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The Fundamental Equations of Gas Dynamics
1 References for astrophysical gas (fluid) dynamics 1. L. D. Landau and E. M. Lifshitz Fluid mechanics 2. F.H. Shu The Physics of Astrophysics Volume II Gas Dynamics 3. L. Mestel Stellar Magnetism

Fundamental equations

é Geoffrey V. Bicknell


2 The equation of continuity 2.1 Derivation of the fundamental equation Define n v = density v i = velocity components S V Mass within volume V is M=

(1)

V



dV

(2)

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Rate of flow of mass out of V is


S

v i n i dS

Therefore, the mass balance within V is given by: d ---- dV = í v i n i dS dt
V




S

(3)

Use the divergence theorem:


S

v i n i dS =

V



v i dV xi

(4)

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and d ---- dV = dt
V



V



dV t

(5)

to give

V



+ v i dV = 0 t xi

(6)

Since the volume V is arbitrary, then + vi = 0 t xi This is the equation of continuity.
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(7)


Aside: Differentiation following the motion Suppose we have a function f x 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 x i = x i t ? We simply calculate f f dx i d f f f x i t t = + ------ = + vi dt t x i dt t xi
(8)

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Useful result from the equation of continuity Write the above equation in the form: v i + vi + =0 t xi xi Going back to the equation of continuity, then v i v i d 1 d ----- = í = í -- -----dt xi xi dt
(10) (9)

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Putting it another way: v i 1 d -- ----- = í - dt xi
(11)

This tells us that in a diverging velocity field v i x i 0 , the density decreases and in a converging velocity field v i x i 0 the density increases.

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2.2 Examples of mass flux Wind from a massive star Massive stars such as O and B stars produce winds with velocities of around 1 000 km s í 1 and mass fluxes of around 10 í 6 solar masses per yr. We can use these facts to estimate the density in the wind as follows. Mass flux = r


S

v i n i dS

(12)

= 4r2 r v r

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where the integral is over a sphere of radius r . We assume that the flow is steady so that the mass flux integrated over any 2 surfaces surrounding the star is the same. Hence, § = 4 r 2 r v r = constant M (13) We shall show in later lectures that far from the star, the velocity is constant, i.e. v r = v = constant Therefore the density is given by: § M r = -----------------4 r 2 v
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(14)

(15)

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For typical parameters: § = 10 í 6 solar masses per yr M r = 0.1 pc = 3.09 10 m v = 1 000 km s í 1 = 10 6 m s í 1 Units 1 solar mass = 2 10
30 16 15 (16)

kg
7

1 pc = 3.09 10

m

(17)

1 year = 3.16 10 seconds The density of the wind at 0.1 pc from the star is: = 5.2 10
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í 22

kg m í 3

(18)
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This is not particularly informative by itself. We usually express the density in terms of particles cm í 3 . Assuming the gas is totally ionised, then = n H m p + n He m He + ...... Now for solar cosmic abundances, n He = 0.085 n H and m He 4 m where m is an atomic mass unit m = 1.66 10
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(19)

(20)

(21)

í 27

kg

(22)
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so that n H m + 4 0.085 n H m = 1.34 n H m Therefore, for the O-star wind: 1.34 n H m 5.2 10
í 22 í 22 (23)

kg m í 3
(24)

5.2 10 n H = ------------------------ í 3 -m 1.34 m = 2.3 10 Hydrogen atoms m í 3
5

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

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3 Conservation of momentum n Consider first the rate of change of momentum within a volume as a re v i v j sult of the flux of momentum. Let i = total momentum , then S § d i = ---- v i dV (27) dt
V

vi V



í p n i is the rate of change of momentum in the volume, V .

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The flux of momentum out of the surface S is


S

v i v j n j dS

(28)

3.1 Surface force There are two complementary ways that we can look at the other aspects of the conservation of momentum. In a perfect fluid there is a force per unit area perpendicular to the surface that exerts a force on the gas inside. This is the pessure p Force on volume = í pn i dS within S


S

(29)

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Therefore the total momentum balance for the volume is: d ---- v i dV = í v i v j n j dS í p n i dS dt
V




S


S

(30)

The surface integrals Using the divergence we can write


S

v i v j n j dS =

V



v i v j dV xj p ij dV xj

(31)

The surface integral of the pressure can be written


S

pn i dS =


S

p ij n j dS =

V



(32)

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Entire momentum equation Again, we take the time derivative inside and obtain:

V



vi + vi vj + p ij dV = 0 t xj xj vi + v i v j + p ij = 0 t xj

(33)

Since the volume is arbitrary, then
(34)

We often write this equation in the form: p vi + vi vj = í t xj xi
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(35)

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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. We write the flux per unit area of n ij n j momentum due to molecular mo v i v j tions as n ij j vi V íp ni S The flux of momentum out of the volume due to molecular motions is then:

V



ij n j dS

(36)

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For a perfect fluid, the relation between ij and p is ij = p ij
(37)

In viscous fluids (to be treated later) the tensor ij is not diagonal.

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Momentum balance When we adopt this approach, the momentum balance of the volume of gas is: d ---- v i dV = í v i v j n j dS í ij n j dS dt
V




S


S



V



vi + v i v j + ij dV = 0 t xj

(38)

The corresponding partial differential equation is: vi + v i v j + ij = 0 t xj
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(39)

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The point to remember from this analysis is that pressure in a fluid is the result of a flux of momentum resulting from the microscopic motions of the particles. Particles of mass m crossing a surface within the fluid with random atomic velocity u i (relative to the bulk velocity) contribute an amount mu i u j to the flux of momentum. A particle 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.

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3.3 Alternate forms of the momentum equations Consider, the isotropic case where ij = p ij . Then, p vi + vi vj = í t xj xi Expand the terms on the left: vi + t vi v i p + vi vj + vj = í t xj xi x j v i v i p + vj =í t xj xi
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(40)

(41)

Using the continuity equation to eliminate the blue terms:
(42)
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In terms of the derivative following the motion: dv i p ------ = í dt xi
(43)

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

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where is the gravitational potential. Therefore the gravitational force acting on volume V is F i = í dV xi
V



(45)

We add this term to the momentum equations to obtain:

V



vi + v i v j + ij dV = í dV t xj xi
V



(46)

implying that: vi + v i v j + ij = í t xj xi
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(47)
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or the alternative form: v i ij v i =í í + vj xj xj xi t When the momentum flux is diagonal ij = p ij p vi + vi vj = í í t xj xi xi v i v i p =í í + vj t xj xi xi
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(48)

(49)

(50)

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3.5 Hydrostatic application of the momentum equations X-ray emitting atmosphere in an elliptical galaxy Elliptical galaxies have extended hot atmospheres extending for 10 í 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 v i = 0 and the momentum equations reduce to 1 p -=í xi xi
(51)

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Let us restrict ourselves to the case of spherical symmetry: 1 p GM r -=í = í ---------------r r r2 G = Newtons constant of gravitation = 6.67 10 SI units M r = Mass within r This can be used to estimate the mass of the galaxy. Put kT p = n k T = --------m
(54) í 11 (53) (52)

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then 1 d kT GM r -- ---- --------- = í ---------------- dr m r2 1 kT d k dT GM r -- ------- ----- + ------- ----- = í ----------------- m dr m dr r2 Rearrangement of terms gives: r 2 kT 1 d 1 dT M r = í ---- ------- -- ----- + -- ----- --- -G m dr T dr
(56)

(55)

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Usually, at large radii, the temperature of the atmosphere is isothermal and the density of the X-ray emitting gas may be approximated by a power-law: T constant r í r d -- ----- = í - dr 0.7
(57)

Therefore, r 2 kT 1 d 1 dT kT M r = í ---- ---------- -- ----- + -- ----- = -------------- r --- -G m p dr T dr mp G
11 T = 3.1 10 ------- 10 7
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r --------------- solar masses 10 kpc

(58)

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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 dynamical system, the conservation of mass, momentum and energy are the fundamental principles to consider. However, before proceeding with the conservation of energy it is necessary to make an excursion into the domain of entropy.

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4.1 Entropy of a fluid Consider an element of fluid with: p s m = = = = = Internal energy density (per unit volume) pressure density entropy per unit mass mass of the element

(59)

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V m s

m Volume = ----- Entropy = s m m Internal energy = --------

(60)

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

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For the infinitesimal volume, above m --------- + pd m kTd s m = d ----- The mass m is constant, therefore kTds = d 1 -- -- + pd
(63) (62)

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Other ways of expressing the entropy relation Expanding the differentials: 1 + p d kTds = -- d í ---------- 2 + p d kTds = d í ---------- Specific enthalpy +p h = ----------- = Enthalpy per unit mass = Specific enthalpy
(65)

(64)

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Expression using the enthalpy 1 + p d kTds = -- d í ---------- 2 1 +p ----------- d í dp = -- d + p í ---- 2 dp = dh í ---- To be symmetric with the previous expression kTds = dh í dp
(67) (66)

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Relation between thermodynamic variables throughout the fluid These relationships have been derived for a given fluid element. However, the equation of state of a gas can be expressed in the form p = p s
(68)

so that any relationship derived between the thermodynamic variables is valid everywhere. Therefore the differential expressions + p d = d í hd kTds = d í ---------- kTds = dh í dp
(69)

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are valid relationships between the differentials of s throughout the fluid. In particular when considering the energy equation, we the changes in these quantities resulting from temporal or changes. We often use the first form for temporal or changes and the second form for spatial changes. Thus consider the change in thermodynamic variables temporal changes alone: The first differential form tells us that: s + p kT = í ---------------= íh t t t t t

p look at spatial spatial due to

(70)

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and for spatial changes s + p kT = í --------------- xi xi xi For spatial changes we often use: s h p kT = í 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 d kTds = d í ----------
Fundamental equations

(71)

(72)

(73)
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then ds d + p d d d kT ---- = ----- í ----------- ----- = ----- í h -----dt dt dt dt dt
(74)

4.2 Adiabatic gas If the gas is adiabatic, then the change of entropy along the trajectory of an element of fluid is zero, i.e. ds ---- = 0 dt and d d ----- í h ----- = 0 dt dt
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(75)

(76)

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This is often re-expressed in a different form using: v i 1 d -- ----- = í - dt xi to give v i d ----- = í + p dt xi
(78) (77)

This equation describes the change in internal energy of the fluid resulting from expansion or compression.

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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 internal energy density are related by: 1 p = í 1 = ---------- p í1 where cp = ---- = Constant r atio of specific heats cv
(80) (79)

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We use this in conjunction with the perfect gas law kT p = n k T = --------m kT = mp Substitute this into: ds d + p d kT ---- = ----- í ----------- -----dt dt dt 1 dp p d ds mp ---- = ---------- ----- í ---------- -- ------- í 1 dt í 1 dt dt
(81)

(82)

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í1 Multiply by ---------- : p 1 dp 1 d ds --- ----- í -- ----- = m í 1 ----p dt dt dt
(83)

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so that d d ds ---- ln p í ---- ln = m í 1 ---dt dt dt d p ds ---- ln ----- = m í 1 ---dt dt ln p ----- = m í 1 s í s 0 p ----- = exp m í 1 s í s 0 We write
(85)
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(84)


K s = exp m í 1 s í s 0 = Pseudo-entropy The equation of state is therefore written: p = K s Adiabatic flow: In adiabatic flow ds ---- = 0 s = constant along a streamline dt
(87) (86)

and K s is a streamline constant, but may differ from streamline to streamline.
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Special cases 5 Monatomic gas (e.g. completely ionised gas) = -3 7 Diatomic gas = -- = 1.4 5

(88)

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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 ---- = ----- í ----------- ----- = ----- í h -----dt dt dt dt dt becomes d + p d ----- í ----------- ----- = í 4 j -dt dt
(90) (89)

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The quantity j represents the total emissivity in units of energy per unit time per unit volume per steradian. The factor of 4 results 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 j = ne np T
(91)

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where n e = electron density n p = proton density T = cooling function This has been the time-honoured way of expressing thermal cooling. A more modern approach is to write 4j = n2 T
(93) (92)

where n is the total number density. The cooling function is calculated using complex atomic physics calculations. Ralph Sutherland has published cooling functions for different plasma conditions.
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5 The energy equations 5.1 Conservation of energy in classical mechanics It's a good idea to look at how we derive the expression for energy in classical mechanics. Suppose we have a particle moving in a time invariant potential field, , with its equation of motion: dv i m ------ = í m dt xi Take the scalar product of this equation with v i . dv i mv i ------ = í mv i dt xi
Fundamental equations

(94)

(95)

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Now dv i d 1d v i ------ = -- ---- v i v i = ----dt dt 2 dt 1 v 2 -2
(96)

Also, the differentiation following the particle motion of d ----- = + vi = vi dt t xi xi since = 0 . Hence t d ---dt
Fundamental equations

(97)

1 mv 2 + m = 0 -2

(98)

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As a consequence, the total energy, 12 E = -- mv + m 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 í 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.
(99)

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We start with the momentum equations in the form v i v i p =í í + vj xj xi xi t and take the scalar product with the velocity: v i v i p vi + vj vi = í vi í vi t xj xi xi
(101) (100)

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Now use: v i vi = t t 1v v = -- i i 2 t 1 v 2 -2 1 v 2 -2

v i vi = xj xj then: t

(102)

1v v = - 2 i i xj

1 v 2 + v -jx 2 j

1 v 2 = í v p í v -ix ix 2 i i

(103)

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We can now take the density and momentum inside the derivatives on the left hand side, using the continuity equation: t 1 v 2 + -2 x j 1 v 2 v = í v p í v -j ix ix 2 i i
(104)

Bring in entropy equations We introduce the entropy equations in order to eliminate the term v i p x i . The first one to use is: s h p = í kT xi xi xi p s h ívi = kTv i í vi xi xi xi
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(105)

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We then manipulate the enthalpy term as follows: h vi = hv i í h v i xi xi xi = hv i + h xi t And now use the other form of the entropy equation s kT = íh t t t s h = í kT t t t
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(106)

(107)

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Putting all of these equations together with appropriate signs: p s h = kTv i í vi ívi xi xi xi h í vi =í hv i í h xi xi t s íh = í + kT t t t
(108)

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Add all of these up: p s s = kT hv i ívi + vi íí xi t xi t xi ds = kT ---- í í hv i dt t x i Substitute in the intermediate form of the energy equation: t 1 v 2 + -2 x j 1 v 2 v = í v p í v -j ix ix 2 i i
(110)

(109)

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gives: 1 2 12 -- v + + -- v v j + hv j t 2 xj 2 ds = í vi + kT --- xi dt

(111)

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Gravitational potential term Tha last term to deal with is í v i x i : í vi =í vi + vi xi xi xi = í vi í t xi =í v i í xi t
(112)

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Final form of energy equation Substitute the last expression into 1 2 12 -- v + + -- v v j + hv j t 2 xj 2 ds = í vi + kT --- xi dt 1 2 12 -- v + + + -- v v j + hv j + v j t 2 xj 2 ds = kT ---dt
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(113)

(114)

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and since j is a dummy repeated subscript, we can write 1 2 -- v + + + t 2 xi 1 v 2 + h + v -2 i

ds = kT ---dt ds When the flow is adiabatic then kT ---- = 0 and dt 1 2 -- v + + + t 2 xi 1 v 2 + h + v = 0 -2 i

(115)

(116)

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When there is radiation ds kT ---- = í 4 j dt and 1 2 -- v + + + t 2 xi 1 v 2 + h + v = í 4 j (118) -2 i
(117)

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Terms in the total energy 12 -- v = Kinetic energy density 2 = Internal energy density = Gravitational energy density 12 -- v + + = E tot Total energy density 2 12 Note analogy with E = -- mv + m for a single particle. 2

(119)

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The energy flux 1 v 2 + h + v Energy flux = F Ei = -2 i 12 -- v v i = Flux of kinetic energy 2 hv i = Enthalpy flux
(120)

v i = Flux of gravitational potential energy An interesting point is that the flux associated with the internal energy is not v i as one might expect but the enthalpy flux hv i = + p v i .
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5.3 Integral form of the energy equation n We can integrate the energy equaF E i tion over volume giving: E tot V S

V



E tot F E i + dV = 0 t xi

and then using the divergence theorem E tot dV + F E i n i dS = í 4 j dV t







(121)

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This says that the total energy within a volume changes as a result 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: FE = 1 v 2 + ----------2 í1 p -- v i n i dS

sphere



(122)

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Since all variables are constant over the surface of the sphere, then 1 v 2 + ---------F E = -2 í1 p --

sphere



v i n i dS
(123)

p § 1 v 2 + ---------- -- = M --2 í 1 We neglect the enthalpy term in comparison to v 2 2 so that 1§ 2 1 F E -- M v = -- 10 í 6 solar masses / yr 10 3 km s í 1 2 2 2 (124) = 3.2 10
Fundamental equations

28

W
68/78

Why do we neglect the enthalpy term?


Consider 2 12 p 12 -- v + ---------- -- = -- v 1 + ----------2 í 1 2 í1 2 12 = -- v 1 + ---------í1 2 where c s = sound speed 2 ---------í1 1 ------- ë 1 M2
2 cs p 12 2 -------- = -- v 1 + ---------- ----2 í 1 v2 v2 (125) 1 ------M2

M = Mach number (126)

In the asymptotic zone the Mach number is high, so that
(127)

We shall justify these statements later.
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Velocity of a jet

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

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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.
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In this case we shall reverse the process to estimate the jet ity. For a radio jet such as we observe in IC4296, once the widened appreciably, then the Mach number is quite low bly of order unity. Hence the energy flux is dominated by thalpy flux. This is the case when 2 ---------í1 1 2 2 ---------- = 6 for = 4 -------- 1 M 3 í1 M2 r z FE

velocjet has probathe en-

(128)

jet



---------- pv z dS í1

(129)

2 = 4 pv z R jet
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Contour image of the inner 150" of IC4296. The diameters of jets can be worked out from images such as this.

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Using the radio data we can estimate the following parameters at a distance 100 from the core: Diameter = 15 = 2.57 kpc Minimum pressure = 5 10
í 13

N m í2

(130)

We also know from an analysis from the radio emission from the lobes of this radio galaxy that F E 10 36 W
(131)

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Since, FE 10 36 W v z = -------------- = --------------------------------------------------------------------2 í 13 2.57 4 pr 2 4 5 10 --------- kpc 2 = 2.5 10 m s í 1 = 25 000 km s í 1 0.08 c This is actually an upper estimate of the velocity since we are using 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.
7

(132)

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6 Summary of gas dynamics equations 6.1 Mass + vi = 0 t xi 6.2 Momentum p vi + vi vj = í í t xj xi xi v i v i p + vj =í í t xj xi xi
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(133)

(134)

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6.3 Internal energy v i d í 4j ----- = í + p dt xi 6.4 Total energy 1 2 -- v + + + t 2 xi 6.5 Thermal cooling 4j = n2 T
(137) (135)

1 v 2 + h + v = í 4 j (136) -2 i

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6.6 Equation of state p = í 1 = K s
(138)

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