## Z

Each positive ion has a mass of Amp, if we ignore the difference between the proton and neutron masses. The total density of the gas is then p = Amp nZ + mene = Amp nz

In going to the second line we have ignored the mass of the electrons relative to the mass of the nucleons. Using equations (10.8) and (10.9), the electron density is related to the total density by ne = (Z/A 2 (P/mp

Substituting this into equation (10.7) and adding the factor of two to account for the difference v y z between our estimate and the detailed calculation, we have

Note that, in a degenerate gas, the pressure depends on the density, but not on the temperature. We have already seen that this point is important in deciding whether the triple-alpha process will take place in a controlled way (normal gas) or in a flash (degenerate gas).

### 10.4.2 Properties of white dwarfs

A star supported by electron degeneracy pressure will be quite small, since it must collapse to a high density before the degeneracy pressure is high enough to stop the collapse. These objects are quite hot, being the remnant of the core of a star. These objects are the stars that appear on the HR diagram as white dwarfs.

Example 10.2 White dwarf density Estimate the density of a white dwarf if it has a solar mass packed into a sphere with approximately 10_ 2 R© (approximately the size of the Earth) as found in Section 3.5.

solution

We find the density by dividing the mass by the volume:

(Remember, the density of water is only 1 g/cm3, so a white dwarf is very dense.)

Example 10.3 White dwarf degeneracy pressure For a white dwarf of density 1.0 X 106 g/cm3, and Z/A = 0.5, estimate the degeneracy pressure and compare it with the thermal pressure of a gas at a temperature of 1.0 X 107 K.

solution

We find the pressure from equation (10.11):

(1.05 X 10"27 erg s)2" 9.11 X 10 "28 g .

For an ideal gas, the pressure is given by

where ne + nZ represents the total density. However, each atom of atomic number Z contributes Z electrons, so ne = ZnZ

We therefore have

We can now relate this to the density p. If A is the mass number of the nuclei, then (ignoring the difference between proton and neutron masses)

This gives

The degeneracy pressure is a factor of about 100 higher than the normal thermal pressure, even at this very high temperature.

We can also use the expression for degeneracy pressure (equation 10.11) to relate the mass and radius of a white dwarf. We saw in Example 9.4 that we could use hydrostatic equilibrium to approximate the central pressure by

If we put this into equation (10.11), and substitute M/4R3 for the density, we find

Rearranging gives h2

4^me

4Amp

3.2 X 1022 dyn/cm2

The right-hand side is all constants, so for a given R a mass can be calculated (Problem 10.12).

A degenerate gas has a very low opacity to radiation. For a photon to be absorbed, an electron would have to jump to a higher energy state.

Path of White Dwarf

Path of White Dwarf

Path of Visible Star

Path of Center of Mass

### Fig 10.13.

Even if we cannot see a white dwarf in a binary system, we can detect its presence.The visible star will appear to wobble as it and the white dwarf orbit their common center of mass.

Path of Visible Star

Path of Center of Mass

### Fig 10.13.

Even if we cannot see a white dwarf in a binary system, we can detect its presence.The visible star will appear to wobble as it and the white dwarf orbit their common center of mass.

However, such a transition would have to be an already empty state, and may require more energy than is carried by an optical photon. In addition, degenerate gases are good heat conductors, which explains why metals are good heat conductors. The low opacity and high thermal conductivity mean that a degenerate gas cannot support a large temperature difference. The internal temperature of a white dwarf is approximately constant across that star, at about 107 K. The outermost 1% is not degenerate, and it is in that thin layer that the temperature falls from 107 K to roughly the 104 K indicated by its spectral type. These conditions make a white dwarf very different from a normal star. In addition, Zeeman splitting measurements suggest very strong magnetic fields, about 107 gauss in some cases.

What happens to a star after it becomes a white dwarf? As it radiates it must get cooler. This is because it is giving off energy, but has no source of new energy. The degeneracy pressure does not depend on the temperature, so the star will maintain its size as it cools. Eventually, it will be too cool to see. It will take tens of billions of years for a star that is now visible as a white dwarf to become that cool. We can understand the long lifetime when we realize that a white dwarf radiates like a 104 K blackbody, but has more energy than a thermal reservoir at 107 K (see Problem 10.13).

Even at their current temperatures, their small sizes make white dwarfs difficult to see. We some times detect their presence in binary systems by measuring their gravitational influence on a star that we can see. We deduce the mass of the white dwarf from its influence on the visible companion's orbit (Fig. 10.13).

### 10.4.3 Relativistic effects

The treatment of degeneracy pressure must be modified by considerations introduced in the special theory of relativity (Chapter 7). This was first realized by S. Chandrasekhar (who shared the 1983 Nobel Prize in physics for his work on stellar structure). Chandrasekhar found that these corrections reduce the degeneracy pressure. This provides an upper limit to the mass that can be supported by electron degeneracy pressure.

The modification arises from the fact that electrons cannot travel faster than the speed of light. In using equation (10.5), we can still say that px is (h/2-n-)(ne)1/3. However, we can no longer say that vx = px/me. We have to use the correct rel-ativistic expression, as discussed in Chapter 7. To find the maximum degeneracy pressure, we take vx = c. This gives

A more detailed calculation gives approximately 0.8 times this. Using this, the expression analogous to equation (10.11) is

If we use the same assumptions to find the mass-radius relation (equation 10.14), we find that the radius drops out, and we simply have an expression for the mass (Problem 10.14):

This mass corresponds to the maximum pressure, so it is the maximum mass that can be supported by electron degeneracy pressure. A more accurate calculation, which takes into account variations in pressure and density with distance from the center of the star, gives a mass that is a factor of about 60 higher. The resulting mass, called the Chandrasekhar limit, is 1.44 M0. A star whose nuclear processes have stopped, and whose mass is greater than 1.44 M0, will continue to collapse beyond the white dwarf phase. The fate of such stars will be discussed in Chapter 11.

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