Minimum value for central pressure of star
We have only two of the four equations, and no knowledge yet
of material composition or physical state. But we can deduce a
minimum central pressure.
2
( ) ( ) ( )dP r GM r r
dr r

= − 2( )
4 ( )
dM r
r r
dr
 =
4
( ) ( )
4
dP GMdP r dM r
dr dr dM r
 = −
40 4
sM
c s
GM
P P dM
r
− = 
2
4 4 40 04 4 8
s sM M
s
s s
GMGM GM
dM dM
r r r  
 = 
Divide these two
equations:
Integrations from center
to surface gives
Lower limit to RHS:
Hence we have
We can approximate the pressure at the surface of the star to be
zero:
For example for the Sun:
𝑃𝑐 = 4.5  1013 Nm-2 = 4.5  108 atmospheres
This seems rather large for gaseous material – we shall see that
this is not an ordinary gas.
2
4
8
s
c s
s
GM
P P
r
− 
2
4
8
s
c
s
GM
P
r

The Virial theorem
Again lets take the two equations of hydrostatic equilibrium and
mass conservation and divide them
Now multiply both sides by 4𝜋𝑟2
And integrate over the whole star

dP(r)
dr
dM(r)
dr

dP
dM
= −
GM
4r4

4r3
dP = −
GM
r
dM

3 V
Pc
Ps
 dP = −
GM
r0
M s
 dM
Where V = vol contained within radius r
Use integration by parts to integrate LHS
3 PV c
s
− 3 P
Vc
Vs
 dV = −
GM
r0
M s
 dM
At centre, 𝑉𝑟 = 0 and at surface 𝑃𝑠 = 0
Hence we have

3 P
0
Vs
 dV +
GM
r0
M s
 dM = 0
Now the right hand term is the total gravitational potential energy
of the star or it is the energy released in forming the star from its
components dispersed to infinity.
Thus we can write the Virial Theorem :
0
3 0
sV
PdV +  =
This is of great importance in astrophysics and has many
applications. We shall see that it relates the gravitational energy of a
star to its thermal energy
The Virial theorem
Minimum mean temperature of a star
We have seen that pressure 𝑃 is an important term in the equation
of hydrostatic equilibrium and the Virial theorem.
What physical processes give rise to this pressure – which are the
most important ?
•Gas pressure 𝑃𝑔
•Radiation pressure 𝑃𝑟
•We shall show that 𝑃𝑟 is negligible in stellar interiors and pressure
is dominated by 𝑃𝑔
To do this we first need to estimate the minimum mean
temperature of a star
Consider the  term, which is the gravitational potential energy:
− =
GM
r0
M s
 dM
We can obtain a lower bound on the RHS by noting: at all points
inside the star 𝑟 > 𝑟𝑠 and hence Τ1 𝑟 > Τ1 𝑟𝑠
Now pressure is sum of radiation pressure and gas pressure: 𝑃 =
𝑃𝑔 + 𝑃𝑟
Assume, for now, that stars are composed of ideal gas with
negligible 𝑃𝑟


GM
r0
M s
 dM 
GM
rs
0
M s
 dM =
GMs
2rs

− = 3 P
0
Vs
 dV = 3
P
0
M s
 dM
Now 𝑑𝑀 = 𝜌𝑑𝑉 and the Virial theorem can be written
k T
P nkT
m

= = The equation of state of ideal gas
𝑘 … Boltzmann’s constant; 𝑚 … average mass of particles
𝑛 … number of particles per m3
Hence we have
And we may use the inequality derived above to write
We can think of the LHS as the sum of the temperatures of all the
mass elements 𝑑𝑀 which make up the star
The mean temperature of the star ത𝑇 is then just the integral divided
by the total mass of the star 𝑀𝑠

− = 3
P
0
M s
 dM = 3
kT
m0
M s
 dM

− = 3
kT
m0
M s
 dM 
GMs
2
2rs
 T
0
M s
 dM 
GMs
2
m
6krs
T
–
 MsT = T0
M s
 dM
T 
GMsm
6krs
Minimum mean temperature of the Sun
As an example for the Sun we have:
Now we know that Hydrogen is the most abundant element in
normal stars and for a fully ionised hydrogen star Τ𝑚 𝑚 𝐻 = Τ1
2 (as
there are two particles, p + e–, for each Hydrogen atom). And for any
other element Τ𝑚 𝑚 𝐻 is greater
ത𝑇𝑆𝑢𝑛 > 2 × 106K
–
–
6 27
4 10 K where 1.67 10 kgH
H
m
T m
m
−
  = 
Physical state of stellar material
We can also estimate the mean density of the Sun using:
3 -3
3
3
1.4 10 kgm
4
Sun
Sun
M
r


= = 
Mean density of the Sun is only a little higher than water and other
ordinary liquids. We know such liquids become gaseous at 𝑇 much
lower than ത𝑇𝑆𝑢𝑛
Also the average K.E. of particles at ത𝑇𝑆𝑢𝑛 is much higher than the
ionisation potential of Hydrogen. Thus the gas must be highly
ionised, i.e. is a plasma.
It can thus withstand greater compression without deviating from an
ideal gas.
Note that an ideal gas demands that the distances between the
particles are much greater than their sizes, and nuclear dimension is
10-15 m compared to atomic dimension of 10-10 m
T
–
–
Now compare gas and radiation pressure at a typical point in the
Sun

Pr
Pg
=
aT4
3
kT
m
=
maT 3
3k
27
6 3 3
4
1.67 10
Taking ~ 2 10 K, ~ 1.4 10 kgm and kg
2
Gives ~10r
g
T T m
P
P
 
−
−
−

=  =  =
Hence radiation pressure appears to be negligible at a typical
(average) point in the Sun. In summary, with now know of how
energy is generated in stars and we have been able to derive a value
for the Sun’s internal temperature with negligible radiation
pressure.
Lets revisit the issue of radiation versus gas pressure. We assumed
that the radiation pressure was negligible. The pressure exerted by
photons on the particles in a gas is:
Where 𝑎 is the radiation density constant

Prad=
aT4
3
Mass dependency of radiation to gas pressure
However we shall later see that 𝑃𝑟 does become significant in higher
mass stars. To give a basic idea of this dependency: replace 𝜌 in the
ratio equation above:

Pr
Pg
=
maT 3
3k
3Ms
4rs
3






=
4ma
9k
rs
3
T3
Ms
And from the Virial theorem : T ~
Ms
rs

Pr
Pg
 Ms
2
i.e. 𝑃𝑟 becomes more significant in higher mass stars.