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Summary of the of Dynamics

Reference: , L.D. Landau & E.M. Lifshitz

1 Introduction

Emission processes give us diagnostics with which to estimate important parameters, such as the , and magnetic , of an astrophysical . provides us with the capability of understanding the transport of mass, and . Normally one spends more than a lecture on Astrophysical Fluid Dynamics since this relates to many areas of . In following lectures we are going to consider one principal application of astrophysical fluid dynamics – accretion discs. Note also that magnetic fields are not included in the following. Again a full treatment of magnetic fields warrants a full course.

2 The fundamental fluid dynamics equations

The equations of fluid dynamics are best expressed via conservation laws for the , mo- mentum and energy.

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2.1 Conservation of mass

n Consider the rate of change of mass within a fixed . This i changes as a result of the mass flow through the bounding .

v ∂ V i ∫ ρdV = –∫ρv n dS ∂t i i S V S Using the , ∂ ∂ Control volume for as- ∫ ρdV +0∫ ()ρv dV = ∂t ∂x i sessing conservation of V V i mass. ∂ρ ∂ ⇒ ∫ ------+ ()ρv dV = 0 ∂t ∂x i V i The continuity Since the volume is arbitrary, ∂ρ ∂ ()ρ -----∂- +0∂ vi = t xi

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2.2 Conservation of momentum Consider now the rate of change of momentum within a vol- ume. This decreases as a result of the of momentum ni through the bounding surface and increases as the result of body (in our case gravity) acting on the volume. Let

V Π ijn j S Π ij = Flux of i component of momentum in the j direction and

f i = Body per unit mass then ∂ ∫ ρv dV = –∫Π n dS + ∫ ρ f dV ∂t i ij j i V S V Π Π th There is an equivalent way of thinking of ij , which is often useful, and that is, ijn jdS is the i component of the force exerted on the fluid exterior to SS by the fluid interior to .

Fluid Dynamics 3/22

Again using the divergence theorem,

∂ ∂Π ∫ ()ρv + ij dV = ∫ ρ f dV ∂t i ∂x i V j V ∂ ∂Π ⇒ρ()ρ ij ∂ vi + ∂ = f i t x j Gravity For gravity we use the gravitational potential ∂φ G f i = –∂ xi For a single gravitating object of mass M GM φ = –------G r

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and for a self-gravitating distribution ∇2φ π ρ G = 4 G ρ()x ′ ⇒ φ = –G ∫ ------i d3x′ G x – x ′ V′ i i where G is Newton’s constant of gravitation. Π Expressions for ij The momentum flux is composed of a bulk part plus a part resulting from the motion of particles moving with () respect to the centre of mass velocity of the fluid vi . For a (an approximation often used in as- trophysics), we take p to be the isotropic , then Π ρ δ ij = viv j + p ij The equations of motion are then:

∂ ∂ ∂φ ()ρ ()ρ δ ρ G ∂ vi + ∂ viv j + p ij = – ∂ t x j xi ∂ ∂ ∂ ∂φ ⇒ ()ρ ()ρ p ρ G ∂ vi + ∂ viv j = – ∂ – ∂ t x j xi xi

Fluid Dynamics 5/22 There is also another useful form for the momentum equation derived using the .

∂ ∂ ∂ρ ∂v ∂ ∂v ()ρ ()ρ ρ i ()ρρ i ∂ vi + ∂ viv j = vi∂ ++∂ vi∂ v j +v j∂ t x j t t x j x j

∂ρ ∂ ∂v ∂v ()ρ ρ i ρ i = vi ∂ + ∂ v j + ∂ + v j∂ t x j t x j ∂v ∂v ρ i ρ i = ∂ + v j∂ t x j Hence, another form of the momentum equation is:

∂v ∂v ∂ ∂φ ρ i ρ i p ρ G ∂ + v j∂ = – ∂------– ------∂ t x j xi xi On dividing by the density ∂ ∂ ∂φ vi vi 1 ∂p G ∂ + v j∂ = – ρ---∂------– ------∂ t x j xi xi

Fluid Dynamics 6/22 Differentiation following the motion (), This is a good place to introduce differentiation following the motion. For a function fxi t , the variation of f following the motion of a fluid element which has coordinates () xi = xi t is given by:

df ∂ f ∂ f dxi ∂ f ∂ f ----- ==∂ + ∂ ------∂ + vi∂ dt t xi dt t xi Hence, the momentum equation can be written compactly as

dv ∂ ∂φ ρ i p ρ G ------= – ∂ – ∂ dt xi xi

Fluid Dynamics 7/22 2.3

m Before going on to consider the consequences of the conservation of S energy, we consider the thermodynamics of a comoving volume ele- U ment. (See the figure at left.) V p

Element of fluid and the variables used to describe its state. Define: m = Mass of element ε = Internal energy density per unit volume P = pressure (as above) s = entropy per unit mass T = (in degrees Kelvin)

Fluid Dynamics 8/22 We have the following quantities for the volume element: mε U ==Total internal energy ------ρ- S ==Entropy ms m V ==volume ---ρ- The second law of thermodynamics tell us that the change in entropy of a mass of is related to changes in other thermodynamic variables as follows: kTdS= dU+ pdV ε ⇒ () m m kTd ms = d------ρ- + pd---ρ-

ε 1 1 ()ε + p ⇒ kTds ==d --- + pd ------dε – ------dρ ρ ρ ρ ρ2 ()ε ⇒ ρ ε + p ρ kTds = d – ------ρ -d

Fluid Dynamics 9/22 Specific A commonly used thermodynamic variable is the specific enthalpy: ε + p h = ------ρ - In terms of the specific enthalpy, the equation ε 1 kTds= dρ--- + pdρ--- becomes ε + p p 1 dp kTds== d------ρ - – d--ρ- + pdρ--- dh – -----ρ-

For a parcel of fluid following the motion, we obtain, after dividing by the time increment of a volume element, ds dε ()ε + p dρ ρkT----- = ----- – ------dt dt ρ dt ds dh 1dp kT----- = ------– ------dt dt ρ dt

Fluid Dynamics 10/22 The fluid is adiabatic when there is no transfer of heat in or out of the volume element: dε ()ε + p dρ ----- –0------= ds dt ρ dt kT----- = 0 ⇒ dt dh 1dp ------–0------= dt ρ dt The quantities ds , dε ,dp etc. are perfect differentials, and these relationships are valid relations from point to point within the fluid. Two particular relationships we shall use in the following are:

∂s ∂ε ∂ρ ρkT = – h ∂t ∂t ∂t ∂ ∂ ∂ ρ s ρ h p kT∂ = ∂ – ∂ xi xi xi 2.3.1 The above equations can be used to derive the equation of state of a gas in which the ratio of specific heats γ ⁄ (= cp cv ) is a constant. Consider the following form of the entropy, internal energy, pressure relation: ()ε ρ ε + p ρ kTds= d – ------ρ -d

Fluid Dynamics 11/22 In a perfect gas, ρkT p = ------µ - mp where µ is the mean molecular weight and p = ()εγ – 1 ⇒γεε + p = Hence, γε µ ()εγ ε ρ mp – 1 ds= d – ----ρ-d ε γ ⇒ µ ()γ d ρ mp – 1 ds = -----ε – ρ---d ⇒εγρµ ()γ () mp – 1 ss– 0 = ln – ln ε ⇒µ----- ==exp[]m ()γ – 1 ()ss– exp[]µm ()γ – 1 s ργ p 0 p

(We can discard the s0 since the origin of entropy is arbitrary.)

Fluid Dynamics 12/22 We therefore have, εµ[]()γ × ργ = exp mp – 1 s ()µγ []ρ()γ × γ p = – 1 exp mp – 1 s = Ks()ργ The function Ks() is often referred to as the pseudo-entropy. For a completely ionised monatomic gas γ = 53⁄ .

2.4 Take the momentum equation in the form:

∂v ∂v ∂ ∂φ ρ i ρ i P ρ G ∂ + v j∂ = – ∂------– ------∂ t x j xi xi and take the scalar product with the velocity:

∂v ∂v ∂ ∂φ ρ i ρ i P ρ G vi∂ + v jvi∂ = – vi∂------– vi------∂ t x j xi xi ∂ ∂ ∂ ∂φ ⇒ ρ 1 ρ 1 P ρ G ∂ ---vivi + v j∂ ---vivi = – vi∂ – vi------∂ t 2 x j 2 xi xi

Fluid Dynamics 13/22 That is, ∂ ∂ ∂ ∂φ ρ 1 2 ρ 1 2 p ρ G ∂ ---v + v j∂ ---v = – vi∂ – vi------∂ t 2 x j 2 xi xi ρρ Before, we used the continuity equation to move the and v j outside the differentiations. Now we can use the same technique to move them inside and we recover the equation: ∂ ∂ ∂ ∂φ 1ρ 2 1ρ 2 p ρ G ∂ --- v + ∂ --- v v j = – vi∂ – vi------∂ t 2 x j 2 xi xi The aim of the following is to put the right hand side into some sort of divergence form. Consider first the term ∂ ∂ ∂ P ρ s ρ h –vi∂ = kTvi∂ – vi∂ xi xi xi ∂ ∂ ρ ds ρ s ρ h = kT----- – kT∂ – vi∂ dt t xi ∂ε ∂ρ ∂ ρ ds ρ h = kT----- – ∂ + h∂ – vi∂ dt t t xi

Fluid Dynamics 14/22 ∂ρ We now eliminate the term using continuity, viz ∂t ∂ρ ∂ ()ρ ∂ = –∂ vi t xi and we obtain ∂ ∂ε ∂ ∂ p ρ ds ()ρ ρ h –vi∂ = kT----- – ∂ – h∂ vi – vi∂ xi dt t xi xi ∂ε ∂ ρ ds ()ρ = kT----- – ∂ – ∂ hvi dt t xi

Fluid Dynamics 15/22 The term

∂φ ∂ ∂ ρ G ()ρφ φ ()ρ – vi∂ = – ∂ Gvi + G∂ vi xi xi xi ∂ ∂ρ ()φρφ = –∂ Gvi – G∂ xi t ∂ ∂ ∂φ ()ρφ ()ρρφ G = –∂ Gvi – ∂ G + ∂ xi t t When the gravitational potential is constant in time,

∂φ ∂φ ∂ ∂ G ⇒ ρ G ()ρφ ()ρφ ∂ = 0 – vi∂ = –∂ Gvi – ∂ G t xi xi t Hence, the energy equation ∂ ∂ ∂ ∂φ 1ρ 2 1ρ 2 P ρ G ∂ --- v + ∂ --- v v j = – vi∂ – vi------∂ t 2 x j 2 xi xi

Fluid Dynamics 16/22 becomes ∂ ∂ ∂ε ∂ 1ρ 2 1ρ 2 ρ ds ()ρ ∂ --- v + ∂ --- v v j = kT----- – ∂ – ∂ hvi t 2 x j 2 dt t xi ∂ ∂ ()ρφ ()ρφ –∂ Gvi – ∂ G xi t Bringing terms over to the left hand side: ∂ ∂ 1ρ 2 ερφ 1ρ 2 ρ ρφ ρ ds ∂ --- v ++ G + ∂ --- v v j ++hvj Gv j = kT----- t 2 x j 2 dt When the fluid is adiabatic ds ρkT----- = 0 dt and we have the energy equation for a perfect fluid: ∂ ∂ 1ρ 2 ερφ 1ρ 2 ρ ρφ ∂ --- v ++ G +0∂ --- v v j ++hvj Gv j = t 2 x j 2

Fluid Dynamics 17/22 The total energy per unit volume is 1 E ==---ρv2 +ερφ + Kinetic + internal + gravitational energy 2 G and the energy flux is 1 1 F ==---ρv2v ++ρhv ρφ v ρ ---v2 ++h φ v Ei, 2 i i G i 2 i = Kinetic + enthalpy + gravitational

3 with

In most astrophysical contexts we do not have to consider molecular viscosity since it is generally small. How- ever, we do need to consider viscosity in circumstances where it is important to discuss the means whereby energy is dissipated in a fluid.

3.1 The momentum flux in a viscous fluid The starting point for considering viscosity is the momentum flux. We put Π ρ δ σ ij = viv j + p ij – ij

Fluid Dynamics 18/22 σ where, the viscous tensor, ij , is given by 2 σ = η v + v – ---δ v + ζδ v ij ij, ji, 3 ij kk, ij kk, η ζδ = 2 sij + ijvkk, The tensor 12 s = --- v + v – ---δ v ij 2ij, ji, 3 ij kk, is the (trace-free) shear tensor of the fluid and

vkk, is the dilation, which is important in compressible fluids. The complete equations of motion are therefore:

∂ ∂ ∂ ∂σ ∂φ ()ρ ()ρ p ij ρ G ∂ vi + ∂ viv j = – ∂ + ∂ – ∂ t x j xi x j xi

Fluid Dynamics 19/22 3.2 Energy conservation

If we now take the scalar product of the momentum equation with vi we obtain

∂ ∂ ∂ ∂φ ∂σ ρ 1 2 ρ 1 2 P ρ G ij ∂ ---v + v j∂ ---v = – vi∂ – vi------∂ + vi∂ t 2 x j 2 xi xi x j that is, the same as before, but with the additional term

∂σ ∂ ij ()σ σ vi∂ = ∂ vi ij –vij, ij x j x j Hence the energy equation becomes ∂ ∂ 1ρ 2 ερφ 1ρ 2 ρ ρφ σ ∂ --- v ++ G + ∂ --- v v j ++hvj Gv j – vi ij t 2 x j 2 ds = ρkT----- – v σ dt ij, ij

Fluid Dynamics 20/22 σ The quantity vi ijn j is interpreted as the work done on the fluid by the viscous force; hence its appearance with terms that we associate with the energy flux. This is not the full story, however. When there is momentum transport associated with viscosity, there is also a heat flux, q which is often represented as being proportional to the temperature with a heat conduction coefficient κ , i.e. ∂ κ T qi = – ∂ xi We then write the full energy equation as ∂ ∂ 1ρ 2 ερφ 1ρ 2 ρ ρφ σ ∂ --- v ++ G + ∂ --- v v j ++hvj Gv j – vi ij +q j t 2 x j 2 ds = ρkT----- – v σ + q dt ij, ij jj, Conservation of energy is expressed by: ∂ ∂ 1ρ 2 ερφ 1ρ 2 ρ ρφ σ ∂ --- v ++ G +0∂ --- v v j ++hvj Gv j – vi ij +q j = t 2 x j 2 and the entropy changes according to ds ρkT----- = v σ – q dt ij, ij jj,

Fluid Dynamics 21/22 σ The term vij, ij represents viscous heating and the term q jj, represents escape of heat from the volume re- sulting from the heat flux. The viscous heating term can be written: σ η ζ 2 ijvij, = 2 sijsij + vkk, remembering the definition of the shear tensor: 12 s = --- v + v – ---δ v ij 2ij, ji, 3 ij kk,

Fluid Dynamics 22/22