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Home , Continuity equation, Fluid dynamics

The Fundamental of Dynamics

1 References for astrophysical gas () dynamics 1. L. D. Landau and E. M. Lifshitz 2. F.H. Shu The of II Gas Dynam- ics 3. L. Mestel Stellar Magnetism

Fundamental equations © Geoffrey V. Bicknell 2 The of continuity 2.1 Derivation of the fundamental equation n Define v  = (1) vi = velocity components S within volume V is V M =  dV (2) V

Fundamental equations 2/78 Rate of flow of mass out of V is

vinidS S Therefore, the within V is given by: d ----- dV = – v n dS (3) dt  i i V S Use the theorem:  v n dS = v dV (4)  i i  xi i S V

Fundamental equations 3/78 and d  ----- dV = dV (5) dt  t V V to give   + v dV = 0 (6)  t xi i V Since the volume V is arbitrary, then   +0vi = (7) t xi This is the equation of continuity.

Fundamental equations 4/78 Aside: Differentiation following the motion

Suppose we have a function fxi t which is a function of both space and time. How does this function vary along the trajectory of a fluid element described by xi = xit ? We simply calculate

d f f dxi f f fxit  t ==+ ------+ vi (8) dt t xi dt t xi

Fundamental equations 5/78 Useful result from the equation of continuity Write the above equation in the form:

  vi ++vi  = 0 (9) t xi xi Going back to the equation of continuity, then

d vi vi 1d ------= –  = –------(10) dt xi xi  dt

Fundamental equations 6/78 Putting it another way:

1d vi ------= – (11)  dt xi

This tells us that in a diverging velocity vi  xi  0 , the density decreases and in a converging velocity field vi  xi  0 the density increases.

Fundamental equations 7/78 2.2 Examples of mass from a massive star Massive stars such as O and B stars produce with veloci- ties of around 1 000 km s–1 and mass of around 10–6 solar per yr. We can use these facts to estimate the density in the wind as follows.

Mass flux = vinidS (12) r S = 4r2r vr

Fundamental equations 8/78 where the is over a sphere of radius r . We assume that the flow is steady so that the integrated over any 2 sur- faces surrounding the star is the same. Hence, · M ==4r2r vr constant (13) We shall show in later lectures that far from the star, the velocity is constant, i.e.

vr==v constant (14) Therefore the density is given by: M· r = ------(15) 2 4r v

Fundamental equations 9/78 For typical parameters: M· = 10–6 solar masses per yr 15 r ==0.1 pc 3.0910 m (16) –1 6 –1 v ==1 000 km s 10 m s Units 30 16 1 solar mass= 210 kg 1 pc= 3.0910 m (17) 7 1year= 3.1610 seconds The density of the wind at 0.1 pc from the star is: –22  = 5.210 kg m–3 (18)

Fundamental equations 10/78 This is not particularly informative by itself. We usually express the density in terms of cm–3 . Assuming the gas is to- tally ionised, then

 = nHmp ++nHemHe ...... (19) Now for solar cosmic abundances,

nHe = 0.085nH (20) and

mHe  4m (21) where m is an atomic mass unit –27 m = 1.6610 kg (22)

Fundamental equations 11/78 so that

  nHm + 4 0.085nHm = 1.34nHm (23) Therefore, for the O-star wind:

–22 –3 1.34nHm  5.210 kg m –22 5.210  n = ------m–3 (24) H 1.34m 5 = 2.310 Hydrogen atoms m–3

Fundamental equations 12/78 Often, instead of the Hydrogen density we use the total number density of ions plus electrons. In this case we put = nm (25)  ==mean molecular weight 0.6156 Hence, –22 5.210 5 n ==------5.110 particles m–3 (26) 0.6156m

Fundamental equations 13/78 3 Conservation of

n Consider first the rate of change of momentum within a volume as a re- vivj sult of the flux of momentum. Let i = total momentum, then S vi · d i = ----- v dV (27) V dt i V is the rate of change of momentum in –pni the volume, V .

Fundamental equations 14/78 The flux of momentum out of the S is

vivjnjdS (28) S 3.1 Surface There are two complementary ways that we can look at the other aspects of the conservation of momentum. In a there is a force per unit area perpendicular to the surface that exerts a force on the gas inside. This is the pes- sure p Force on volume = – pn dS (29) within S  i S

Fundamental equations 15/78 Therefore the total momentum balance for the volume is: d ----- v dV = – v v n dSpn– dS (30) dt i  i j j  i V S S The surface Using the divergence we can write  vivjnjdS =  vivj dV (31) xj S V The of the can be written  pnidS ==pijnjdS  pij dV (32) xj S S V

Fundamental equations 16/78 Entire momentum equation Again, we take the time derivative inside and obtain:     vi + vivj + pij dV = 0 (33) t xj xj V Since the volume is arbitrary, then   vi +0vivj + pij = (34) t xj We often write this equation in the form:   p vi + vivj = – (35) t xj xi

Fundamental equations 17/78 3.2 2nd way - momentum flux Across any surface in a gas there is a flux of momentum due to the random motions of atoms in the gas.  n n We write the flux per unit area of ij j momentum due to molecular mo- v v i j tions as ijnj The flux of momentum out of the S vi volume due to molecular motions is V then:

 n dS (36) –pni  ij j V

Fundamental equations 18/78 For a perfect fluid, the relation between ij and p is

ij = pij (37)

In viscous (to be treated later) the tensor ij is not diago- nal.

Fundamental equations 19/78 Momentum balance When we adopt this approach, the momentum balance of the vol- ume of gas is: d ----- v dV = – v v n dS–  n dS dt i  i j j  ij j V S S (38)     vi + vivj + ij dV = 0 t xj V The corresponding partial is:   vi +0vivj + ij = (39) t xj

Fundamental equations 20/78 The point to remember from this analysis is that pressure in a flu- id is the result of a flux of momentum resulting from the micro- scopic motions of the particles. Particles of mass m crossing a surface within the fluid with ran- dom atomic velocity ui (relative to the bulk velocity) contribute an amount muiuj to the flux of momentum. A with an equally opposite velocity contributes exactly the same amount to the momentum flux. Hence, atomic motions in both directions across the surface contribute equal amounts to the pressure.

Fundamental equations 21/78 3.3 Alternate forms of the momentum equations

Consider, the isotropic case where ij = pij . Then,   p vi + vivj = – (40) t xj xi Expand the terms on the left:

 vi  vi p vi ++ vi vj + vj= – (41) t t xj xj xi Using the to eliminate the blue terms:

vi vi p  + vj = – (42) t xj xi Fundamental equations 22/78 In terms of the derivative following the motion:

dvi p ------= – (43) dt xi

3.4 Additional We can in general have additional forces acting on the fluid. In particular, we can have a gravitational force derived from a grav- itational potential. This is not a surface force like the pressure but a which is proportional to the volume of the re- gion. We write:  Gravitational force per unit mass = – (44) xi

Fundamental equations 23/78 where  is the gravitational potential. Therefore the gravitational force acting on volume V is  Fi = –  dV (45) xi V We add this term to the momentum equations to obtain:

    vi + vivj + ij dV = –  dV (46) t xj xi V V implying that:

   vi + vivj + ij = – (47) t xj xi Fundamental equations 24/78 or the alternative form:

vi vi ij   + vj = – –  (48) t xj xj xi When the momentum flux is diagonal

ij = pij (49)   p  vi + vivj = – –  t xj xi xi (50) vi vi p   + vj = – –  t xj xi xi

Fundamental equations 25/78 3.5 Hydrostatic application of the momentum equations X-ray emitting in an elliptical galaxy Elliptical galaxies have extended hot extending for 10– 100 kpc from the centre of the galaxy. Our first approach to understanding the distribution of gas in such an atmosphere is to consider an hydrostatic model.

In this case vi = 0 and the momentum equations reduce to 1p  --- = – (51) xi xi

Fundamental equations 26/78 Let us restrict ourselves to the case of spherical : 1p  GM r --- ==– –------(52) r r r2 G = Newtons constant of gravitation –11 = 6.6710 SI units (53) Mr= Mass within r This can be used to estimate the mass of the galaxy. Put kT pnkT==------(54) m

Fundamental equations 27/78 then 1 d kT GM r ------= –------dr m r2 (55) 1 kT d k dT GM r ------+ ------= –------ m dr m dr r2 Rearrangement of terms gives: r2 kT 1d 1dT Mr= –------+ ------(56) G m  dr T dr

Fundamental equations 28/78 Usually, at large radii, the of the atmosphere is iso- thermal and the density of the X-ray emitting gas may be approx- imated by a power-law: T constant r–  0.7 r d (57) ------= –  dr Therefore, r2 kT 1d 1dT kT Mr==–------+ ------r G mp  dr T dr mpG (58) 11 T r = 3.110 ------------solar masses 107 10 kpc

Fundamental equations 29/78 The amount of matter implied by such observations is larger than can be accounted for by the stellar light from elliptical galaxies and implies that elliptical galaxies, like spiral galaxies, have large amounts of dark matter.

4 Entropy In any , the , momentum and are the fundamental principles to consider. However, before proceeding with the it is necessary to make an excursion into the domain of entropy.

Fundamental equations 30/78 4.1 Entropy of a fluid Consider an element of fluid with:  = Internal (per unit volume) p = pressure  = density (59) s = entropy per unit mass m = mass of the element

Fundamental equations 31/78 m Volume = ------V  m Entropy = sm (60)  m Internal energy = ------s 

The relationship between internal energy U , pressure, volume V and entropy S of a gas is kTdS= dU+ pdV (61)

Fundamental equations 32/78 For the volume, above m m kTd sm = d------+ pd------(62)   The mass m is constant, therefore  1 kTds= d--- + pd--- (63)  

Fundamental equations 33/78 Other ways of expressing the entropy relation Expanding the differentials: 1  + p kTds = ---d – ------d  2 (64)  + p kTds= d – ------d  Specific  + p h ==------Enthalpy per unit mass (65)  = Specific enthalpy

Fundamental equations 34/78 Expression using the enthalpy 1  + p kTds = ---d – ------d  2 1  + p dp = ---d + p – ------d – ------(66)  2  dp = dh – ------ To be symmetric with the previous expression kTds = dh– dp (67)

Fundamental equations 35/78 Relation between thermodynamic variables throughout the fluid These relationships have been derived for a given fluid element. However, the of a gas can be expressed in the form pp=  s (68) so that any relationship derived between the thermodynamic var- iables is valid everywhere. Therefore the differential expressions  + p kTds== d – ------d d – hd  (69) kTds = dh– dp

Fundamental equations 36/78 are valid relationships between the differentials of sp throughout the fluid. In particular when considering the energy equation, we look at the changes in these quantities resulting from temporal or spatial changes. We often use the first form for temporal or spatial changes and the second form for spatial changes. Thus consider the change in thermodynamic variables due to temporal changes alone: The first differential form tells us that: s   + p    kT ==– ------– h (70) t t  t t t

Fundamental equations 37/78 and for spatial changes s   + p  kT = – ------(71) xi xi  xi For spatial changes we often use: s h p kT =  – (72) xi xi xi Derivation along a fluid trajectory Another use for the entropy equation is to consider the variation of the entropy along the trajectory of a fluid element. Take  + p kTds= d – ------d (73) 

Fundamental equations 38/78 then ds d  + p d d d kT----- ==----- – ------– h------(74) dt dt  dt dt dt

4.2 Adiabatic gas If the gas is adiabatic, then the change of entropy along the tra- jectory of an element of fluid is zero, i.e. ds -----= 0 (75) dt and d d ----- –0h------= (76) dt dt

Fundamental equations 39/78 This is often re-expressed in a different form using:

1d vi ------= – (77)  dt xi to give

d vi ----- = – + p (78) dt xi This equation describes the change in internal energy of the fluid resulting from expansion or compression.

Fundamental equations 40/78 4.3 Equation of state The above equations can be used to derive the equation of state for a gas, given relationships between  and p . One important case is the  -law equation of state where the pressure and inter- nal energy density are related by: 1 p =  – 1   = ------p (79)  – 1 where

cp  ==----- Constant ratio of specific heats (80) cv

Fundamental equations 41/78 We use this in conjunction with the perfect gas law kT pnkT==------m (81) kT = mp Substitute this into: ds d  + p d kT----- = ----- – ------dt dt  dt (82) ds 1 dp  pd  mp----- = ------– ------dt  – 1 dt  – 1 dt

Fundamental equations 42/78  – 1 Multiply by ------: p 1dp 1d ds ------– ------= m – 1 ----- (83) p dt  dt dt

Fundamental equations 43/78 so that d d ds -----lnp – ----- ln = m – 1 ----- dt dt dt d p ds ----- ln----- = m – 1 ----- dt  dt (84) p ln----- = m – 1 ss–  0 p ----- = expm – 1 ss–  0 We write (85)

Fundamental equations 44/78 Ks==exp m – 1 ss– 0 Pseudo-entropy

The equation of state is therefore written:

pKs=  (86) Adiabatic flow: In adiabatic flow ds -----= 0  s = constant along a streamline (87) dt and Ks is a streamline constant, but may differ from stream- line to streamline.

Fundamental equations 45/78 Special cases 5 Monatomic gas (e.g. completely ionised gas)   = --- 3 (88) 7 Diatomic gas   ==--- 1.4 5

Fundamental equations 46/78 4.4 Radiating gas In astrophysics we often have to take account of the fact that the gas radiates energy at a sufficient rate, that we have to take into account the effect on the internal energy. Let the energy radiated per unit volume per unit time per steradian be j then the internal energy equation ds d  + p d d d kT----- ==----- – ------– h------(89) dt dt  dt dt dt becomes d  + p d ----- –4------= – j (90) dt  dt

Fundamental equations 47/78 The quantity j represents the total emissivity in units of energy per unit time per unit volume per steradian. The factor of 4 re- sults from integrarting over 4 steradians. Thermal gas When we are dealing with a thermal gas, that is one in which the ions and electrons are more or less in thermal equilibrium, then the total emissivity may be expressed in the form:

4jn= enpT (91)

Fundamental equations 48/78 where

ne = electron density (92) np = proton density T = cooling function This has been the time-honoured way of expressing thermal cooling. A more modern approach is to write

4jn= 2T (93) where n is the total number density. The cooling function is calculated using complex calculations. Ralph Sutherland has published cooling functions for different conditions.

Fundamental equations 49/78 5 The energy equations 5.1 Conservation of energy in It’s a good idea to look at how we derive the expression for en- ergy in classical mechanics. Suppose we have a particle moving in a time invariant potential field,  , with its equation of motion:

dvi  m------= –m (94) dt xi

Take the scalar product of this equation with vi .

dvi  mvi------= –mvi (95) dt xi

Fundamental equations 50/78 Now

dvi 1 d d 1 v ------==------v v --------v2 (96) i dt 2dt i i dt2 Also, the differentiation following the particle motion of  d    ------==+ vi vi (97) dt t xi xi  since = 0. Hence t d 1 --------mv2 + m = 0 (98) dt2

Fundamental equations 51/78 As a consequence, the total energy, 1 E = ---mv2 + m (99) 2 is a constant of the motion, i.e. it is conserved.

5.2 Conservation of energy in gas dynamics The consideration of energy in gas dynamics follows a similar line – We start by taking the scalar product of the momentum equations with the velocity. The end result is not as simple but has some interesting and useful consequences.

Fundamental equations 52/78 We start with the momentum equations in the form

vi vi p   + vj = – –  (100) t xj xi xi and take the scalar product with the velocity:

vi vi p  vi + vjvi = – vi – vi (101) t xj xi xi

Fundamental equations 53/78 Now use:

vi  1  1 v ==---v v ---v2 it t2 i i t2 (102) vi  1  1 v ==---v v ---v2 i i i  xj xj 2 xj 2 then:  1  1 p   ---v2 + v ---v2 = – v – v (103)  j  i i t 2 xj 2 xi xi

Fundamental equations 54/78 We can now take the density and momentum inside the deriva- tives on the left hand side, using the continuity equation:  1  1 p  ---v2 + ---v2v = – v – v (104)  j i i t 2 xj 2 xi xi Bring in entropy equations We introduce the entropy equations in order to eliminate the term vip  xi . The first one to use is: s h p kT =  – xi xi xi (105) p s h –vi = kTvi – vi xi xi xi Fundamental equations 55/78 We then manipulate the enthalpy term as follows:

h   vi = hvi – h vi xi xi xi (106)   = hvi + h xi t And now use the other form of the entropy equation

s   kT = – h t t t (107)   s  h = – kT t t t

Fundamental equations 56/78 Putting all of these equations together with appropriate signs: p s h –vi = kTvi – vi xi xi xi

h   (108) –vi = – hvi – h xi xi t   s –h = – + kT t t t

Fundamental equations 57/78 Add all of these up:

p s s   –vi = kT + vi – – hvi xi t xi t xi (109) ds   = kT----- – – hvi dt t xi Substitute in the intermediate form of the energy equation:  1  1 p  ---v2 + ---v2v = – v – v (110)  j i i t 2 xj 2 xi xi

Fundamental equations 58/78 gives:

 1 2  1 2 ---v +  + ---v vj + hvj t 2 xj 2 (111)  ds = – vi + kT----- xi dt

Fundamental equations 59/78 Gravitational potential term

Tha last term to deal with is –vi  xi :    –vi = – vi +  vi xi xi xi   (112) = – vi – xi t   = – vi –  xi t

Fundamental equations 60/78 Final form of energy equation Substitute the last expression into

 1 2  1 2 ---v +  + ---v vj + hvj t 2 xj 2 (113)  ds = – vi + kT----- xi dt

 1  1 ---v2 +++ ---v2v ++hv v t 2 x 2 j j j j (114) ds = kT----- dt

Fundamental equations 61/78 and since j is a dummy repeated subscript, we can write  1  1 ---v2 +++ ---v2 ++h  v t 2 x 2 i i (115) ds = kT----- dt ds When the flow is adiabatic then kT-----= 0 and dt  1  1 ---v2 +++0---v2 ++h  v = (116) i t 2 xi 2

Fundamental equations 62/78 When there is radiation ds kT-----= – 4 j (117) dt and  1  1 ---v2 +++4---v2 ++h  v = – j (118) i t 2 xi 2

Fundamental equations 63/78 Terms in the total energy 1 ---v2 = density 2  = Internal energy density (119)  = Gravitational energy density 1 ---v2 ++= E Total energy density 2 tot 1 Note analogy with E = ---mv2 + m for a single particle. 2

Fundamental equations 64/78 The energy flux 1 Energy flux = F = ---v2 ++h  v Ei 2 i 1 ---v2v = Flux of kinetic energy 2 i (120)

hvi = Enthalpy flux

vi = Flux of gravitational potential energy An interesting point is that the flux associated with the internal energy is not vi as one might expect but the enthalpy flux hvi =  + p vi.

Fundamental equations 65/78 5.3 Integral form of the energy equation n We can integrate the energy equa- F Ei tion over volume giving:

S Etot FEi Etot + dV = 0  t xi V V and then using the divergence theo- rem  E dV+4 F n dS = – jVd (121) t tot  Ei i 

Fundamental equations 66/78 This says that the total energy within a volume changes as a re- sult of the energy flux out of that volume and the radiative losses from the volume.

5.4 Examples of energy flux Wind from O star Suppose that the wind is spherically symmetric: 1  p F = ---v2 + ------v n dS (122) E  2  – 1 i i sphere

Fundamental equations 67/78 Since all variables are constant over the surface of the sphere, then 1  p F = ---v2 + ------ v n dS E 2  – 1  i i sphere (123) 1  p = M· ---v2 + ------2  – 1

We neglect the enthalpy term in comparison to v2  2 so that

1 · 2 1 –6 3 –1 2 FE  ---Mv = ---10 solar masses / yr  10 km s 2 2 (124) 28 = 3.210 W Why do we neglect the enthalpy term? Fundamental equations 68/78 Consider 2 1  p 1 2 p 1 2 cs ---v2 + ------==---v2 1 + ------v2 1 + ------2  – 1 2  – 1 2 2  – 1 2 v v (125) 1 2 1 = ---v2 1 + ------2  – 1M2

where cs = speed M = (126) In the asymptotic zone the Mach number is high, so that 2 1 ------1« (127)  – 1M2 We shall justify these statements later.

Fundamental equations 69/78 Velocity of a jet

20 cm image of the radio gal- axy, IC4296. This image shows the jets (unresolved close to the core) and the lobes of the radio source.

Fundamental equations 70/78 Images of the jets in the radio galaxy IC4296 close to the core. The resolutions are 1" and 3.2" on the left and right respectively.

Fundamental equations 71/78 In this case we shall reverse the process to estimate the jet veloc- ity. For a radio jet such as we observe in IC4296, once the jet has widened appreciably, then the Mach number is quite low proba- bly of order unity. Hence the energy flux is dominated by the en- thalpy flux. This is the case when 2 1 2 4 ------1  M2  ------==6 for  --- (128)  – 1M2  – 1 3  F  ------pv dS z E   – 1 z r jet (129) 2 = 4pvzRjet

Fundamental equations 72/78 Contour image of the inner 150" of IC4296. The diame- ters of jets can be worked out from im- ages such as this.

Fundamental equations 73/78 Using the radio data we can estimate the following parameters at a distance 100 from the core: Diameter== 15 2.57 kpc (130) –13 Minimum pressure= 510 N m–2 We also know from an analysis from the radio emission from the lobes of this radio galaxy that 36 FE  10 W (131)

Fundamental equations 74/78 Since,

FE 1036 W v ==------z 2 –13 2.57 2 4pr 4  510  ------k p c 2 (132) 7 =2.510 m s–1 = 25 000 km s–1  0.08c This is actually an upper estimate of the velocity since we are us- ing a minimum estimate of the pressure. However, it is unlikely that the pressure is too far from the minimum value, so that this is a reasonable estimate of the velocity in the western jet of IC4296 at this distance.

Fundamental equations 75/78 6 Summary of gas dynamics equations 6.1 Mass   +0vi = (133) t xi

6.2 Momentum   p  vi + vivj = – –  t xj xi xi (134) vi vi p   + vj = – –  t xj xi xi

Fundamental equations 76/78 6.3 Internal energy

d vi ----- = – + p – 4j (135) dt xi

6.4 Total energy  1  1 ---v2 +++4---v2 ++h  v = – j (136) i t 2 xi 2

6.5 Thermal cooling 4jn= 2T  (137)

Fundamental equations 77/78 6.6 Equation of state p == – 1 Ks (138)

Fundamental equations 78/78