Chapter 4 Hydrostatic Equilibrium - University of...

Chapter 4

Hydrostatic Equilibrium

A fundamental property of main sequence stars like ourSun is their stability over long periods of time.

• In the case of the Sun, the geological record indicatesthat it has been emitting energy at its present rate forseveral billion years,with relatively small variation.

• The key to this stability is that main sequence starsare in a state ofnear perfect hydrostatic equilibrium.

• In hydrostatic equilibrium the pressure gradients pro-duced by thermonuclear fusion and internal heat al-most exactly balance the gravitational forces.

Thus the starting point for an understanding of stellarstructure is an understanding of hydrostatic equilibriumand departures from that equilibrium.

67

68 CHAPTER 4. HYDROSTATIC EQUILIBRIUM

4.1 Newtonian Gravity

The Newtonian gravitational field is derived from a gravi-tational potentialΦ that obeys the Poisson equation,

∇∇∇2Φ = 4πGρ ,

∇∇∇ ≡ ∂∂x

iii +∂∂y

jjj +∂∂z

kkk

whereρ is the mass density. For the special case of spher-ical symmetry, this may be written as

1r2

∂∂ r

(

r2∂Φ∂ r

)

= 4πGρ .

The gravitational acceleration is given by

ggg = −∇∇∇Φ.

For the particular case of spherical symmetry,

ggg = (gr ,gθ ,gϕ) = (−g,0,0),

whereg≡ |ggg| > 0, so only the radial component is nonva-nishing and

g =∂Φ∂ r

=Gmr2 ,

wherem= m(r) is the mass contained within the radiusr .Hence, for spherical geometry

Φ(r) =∫ r

0g dr+ constant=

∫ r

0

Gmr2 dr + constant.

The constant is fixed by requiring thatΦ → 0 asr → ∞.

4.2. CONDITIONS FOR HYDROSTATIC EQUILIBRIUM 69

r

r+dr

r

drP(r)

P(r+dr)

Unit area

(a) (b)

Figure 4.1:Spherical mass shells. In (b) the small shaded volume has height drand unit area on its inner surface. Therefore its volume is 1×dr = dr.

4.2 Conditions for Hydrostatic Equilibrium

The local gravitational acceleration at a radiusr is givenby

g =∂Φ∂ r

=Gmr2 ,

wherem(r) is the mass contained within a radiusr . Themass contained in a thin spherical shell is (see Fig. 4.1)

dm= m(r +dr)−m(r) = 4πr2ρ(r)dr.

Integrating this from the origin to a radiusr yields themass functionm(r),

m(r) =

∫ r

04πr2ρ dr.

(Total mass contained within the radiusr .)


r

r+dr

r

drP(r)

P(r+dr)

Unit area

(a) (b)

Now consider the total gravitational force acting on avolume of unitarea in the concentric sphere of radiusr and depthdr.

• The magnitude of this force (per unit area) will be

Fg = −g(r)︸︷︷︸

a

ρdr︸︷︷︸

m

= −ρGm(r)

r2 dr, (Gravity)

Negative sign→ directed toward the center of the sphere.

• The force per unit area resulting from the pressure differencebetweenr andr +dr is

P(r)−P(r +dr) = −∂P∂ r

dr (Pressure Gradient)

Negative sign→ directed outward.

• The inwardly directed gravitational force is counterbalanced bya net outward force arising from the pressure gradient of thegasand radiation that has a magnitude

Fp = P(r)−P(r +dr) = −∂P∂ r

dr.


r

r+dr

r

drP(r)

P(r+dr)

Unit area

(a) (b)

• The total force acting on this volume of unit surface area isthen

F = Fg +Fp = −∂P∂ r

dr︸︷︷︸

Pressure Gradient

−Gm(r)r2 ρdr

︸︷︷︸Gravity

,

• by Newton’s 2nd law the equation of motion is (remember: unitarea)

F = ma= ρdr︸︷︷︸

Mass

× ∂ 2r∂ t2︸︷︷︸

Acceleration

,

• This leads to

ρ∂ 2r∂ t2 = −∂P

∂ r− Gm(r)

r2 ρ .

• For hydrostatic equilibriumthe left side vanishes because theacceleration∂ 2r/∂ t2 = 0 and we obtain

dPdr

= −Gm(r)r2 ρ = −gρ ,

where partial derivatives have been replaced with derivatives be-cause by our assumption there isno longer any time dependence.


Hydrostatic Equilibrium and Stellar Interiors

In the equation

dPdr

= −Gm(r)r2 ρ = −gρ ,

bothρ andGm(r)/r2 are positive.

1. ThusdP/dr ≤ 0 and pressure must decrease outwardeverywhere for a gravitating system to be in hydro-static equilibrium.

dP/dr is always negative under conditions ofhydrostatic equilbrium.

2. This will in turn imply that density and temperaturemust increase toward the center of a star.

Thus, the conditions of hydrostatic equilibrium are suffi-cient to ensure thatstars must be much more dense andhot near their centersthan near their surfaces.


The equations

dPdr

= −Gm(r)r2 ρ = −gρ ,

dm= 4πr2ρ(r)dr.

are our first two equations of stellar structure.

• They constitutetwo equations in three unknowns(P,m, andρ as functions ofr).

• This system of equations may beclosed by specifyingan equation of staterelating these quantities.

Before considering that, we explore some consequencesthat follow from these equations alone.


4.3 Lagrangian and Eulerian Descriptions

In studying fluid motion, there are two basic computational points ofview that we can take.

1. We can fix a grid and watch the fluid flow through the grid; thisis calledEulerian hydrodynamics.

2. Alternatively, we can construct coordinates that are attached tothe mass elements and move with them; this is calledLagrangianhydrodynamics.

To appreciate the difference, consider determining thetemperature of the atmosphere over time either fromweather balloons drifting with the wind, or from fixedpoints on the ground.

• The first is aLagrangianpoint of view, since the co-ordinates of a balloon move with the fluid.

• The second isEulerian, since one observes the airfrom fixed observation points as it flows by.

3. In the limit thataccelerations of the fluid can be neglected,La-grangian and Eulerian descriptions of hydrodynamics reduce toLagrangian and Eulerian descriptions ofhydrostatics.

4.3. LAGRANGIAN AND EULERIAN DESCRIPTIONS 75

4.3.1 Lagrangian Formulation of Hydrostatics

Let us illustrate the Lagrangian approach by reformulatingthe pre-ceding equations withm(r) rather thanr as the independent variable.

• For the change of variables between Eulerian and Lagrangianrepresentations(r, t) → (m,t), we can use

∂∂m

=∂ r∂m

∂∂ r

·

• Sincedm= 4πr2ρ(r)dr. we have

∂ r∂m

=1

4πr2ρ,

and in operator form the transformation between the two repre-sentations is

∂∂m

=1

4πr2ρ∂∂ r

.

• Now we convert the Eulerian equation

ρ∂ 2r∂ t2 = −∂P

∂ r− Gm(r)

r2 ρ .

to Lagrangian coordinates by using∂P∂ r

=∂m∂ r

∂P∂m

= 4πr2ρ∂P∂m

14πr2

∂ 2r∂ t2

︸︷︷︸∝ Acceleration

= −∂P∂m

− Gm(r)4πr4 .

• For the case of hydrostatic equilibrium, the accelerationtermmay be neglected and we obtain theLagrangian equation forhydrostatic equilibrium

dPdm

= − Gm4πr4.


Table 4.1: Equations of hydrostatics

Eulerian coordinates(r,t) Lagrangian coordinates(m,t)

dmdr

= 4πr2ρdrdm

=1

4πr2ρdPdr

= −Gmρr2

dPdm

= − Gm4πr4

In Table 4.1 we summarize the equations of spherical hy-drostatics in Eulerian and Lagrangian form.

4.3. LAGRANGIAN AND EULERIAN DESCRIPTIONS 77

4.3.2 Contrasting Lagrangian and Eulerian Descriptions

Eulerian and Lagrangian representations have advantages and disad-vantages in a particular context.

• Our observational mindset is often Eulerian:we tend to think ofmonitoring a river by placing a measuring device at a fixed pointon the river rather than imagining a measuring device floatingdown the river with a given packet of water.

• We tend to formulate microscopic laws of physics in a Lagrangianway: for the collision of billiard balls, we normally imagine fol-lowing each ball. We seldom imagine staking out points on thetable and asking how balls move past those fixed points (a clearlyEulerian point of view).

• Because the Lagrangian point of view is often more simply tiedto the underlying physical laws,the Lagrangian formulation isoften preferred when there are clear symmetries and conserva-tion lawsthat play significant roles in the system.

Example: Imagine a spherical star that is neither gainingnor losing mass, but is pulsating radially in size.

– Theradial distance to the surface(an Eulerian coor-dinate) is changing with time.

– The mass contained within the outermost radius(aLagrangian coordinate) is constant in time.

• On the other hand, if spherical symmetry is broken and thereis convective and turbulent motion of the fluid, the Euleriande-scription is often simpler than the Lagrangian description.


4.4 Dynamical Timescales

A particularly important concept in astrophysics is that ofadynamicaltimescale, because a dynamical timescale sets the order of magnitudefor the time required for a system to respond to a perturbation.

• The dynamical response of stars to perturbations of their hy-drostatic equilibrium is of obvious significance in understandingstars and their evolution.

• Consider thefree-fall timescaletff

tff ≃√

1Gρ

≃√

Rg

whereρ = M/(43πR3) is the average density andg = GM/R2 is

the gravitational acceleration.

• This defines atimescale for collapseof a gravitating sphere if itsuddenly lost all pressure support.

4.4. DYNAMICAL TIMESCALES 79

• We may introduce a second dynamical timescale by consideringthe opposite extreme: if gravity were taken away, how fast wouldthe star expand by virtue of its pressure?

• This timescale can depend only onR, ρ , and P, and the onlycombination of these quantities having time units is

texp ≃ R

√

ρP≃ R

vs

,

This characteristic expansion timescale has a simple phys-ical interpretation:

1. (ρ/P)1/2 is approximately theinverse of the meansound speedvs for the medium.

2. This implies thattexp is approximately thetime for asound wave to travel from the center to the surface ofthe star.

This intepretation makes sense because pressure wavespropagating outward should be characterized by thattimescale.

• Hydrostatic equilibrium will clearly be precarious unless the twodynamical timescales are comparable with each other; therefore,we define ahydrodynamical timescalefor the system through

τhydro ≃ texp ≃ tff ≃√

1Gρ

.


Table 4.2: Hydrodynamical timescales

Object ∼ M/M⊙ ∼ R/R⊙ τhydro

Red Giant 1 100 36 days

Sun 1 1 55 minutes

White Dwarf 1 1/50 9 seconds

Example: For the Sunρ = 1.4 g cm−3 and

t⊙hydro = τhydro ≃√

1Gρ

≃ 55 minutes.

• If hydrostatic equilibrium were not satisfied wewould expect to seechanges in a matter of hours,but the fossil record indicates that the Sun has beenextremely stable for billions of years.

• We conclude thatthe Sun is in very good hydrostaticequilibrium.

In Table 4.2 we illustrate the hydrodynamical timescalefor several kinds of stars calculated using this formula.

4.5. VIRIAL THEOREM 81

4.5 Virial Theorem

Stars generally have at their disposal two potentially largesources of energy:

1. Gravitational energy,which can be released by con-traction.

2. Internal energy,which can be produced both by con-traction and by fusion and other internal processes.

We now derive an important relationship between internaland gravitational energy for objects in approximate hydro-static equilibrium called thevirial theorem.


We may multiply both sides of the Lagrangian equation

dPdm

= − Gm4πr4.

by 4πr3 and integrate overdmfrom 0 to M ≡m(R) to give∫ M

0

Gmr

dm=−4π∫ M

0r3 ∂P

∂mdm

︸︷︷︸

Integrate by parts

=− 4πr3P

∣∣∣∣

m=M

m=0︸︷︷︸

identically zero

+12π∫ M

0r2P

∂ r∂m

dm

= 12π∫ M

0r2P

14πr2ρ

dm

=∫ M

0

3Pρ

dm,

• ρ , r , andP are functions of independent variablem

• An integration by parts was used to obtain line 2

• In the first term of line 2

1. r vanishes whenm= 0 (center of star)

2. P vanishes whenm= M (surface of star).

Thus this term is identically zero.

•drdm

=1

4πr2ρwas used in going from line 2 to line 3


The equation∫ M

0

Gmr

dm=∫ M

0

3Pρ

dm,

may be given a simple interpretation (Exercise 4.5). Firstconsider the right side:

• P/ρ = kT/µ for an ideal monatomic gas

• Thus the right side is twice the internal energyU be-cause

∫ M

0

3Pρ

dm=3kT

µ

∫ M

0dm=

3MkTµ

= 3

(Mµ

)

︸︷︷︸

N

kT = 3NkT = 2U.

since for an ideal monatomic gasU = 32NkT.

Hence the virial theorem (for an ideal gas) is equivalent to∫ M

0

Gmr

dm= 2U.

Let us now interpret the integral on the left side by askingthe question

What is the total gravitational energy releasedin forming a star?


r

m(r)

∆m = 4π r 2ρdrs = ∞

Figure 4.2:Gravitational assembly of a star by the accretion of mass shells.

Consider Fig. 4.2, where we allow a shell of mass∆m tofall from infinity onto the surface of a spherical mass ofradiusr and enclosed massm(r). The gravitational energyreleased in this process is

dΩ =

∫ r

∞Fg ds=

∫ r

∞g(s)∆m ds

=

∫ r

∞

Gm(r)s2

︸︷︷︸gs

4πr2ρdr︸︷︷︸

∆m

ds

= −Gm(r)s

∣∣∣∣

r

∞×4πr2ρdr = −4πr2ρdr

Gm(r)r

,

and the total energy released in assembling a star of radiusR and massM from such mass shells is

Ω =

∫

dΩ = −4π∫ R

0r2ρ

Gm(r)r

dr = −∫ M

0

Gm(r)r

dm,

wheredm/dr = 4πr2ρ was used andM ≡ m(R).


Thus we have obtained

∫ M

0

Gm(r)r

dm= −Ω (Gravitational energy of star)

and from the previous

∫ M

0

Gmr

dm= 2U.

we see that

Virial Theorem for ideal gas: 2U +Ω = 0,

whereU is the internal energy of the star andΩ is its gravitational energy.


The result

2U +Ω = 0 (or in the formU = −12Ω )

is termed thevirial theoremfor an ideal, monatomic gas.

1. It establishes an important general relationship be-tween the internal energy and gravitational energy ofa star in approximate hydrostatic equilibrium.

2. The virial theorem is of broad applicability becauseof

• The very general conditions under which it wasderived

• Because it relates the two most important energyreserves for a star, gravitational and internal en-ergy.

We shall often use the virial theorem and concepts derivedfrom it in discussions of stellar structure and evolution.


As a star forms, gravitational contraction releases anamount of energy∆Ω.

• As stars form they go through a sequence of stagesthat are often nearly in hydrostatic equilibrium.

• Sincethe virial theorem must be satisfied for hydro-static equilibrium to hold,as a newly-forming starcontracts the thermal energy must change by

∆U = −12∆Ω

and the excess energy must be radiated into space.

A cloud of gas and dust collapsing to forma star cannot be in hydrostatic equilibrium.However, through much of the collapse thenascent star is only slightly out of equilib-rium, so we may expect the virial theorem tobe approximately satisfied.

Thus, gravitational contraction has three consequences:

1. The star heats up,

2. Some energy is radiated into space,

3. The total energy of the star decreases and it becomesmore bound.

This leads to the interesting consequence that the star“heats up while it cools”.


4.6 Kelvin–Helmholtz Timescale for the Sun

If approximate hydrostatic equilibrium is to be main-tained,

1. At each infinitesimal step of the contraction the starmustwait until half of the released gravitational en-ergy is radiatedbefore it can continue to contract.

2. This implies that there is atimescale for contractionin near hydrostatic equilibrium that is set by the timerequired to radiate the excess energy.

This contraction timescale is called theKelvin–Helmholtztimescalefor the system.

4.6. KELVIN–HELMHOLTZ TIMESCALE FOR THE SUN 89

Estimate theKelvin–Helmholtz timescaleby assuming

1. constant densityρ and

2. a corresponding massm(r) = 43πr3ρ

Then the gravitational energy released in collapsing theinitial cloud of gas and dust to a star of radiusR is

Ω =−∫ R

04πr2ρ

Gm(r)r

dr

=−163

π2ρ2G∫ R

0r4 dr

=−1615

π2ρ2GR5

=−35

GM2

R,

whereM = 43πR3ρ .

Taking M = M⊙ andR = R⊙, we find thatΩ⊙ = 2.3×1048 erg of gravitational energy was released in formingthe Sun.

By the virial theorem, half of this must be radiated whilethe Sun contracts:

E⊙rad = 1

2Ω⊙ ≃ 1048 erg.

The Kelvin–Helmholtz timescaletKH sets thetime required to radiate this energy.


We may make a rough estimate of the Kelvin–Helmholtztimescale for the Sun by assuming that it has radiated at itspresent luminosity ofL⊙ = 4×1033 erg s−1 for its entirelife. Then

tKH ≃ E⊙rad

L⊙≃ 107 years,

and we conclude that the Sun contracted to the mainsequence on a Kelvin–Helmholtz timescale of approxi-mately10 million years.

Generally, we shall define a Kelvin–Helmholtz timescalefor a star by the relation

tKH =ΩL≃ GM2/R

L,

whereR is the radius,M the mass, andL the luminosity.

4.7. TIMESCALE SET BY RANDOM WALK OF PHOTONS 91

4.7 Timescale Set by Random Walk of Photons

More microscopically, we may view the contractiontimescale as being set by the time for photons producedin the core of the star to make their way by a random walkto the surface of the star.

• For a random walk, the distance traveled afterZ scat-terings is (see Exercise 4.3)

∆x≃ λ√

Z,

whereλ is the average distance (mean free path) tra-versed by the photon before being scattered.

• To escape, a photon must undergo approximately

Z =

(∆xλ

)2

=

(Rλ

)2

scatterings.

• For the Sun, this corresponds to 1022 scatterings be-fore reaching the solar surface if we assume an aver-age mean free path of 0.5 cm.

• We may attach a timescale to this random walk byestimating the average lifetime of the state formedwith each scattering.

• Taking a characteristic lifetime of 10−8 seconds forsuch states, we again find approximately 107 yearsfor contraction of the Sun to the main sequence.


4.8 Kelvin–Helmholtz Timescale for Other Stars

We may relate the Kelvin–Helmholtz timescale for otherstars to that of the Sun by the following considerations.

• As a very rough approximation, we may assume thatfor main sequence starsM/R≃ constant(good toabout a factor of two—see table in Ch. 1).

• Then, if we assume the photon absorption cross sec-tion to be proportional to the density and also to beapproximately independent of the temperature whenaveraged over the star,

ρ ≡ M

(43πR3)

=1

4πR2

MR︸︷︷︸

constant

≃ R−2 λ ≃ 1ρ

∝ R2,

and the number of random walk scatterings for a pho-ton to reach the surface of the star may be estimatedas

Z ∝(

Rλ

)2

≃(

RR2

)2

≃ R−2.

• Therefore, we conclude that the contraction time fora main sequence star of radiusR behaves approxi-mately as

tcon ∼ Z ∼ R−2.

4.8. KELVIN–HELMHOLTZ TIMESCALE FOR OTHER STARS 93

Suppose that for some starR≃ 10R⊙.

• Then the contraction time would be(10R⊙/R⊙)2 ≃100times shorter than that of the Sun.

• This corresponds to a time of about107 × 10−2 ≃105 yr to reach the main sequence.

This is one of many examples that we shall encounterillustrating that more massive stars evolve more rapidlythrough all phases of their lives.

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