Download Handout 15: Virial Theorem E = P.E. + K.E = (1/2)P.E. = -K.E.

Handout 15: Virial Theorem
E = P.E. + K.E = (1/2)P.E. = -K.E.
„
The virial theorem is crucial for an overview of
the stellar interior
… See
„
Holds when P.E. is from a force ~ 1/r2
… i.e.
„
discussion Chap. 2 of C&O
gravity and electrostatic forces
Consider orbit of satellite around Earth
m⋅ v
2
Fgravity
r
KE
G⋅ m⋅ M
2
r
1
2
⋅ m⋅ v
2
1 G⋅ m⋅ M
⋅
2
r
−1
2
⋅ PE.
Drag causes increase in speed
„
Non-intuitive consequences
… Atmospheric
drag causes r to decrease
… P.E. decreases
… |P.E.| increases
… K.E. increases
… Satellite speeds up!
Gas drag causes speed to increase
„ Half the decrease in P.E. goes into heating up the
atmosphere
„
Virial theorem in globular clusters,
clusters of galaxies
„
Assuming equilibrium has been reached
K.E. Î high total mass
… Missing mass discovered in galaxy clusters
… High
„
„
Zwicky, 1930’s
For the stellar interior, a virial theorem results if
… Hydrostatic
equilibrium dP/dr = -gρ
… Pressure from ideal gas E.O.S P = nkT
„
And <K.E.> = (3/2)kT
… Massive
particles, non-relativistic speeds
Derivation of Virial Theorem for stars
d
P
dr
−ρ ⋅
G⋅ M r
2
r
multiply by 4π r3 and integrate from r = 0 to R
Right Hand Side
RHS
⌠
−⎮
⎮
⌡
R
ρ⋅
G⋅ M r
r
2
⋅ 4⋅ π⋅ r dr
0
2
4⋅ π⋅ r ⋅ ρ ⋅ dr
dm
−dm⋅
RHS
G⋅ M r
r
Gravitational_PE_of_star
dPE
The virial theorem for stars
RHS
⌠
⎮
⎮
⌡
LHS
Gravitational_PE_of_star
R
3
d
P⋅ 4⋅ π⋅ r dr
dr
0
(
P( r
LHS
⌠
−3⋅ ⎮
⌡
R)
R
2
P⋅ 4⋅ π⋅ r dr
0
KE
)
3
d
P⋅ 4⋅ π⋅ r
dr
integrate by parts
and
0
⌠
−3⋅ ⎮
⌡
3
2
d
P⋅ 4⋅ π⋅ r + P⋅ 12⋅ π⋅ r
dr
r( r
R
2
n ⋅ k⋅ T⋅ 4⋅ π⋅ r dr
0
−1
2
⋅ PE
the virial theorem
0)
so
0
2 ⌠
−3⋅ ⋅ ⎮
3 ⎮
⌡
R
0
3
2
n ⋅ ⋅ k⋅ T⋅ 4⋅ π⋅ r dr
2
−2⋅ KE..
Consequences for stellar evolution
„
Consider collapse of a star
… As
R decreases, PE decreases
PE
⌠
−⎮
⎮
⌡
R
G⋅ M r
r
(
2
)
⋅ 4⋅ π⋅ r ⋅ ρ dr
0
PE will be of order
i.e. for constant density
−G⋅ M
2
R
PE
−3 G⋅ M
⋅
R
5
2
For a star, gas is compressible, therefore density increases as r decreases
Therefore more tightly bound than constant density
Therefore PE is more negative than this
Kelvin-Helmholtz contraction
„
Contract from 100 to 1 Rsun
… PE
„
goes from near zero to –GMsun2/Rsun
KE increases by –(1/2)∆PE
…
„
Star gets hotter inside
Where did the other half of the PE go?
…
It was radiated away as photon luminosity
… Kelvin-Helmholtz
„
For the sun at R = Rsun, timescale = 107 yr
… This
„
„
timescale = -P.E./2*L
is
Timescale for contraction to the Main Sequence
Age of the sun if only gravitational energy available
Nuclear energy sources
„
Life on earth is at least 3 to 4 billion years old
… Age
of sun much longer than 10 million years
… Nuclear energy sources crucial
„
Thermal equilibrium
…
…
…
Energy flux radiated from the surface exactly equal to the
energy generated by nuclear fusion reactions in the interior
This condition defines the locus of the Main Sequence on
the H-R diagram
ε(ρ,T) is the rate of energy generation, power/mass
Thermal equilibrium
Lradiant
⌠
⎮
⌡
R
0
2
4⋅ π⋅ r ⋅ ρ ⋅ ε dr..
Temperature in the sun
„
The virial theorem gives the average
temperature of the sun
3
2
⋅ k⋅ T⋅ N
N
Tave
−1
2
⋅ PE
M sun
µ ⋅ mH
G⋅
M sun
2
Rsun
µ
2 µ ⋅ mH⋅ G ⎛ M sun ⎞
⋅
⋅⎜
k
3
⎝ Rsun ⎠
0.6
7
10 ⋅ K
amazing
Just by looking at the sun for ~ 1 hour Î hydrostatic equilibrium
Î Virial theorem Î interior is 10 million K!
Sun temperatures of 10 million K
Î nuclear fusion reactions
High velocities necessary to overcome
coulomb barrier
„ 4 H Î eventually, He + energy
„
… And
„
energy ~ 1% mc2
Very significant, c.f. (1/2)PE
1% Mc2 = 1052 ergs
… GM2/2R = 1048 ergs
…
„
Rather than the K-H time of 107 yr, we have 1011
years for nuclear, Main Sequence lifetime
…
Stellar evolution Î only 10% of H burnt on M.S.