From Newtons law to hydrodynamic equations 18.354 - L14 Goal: - - PowerPoint PPT Presentation

From Newton’s law to hydrodynamic equations

18.354 - L14

∂ρ ∂t + r · (ρu) = 0.

D Dt = ∂ ∂t + (u · r)

Du Dt = rp ρ + g.

Goal: derive

13.2 From Newton’s laws to hydrodynamic equations

a many-particle system governed dxi dt = vi , mdv dt = F i,

• F (x1, . . . , xn) = G(xi) +

X

j6=i

H(xi xj) = rxiΦ(x1, . . . , xn)

s H(r) = H(r)

X

6

We define the fine-grained phase-space density f(t, x, v) =

N

X

i=1

δ(x xi(t))δ(v vi(t))

X where δ(x xi) = δ(x xi)δ(y yi)δ(z zi)

X

6

We define the fine-grained phase-space density f(t, x, v) =

N

X

i=1

δ(x xi(t))δ(v vi(t))

∂ ∂tf =

N

X

i=1

d dt [δ(x xi)δ(v vi)] =

N

X

i

{δ(v vi)rxiδ(x xi) · ˙ xi + δ(x xi)rviδ(v vi) · ˙ vi} = rx

N

X

i=1

δ(v vi)δ(x xi) · vi rv

N

X

i=1

δ(x xi)δ(v vi) · F i m where, in the last step, we inserted Newton’s equations and used that ∂ ∂xi δ(x xi) = ∂ ∂xδ(x xi)

X Writing r = rx and inserting (309) for the forces, we may rewrite m ✓ ∂ ∂t + v · r ◆ f = rv

N

X

i=1

δ(x xi)δ(v vi) · 2 4G(xi) + X

j6=i

H(xi xj) 3 5 = rv

N

X

i=1

δ(x xi)δ(v vi) · 2 4G(x) + X

xj6=x

H(x xj) 3 5 =

• 2

4G(x) + X

xj6=x

H(x xj) 3 5 · rvf (

∂ ∂tf = v · rx

N

X

i=1

δ(v vi)δ(x xi) rv

N

X

i=1

δ(x xi)δ(v vi) · F i m = v · rxf 1 mrv

N

X

i=1

δ(x xi)δ(v vi) · F i. (313)

To obtain the hydrodynamic equations from (314), two additional reductions will be necessary:

• We need to replace the fine-grained density f(t, x, v), which still depends implicitly
• n the (unknown) solutions xj(t), by a coarse-grained density hf(t, x, v)i.
• We have to construct the relevant field variables, the mass density ρ(t, r) and velocity

field u, from the coarse-grained density ¯ f.

r r m ✓ ∂ ∂t + v · r ◆ f X 4 X

6

5 =

• 2

4G(x) + X

xj6=x

H(x xj) 3 5 · rvf (314)

hf(t, x, v)i = Z dP(Γ0) f(t, x, v). (315)

{

1 N

} s {x1(0), . . . , xN(0); v1(0), . . . , vN(0)} =: Γ0.

e {x1(t), . . . , xN(t)}

m ✓ @ @t + v · r ◆ hfi = rv · [G(x)hfi + C] (316) where the collision-term C(t, x, v) := X

xj6=x

hH(x xj)f(t, x, v)i (317)

We now define the mass density ⇢, the velocity field u, and the specific kinetic energy tensor Σ by ⇢(t, x) = m Z d3v hf(t, x, v)i, (318a) ⇢(t, x) u(t, x) = m Z d3v hf(t, x, v)i v. (318b) ⇢(t, x) Σ(t, x) = m Z d3v hf(t, x, v)i vv. (318c)

Z h i The tensor Σ is, by construction, symmetric as can be seen from the definition of its individual components ⇢(t, x) Σij(t, x) = m Z d3v hf(t, x, v)i vivj, Z and the trace of Σ defines the local kinetic energy density ✏(t, x) := 1 2Tr(⇢Σ) = m 2 Z d3v hf(t, x, v)i |v|2. (319)

We now define the mass density ⇢, the velocity field u, and the specific kinetic energy tensor Σ by ⇢(t, x) = m Z d3v hf(t, x, v)i, (318a) ⇢(t, x) u(t, x) = m Z d3v hf(t, x, v)i v. (318b) ⇢(t, x) Σ(t, x) = m Z d3v hf(t, x, v)i vv. (318c)

m ✓ @ @t + v · r ◆ hfi = rv · [G(x)hfi + C] (316)

Z Integrating Eq. (316) over v, we get @ @t⇢ + r · (⇢u) =

• Z

dv3 rv · [G(x)hfi + C] , (320) but the rhs. can be transformed into a surface integral (in velocity space) that vanishes since for physically reasonable interactions [G(x)hfi + C] ! 0 as |v| ! 1. We thus recover the mass conservation equation @ @t⇢ + r · (⇢u) = 0. (321)

Mass conservation

To obtain the momentum conservation law, lets multiply (316) by v and subsequently integrate over v, Z dv3 m ✓ @ @t + v · r ◆ hfiv =

• Z

dv3 vrv · [G(x)hfi + C] . (322)

Momentum conservation

m ✓ @ @t + v · r ◆ hfi = rv · [G(x)hfi + C] (316)

The lhs. can be rewritten as Z dv3 m ✓ ∂ ∂t + v · r ◆ hfiv = ∂ ∂t(ρu) + r · Z dv3 mhfivv = ∂ ∂t(ρu) + r · (ρΣ) = ∂ ∂t(ρu) + r · (ρuu) + r · [ρ(Σ uu)] = ρ ∂ ∂tu + u ∂ ∂tρ + ur · (ρu) + ρu · ru + r · [ρ(Σ uu)]

(321)

= ρ ✓ ∂ ∂t + u · r ◆ u + r · [ρ(Σ uu)] (323)

Z dv3 m ✓ @ @t + v · r ◆ hfiv =

• Z

dv3 vrv · [G(x)hfi + C] . (322)

✓ ◆ The rhs. of (322) can be computed by partial integration, yielding

• Z

dv3 vrv · [G(x)hfi + C] = Z dv3 · [G(x)hfi + C] = ρg + c(t, x), (324) where g(x) := G(x)/m is the force per unit mass (acceleration) and the last term c(t, x) = Z dv3C = Z dv3 X

xj6=x

hH(x xj)f(t, x, v)i (325)

X

6

encodes the mean pair interactions. Combining (323) and (324), we find ρ ✓ ∂ ∂t + u · r ◆ u = r · [ρ(Σ uu)] + ρg(x) + c(t, x). (326) The symmetric tensor Π := Σ uu (327)

• measures the covariance of the local velocity fluctuations of the molecules w

related to their temperature. To estimate c, let us assume that the pair interaction force H can be derived from a pair potential ϕ, which means that H(r) = rrϕ(r). Assuming further that H(0) = 0, we may write c(t, x) = Z dv3 X

xj(t)

h[rxϕ(x xj)]f(t, x, v)i (328) X Replacing for some function ζ(x) the sum over all particles by the integral X

xj

ζ(xj) ' 1 m Z d3y ρ(t, y) ζ(y) (329)

X we have c(t, x) ' 1 m Z dv3 Z d3y ρ(t, y) h[rxϕ(x y)]f(t, x, v)i = 1 m Z dv3 Z d3y ρ(t, y) h[ryϕ(x y)]f(t, x, v)i = 1 m Z dv3 Z d3y [rρ(t, y)] hϕ(x y)f(t, x, v)i (330) X Replacing for some function ζ(x) the sum over all particles by the integral X

xj

ζ(xj) ' 1 m Z d3y ρ(t, y) ζ(y) (329)

In general, it is impossible to simplify this further without some explicit assumptions about initial distribution P that determines the average h · i. There is however one exception, namely, the case when interactions are very short-range so that we can approximate the potential by a delta-function, ϕ(r) = φ0a3δ(r), (331)

c(t, x) '

• Z

Z 1 m Z dv3 Z d3y [rρ(t, y)] hϕ(x y)f(t, x, v)i

where ϕ0 is the interaction energy and a3 the effective particle volume. In this case, c(t, x) = ϕ0a3 m Z dv3 Z d3y [rρ(t, y)] hδ(x y)f(t, x, v)i = ϕ0a3 m [rρ(t, x)] Z dv3hf(t, x, v)i = ϕ0a3 m2 [rρ(t, x)]ρ(t, x) = ϕ0a3 2m2 rρ(t, x)2 (332)

• r

Inserting this into (326), we have thus derived the following hydrodynamic equations ∂ ∂tρ + r · (ρu) = (333a) ρ ✓ ∂ ∂t + u · r ◆ u = r · Ξ + ρg(x), (333b) where Ξ :=  ρ(Σ uu) + ϕ0a3 2m2 ρ2I

• (333c)

is the stress tensor with I denoting unit matrix. a commonly adopted closure condition is the ideal isotropic gas approximation Σ uu = kT m I, (334) where T is the temperature and k the Boltzmann constant. For this closure condition,

• Eqs. (333a) and (333b) become to a closed system for ρ and u.

Closure problem

Traditionally, and in most practical applications, one does not bother with microscopic derivations of Ξ; instead one merely postulates that Ξ = pI + µ(r>u + ru>) 2µ 3 (r · u), (335) where p(t, x) is the pressure field and µ the dynamic viscosity, which can be a function

• f pressure, temperature etc. depending on the fluid. Equations (333a) and (333b) com-

bined with the empirical ansatz (335) are the famous Navier-Stokes equations. The second summand in Eq. (335) contains the rate-of-strain tensor E = 1 2(r>u + ru>) (336) and (r · u) is the rate-of-expansion of the flow.

Ξ :=  ρ(Σ uu) + ϕ0a3 2m2 ρ2I

• Σ uu = kT

m I,

r · For incompressible flow, defined by ρ = const., the Navier-Stokes equations simplify to r · u = (337a) ρ ✓ ∂ ∂t + u · r ◆ u = rp + µr2u + ρg. (337b) In this case, one has to solve for (p, u).

14 The Navier-Stokes Equations

ρDui Dt = ∂p ∂xi + 2µ X ∂ ∂xj 1 2 ✓∂uj ∂xi + ∂ui ∂xj ◆ = rip + µri(r · u) + µr2ui.

When the fluid density doesn’t change very much we have seen that r · u = 0, and under these conditions the Navier-Stokes equations for fluid motion are ρDu Dt = rp + µr2u. (347)

An important parameter that indicates the relative importance of viscous and inertial forces in a given situation is the Reynolds number. Suppose we are looking at a problem where the characteristic velocity scale is U0, and the characteristic length scale for variation

• f the velocity is L. Then the size of the terms in the equation are

∂u ∂t ⇠ U 2 L , u · ru ⇠ U 2 L , µr2u ⇠ µU0 L2 . (349)

The ratio of the inertial terms to the viscous term is ρU 2

0 /L

µU0/L2 = ρU0L µ = Re, (350)

14.2 The Reynolds number

For an incompressible flow, we have established that the equations of motion are ρ∂u ∂t + ρu · ru = rp + µr2u + fext, (348)