The interior of the stars: Polytropic Gaseous Spheres

Ilídio Lopes

The interior of the stars: Polytropic Gaseous Spheres

HST image of AG Carinae (55 Msun, 50-500 Rsun, 600000 – 900000 Lsun): a difficult balance between gravity and radiation to avoid self-destruction.

Last Lecture Questions

Ilídio Lopes 3

4

Nuno’s Question (Equation of State)

Kippenhahn (15.9)

Astrophysics · Engineering Physics2019-2020 · Ilıdio Lopes

The Degenerate Electron Gas (Lecture 12) 10

xvi. For the equation of state we need the pressure, which by definition is the flux of momentumthrough a unit surface per second. We consider a surface element dσ having a normal vector "n,as indicated in Fig. 3 [15.3]. An arbitrary unit vector "s, together with "n, defines an angle ϑ.

xvii. Let us determine the number of electrons per second that go through the area dσ into a smallsolid angle dΩs around the direction "s. We restrict ourselves to electrons for which the absolutevalue of their momentum lies between p and p+ dp.

xviii. Since the momentum distribution is isotropic, we can carry the integration over angles di-rectly, and the RHS of equation (13), reads

Pe =

∫

©

∫ ∞

0f(p)v(p)p dp

[

(cosϑ)2

4πdΩs

]

(14)

∫

©

(cosϑ)2

4πdΩs = 2

∫

©/2

(cosϑ)2

4πdΩs =

1

2π

∫ 2π

0dφ

∫ π/2

0(cosϑ)2 sinϑdϑ (15)

where© is the sphere and©/2 is the hemisphere (half of a sphere). It follows that∫ π/2

0(cosϑ)2 sinϑdϑ =

∫ 0

1µ2(−dµ) =

∫ 1

0µ2dµ =

[µ

3

]

=1

3(16)

µ = cosϑ dµ = − sinϑdϑ (17)

Pe =

∫ ∞

0f(p)v(p)p

1

3dp Pe =

1

3

∫ pF

0

8πp3

h3v(p)dp (18)

where f(p) = 8πp2

h3 .

xix. Alternative expression:We obtain a very general equation for the pressure as the integral over the momentum p (in thephase space 6 dimensions):

Pe =1

3

∫ ∞

0n(p) p v(p)4πp2dp, (19)

where 1/3 integration over the solid angle and 4πp2dp corresponds to a the shell [p, p+dp]. Sincethere are two spin states, two electrons can occupy each cell in momentum space. Thus we havefor our Fermion gas

n(p) = 2×1

h31

e(E−µ)/kT + 1(20)

The number density of electrons ne reads

ne =

∫ ∞

0n(p)4πp2dp =

8π

h3

∫ ∞

0

p2

e(E−µ)/kT + 1dp (21)

Astrophysics · Engineering Physics2019-2020 · Ilıdio Lopes

The Degenerate Electron Gas (Lecture 12) 10

xvi. For the equation of state we need the pressure, which by definition is the flux of momentumthrough a unit surface per second. We consider a surface element dσ having a normal vector "n,as indicated in Fig. 3 [15.3]. An arbitrary unit vector "s, together with "n, defines an angle ϑ.

xvii. Let us determine the number of electrons per second that go through the area dσ into a smallsolid angle dΩs around the direction "s. We restrict ourselves to electrons for which the absolutevalue of their momentum lies between p and p+ dp.

xviii. Since the momentum distribution is isotropic, we can carry the integration over angles di-rectly, and the RHS of equation (13), reads

Pe =

∫

©

∫ ∞

0f(p)v(p)p dp

[

(cosϑ)2

4πdΩs

]

(14)

∫

©

(cosϑ)2

4πdΩs = 2

∫

©/2

(cosϑ)2

4πdΩs =

1

2π

∫ 2π

0dφ

∫ π/2

0(cosϑ)2 sinϑdϑ (15)

where© is the sphere and©/2 is the hemisphere (half of a sphere). It follows that∫ π/2

0(cosϑ)2 sinϑdϑ =

∫ 0

1µ2(−dµ) =

∫ 1

0µ2dµ =

[

µ3

3

]

=1

3(16)

µ = cosϑ dµ = − sinϑdϑ (17)

Pe =

∫ ∞

0f(p)v(p)p

1

3dp Pe =

1

3

∫ pF

0

8πp3

h3v(p)dp (18)

where f(p) = 8πp2

h3 .

xix. Alternative expression:We obtain a very general equation for the pressure as the integral over the momentum p (in thephase space 6 dimensions):

Pe =1

3

∫ ∞

0n(p) p v(p)4πp2dp, (19)

where 1/3 integration over the solid angle and 4πp2dp corresponds to a the shell [p, p+dp]. Sincethere are two spin states, two electrons can occupy each cell in momentum space. Thus we havefor our Fermion gas

n(p) = 2×1

h31

e(E−µ)/kT + 1(20)

The number density of electrons ne reads

ne =

∫ ∞

0n(p)4πp2dp =

8π

h3

∫ ∞

0

p2

e(E−µ)/kT + 1dp (21)

Astrophysics · Engineering Physics2019-2020 · Ilıdio Lopes

The Degenerate Electron Gas (Lecture 12) 10

xvi. For the equation of state we need the pressure, which by definition is the flux of momentumthrough a unit surface per second. We consider a surface element dσ having a normal vector "n,as indicated in Fig. 3 [15.3]. An arbitrary unit vector "s, together with "n, defines an angle ϑ.

xvii. Let us determine the number of electrons per second that go through the area dσ into a smallsolid angle dΩs around the direction "s. We restrict ourselves to electrons for which the absolutevalue of their momentum lies between p and p+ dp.

xviii. Since the momentum distribution is isotropic, we can carry the integration over angles di-rectly, and the RHS of equation (13), reads

Pe =

∫

©

∫ ∞

0f(p)v(p)p dp

[

(cosϑ)2

4πdΩs

]

(14)

∫

©

(cosϑ)2

4πdΩs = 2

∫

©/2

(cosϑ)2

4πdΩs =

1

2π

∫ 2π

0dφ

∫ π/2

0(cosϑ)2 sinϑdϑ (15)

where© is the sphere and©/2 is the hemisphere (half of a sphere). It follows that∫ π/2

0(cosϑ)2 sinϑdϑ =

∫ 0

1µ2(−dµ) =

∫ 1

0µ2dµ =

[

µ3

3

]

=1

3(16)

µ = cosϑ dµ = − sinϑdϑ (17)

Pe =

∫ ∞

0f(p)v(p)p

1

3dp Pe =

1

3

∫ pF

0

8πp3

h3v(p)dp (18)

where f(p) = 8πp2

h3 .

xix. Alternative expression:We obtain a very general equation for the pressure as the integral over the momentum p (in thephase space 6 dimensions):

Pe =1

3

∫ ∞

0n(p) p v(p)4πp2dp, (19)

where 1/3 integration over the solid angle and 4πp2dp corresponds to a the shell [p, p+dp]. Sincethere are two spin states, two electrons can occupy each cell in momentum space. Thus we havefor our Fermion gas

n(p) = 2×1

h31

e(E−µ)/kT + 1(20)

The number density of electrons ne reads

ne =

∫ ∞

0n(p)4πp2dp =

8π

h3

∫ ∞

0

p2

e(E−µ)/kT + 1dp (21)

Astrophysics · Engineering Physics2019-2020 · Ilıdio Lopes

The Degenerate Electron Gas (Lecture 12) 10

xvi. For the equation of state we need the pressure, which by definition is the flux of momentumthrough a unit surface per second. We consider a surface element dσ having a normal vector "n,as indicated in Fig. 3 [15.3]. An arbitrary unit vector "s, together with "n, defines an angle ϑ.

xvii. Let us determine the number of electrons per second that go through the area dσ into a smallsolid angle dΩs around the direction "s. We restrict ourselves to electrons for which the absolutevalue of their momentum lies between p and p+ dp.

xviii. Since the momentum distribution is isotropic, we can carry the integration over angles di-rectly, and the RHS of equation (13), reads

Pe =

∫

©

∫ ∞

0f(p)v(p)p dp

[

(cosϑ)2

4πdΩs

]

(14)

∫

©

(cosϑ)2

4πdΩs = 2

∫

©/2

(cosϑ)2

4πdΩs =

1

2π

∫ 2π

0dφ

∫ π/2

0(cosϑ)2 sinϑdϑ (15)

where© is the sphere and©/2 is the hemisphere (half of a sphere). It follows that∫ π/2

0(cosϑ)2 sinϑdϑ =

∫ 0

1µ2(−dµ) =

∫ 1

0µ2dµ =

[

µ3

3

]

=1

3(16)

µ = cosϑ dµ = − sinϑdϑ (17)

Pe =

∫ ∞

0f(p)v(p)p

1

3dp Pe =

1

3

∫ pF

0

8πp3

h3v(p)dp (18)

where f(p) = 8πp2

h3 .

xix. Alternative expression:We obtain a very general equation for the pressure as the integral over the momentum p (in thephase space 6 dimensions):

Pe =1

3

∫ ∞

0n(p) p v(p)4πp2dp, (19)

where 1/3 integration over the solid angle and 4πp2dp corresponds to a the shell [p, p+dp]. Sincethere are two spin states, two electrons can occupy each cell in momentum space. Thus we havefor our Fermion gas

n(p) = 2×1

h31

e(E−µ)/kT + 1(20)

The number density of electrons ne reads

ne =

∫ ∞

0n(p)4πp2dp =

8π

h3

∫ ∞

0

p2

e(E−µ)/kT + 1dp (21)

Last Lecture (summary)

6

These basic equations supplemented with • Equation of state (pressure of a gas as a function of its density and

temperature)• Opacity (how opaque the gas is to the radiation field)• Core nuclear energy generation rate

For our stars – which are isolated, static, and spherically symmetric – there are four basic equations to describe structure. All physical quantities depend on the distance from the centre of the star alone

1) Equation of hydrostatic equilibrium: at each radius, forces due to pressure differences balance gravity

2) Conservation of mass3) Conservation of energy : at each radius, the change in the energy flux = local

rate of energy release4) Equation of energy transport : relation between the energy flux and the local

gradient of temperature

Equations of stellar structure(reminder)

Equations of stellar structure:“Simple stellar model” (Polytropes and simple

models)

8

Deduction of Polytropic models (blackboard)

Astrophysical Notes: Application to main sequence stars, cluster of stars and cluster of galaxies (isothermal spheres). See notes on page 287 from Peter Schneider, ”Extragalactic Astronomy and Cosmology -An Introduction”Second Edition, Springer.

Astrophysical Notes: Stellar Structure and Evolution, Kippenhahn, Rudolf; Weigert, Alfred; Weiss, Achim, , 2012. ISBN: 978-3-642-30255-8

Compute Polytropic models. See Chapter 19 Polytropic Gaseous Spheres from ”Stellar Structure andEvolution” (page:213 , Second Edition, Rudolf Kippenhahn, Alfred Weigert, Achim Weiss) for details.

10

Polytropic models Take the equation for hydrostatic support (in terms of the radius variable r),

Multiply by r2/𝜌 and differentiating with respect to r, gives

Now substitute the equation of mass-conservation on the right-hand side, and we obtain

Let us now adopt an equation of state of the form (where is it customary to adopt 𝛾= 1+1/n) . K is a constant and n is known as the polytropic index.

€

P = Kργ = Kρn+1n

€

dP(r)dr

= −GM(r)ρ(r)

r2

€

ddr

r2

ρdPdr

#

$ %

&

' ( = −G

dMdr

€

1r2

ddr

r2

ρdPdr

#

$ %

&

' ( = −4πGρ

11

Solving the Lane-Emden equationIt is possible to solve the equation analytically for only three values of the polytropicindex n

€

n = 0, θ =1− ξ 2

6%

& '

(

) *

n =1, θ =sinξξ

n = 5, θ =1

1+ ξ2

3% & '

( ) *

0.5

Solutions for all other values of n must be solved numerically i.e. we use a computer program to determine 𝜃 for values of ξ. Solutions are subject to boundary conditions:

€

dθdξ

= 0, θ =1 at ξ = 0

Polytropic models (examples)

12

For n < 5 polytropes, the solution for θ drops below zero at a finite value of ξ and hence the radius of the polytrope ξ R can be determined at this point. In the numerically integrated solutions, a linear interpolation between the points immediately before and after θ becomes negative will give the value for ξ at θ=0. The roots of the equation for a range of polytropic indices are listed below. In the two cases where an analytical solution exists, the solutions are easily derived.

n xR0 2.45 3.33 ´ 10-1

1 3.14 1.01 ´ 10-1

2 4.35 2.92 ´ 10-2

3 6.90 6.14 ´ 10-3

4 15.00 5.33 ´ 10-4

€

−dθdξ%

& '

(

) * ξ =ξ R

€

ρ = ρcθn

€

r =αξRecall:

Polytropic models (Comparison with real models)

13

Numerical solutions to the Lane-Emden equation for (left-to-right) n = 0,1,2,3,4,5

Compare with analytical

€

n = 0, θ =1− ξ 2

6%

& '

(

) *

n =1, θ =sinξξ

n = 5, θ =1

1+ ξ2

3% & '

( ) *

0.5

Solutions decrease monotonically and have θ=0 at ξ= ξR (i.e. the stellar radius)

With decreasing polytropic index, the star becomes more centrally condensed. What does a polytrope of n=5 represent ?

Polytropic models (Examples)

10-min Pause

chill out …relax …

have a break . . .

Thank you for your attention!

Any Questions?