university of naples federico ii, academic year 2011-2012 istituzioni di astrofisica · ·...
TRANSCRIPT
Lecture 6
The equations of stellar structure
University of Naples Federico II, Academic Year 2011-2012Istituzioni di Astrofisica, read by prof. Massimo Capaccioli
Karl &
Martin Schwarzschild
Learning outcomes
The student will :
• see the first two basic equations of stellar structure (the hydrostatic
equation and the conservation of the mass);
• analyze the hydrostatic and the spherical symmetry assumptions;
• get in touch with the Virial Theorem;
• evaluate the minimum pressure and temperature at the Sun’s center;
• discuss the mean molecular weight;
• consider the physical state of the stellar material;
• know pressures other then thermal.
A star is: a huge mass of gas
self-gravitating
self-luminous
supported by internal pressure
supplied by central energy source or by its own potential
Some questions:source of pressure ?
source of energy ?
why a gas ?
how long do they live ?
What is a star ?
Ideal stars are:
– isolated
– spherical
– non-rotating
– steady
Real stars may be:
– embedded in gas and/or dust
– in a double or multiple star system
– connected to surrounding gas by magnetic field lines
– rotating rapidly
Real versus ideal star ?
5
To get started
What are the main physical processes determining the structure of stars ?
• Stars are held together by their gravitation: each mass element of a star exerts
an attraction on all the other mass elements [Why so?].
• The gravitational collapse is opposed by the internal (thermal) pressure.
• Being the principal forces acting on the stellar structure, these two forces must
be in balance. But how finely? And for how long?
• Stars are continually radiating into space, thus they loose energy.
• Thus, if thermal properties are constant, a continual energy source must exist.
Theory must describe: 1. the origin of this energy &
2. the way it is transported to the surface.
6
Basic (temporary) assumptions
two fundamental assumptions
rate of change constant
For now we make :
1) the of properties is assumed : ;
2) al
with time
spherical symmetrl stars are & about their centers: .
We will
0ic
0
t
θ φ
∂ ≡∂
∂ ∂= ≡∂ ∂
start with these assumptions and later reconsider their validity.
7
For our star models, which are isolated, static, andspherically symmetric, there are four basic equations to describe the structure.
Spherical symmetry implies that all physical quantities depend on the distancerfrom the center of the star alone.
1) Equation of hydrostatic equilibrium: at each radius r, the forces due to
pressure differences balance the gravity.
2) Conservation of mass: in particular, since the models are static, ∂ρ(r)/∂t ≡ 0.
3) Conservation of energy: at each radius, the change in the energy flux equals
the local rate of energy release.
4) Equation of energy transport: relation between the energy flux and the local
gradient of temperature.
Four basic equations
8
These era three additional equations to be considered:
1) Equation of state: pressure of a gas as a function of its density and temperature;
2) Opacity: how hard is for the radiation field to travel across the stellar material;
3) Core nuclear energy generation rate: rate at which energy is generated at the center in oder to compensate the energy losses.
Four basic equations’ supplements
9
Equation of hydrostatic support
DomenicoFetti, 1620 Archimedes Thoughtful
The balance between gravity and internal pressure is known as
. Consider the forces acting on
an elementary
hydrostatic
equilibrium radially
density
outwar
mass: ,
where is t( h
d
e t :)
a
m (r) s r
r r
δ ρ δ δρ
⋅
=
: pressure exerted by the stellar
material on the : .
: pressure exerted by the stellar materi
force
lower face
inward force
upper face gravitational
al
attron the plu actions of
(
all t
)
h
P r sδ
⋅
2 2
e stellar material lying within .
By equation the inward and outward components, one get
( ) ( )( ) ( )
:
( )
s
.
r
GM r GM rP r r s m P r r s r s r
r rδ δ δ δ δ ρ δ δ+ + = + +
r
rδsδ
10
2
2
In hydrostatic equilibrium:
For a infinitesimal elemn it is:
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( )
e
l
nt
etting
Hence, rearrangin
( ) 0
g ab
GM rP r r s r s r P r s
rGM r
P r r P r r rr
P r r P r dP rr
r dr
δ δ ρ δ δ δ
δ ρ δ
δ δδ
+ + =
⇒ + − = −
+ − = →
2
hydrostatic support
ove we get:
the equa
tion
( ) ( ) (
of
.
)
dP r GM r r
dr r
ρ= −
2
( )( ) ( )
GM rP r r s r s r
rδ δ ρ δ δ+ +
( )P r sδ
( )
( )
r V
r s r
ρ δρ δ δ
=
r
rδ
Why d ?
Equation of hydrostatic support
Star centre
11
2
Inside the star consider a thin shell with radius and outer radius .
The contained within the is
determined by th
( )
( )e density .
mass
4
stellar radius
r r r
M r r
r
V r r
M
δ
ρ
δ π δδ δ⇒
+
== 2
2
in the limit where
mass con
.
T
( ) 4 ( )
( )4 (
his the e servatquatio ionn of .
0
)
V r r r r
dM rr r
dr
r
ρ π δ ρ
π ρ
δ
=
=
→
⇒
Equation of mass conservation
r
r rδ+
2
0
2 20
What is the total mass, and what the average den
sphere of gas outer radius densi
sity within r
4
a
Imagine a with and e:
.
d
iu
4
t
y p
s ?
rof
il
r
r
RR
r
r
dMr R
dr
ρ ρ
ρ
π ρ πρ
=
= =
2 30 0 00
3
22 3
0 00
,
where is the .
For the within
total volume
average density
decrease in the local densi
4 4 3
4
3
( ) 4 4 3 ( )
:
.
that is, the t
R
R R
R
r
r
M R dr R V
V R
r
RM r R dr r r
r
πρ πρ ρ
π
πρ πρ ρ ρ
= = =
=
= = =
→
∫
∫
is compensated exactly by increase in vo
y the lume.
Example: compute mass from density
( ) ( )
33
2 20 0
0 20
central solar pressure the density
is constan3
1410 kg m
Make a crude estimate of the , assumi
4
( ) ( ) ( ) ( ) ( )
ng tha
( )
( ) (
t
:t
)
.-
R R
R
M
R
dP r GM r r GM r rdP r dr
dr r r
GM r rP R P dr
r
ρ ρπ
ρ ρ
ρ
= = =
= − ⇒ = −
− = −
∫ ∫⊙ ⊙
⊙
⊙
⊙
⊙
⊙ ⊙⊙
( )
( )
2 2 2
0
2 2 100
4 2
3 3
0
21.34 1
Assume the to be , :
.
Thi
outer pressure negligible
grosss is a since t underestima he density increases strongly toward the centre. The
0 Pa
pre
3
t
RG rdr G R
P R
P G R
πρ πρ
πρ
= − = −
=
= = ×
∫ ∫⊙
⊙ ⊙ ⊙
⊙ ⊙
⊙ ⊙⊙
( ) 16
5
1160 2.5 10 Pa 2.5 10 at
[
oper value is 10 times
Remember that: 1 Pa 10
hi
bar 101.325 at
mgher:
]
.m
ρ = × = ×= =
⊙
Important example: the pressure at Sun’s center
34
3Rπρ⊙ ⊙
14
We have assumed that the d.
Consider the cas an
gravity and pressure forces are balance
the outward inward forces not equal.resulta
How v
nt fo
alid is tha
e where The acting on the ele
d amerce nt
r
e
t ?
of s
( ) ( )
( )
( )
2
2
2
and will give rise to an :
urface thicknessacceleration
acceleration due to gravity
Since is ,
( )
( )(
( ) ( ) ( )
( )
) ( ) ( ) ( )
t
GM rP r r s r s r P r s r s r a r
r
s ra
dP r GM rr r a r
dr r
g r GM r r
δ δ ρ δ δ δ
δ δ
ρ ρ
ρ δ δ+ + −
+ =
=
⇒
=
( ) ( )
hen:
whi generalised fch is the of the equati
( )( )
on oorm hydrostatic suppo
( )
f r . t
,dP r
g r r r a rdr
ρ ρ+ =
Accuracy of hydrostatic assumption
Accuracy of hydrostatic assumption
15
( ) ( )
a small fraction of gravitatiAssume that the resultant force on a mass element of the star is not zero, but rather ( ):
onal term
0 1
( ) ,
so that: )
( )
(
r g r r a r
dP r d
β
βρ ρ
<
=
≪
( )
( ) ( )2 2
,
Thus there is an : .
Assuming the motion begins , then
( ) ( ) 0
the spatial displacement after a time is
inward acceleration
at rest
1 1
:
2 2
r g r r a
a g
dt
d a t g t t
ρ ρ
β
β
+ = ≠
=
∆∆
∆ →∆ = ∆ = ∆ 2 d
gβ∆=
16
Class Tasks
8
1. Estimate the timescale for the by an observable amount (as a function of ). Assume is small. Is the timescale likely ? (for the Sun
Sun's
: 7 10 m 2.
radius to c a
n
;
ge
h
R g
ββ
= × =⊙ ⊙
2 2
9
)
2. We know from geological and fossil records that it is unlikely to have changed its
5 10 m s
10 y flux output significantly over the last . Hence find an upper limit for . What doe
r
β
−×
s this imply about the assumption of hydrostatic equilibrium ?
( ) ( )2 2
3
4
9 9 15
21 2
2
1 2.4 10 s
1" 1 5.6 10 56.6
30 60"
1 10 10 56.6
Assu
me:
I =3.f: 2 0 1
R R Rd g t R t R t
R g g R
Rt
R
Rt
R
t
ββ β
β
β
ββ
−
−
∆∆ = ∆ = ∆ ∆ = ∆ ∆ =
∆∆ = ×
∆
⇒ ⇒ ⇒
⇒
⇒
⇒
= = × ∆ =×
∆ = = ×⇒
⊙ ⊙ ⊙
⊙ ⊙
⊙ ⊙
⊙
⊙
⊙
⊙
17
The dynamical timescale
2* *
13 2
*
13 2
If we allow the star to collapse, set and substitute
in .
.
is known as the What is
(no pressure
Assuming :
22 1
2=
dynamical tim
e.
1 )
d
d
d r g GM r
rdt t
g GM
rt
GM
i e.
t
.
β β
β
∆ = =
∆∆ = ∆ =
=
→
( )8
30
11
7 10 m
1.99 10 kg 38 hours
6.660 10 MKS
it a measure of ?
For the Sun
:
d
r
M t
G −
= × =
×
⇒× = =
⊙
⊙ ⊙
18
Accuracy of spherical symmetry assumption
To what extent are they flattened at the poles ? Shall we account for the departures from the spher
Stars are rotating gaseous bodies.
Consider a star of mass radius,
ic
,
al symmetr
and angul ve
y ?
ar M r
2
locitymass
no departurespherical symmetry
gravitational force larg
. A
.There will be f
near the surface at the equat
rom provide
or feel
d thatt
s a centrifugal force (where it is largest):
he e
mm r
ωδ
δ ω
( )2 2 2 3
2
:
1
ly exceed the centrifugal force
dimesions of a tid
.
Notice that has the .e
GMGM mm r
r r
δδ ω ω
ω
→≪ ≪
mδr
ω
19
32 2
3
23 2 2
2
2 2
2
2 2
,
spherical
, ,
.
Note the RHS of equation: is similar to the inverse of:
or:
a 2 = rotation pe
nd, a riods4 2
where
I
sy
:
2
m
f
d
d d
dd
GM rt
r GMGM
r t t
P P
P tP t
ω
ω
ω ππ π
=
=
=
→
→
≪
≪
≪ ≫
6
is to hold, then .
For the majo
metry
departure
~ 2000s
s from
For example
spherical sy
~ 3 10 s 1 mont
mmetry can be rity of stars,ignored
.
Some s
for the
tarsd
Sun, wh
o rotat
ile .h
e an rapidly d
d
d
P t
t P ×
≫
≃
rotational effects must be included in the structure equations; this can change the output of models.
Accuracy of spherical symmetry assumption
dynamical time
20Equatorial velocity (lower limit) for giant stars
Stellar rotation
rotational ax
is the angle between the theand
isline of sigh
Why we use
th
vsie .
t
n ?
i
i
21
Minimum value for the central pressure of star
2 of the 4 equations material
composition physic
We have only , and no knowledge yet of the
and of the of the star.
T
al state
minimum central press
Wh
hough we can deduce a
y, in principle, do y
.
o
ure
u thi
22
Divide the
nk there need
hydrostatic e
s to be a minimum value ?
Given what we know, what is thi
quation:
s likel
( ) ( ) ( ) ( ) 4 ( )
( )
by ,
to ge( )
y to depend upon
t
?
:
dP r GM r r dM rr r
dr r drdP GdP r dM r
dr dr dM
ρ π ρ= − =
≡ = −4
40
2
4 4 40 0
.
Integrating over the whole star it is: .
The lower limit to
stell
4
4
4
ar radius
the RHS is: ,
where is
4 8
the
.
s
s s
M
c s
M MS
s s
s
M
rGM
P P dMr
GMGM GMdM dM
r r r
r r
π
π
π π π
− =
> =
≥
∫
∫ ∫
max of r
22
( )
( )
( )
2
4
2
4
pressure at the surface of the star to be zero
.Hence it is:
A
For example,
pproxima
ting the
,
for the S
8
0
un:
8
:
sc s lower
s
s
sc lower
s
c lower
GMP P
r
P
GMP
r
P
π
π
− >
=
>
13 8
to be a gaThis value seems rather large .We shall see that the material of stars
= 4.5 10 Pa = 4.5
is not
s
ord
10
inaan . ry
t .a m
gas
× ×
Minimum value for central pressure of star
Hannes Alfvén (1908–1995)
23
The Virial Theorem
4
33
mass
conservation
( ) ( )
hydrostatic eqAgain le
ts take the two equations of and
and divide them:
.
Now multiply both sides by :
uilibrium
4
4 34
dP GMdP r dM r
dr dr dM
r
r
r dPπ
π
π
≡ = −
=
3
0
where contained within radius 4
= volume 3
and integrateover the whole sta :
.
Use integration by parts to integ
r
rate S
,
3
LH
s s
c
P M
P
V r r
GMVdP dM
r
GMVdP dM
r
π=
= −
= −∫ ∫
[ ]0
at centre , and at surfa
:
3 3
0 0e c .
s s
c
V Ms
c V
c s
GMPV PdV dM
rV P
− = −
= =
∫ ∫
24
0 0
RHS = total gravitational potential energy of the star
Hence we have:
.
Now the or energy
released
it is the
in forming the star from its components dis
3 0s sV M GMPdV dM
r− =∫ ∫
0
Class task:
Can you show this to be true ? Note that
work done = force distance = mass a
persed to infinity.
Thus we can write t Virial Th
ccel. d
eorehe : 3
s .
m
i t
sVPdV
× × ×
+∫ .
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
ener .
0
gy
Ω =
The Virial Theorem
Rudolf Clausius (1822–1888)
25
( ) 84.5 10 atm
We have seen that the is an important term in the equation of the
hydrostatic equilibrium and in the Virial Theorem.
We derived a minimum value for the Solar central pr
pr
essur
essure
e c
P
P
> ×⊙
What physical processes giv
gas press
e rise to this pressure? Which is th
e most importan
ure
radia
.
tion pressure
t ?
gP
P
We shall show that is in the stellar interiors and that pressure is
dominated by .
To do this we first need to estimate the minimum mean temperature of a star.
Consider t
negl
he
igible
,
whicterm
r
r
g
P
P
Ω
0
h is the :
gravitation
al po
tential energ
.
y
sM GMdM
r− Ω = ∫
Two types of pressure
Minimum mean temperature of a star
26
2
0 0
We can obtain a lower bound on the RHS by noting that at all points inside
the star it is , and hence ,
.
Now , and the can be writ
1 r 1
2
Virial Theor m te
s s
s s
M Ms
s s
r r r
GMGM GMdM dM
r r r
dM dVρ
< >
− Ω =→ > =
=
∫ ∫
0 0
sum of radiation pressure gas pr
en
.
The pressure is and : .
Assume, for now, that stars are composed of an
essure
ideal gas
3 3
and that i
negligi ,ble
s
s sV
r
r
M
g
PPdV dM
P P P
P
ρ− Ω = =
= +
∫ ∫
3number of particles per m average mass of particles, Boltzmann'
w
s constant.
here
equation of state of
,
anso that: , , ideal ga s
nmk
k TP nkT
m
ρ
===
= =
Minimum mean temperature of a star
Note that this is the minimum value for the potential (maximun for the potential energy)
27
For useful pages on the ideal gas laws, consult the links listed at:
http://heybryan.org/~bbishop/search/AP_Physics_C/ideal%20gas%20law.html
Aid
28
0 0
2 2
0 0
Hence it is:
.
We may use the inequality derived above to
3 3
3
write:
The LHS may be thought a um
2 6
s the s o
s s
s s
M M
M Ms s
s s
P kTdM dM
m
GM GM mkTdM TdM
m r kr
ρ− Ω = =
− Ω = > >→
∫ ∫
∫ ∫
of all the mass
elements which make up the star.
The of the star is then just the integral divided by the
f the temperatures
mean temperature total
m of the star :
ass
s
s
dM
T
M
M T T→ =0
6
sMs
s
GM mdM T
kr>→∫
Minimum mean temperature of a star
29
Minimum mean temperature of the Sun
6 27
As an example, for the Sun we have:
, where .
Now we know that H is the most abundant element in stars and for a fully
ionised hydrogen star (as there are tw
4 10 K 1.6
o part
7 10 kg
1 2 i
HH
H
mT m
m
m m
−> × = ×
= +
6
numb
cles,
er of
and ,
f
e 1
or each H atom).
And for is 1 ion and
Thus:
.
p e
any other elem
e
nt greater
2 10 K
. Hm m
T
−
−
≥
> ×
⊙
The mean molecular weight is an important quantity, because
the pressure support against gravity
A sudden changes in the ionization state or chemical c
number of depends on the .
free particl
esH
m
mµ ≡
In general, the value of
omposition of the star
solv
can lead t
ing the Saha eq requires to determin
o sudden chang
uationionizati
es in the pre
on state
the of everye
ssure
ato
.
m.
m
We can derive two useful expressions for the cases of or
gasses.Define: , are the s of H, He, and metals, resp
fully neutralfully ionized
mass fraction
Neutral:
ectiv
ely1 1
.
1
,
4
,nn
X Y Z
X Y ZAµ
≅ + +
1 1For So
Ioni
lar
zed:
abundances: 1
.
5
1 32
4
5
1
2
..
i
n
X
A
Y Zµ
≈
≈ + +
Minimum molecular weight
By how much does the pressure increase following complete ionization, for a neutral gas with the foll
owing compos
ition (typical of young
0.70
star
s) ?
X =
Remember
1 1
N
eutral:
0.28 = 15.5
0.02
1 Ionized:
1 1 1 3 1 2
4 4 2
1 0.28 0.020.70 0
4 15
tha
.5
so:
t:
, ,
n
nn i
n
YA
Z
X Y Z X Y ZAµ µ
µ
=
=
≅ + + ≈ + +
≅ + + =
1
3 0.02.771
2 0.70 0.28 = 1.624 2
= =2. 0
,
.1
,i
i n
n i
P
P
µµµ
≈ × + × +
Example
32
( )
For a fully neutral gas: .
For a fully ionized gas: .
1
total mass of Hydrogen
total masstotal mass of Helium
to
Let's define:
j j j jj j
n nj j
j j
j jj
n
j jj
N m N A
mN N
N A
N z
X
Y
µ
µ
= =
=+
→
≡
≡
∑ ∑
∑ ∑
∑
∑
( )n
n
For a gas:neutr
1tal mass
total mass of metals
total mass1 1 1
al 1 4
1 3
1/15.
For
fully ionized 1
2a ga s 4
:
5
nn
n
X Y Z
Z
X Y ZA
ZX Y
A
ZA
µ
µ
+ + =
≡
+ +
++ +
⊙∼≃
≃ ( )1 1/ 2n
A⊙∼
number of free electron resulting from the complete ionization of the atomic species j
j j HA m m=
Mean molecular weights: recap
33
Physical state of stellar material
3 33
3 1.4 10 k
Th
mean densit
e mean dens
The is simply
ity of the Sun
computed:
is only a little higher tha
n water and o
y of t
ther
o
he Su
rdina
g m4
ry l .iquids
ow
n
H
M
rρ
π−= = ×⊙
⊙
⊙
( )
ever, we know such liquids become gaseous at much lower than .
Also the of particles at is much larger than thaverage kinetic energy
ionisation pote 13.6 eV= 3 2 kTn
e
of H .
Thus the gas
tia
co
l
T T
T
⊙
⊙
highly intitutin onisedg the Sun must be , a . p. l i as .smai e
34
A stellar plasma can therefore withstand greater compression
.
There is in fact ample room for individual particles when atoms feel
without d
already
eviating
from an i
too crowded: remem
deal
ber
gas
tha
15 10
an ideal gas distances between particles
the order of their sizes nuclear di
t demands
to be of . Now, are of the order of
, and those of of the o
10 m 10
me
mrder of .
Let's re
n
v
s
i
ions
atom
sit th s
s
e i
− −
sue of radiation gas pressure.
We assumed that the .
The pressure exert
radiation pressure was
ed by photons on the pa
negligible
rticles in
vs
4
16 3 4
3
7.55 10 J m K
a gas is: ,
radiatiwhere on densit .y constant
rad
aTP
a − − −
=
= × =
Physical state of stellar material
35
A
p
dx
z
yx θ
θ
m
mfp
ip
Let's consider a box (conveniently oriented) containing a system of equal
particles subject to
mass
elastic collisions
normal to the -axi
only
Bouncing on the wall , each particle s momentumof
.
c m
o
x
m
p
municates an : , where .
Therefore, the applied by th
impulse
force collie of one parsio
2 2v
2 vn ticle is: .
xx
x x x
xf t p p i t
p pf
t x
∆∆ = −∆ = ∆ =
= =∆ ∆
Ideal gas: state equation review
36
2 2 2
contribution to the pressure
Since is proportional to and, on average: , then .
The by a momll the particles with
between and is thus
e
1 v v v v
ntum
:
v3
1
3
x x x y z P
p
pP p p
pp f
xN dN dp
Np p dp dF f N dp
x
= = =∆
=
+ = =∆
0
0
v
1v
3
1v
3
surf
.
By integr
ace
number of particles per unit
momentum
ating over all particles: .
Dividing each member by the : ,
and noting that: , where is a
,
sinc
p
p
pp p
p dp
NF p dp
x
NFA p dp
A A xN
n dp dp nV
∞
∞
=∆
=∆
=
∫
∫
0
e and , then:
, , valid for mass or masintegral so
the pressure the volume eleme
l
n
f e p ss particleressure s1
v3
.
t
p
F A P A x dV
P n p dp∞
= ∆ =
= ∫
Ideal gas: state equation review
37
2
2
2v0
v
3 2 5 2v
v
2 43 20
3 2v 2 2
v
In the classical case: , and .
Taking from the Maxwellin distribut
v v
v 4 v v2
1v v
3
4 4 2v v
io
3 2 3 2
n: ,
m k
p
m T
T
k
P mn d
n
m
p m n dp n d
mn d n e d
kT
kT mP mn e d mn
kT m kTπ π
π π
ππ
∞
∞
−
−
=
= =
= =
=
→
∫
∫2
2
2
v 2 4
0
1 24
0
27
41 2 0
.
Since (integrate by parts): ,
where ( is the , that is
3
8
1.6
v
7
the mean mass of all pa
4 10 kg mean molecola
r
r weig
v
8
)
t
3
ht
m kT
x
H
H
x
H
e d
nkT e x dx
P nkT kT kTmm
m m
e x dx
m
πρ ρ
µµ
π
∞ −
∞ −
∞ −
−
=
=
=
= = =
= = ×
∫
∫
∫
This is a very important fact to be kept well in mind!
For instance, if a neutral
icles present,includi
gas of hydrogen is f
ng electron
ully ionize
s if the matter is io
d and for some reaso
niz
n
ed.
its
temperature remains constant, then the pressure must reduce by 1 2.
2
2
2v v0
3 2v 2 2
v v
3 2 5 2v 2 4
3 20
Maxwellin distribut
In the classical case: ,
ion
and .
Taking from the :
1v v v v
3
v 4 v v2
4 4 2v v
3 2 3 2
,
p
m kT
m kT
p m n dp n d P mn d
mn n d n e d
kT
m kT mP mn e d mn
kT m kT
ππ
π ππ π
∞
−
∞ −
= = =
=
= =
→
∫
∫2
2
2
v 2 4
0
41 2 0
1 2
27
4
0
.
Since ( ):
mean molecolar
,
where ( ) is the ,
i
that is
the mean mass of all par
ntegrate by parts
1.674 10
v v
8
3
3
8
weig th
t
gk
m kT
x
x
H
HH
e d
nkT e x dx
e x dx P nkT kT kTmm
m m m
ππ ρ ρ
µµ
∞
∞ −
∞
−
−
−
=
=
= =
= ×
= =
=
∫
∫
∫
This is a very important fact to be kept well in mind!
For instance, if a neutral
icles present,includi
gas of hydrogen is f
ng electron
ully ionize
s if the matter is io
d and for some reaso
niz
n
ed.
its
temperature remains constant, then the pressure must reduce by 1 2.
Ideal gas: state equation review
38
Let's consider a . Being , there is no limit
to their . The energy of a single photon with
is , th
gas
e
of photons bosons
density frequency
m , the velocity .
Replacing these valu
omentum vE h p E c h c cν
νν ν= = = =
3
3
total pressure momentum distribution fun
es into the general equation giving the by a system of particles with ,
where is given by th
ction
Planck equatie : , we g8
n t:o e1
p
h kT
rad
n h u
h du u d
c e
P
ν
ν ν ν
νπ ν νν
=
=−
34
30 0 0 0
15 3 4
1 1 1 1 8 1v
3 3 3 3 1 3
47.566 10 erg cm
,
with .K
p h kT
h h dn p dp n cd u d aT
c c e
ac
ν ν νν π ν νν ν
σ
∞ ∞ ∞ ∞
− − −
= = = = =−
= = ×
∫ ∫ ∫ ∫
4The total pressure (gas + radiation): 3
.1
totH
kTP aT
m
ρµ
= +
Radiation pressure review
no need to integrate: use Stafan-Boltzmann law.
39
Physical state of stellar material
4 3
276 3 3
Now compare gas and radiation pressure at a typical point in the Sun:
.
With ,
3 3
1.67 10~ 2 10 K ~ 1.4 10 kg m kg , and ,
on
2
r
g
P aT kT maT
m kP
T T m
ρρ
ρ ρ−
−
= =
×= × = × =⊙ ⊙
4
averag
e gets: for the Sun.
Hence radiation pressure appears to be negligible at a typical ( )
In summary, with no knowledge of how energy is gene
~ 1
ra
point
ted in
in the Sun.
stars we h
ave
e n
e
b e
0 r
g
P
P−
able to derive a value for the Sun's internal temperature and deduce that
it is composed of a near ideal gas plasma with negligible radiation pressure.
40
Mass dependency of radiation to gas pressure
3
3
However we shall later see that does become
.
To give a basic idea of this dependency, replace in the ratio equation
abov
significant in highe
e :
r
mass stars
r
g
r
P maT
P k
P
ρ
ρ =
3 33
3
2
and from the Virial theorem:
more significant
higher mass
4
be sco tm a
933
4
~
. .
s n rsie .
sr
g ss
s
s rs
s g
r
r TP maT ma
P k MMk
r
M PT M
r P
Pi e
π
π
= =
∝→
The ideal gas law neglects both special relativity and quantum mechanics. It therefore breaks down at high velocities (temperatures) and at high densities.
The Fermi-Dirac distributionis a modification of the Maxwell-Boltzmann distribution, accounting for the Heisenberg uncertainty principleand the Pauli exclusion principle.
Essentially, the Pauli exclusion principlestates that no two fermions (e.g. electrons and protons) can occupy the same quantum state.
This provides an extra source of pressure when densities get high.
Non-ideal gases
The Bose-Einstein distributionfunction applies to bosons(such as photons).
In this case, the presence of a particle in a quantum state enhancesthe probability that another particle will occupy the same state
Both the Fermi-Dirac and the Bose-Einstein distribution functions approach the Maxwell-Boltzmann distribution at high energies.
Non-ideal gases
The ideal gas law neglects both special relativity and quantum mechanics. It therefore breaks down at high velocities (temperatures) and at high densities.
43
Electrons degeneracy
It occurs when the temperature , and is a of:
, stating th only one fermion may coexist within a given quant
consequence
Pauli's exclusion principle
Heisenber
um
g'
sta
s indetermination p
at , dt ane
0T →
⊳
⊳ : , which requires to particles densely confined ( ) to have a high momentum: .
Let's concentrate on a .
r
inciplevery small
gas of electro
ir
ns
all parealistiAssu call ry ticleming ( ) that
x px
p p
∆ ∆ ≈∆
> ∆
ℏ
0
1 1v v
3
have s sam
3electron
e momentu
densit
the ,
from the : , on
m
pressure int e gets: ,
with
e ral
y
g
.
p e
e
P n p dp P n p
n
∞= ≈
=∫
44
Electrons degeneracy: first approach
1
1 3 1 3
completely ionized system
mean volumes
average separ
In , electrons are highly packed
within .
The is thus: , implying: .
For complete ionizati proton density elec
ation
o
tn:
e
e e
e
n
x n p p nx
n
−
−
≈
∆ ≈ ≈ ∆ ≈ ≈∆
= ×
ℏℏ
1 3
1 3
1 3non relati
rons number per
vistic electrons
then: .
For (
proton
,
v): .v
H H
ee e H
ee
Z Zp
A m A m
p Zn
m m m A mp m
ρ ρ
ρ
= = ≈
= = =
=
ℏ
ℏ ℏ
x∆
e
45
( )
5 32
2 3 5 32 2
pressure complete degenera
By combining the various parts:
we obtain the approximate expression: .
The exact e
1v
3
1
3
3
5xpression is: .
The f t o a
e
e H
e H
P n p
ZP
m A m
ZP
m A m
ρ
π ρ
≈
≈
≈
ℏ
ℏ
5 3
ed electron gas density
not on tempe
Why protons degenerate
ratur
at hi
depends on
but
gher temperatur ?
.e
es
ρ
eH
Zn
A m
ρ =
1 3
H
Zp
A m
ρ ≈
ℏ
1 3
ve H
Z
m A m
ρ =
ℏ
Electrons degeneracy