Hydrodynamical equations
Derivation and simple solutions
Jiří Krtička
Masaryk University
Derivation of hydrodynamical equations
Boltzmann equation
Particle distribution function F(t,       gives the number of particles in the element of the phase space dx d£ = dxi 6x2 dx$ d£i d^2 d^3 with coordinates x and momenta £ as
F(t,x, £)dxd£.
The time evolution of the particle distribution function under the influence of external force f acting on partice with mass m and taking into account particle collisions is
DF_        OF       df_ _ /dF\ dt     mdxh     hd^h     \ dt / con'
which is the Boltzmann equation. Here used the Einstein summation convention for index h.
1
Boltzmann equation
Using the Poisson bracket
dH OF    dH OF
F} =
dxh d£h    d£h dxh
the Boltzmann equation for the system that obeys the Hamilton equation can be rewritten as
9F-{H,F) <iF
dt V dt
coll
For stationary collisionless system the distribution function depends on the particle energy only,
{H, F} = 0.
2
Momentum equations
The Boltzmann equation can be solved numerically to derive the particle distribution function. However, for most of practical applications, the distribution function is very close to the Maxwelian distribution expressed at given location in the frame comoving with the fluid. In such a case, just mean quantities are of real importance for the description of the flow. These are moments of the Boltzmann equation
These can be derived by multiplying the Boltzmann equation by m and £/m and by integrating. However, the equation for n-th moment contains n + 1-th moment. Consequently, we shall close the equations somehow to avoid obtainig infinite set of equations. This is done for the equation for the 2nd moment using thermodynamical relations for pressure.
(Oth moment, flow density)
(1st moment, flow velocity).
3
The continuity equation
Multiplicating the Boltzmann equation by particle mass m and integrating over the velocity space
dF ^     f    &dF f    rdF f    fdF . A.
An — d§ + / m — -— d£ + / mth——d$> = / m   — d§ dt   S    J     mdxh        J       d^h        J \dt
-v-'       V-v-'      V-v-'        >-^
12 3 4
1 =      —    r at = m— = —-,
aty dt dt1
2=7—/ ^'      = at?-— ni//, = —-,
3 =     f ^[H^oo d^' = 0 is //, does not depend on £,
4 = 0 for conserved quantity (m),
where
n = J Fd£ is number density of particles,
p = Ann is the density,
Vh = ~k f      d£ is the mean speed.
coll
The continuity equation
This gives
+ d(pvh) dt dxh
t + v.„„,
which is the continuity equation.
The continuity equation: interpretation
Integration over volume fixed in space gives
or, using the Stokes theorem
4~ / pdV = (f pvdS, dtjv Jdv
which is the expression of the law of conservation of mass.
The continuity equation: Lagrangian picture
Introducing the Lagrangian derivative, describing the time change of any quantity q(t,x) following a moving fluid particle,
Dq(£,x) _ dq(t,x)    dq(t,x) dxh Dt     ~     dt dxh    dt '
D      d ^ = — + v • V
Dt dt
the continuity equation can be rewritten as
^ + pV • v = 0. Dt
which for incompressible fluid (p = const.) is
V • v = 0.
7
Equation of motion
Multiplicating the Boltzmann equation by £/ and integrating
dt
m dxh
1
2
3
dt
4
coll
1 =
2= —
d_
di 1 9
£iFdZ = mw(nvi) = -5r
m dxh
d
dxh d
UhFd£ = m— J (q + Vj){ch + vh)Fd£
ViVh [ Fd£ + vh [ c;Fd£ + v; [ chFd£ + / c;chFd£
d
(mnviVh + 0 + 0 + p/,/)
(pVjVh + phi)
3
4
dxh dxh Eh I U&F]^ d? -J Eh SihfhFdd = -nfi = -pgi, 0 for conserved quantity (£), where Ch = Ch/m — Vh is the thermal speed, Phi = m J CjChF d£ is the pressure tensor, phi = p5/i/, gi = />■/AT7 is force per unit of mass (acceleration).
8
Equation of motion
This gives
—^I--1-      (pviVh + pohi) = pgi
at       dxh v-v-/
which is, after differencing and using the continuity equation,
dvi dvi dp
at ax/, ox-,
where V\,k is the momentum flux density tensor, or
P~dt + PV'^V = ~VP + PS,
the momentum equation. Introducing the Lagrangian derivative the momentum equation has a form of Newton's second law
Dv
Energy equation
Multiplicating the Boltzmann equation by CiCj/m ar|d integrating
m   J dt        J m2        dxh       J       m d£h        J ™      \dt J co\\
V-v-'    V-v-'    V-v-'     >-v-_—/
12 3 4
1 d  f d  f d
1= möt J ^JFd^ = mö~t J (a + Vi)(cj + Vj)Fd£ =— (pviVj +Pij),
0, terms with h ^ i and /? 7^7 (direct integration), — finvj — fj-nv;, terms with h = / or h = j (per-partes),
0 when contraction is performed, where
Phij = J ChCjCjF d£/A77 is p/,// = 0 when neglecting viscosity.
3 =
10
Energy equation
After the contraction and multiplication by \ we derive
d n  2 3
d
dt\2pV +2P) + d^\2
-pvhv + -pvh ) - pvigi = 0,
or, introducing the specific energy pe = |p,
d pv
pe + Hr- ) + V
0t
pv   e —
+ pi/
- pvg = 0,
which is the energy equation
li
Energy equation: some manipulations
Multiplication of momentum equation by v, and summation gives
dv-, dv; dp
+ PViVh— = ~Vj— + Vjpgj, Ot OXh OX;
or
dfl2\l2dpdfl2   \     1 2d(pvh) dp
i_i
=0 (continuity equation)
Substracting this from the energy equation yields equation for the internal energy
+ V • (pev) = -pV • v, which can be rewritten using the continuity equation as
p— = -pV • v. HDt H
12
Energy equation: second law of thermodynamics
The conservation of entropy for isentropic flow requires that
d7 = 0'
which for the specific entropy of ideal gas s = cy In(pv^) + const. = cy \n(pp~") + const, is (using p = |pe)
+   v a
Derivating and multiplying by p™ we arrive at
-^ll + v • (pei/) - peV • v - xe^- - xev • Vp = 0. dt v     J dt
Eliminating the last two terms using the equation of continuity and
noting that h — 1 = | for ideal gas we derive the equation for the
internal energy once again
Q-^- + V • (pel/) = -pV • v.
13
Many faces of the beast
Collecting the nuggets: the hydrodynamical equations
dp dt
+ V • (pv) = 0
P
d_ dt
dv
dt
2
pe + ^- ) + V
+ pv Vi/ = -Vp + pg.
2
PV    € +
V
- system of nonlinear first-order partial differential equations
- unknowns p, v, p, and e (+equation of state)
- initial and boundary conditions crucial
- inviscid flow, no magnetic field
- some special analytic solutions, general solution only numerically
- stationary solutions are important (d/dt = 0, but v ^ 0)
14
The hydrodynamical equation in planar symmetry
In a planar symmetry the hydrodynamic quantities do not depend on x and y coordinates, there is no flow in x and y directions (v = v(z)z) and the hydrodynamical equations are
The hydrodynamical equations in spherical coordinates
In spherical coordinate system, the components of the velocity vector are v = (vr, vq, v^) and the components of force are g = (gy, go, g^). The equation of continuity is
dp     1 9, 2    .       1    9 , . *    \       1    ^ f \ at     rz or rs\nOoO rsmOocp
and the components of equation of motion take the form of
dvr       dvr    ve dvr dvr    ve + vl       1 dp
- \ vr--\----\- —---—-— =----\- gr,
dt        dr      r d6     rs\n9 dcj) r pdr '
dvp    ^ dvo_    vo_dvo_ dvp     vrve _ vlCQt # _    1 dp
dt      r dr      r d6     rsin 9 dcf)       r r rpdO
dvcfi | ^ dv<f, | vodVff, |   v<f,   dv<f, | vrv^ | ifr vfr cot 0 _       1 Op dt     r dr     r d6    rsmO dcf)      r r rpsmOdcj) ^
16
The hydrodynamical equations in spherical symmetry
In a spherical symmetry the hydrodynamic quantities do not depend on 6 and cj) coordinates, there is no flow in 6 and <fi directions (v = v(r)r) and the hydrodynamical equations are (v = vr)
The hydrodynamical equations in cylindrical coordinates
In cylindrical coordinate system, the components of the velocity vector are v = (vr, v^,, vz) and the components of force are g = (gR, g^, gz). The equation of continuity is
dp     1 d 1^/    \     d . .
a?+ RdR{RpVR) + + d~z{pVz) =0
and the components of equation of motion take the form of
9vr        Ovr_    v^8vr_       dvR     v% _    1 dp dt     VR OR     R dcj)      z dz     R~   pdR gR'
dvcfi        dv^_    v^dv^       dv<f,     vrv^ _     1 dp dt     VR dR     R dcj)        dz       R   " pRdcj)^g^
dvz        dvz        dvz       dvz       1 dp
~dF + VRdR +~R^ + Vz^I = ~~pd~z+gz-
18
Hydrostatic equilibrium
Static case: d/dt = 0, v = 0
In a static case the equation of continuity is fulfilled identically and the momentum equation leads to
Vp = pg
the equation of hydrostatic equilibrium. The energy equation Qracj = 0 gives the radiative equilibrium equation.
19
Atmosphere in hydrostatic equilibrium
The equation of hydrostatic equilibrium in homogeneous gravitational field directed along the z-axis is
dp dz
which, using the ideal gas equation of state p = pkT/(pm^), leads to
d(p"0 pgmH
—A-=--j—p.
dz k
In isothermal atmosphere T = const, this has the solution
p = poe"z/", H=-^—,
pmug
where H is the atmospheric scale-height. For z —>► oc is p —> 0, as it should be.
20
Atmosphere in hydrostatic equilibrium: spherical symmetry
The equation of hydrostatic equilibrium in spherically symmetric isothermal case is
There are two problems with this solution applied for gas spheres. For
r —)► 0 the equation is not applicable, because one should insert
M = M(r): Lane-Emden equation. Moreover, for r —>* oo is
P —^ Po 7^ Pism- Solution: Bonnor-Ebert spheres with external pressure.
Matter may escape from the regions, where the thermal speed is higher
than the escape speed: atmospheric escape: loss of planetary
atmospheres, solar-type (coronal) winds.
dp
which, with g = GM/r
2, has the solution
P
21
Lane-Emden equation
Consider a spherical mass in equilibrium. The hydrostaic equilibrium equation is
dp pGM(r)
dr
Pg
The polytropic relation p = Cp1+1/n with the definition of mass inside radius r, which is M(r) = Att Lr pr'2 dr', gives after differentiation
1 d
r2 dr   p dr
r2 d
P
l+l/n
4ttG
C
p.
Introducing new variables 9 and £ via p = AO" and £ = r/a, where A is arbitrary dimensional constant and
a
C(l + n)
47tGA1+1/"
we arrive at Lane-Emden equation
1 d ^2d9
e2 de r de
-en.
22
Hydrostatic atmospheres with radiative force
The equation of hydrostatic equilibrium in spherically symmetric atmosphere in radiative equlibrium is
dp
= -Pg + Pgrad,
with the radiative force g"racj = \ § KpFvdv = j^i- The temperature given by the energy transport equation
d7" 3 npL
dr        AaT3 Aircr2 This can be rewritten in terms of pracj = (a/3)7~4 as
dprad     4a7~3d7~ KpL
dr 3    dr Aivcr2
Therefore, the equation of hydrostatic equilibrium is
dptot _ d(p + prad) _ _ dr dr ^
Atmospheres close to the Eddington limit
Dividing the last two equations
dptot      dp g      1 dp 1
a-      = a-+ 1 = -       = ?'       or   a- =r-1'
dPrad       dprad gra(j       I dprad I
where the generalized Eddington factor is
r =
AttcGM '
The derivative dp/dpracj is a function of l~ only. Moreover, the the point at which the envelope solution crosses the Eddington Imimit l~ = 1 needs to me an extremum in p (Gräfener et al. 2012).
24
Envelope inflation close to the Eddington limit
6 5 logio(Prad/dyn cm2)
Logarithm of the Eddingtor factor l~ (colors) in the pracj — p plane with Iglesias & Rogers (1996) opacities. Black arrows denote slopes
dp/dprad-
The numerical solution for 23 M@ star almost precisely follows a path with T = 1 and crosses the Eddington limit at the lowest gas pressure (corresponding to the Fe-opacity peak). The gas density increases outwards leading to density inversion. This explains why WR and LBV stars have extended envelopes.
(Grafener et al. 2012)
25
Suggested reading
G. K. Batchelor: An Introduction to Fluid Dynamics
D. Mihalas & B. W. Mihalas: Foundations of Radiation Hydrodynamics
F. H. Shu: The physics of astrophysics: II. Hydrodynamics
A. Feldmeier: Theoretical Fluid Dynamics
26