Notes for stellar-structures, J. Hedberg © 2018

Stellar Interiors1. Hydrostatic Equilibrium2. Mass continuity3. Equation of State

1. The pressure integral4. Stellar Energy Sources

1. Where does it come from?5. Intro to Nuclear Reactions

1. Fission2. Fusion3. Binding Energies4. Bringing them together5. Energy / Tunneling6. Cross Section7. p-p chain8. CNO Cycle

6. Luminosity Gradient7. Energy Transport

1. Radiative2. Convective Transport

8. Stellar Model Building

Figure shows the basic structure of a star. (It is not scale!) The light we seecomes from photons in the stellar atmosphere.

Hydrostatic Equilibrium

Fig.1The basic structure of a star.

PHY 454 - stellar-structures - J. Hedberg - 2018

updated on 2018-10-09

Let's imagine a small cylindrical volume of gas that is located in a larger region.There will be three forces acting on it: is the force of gravity, and twoforces from the pressure, one from the pressure above: and onefrom pressure below: . The net force from the pressure will be :

where is the area (whose surface normal points in the y direction) of the gasvolume. The gravitation force will be given by:

where is the mass enclosed by a sphere of radius centered about theearth's center.

If this volume of gas is in equilibrium, then the pressure forces and the gravitational forces will be equal:

Solving for :

If we consider this cylinder to be infinitely thin: , then we have a differential form:

Eq. is called the equationofhydrostaticequilibrium (for a spherical symmetric body).

To apply this to a star, we should acknowledge the limiting assumptions we made:

1. The star is spherical2. The star is not rotating3. The star is not expanding or contracting4. Gravity and Pressure gradients are the only forces acting.

A better way to write this would be to include the dependencies:

We can see that there are 3 quantities that depend on : Pressure ( ), Mass ( ) and Density ( ). This makes a uniquesolution not possible. We need more equations!

Mass continuity

∆ r

P

Fgrav

P+∆ P

Fig.2A small volume of gas has threeforce vectors acting in the verticaldirections: gravity, pressure frombelow and pressure from above.

FgravP +ΔP

P

= A[P − (P +ΔP )]Fpres

A

= −FgravG ρAΔrMr

r2

Mr r

A[P − (P +ΔP )] − = 0G ρAΔrMr

r2(3)

ΔP

ΔP = − ΔrG ρMr

r2(4)

Δr→= 0

= −dP

dr

G ρMr

r2(5)

(5)

= −dP

dr

GM(r)ρ(r)r2

(6)

r P M ρ



The equationofmasscontinuity relates density and radial position and total mass. In short it says that the totalmass is just the sum of all the infinitesimal spherical shells (i.e. onion skins)

Equation of State

The equationofstate relates density, temperature, and pressure for a gas. For many stars, the ideal gas law willwork just fine.

Mean Molecular Weight

where , and is the average mass of a gas particle in kg. therefore gives the theaverage mass of a free particles in the gas, in units of hydrogen. And, let's say that .

This allows us to write the ideal gas law as:

The pressure integral

Derivation of the pressure integral: Assuming a particle in the x direction, normal to a surface:

Since it must travel the distance twice in the time :

Thus the average force:

We can sub:

The number of particles with a given momentum between and is , then the total:

= 4π ρ(r)dM

drr2 (7)

PV = nRT (8)

(8)

μ≡m

mH

= 1.6735... × kgmH 10−27 m μ

n = ρ/m

=PgρkT

μmH

(9)

P = pvdp13

∫ ∞

0np (10)

fΔt= −Δp = 2px i

Δx Δt

Δt= 2Δxvx

f = =2pxΔt

pxvx

Δx

pv ≈13 pxvx

f(p) =13pv

Δx

p p+ dp dpNp

N = dp∫ ∞

p



Contribution to the total force by all particles within that momentum range:

Divide by

For non-relativistic particles, :

Recall the Maxwell-Boltzmann distribution:

Putting this into Eq. we'll get a simple expression:

(which is just the ideal gas law since )

Stellar Energy Sources

Where does it come from?The first suggestion to explain where stars got the energy was to look at the gravitational potential. Let's say we havetwo particles, then the gravitational potential energy between them is:

If the distance between these particles decreases, then the potential energy must get more negative, and we wouldexpect to find energy in other forms, kinetic for example.

How much energy is there in gravity?Of course there are more than two particles in a star, so we'll have to expand that equation a bit. If a point mass is located outside a spherically symmetric mass , then the force on that point mass will be directed to the centerand will have a magnitude:

N = dp∫ ∞

0Np

dF (p) = f(p) dp = pv dpNp13Np

Δx

A

dp ≡ dpnpNp

ΔV

P = pv dp13

∫ ∞

0np (11)

p = mv

P = m dv13

∫ ∞

0nvv

2 (12)

dv = n 4π dvnv ( )m

2πkT

3/2v2e−

mv2

2kT

(12)

= nkTPg (13)

n = N/V

U = −GMm

r(14)

r

dm

Mr

d = GFg,idMr mi

r2



The potential energy of the point mass is thus:

Now we'll consider a shell of thickness and mass :

where is the density and is the volume of the shell. Thus, eq becomes

We can integrate over the radius of the star like this:

Of course we probably don't know very well at this point, but we can guess based on it's average value:

with being the total mass of the star. Also:

Putting this into eq. , we get:

Applying the virial theorum, we have:

If we plug some sun values into this:

Now if the sun luminosity was constant at today's value: , then this would mean there is enoughenergy to last for about years (about 10 million years). This is notably too short given other observations fromgeology that suggest the earth and moon and things have been around for billions of years. There must be anothersource of energy contributing to the star's luminosity.

Intro to Nuclear Reactions

d = −GUg,idMr mi

r(15)

dr dm

dm = 4π ρdrr2

ρ 4π drr2 (15)

d = −G drUg4π ρMr r2

r

R

= −4πG ρr drUg ∫ R

0Mr (16)

ρ

ρ ∼ =ρM

π43 R3

M

∼ πMr

43r3 ρ

(16)

∼ G ∼ −Ug16π2

15ρ2R5 3

5GM 2

R(17)

E ∼ −310GM 2

R(18)

≈ 1.1 × J310

GM 2☉

R☉1041

3.84 × W1026

1 × 107



FissionWhen an atom breaks apart. Power plants...

FusionWhen you smash two atoms together. Suns...

Binding Energies

Take four Hydrogen atoms (1 proton & 1 electron), each have a mass of 1.00782503214 u. Together, they thereforehave a combined mass of 4.03130013 u. One Helium-4 atom has a mass of 4.002603 u. This is obtained by addingthe mass of the neutrons and the protons together. Thus, let's say we were able to mash 4 Hydrogen atoms together

Fig.3

http://www.nuclear-power.net/nuclear-power/fission/

Fig.4

http://www.nuclear-power.net/nuclear-power/nuclear-fusion/

Fig.5

Carroll & Ostlie, figure 10-9.



and make a Helium. Now there would be a change in mass between the initial and final constitutions: .

Since , we can see that this amount of mass is associated with an energy of 26.731 MeV. This is essentiallythe amount of energy in the formation of a Helium nucleus, aka the bindingenergy. (This is also the amount ofenergy that would be required to take apart a Helium nucleus.) The trick is to figure out exactly how we can combine4 Hydrogen atoms to make a Helium. (that's the study of High energy physics earlier last century)

Bringing them togetherOne impediment is the Coulomb repulsion between two protons, both of whichare positively charged.

As the separation distance between two charged (alike) particles decreases,the force of repulsion increases. The Coulomb potential between two protonslocated 1 fm apart would be:

If two protons wanted to get closer than 1 fm, they would need to have akinetic of energy greater than this value to overcome the potential barrier. Theenergy of a proton at the center of the sun is about:

This would be insufficient energy to get over the barrier. This is where quantumtunneling comes in.

Tunneling

Fig.7A 1-d simulation of a wave packet incident on a potential barrier. Some of the wave tunnels through the barrier.

Simulation based on code from https://github.com/jakevdp/pySchrodinger

Criteria for tunneling: a) the distance between protons should be comparable to the de Broglie wavelength:

and b) the kinetic energy ( ) of the proton needs to be comparable to the electrostatic potential energy at thatseparation.

Solving for the kinetic energy:

Δm = 4.03130013 − 4.002603 = 0.028697 u

E = mc2

F(r)

U(r)

r

r

Fig.6The plots of the Coulomb force(top) and the Coulomb potential(bottom).

≈ (1.60× C10−19 )2

4π(8.8× )(1.60× J )( m)10−12 C2 J−1 m−1 10−19 eV−1 10−15

≈ 1.4 × eV106

≈ 1.4 MeV

⟨E⟩ = ≈3kTc2

3(1.38 × J )(1.47 × K10−23 K−1 107

2(1.6 × J )10−19 eV−1

= = 7 × mλDBh

p10−13 (20)

E

U = = ≈ Ee2

4πϵ0

1λDB

e2

4πϵ0

(2 Emp )1/2

h(21)

E ≈ ( ) ≈ α ≈ 3keV2

p2



https://github.com/jakevdp/pySchrodinger

Since this energy is comparable to the kinetic energy of a proton in the core of the sun, we can expect tunnelingthrough the Coulomb barrier to be possible. The next step is to figure out just howlikely such events are.

Energy / Tunneling

Cross SectionThe crosssection gives a probability of interaction. It is the number of reactions per target nucleus per time dividedby the flux of incident particles.

Cross SectionIf a particle hits the wall, then there is a reaction. This is an abstraction of thenuclear collision process. (there is no cylinder, no wall...)

: just a particle

: an incident particle

: # of incident particles having energies between and per unit volume

: # of particles that "hit the wall" / time or # of reactions / time

: # of particles with an energy between E and , (might not be incident)

E ≈ ( ) ≈ α ≈ 3keVe2

4πϵ0

2mp

h21

2π2mpc

2 (22)

2 4 6 8 10

Coulomb repulsion (1/r)

Strong Nuclear potential well

r (fm)

U (r

) [A.

U]

Fig.8A qualitative graph showing thewhat happens when two protons getclose engouh for the strong force todominate.

σ(E) ≡numberofreactions/nucleus/time

numberofincidentparticles/area/time

ds = v(E) dt

σ (E)

Fig.9Imagine a cylinder of length and area . All the

particles within the volume of thecylinder will reach the wall in the time

.

ds = v(E)dt σ(E)

dt

n

ni

dEniE E E+ dE

dNE

dEnE E+ dE



We have a target particle ( ) and an incident particle ( ). If the incident particle strikes the area , then there willbe a nuclear reaction.

Take the number of incident particles per unit volume having energy between and to be: . (This islike a density) To find the number of such particles inside the cylinder:

The number of particles with the appropriate kinetic energy is some fraction of the whole:

The number of reactions per target nucleus per time having energies between and is

Lastly, if there are targets per unit volume, then the total number of reactions per unit volume per unit time, givenall possible energies is:

But, what exactly is ?

(Since )

Also, for a barrier and particle with energy , the probability for tunneling through will go as:

In this case the is the Coulomb potential: and is just the kinetic energy: . We canassume the distance is basically the de Broglie wavelength: .

In which case we obtain:

where is given by

x i σ(E)

E E+ dE dEniE

d =NE dEniE density

σ(E)v(E)dt volume

dE = dE = dEniEdE∫ ∞

0 niE

dE∫ ∞0 nE

nEni

nnE

dt E E+ dE

= = σ(E)v(E) dEreactionspernucleus

timeintervaldNE

dt

ni

nnE

nx

= σ(E)v(E) dErix ∫ ∞

0nxni

nE

n(23)

σE

σ(E)

σ(E) ∝ π ∝ π ∝λ2DB

( )hp

2 1E

(24)

K = E = /2 = /2μmv2 p2 μm

U E

σ(E) ∝ e−2 U/Eπ2(25)

U =UcZ1Z2e

2

4π rϵ0E E = /2μmv

2

r r ∼ λ = h/p

σ(E) ∝ e−bE−1/2

(26)

b

b≡πμ

1/2m Z1Z2e

2

h21/2ϵ0



We can combine eq and and define a general function to get a functional form for :

Finally, we can write a reaction rate as:

The peak will be at

p-p chain

One way to convert Hydrogen to Helium is through the proton-proton chain(PP1). The overall reaction combines 4 protons to create 1 Helium-4 plus someother stuff.

The first step in this reaction take two protons and produces a deuterium, apositron, and an electron neutrino.

Then the deuterium combines with another proton to make a Helium-3 atomplus a gamma ray.

Lastly, two Helium-3s combine to create Helium-4 and 2 protons

CNO CycleIn stars that more than , a different nuclear fusion process will dominate. This is the CNOcycle. It's morecomplicated, but the basic idea is that Carbon, Nitrogen, and Oxygen act a catalysts to speed the fusion of hydrogen.The net result is:

(24) (26) S(E) σE

σ(E) =S(E)E

e−bE−1/2

(27)

= S(E) dErix ( )2kT

3/2ninx

( πμm )1/2∫ ∞

0e−bE

−1/2e−E/kT (28)

2 4 6 8 10

0.2

0.4

0.6

0.8

1.0

2 4 6 8 10

0.2

0.4

0.6

0.8

1.0

Energy (keV)

e-E/k T

e-b E -1/2(× 103)

e-E/k Te-b E -1/2(× 106)

likel

ihoo

d [A

.U.]

Fig.10The likelihood that a nuclearreaction will occur.

=E0 ( )bkT

2

2/3

1H 1H

2H 1H

3He

νe

e+

γ

1H1H

2H1H

1H 1H

3He

νe

e+

γ

4He

Fig.11The full p-p chain

4 H → He + 2 + 2 + 2γ11

42 e+ νe

H+ H → H+ +11

11

21 e+ νe

H+ H → He + γ21

11

32

He + He → He + 2 H32

32

42

11 (32)

1.3M☉



Luminosity Gradient

Here, is the total energy released per kilogram per second by all the star stuff: gravity and nuclear reactionstogether.

For spherically symmetric star, the mass of a thin shell of thickness will be:

which leads to:

Energy Transport

Radiative

Radiation PressureEnergy density of Blackbody radiation:

We can integrate this over all wavelengths/frequencies

Thus we obtain:

1H1H

1H

νe

e+

νe

e+

15N

12C

4He

13N

13C

14N

15O

γ

γ1Hγ

Fig.12The CNO cycle

4p→ HE+ 2 + 2 + 3γ4 e+ νe

dL = ϵ dm (34)

ϵ

dr

dm = d = ρdV = 4π ρdrMr r2 (35)

= 4π ρϵdLr

drr2 (36)

dv = dv =uv4πcBv

8πh /ν3 c3

− 1ehν/kT(37)

u = dλ = dν∫ ∞

0uλ ∫ ∞

0uν

u = (T)dλ = = a4πc

∫ ∞

0Bλ

4σT 4

cT 4 (38)

7.565767 × J−16 −3 −4



( is the radiation constant given by: )

The radiationpressure can be shown to be one-third of the energy density:

The radiation pressure on the inner surface:

The radiation pressure on the outer surface:

(assuming )

Thus, the net radiation force will be given by:

which using eqs. and , we can get:

The optical depth is given by:

where is the opacity. The probability that a photon will be absorbed while passing through the shell is .

The rate at which photons carry energy through the shell is the luminosity, . Momentum is carried through theshell is just , since .

The force on the shell, i.e. the rate at which photon momentum is transferred to the shell will be

a 7.565767 × J10−16 m−3 K−4

=PradaT 4

3(39)

r

r + dr

starcenter

sphericalshell

T

T + dT

Fig.13A thin spherical shell centeredon the star center. Let dr≪ r

(r) =PradaT 4

3(40)

(r+ dr) = = ≈ (1 + 4 )Prada(T + dT)4

3a

3T 4(1 + )dT

T

4a

3T 4 dT

T(41)

dt/T ≪ 1

= [ (r) − (r+ dr)]4πFrad Prad Prad r2 (42)

(40) (41)

≈ − 4 4π = − a dTFrada

3T 4 dT

Tr2

16π3

r2T 3 (43)

dτ = −ρ(r)κ(r)dr (44)

κ dP ≈ dτ

L(r)L(r)/c p = E/c

= dτ = − ρ(r)κ(r)drradL(r) L(r)



Combing eq. equation with eq. , we get:

This can be rearranged to yield the equationofradiativeenergytransport :

This equation related the luminosity to the temperature gradiant of a star.

Convective TransportThis is the equationofconvectiveenergytransport .

It shows that the temperature gradient is proportional to the pressure gradient. ( is the adiabaticindex and dependson the gas in question. )

Stellar Model Building

= dτ = − ρ(r)κ(r)drFradL(r)c

L(r)c

(45)

(45) (43)

− a T(r dT = − ρ(r)κ(r)dr16π3

r2 )3L(r)c

(46)

= −dT

dr

3ρ(r)κ(r)L(r)16πac T(r)3r2

(47)

= (1 − )dT

dr

1γ

T(r)P (r)

dP

dr(48)

γ

Fig.14The different transport modesare separated into regions.

dP= −

G ρMr



These are the basic equations that show the relationship between the main attributes of a star. They are 4 differentialequations that relate: pressure, mass, density, temperature, energy generation, luminosity, and the radius of the star.

Boundary ConditionsIn order to solve these equations, we need boundary conditions: At the center we should expect:

and at the surface of the star:

1. Build a star mesa http://mesa-web.asu.edu/index.html

dP

dr

dM

dr

dL

dr

dT

dr

= −G ρMr

r2

= 4π ρ(r)r2

= 4π ρ(r)ϵ(r)r2

= − [forradiative]3ρ(r)κ(r)L(r)16πac T(r)3r2

= (1 − ) [forconvective]1γ

T(r)P (r)

dP

dr

} as r→ 0Mr

Lr

→ 0→ 0

as r→T

P

ρ

→ 0→ 0→ 0

⎫⎭⎬⎪⎪ R∗



http://mesa-web.asu.edu/index.html

Notes for stellar-structures, J. Hedberg © 2018

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