ASTR 498N Lecture 5 So What’s a Virial?
(online at www.astro.umd.edu/~drabin/)
Gravity is nothing if not persistent.
Charles Bogle
We’re sneaking up on stellar structure. Much of a star’s life (like ours, as you will eventually discover) consists of a valiant struggle to resist the tireless force of gravity.
Free
fall
(O&C §2.2)
What Newton says, goes:
As you know, the gravitational field at a point outside (or, in fact, within) a spherically symmetric shell or ball of mass is the same as if all the mass interior to that point were concentrated at the center. Consider the free-fall of the mass m from rest at initial radius R0 (we’ll assume M>>m for simplicity, although everything goes through the same in the center-of-mass frame).
Acceleration of m:
We already know a first integral,
where K is the kinetic energy and U the gravitational potential energy of the system,
from which we can write
and immediately integrate:
We
can write this in terms of the mean mass density inside R0, , as
Apart from the numerical factor, this is an equation worth remembering! “One over root G rho” pops up again and again as a characteristic dynamical time in a gravitational system; for example, it applies to the period of a radially pulsating star. Such a general relationship must be derivable from order-of-magnitude reasoning.
ROM (Rough Order of Magnitude estimate)
If you try to construct a quantity with the dimensions of time from G, M, and R (the only primary quantities available), you’ll get the same estimate.
Hydrostatic equilibrium
(O&C §10.1)
What can resist gravity? Pressure! Consider a shell of mass within a spherical star. The shell is at radius r with thickness dr, local mass density ρ(r), and mass Mr interior to r. The downward gravitational force on the shell is
while the pressure force supporting the shell is
In hydrostatic equilibrium, these forces balance:
This is a first-order differential equation and therefore requires a boundary condition, such as the central pressure Pc.
We can write this in another useful way. From
we have
Dividing the first boxed equation by the second gives
where the notation stresses that, in this form, the interior mass Mr is the independent variable and r is dependent. This so-called Lagrangian (“follow the mass”) form of the equation is often convenient.
Virial theorem
(O&C §2.4)
Start from the Lagrangian form of the hydrostatic equilibrium equation, written as
and integrate over the whole star:
(*)
where Vr is defined analogously to Mr, and Ps and Ms refer to the surface of the star. The integrand on the right hand side is just the gravitational potential energy of a shell with mass dMr at a distance r from the center of a spherical mass distribution of total mass Mr. The integral is therefore the gravitational potential energy U of the star as a whole.
Integrate the left hand side by parts:
The integrated part on the right vanishes strictly at the lower limit (because Vc = 0) and to a very good approximation at the upper limit (because Ps << Pc). For a nonrelativistic gas, 3P is equal to twice the thermal (kinetic) energy density, 2K.
Altogether then, we can write equation (*) above as
For a relativistic (e.g., photon) gas, pressure is equal to one third (not two thirds) of the energy density, so instead we would have
Now the virial theorem (from L. vis
vires: force, power, strength) is
more general than the above derivation would suggest and applies to a wide
variety of gravitating systems in equilibriumfor example, binary star orbits or a cluster of
galaxies. However, the common
derivation in terms of point masses (as in O&C §2.4) makes the result as it
applies to stars appear unnecessarily mysterious. [For the record, the virial of Clausius for a system of
point masses is
so there.]
Consequences of the virial theorem
For now, we’ll restrict attention to nonrelativistic systems. Since the total energy is E = K + U, we can also write the virial theorem as E + K = 0.
Consider a slowly contracting star, nearly in
equilibrium so that the virial theorem is satisfied at every stage. If the gravitational potential energy
changes by ΔU (negative), the thermal energy must change
by ΔK = ΔU
(positive). The star heats up! The change in the total energy is ΔE
= ΔK + ΔU =
ΔU
(negative). The star as a whole must
lose energy in some form (usually, radiation).
Thus we have the Three-Fold Way of quasistatic gravitational contraction:
1. The star gets hotter.
2. Energy is liberated from the system.
3.
The
total energy of the system decreasesi.e., the star becomes more tightly bound.
In thermodynamic terms, a star has negative specific heat.
Jeans mass (O&C §12.2)
Consider
a spherical cloud (protostar) of constant density. If 2K < U, the kinetic agitation in the cloud will
not suffice to overcome gravity, and the cloud will collapse. Using the dimensional estimate
and the
internal kinetic energy of a perfect gas,
, the condition for collapse is
If we eliminate the mass through , we estimate the Jeans length,
such that a cloud of density ρ and temperature T will collapse if R > RJ. Note our old friend “one over root G rho.”
The corresponding Jeans mass is
Stars by ROM
(O&C §10.3)
Central pressure
From the Lagrangian form of the hydrostatic equilibrium equation, we estimate
This expression underestimates the central pressure of the Sun by about two orders of magnitude; it’s closer to the pressure at the half-radius point (as we’ll see, stars are centrally concentrated). In general, we must remember that ROM means what it says.
Central temperature
Temperature is related to pressure through an equation of state, P = P(T,ρ,C) where C represents chemical composition. For pressure arising from a perfect gas and radiation (which applies in most parts of most stars),
where a = 4σ/c
and μ is the mean
molecular weight. Neglect radiation and
use the mean density and our
earlier estimate for Pc to derive
For a gas of
pure ionized hydrogen, μ = . For the
Sun, this estimate gives about 4e6 K, a factor of four lower than the
standard model.
P Repeat this calculation for pure radiation pressure. Is radiation pressure more important for high or low-mass stars?
Kelvin-Helmholtz timescale
(O&C §10.3)
The present gravitational potential energy of the
Sun is, dimensionally, . We know
from the virial theorem that the Sun had to release half this much energy
during contraction. The present luminosity
of the Sun is 4e26 W. If that
luminosity also applied in the past, the Sun would have radiated away the
necessary amount of energy in 1e15 sec or 30 million years. This is known as the Kelvin-Helmholtz time
scale.
At the time Darwin published On the Origin of Species in 1859, most astronomers believed that the Sun was in fact powered by gravitational contraction. Kelvin and Darwin had several public disagreements about the extended timescales required for biological evolution in Darwin’s picture. The advent of radioactive dating settled the issue (for scientists), showing that the Earth is several billion years old.
In 1929, Eddington noted that, if the Sun could
convert all its rest mass to energy according to , the Sun could shine at its present rate for ~1e13
y. Thus, a subatomic process that
converts 10% of the mass of the Sun to energy with about 0.1% efficiency would
suffice to power the Sun for a geological time scale.