inserting this into (2), we haveVirial TheoremdU =  GM_{r}4pr^{2}r dr / rwhich can be integrated over the radius of the star, from the center (r = 0) to the surface (r = R) to give the total potential energy of the star:In this expression, note that M_{r} is itself an integral, since the total mass interior to the radius r requires integrating the density from the center to the point r:
RU =  4pG M_{r} rrdr (3) 0 so the entire expression requires the change of density with radius in the star. However, we can make some simplifying assumptions to see if gravitational potential energy can be the energy source for the Sun. Let's assume that the density is constant, with the value of the average density <r> = M_{r} / (4/3 pR^{3}), so that (4) becomes M_{r} ~ 4/3 pr^{3}<r>. Then the potential energy, (3), becomes
rM_{r} = 4p rr^{2}dr (4) 0
RU ~  4pG <r>^{2 } 4/3 pr^{4}dr 0 or
16p^{2} 3GM ^{2} U ~  G <r>^{2}R^{5} ~  15 5R We will discuss in a moment the virial theorem, which says that a gravitational system moving from one equilibrium state to another will release 1/2 its available energy as internal kinetic energy, and the other 1/2 is available to be radiated away. Thus, if in the initial configuration all of the mass is at infinity (zero potential energy) then the amount that can be radiated in getting to the final (current) configuration is U/2, or
DE = E_{f}E_{i} = U/2 = 3GM^{2} / 10RIn the case of the Sun, this is DE =3GM^{2} / 10R ~ 1.1 x 10^{41} J. Again, since the Sun shines at the rate L = dE/dt = 3.826 x 10^{26} J/s, it will take onlyDt = t_{KH }= DE / L = 1.1 x 10^{41} J / 3.826 x 10^{26} J/s ~ 10^{7} years.This time, t_{KH} is called the KelvinHelmholtz time scale, and was calculated near the turn of the century. At the time, this was considered an ironclad argument that the Earth and Sun coud not be much older than this, although geologists and paleontologists were finding that the Earth must be much older. We now know from radioactive dating techniques that the Earth and Moon are at least 4.5 billion years old. Clearly there must be another source of energy for the Sun. We have looked at gravitational energy in some detail, however, because for some objects and at some parts of their lifecycle gravitational energy is the dominant energy source.
As we mentioned, the virial theorem states that when a system is in one equilibrium state and changes to another equilibrium state, the difference in energy goes equally into two parts: (1) the internal energy of the system and (2) radiation or other loss mechanisms. What do we mean by equilibrium? An equilibrium is a situation in which the configuration of the system is not changing. However, by the phrase "not changing" we do not mean that the system cannot evolve. Rather, the system is allowed to change only slowly over some relevant timescale. Consider a planet orbiting around a star. If the planet orbit does not change appreciably over many orbits, then the system can be said to be in equilibrium. Consider a collapsing cloud of gas. If many collisions between particles occur before the cloud changes radius appreciably, the system can again be said to be in equilibrium.Nuclear EnergyIn the collapsing cloud example, above, we used the virial theorem to find the amount of energy radiated away. The virial theorem also says that this same amount of energy (3GM^{2} / 10R) goes into internal energy (kinetic energy, or heat in this case). It the cloud starts out at temperature 0 K, what will be its temperature at radius R? The number of particles in the cloud is N = M / mm_{H}. and each has thermal energy is 3/2 kT, so the total thermal energy is
3/2 NkT = 3GM^{2} / 10Rso the temperature change isDT = T_{f}  T_{i} = GM ^{2} / 5RNk = GM mm_{H }/ 5kR ~ 2.78 x 10^{6} Kwhere the numerical value is found using solar values for the mass and radius. In fact, the internal temperature of the Sun is higher than this, as we can see by simple application of hydrostatic equilibrium and the ideal gas law:The virial theorem has a wide range of validity and usefulness. We can use it in problems concerning planetary orbits, stellar collapse, accretion disks around black holes, and motions in clusters of galaxies.
 Average density of Sun: <r> = M_{sun}/4/3pR^{2}_{sun} ~ 1410 kg m^{3}
 Core pressure of Sun (Chapter 16, pg 311): dP/dr =  rg =  GM r/ R^{2} which using scaling arguments is roughly P_{c} = GM <r>/ R ~ 2.7 x 10^{14} N m^{2}
 Core temperature of Sun (Chapter 16, pg 312): P = rkT / mm_{H} =>
T_{c} = mm_{H}P_{c }/ <r>k ~ 1.4 x 10^{7} K.
The energy contained
in the nucleus is given by the rest mass of the nucleus,
E = mc^{2}.
Rest masses of an individual proton and neutron are very nearly the same:
Example:
How long will the Sun last if it shines at its current rate and converts
10%
of its mass from hydrogen to helium in nuclear fusion?
which is more than enough
time to account for the age of the solar system.