ultra-relativistic case
ETOT(UR) = 0. (2.12)
That is, in the ultra-relativistic case, the system is at the boundary of being bound (i.e. ETOT < 0) and unbound (i.e. ETOT> 0). If a bound system approaches the
ultra-relativistic limit, it moves closer towards becoming unbound.
2.3 The lower mass limit for stars
In Chapter 1 we discussed hydrogen burning in stars, but did not consider which self-gravitating objects attain stardom and which do not. If the temperature in the core of a protostar continued to increase as it contracted, ultimately the ignition temperature for hydrogen burning would be reached in every case. However, the temperature cannot always increase, due to the onset of electron degeneracy (see the box below). Unlike the case of an ideal gas, in a degenerate electron gas, an increase in pressure does not lead to an increase in temperature. If a contracting protostar becomes degenerate before it becomes hot enough for hydrogen fusion to begin, it will instead become a brown dwarf. The situation under which this can occur is explored in this section.
Electron degeneracy
Electron degeneracy is a condition in which the quantum nature of electrons cannot be ignored. Although it is often the case that electrons can be treated as isolated, classical particles, like all matter they also have wave-like properties. This is one example of the wave–particle duality that pervades a quantum view of the Universe, and which is beautifully demonstrated in an experiment showing that a beam of electrons can be diffracted.
If electrons are packed so densely that their separations are comparable to their de Broglie wavelengths, then the full range of their quantum properties has to be considered, including the Pauli exclusion principle which prohibits the overlapping of electrons having the same energy. This leads to some unexpected but well understood consequences, such as the decoupling of pressure and temperature in a degenerate electron gas. Other consequences of electron degeneracy will be encountered later in this book, as will consequences of neutron degeneracy.
In a cold, dense gas, the behaviour of the electrons is controlled by the laws of quantum mechanics once the typical separation between the electrons becomes comparable to their de Broglie wavelength. The de Broglie wavelength is given by λdB= h/p where h is Planck’s constant and p is the electron’s momentum.
Now, in the core of a star, an electron’s kinetic energy can be written as
EK= 32kTc, where Tcis the core temperature of the star. Another expression for
the kinetic energy of an electron is simply EK = 12mev2. So equating these two
we obtain 3kTc= mev2 or v = (3kTc/me)1/2. (Strictly, electrons will exist with
a range of speeds, but (3kTc/me)1/2is close to the average.) The momentum of
wavelength of an electron in the core of a star is
λdB = h
(3mekTc)1/2
. (2.13)
The core of a star consists of an ionized plasma containing equal numbers of protons and electrons. The mean separation l between the protons in the core is given by ρc= mp/l3where ρcis the density of the plasma in the core and mpis
the mass of a proton. The mean separation between the electrons will also be l (because of overall charge neutrality), hence the plasma will become degenerate when λdB≈ l, or when , mp ρc -1/3 ≈ h (3mekTc)1/2 .
So the limiting density for the core, below which electron degeneracy can be avoided is
ρc< mp(3mekTc) 3/2
h3 . (2.14)
In order to find out what mass of star corresponds to this density limit, we must find an expression linking the star’s mass to its internal temperature. We can do that by making use of the result obtained in the last section found by combining the virial theorem with the ideal gas law.
As you saw in Section 2.2, the condition for hydrostatic equilibrium inside a star is 2EK+ EGR = 0. For an ideal gas containing N protons and N electrons, the
kinetic energy of the gas is EK = 32NkTI+ 32NkTI = 3NkTIwhere TIis a
typical internal temperature of the star. The gravitational potential energy of the gas is EGR = −3GM2/5R (for a spherical region of uniform density). Hence we
can write 2 × 3NkTI= 3GM2/5R and so the typical thermal energy inside the
star is
kTI= GM 2
10NR.
Now, we note that the total mass of the star can be written as M = Nmp+ Nme
where mpand meare the masses of a proton and electron respectively. Since mp >> me, we have M ≈ Nmpand the number of protons (or electrons) can be
written as N = M/mp, so the typical thermal energy becomes kTI= GMm10Rp
and since the average density for a sphere of mass M and radius R is
$ρ6 = M/43πR3, we can replace the radius R by (3M/4π $ρ6)1/3. Therefore the
expression for the typical thermal energy becomes
kTI= 101 , 4π 3 -1/3 GmpM2/3$ρ61/3. (2.15)
So, for an ideal gas, the temperature inside a star is proportional to the two-thirds power of the mass multiplied by the one-third power of the average density. We may therefore combine Equation 2.15 with Equation 2.14 to calculate the limiting mass below which the core of a star will be degenerate.
2.3 The lower mass limit for stars
Worked Example 2.3
What is the mass limit between objects which become stars and those which become brown dwarfs?
(a) Combine the condition to avoid degeneracy from Equation 2.14 with the expression for the typical internal thermal energy from Equation 2.15 to find an expression for the limiting core thermal energy kTcto avoid
degeneracy, in terms of the mass of the star. As an order of magnitude approximation, you should assume that the core of the star is about 10 times hotter than its typical internal temperature TIand that the core density is
about 100 times the average density, as is the case for the Sun.
(b) Compute the minimum mass for a star to ignite hydrogen burning in its core and avoid becoming a degenerate brown dwarf. Adopt
Tign = 1.5 × 106 K as the minimum ignition temperature for hydrogen
burning.
Solution
(a) Assuming Tc= 10TIand ρc= 100 $ρ6, Equation 2.15 becomes
1 10kTc= 1 10 , 4π 3 -1/3 GmpM2/3(ρc/100)1/3.
Substituting into this the limiting core density from Equation 2.14, canceling the factors of 1/10 and rearranging slightly gives
kTc< , 4π 300 -1/3 GM2/3m4/3p (3mekTc)1/2/h.
Collecting the terms in kTcand rearranging this further gives the condition
to avoid degeneracy as kTc< , 4π 300 -2/3 3G2m8/3ph2meM4/3. (b) If Tc= Tign, we require M4/3 > kTign '4π 300 +2/3 3G2 m8/3p me h2 .
Putting the numbers into the right-hand fraction in the denominator, we have
m8/3p me h2 = ' 1.673 × 10−27kg+8/3× 9.109 × 10−31kg (6.626 × 10−34J s)2 = 8.184 × 10−36kg11/3J−2s−2. Therefore M4/3> ' 1.381 × 10−23J K−1× 1.5 × 106K 4π 300 +2/3 × 3 ×'6.673 × 10−11N m2kg−2+2× 8.184 × 10−36kg11/3J−2s−2 > 1.571 × 1039J3N−2m−4kg1/3s2.
Taking the (3/4)-power of both sides gives M > (1.571 × 1039)3/4[(N m)3N−2m−4kg1/3s2]3/4 > (2.5 × 1029)(N m−1kg1/3s2)3/4 > (2.5 × 1029)(kg m s−2m−1kg1/3s2)3/4 > (2.5 × 1029)(kg4/3)3/4 > 2.5 × 1029kg.
Since 1 M& = 1.99 × 1030kg, we can also express the limiting mass as
M > 2.5 × 1029kg/1.99 × 1030kg M−1
& ∼ 0.1 M&, to the nearest order of
magnitude.
The example above shows that the temperature at which degeneracy could set in depends on the mass of the star, according to T ∝ M4/3. Successful stars reach
the hydrogen ignition temperature before degeneracy sets in, but objects of sufficiently low mass become degenerate whilst cooler. For the latter objects, further contraction does not increase the temperature, so they fail to initiate thermonuclear reactions at a sufficiently high rate. Detailed computations show that the limit between successful stars and unsuccessful brown dwarfs occurs at
M = 0.08 M&.