Variational formulation of an ideal uid and uid … · Variational formulation of an ideal uid and uid Maxwell equations Tsutomu KAMBE Tokyo,1 Japan International Workshop on Mathematical

Variational formulation of an ideal fluidand fluid Maxwell equations

Tsutomu KAMBE

Tokyo,1 Japan

International Workshop on Mathematical Fluid Dynamics

Waseda University 早稲田大学

March 10, 2010

[email protected] http://www.purple.dti.ne.jp/kambe/

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Outline

Part I. Variational formulation

There are two ways in description of fluid flows:Lagrangian description and Eulerian description.

• Variational formualtions and Lagrangian functionals are recon-sidered from the viewpoint of guage theory(expressed in a form independent of any particular coordinate system).

• Transformation between the two spaces are examined, and itsuniqueness is considered.

• Importance of vorticity equation is emphasized in this regard.

Part II. Fluid Maxwell Equations

There is analogy between the equations ofFluid mechanics and Electromagnetism.

• Defining two vector fields analogous to the electric and magneticfields, fluid Maxwell equations are proposed.

• Vorticity corresponds to the magnetic field.

• Equation of sound wave is derived. Hence, the sound wave isanalogous to the Electromagnetic wave.

• Forces acting on a test particle moving in a flow field are expressedwith the same form as of the electromagnetism.

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Part I. Variational formulation

1. Definitions

(a) Two descriptions

• Lagrangian particle coordinates:

a = (a1, a2, a3) = (a, b, c), time : τ = t = a0,

aµ = (τ, a1, a2, a3), µ = 0, 1, 2, 3

d3a = da db dc

• Eulerian physical space coordinates:

x = (x1, x2, x3) = (x, y, z), time : t

volume element : d3x = dx dy dz

Each particle is identified by a = (a, b, c). Its position is expressed with

X = X(τ,a) = (X1, X2, X3) = (X, Y, Z), a = X(0,a)

Mass density ρ：dm = d3a = ρ(t,x) d3x,

From dm = da db dc = ρ dX dY dZ,

ρ =1

J, J =

∂(Xk)

∂(al)=∂(X1, X2, X3)

∂(a1, a2, a3)=∂(X,Y, Z)

∂(a, b, c)(Jacobian)

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(b) Ideal Fluid

The mass dm of a fluid particle is invariant during its motion:

∂τ (dm) = 0 ,

∂τ ≡ ∂/∂τ, for fixed a.

There is no dissipation of kinetic energy. Therefore,

⇒ no heat generation, ⇒ no entropy change:

∂τ s = 0 , (Definition of an ideal fluid).

⇒ s = s(a) = s(a, b, c).

(c) Thermodynamic relations: Setting as δs = 0 (vanishing variation),

δϵ = (δϵ)s =p

ρ2δρ, δh = (δh)s =

1

ρδp,

( · )s denotes variation with s fixed.

p : pressure

ϵ, h (= ϵ+ p/ρ) : internal energy, enthalpy, (per unit mass).

For small changes of δρ (density) and δs (entropy):

δϵ = (p/ρ2)δρ+ Tδs , δh = (1/ρ)δp+ Tδs.

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2. Lagrangian description

(a) Lagrangian functional L is defined by

L =

∫Ma

Λ(Xkµ, X

k) d3a

Λ(Xkµ, X

k) = 12 X

kτ X

kτ − ϵ(ρ, s) , (1)

(kinetic energy) − (internal energy)

Xkτ : k-component velocity, ϵ : internal energy.

Action : I =

∫ τ2

τ1

L dτ =

∫ τ2

τ1

dτ

∫Ma

Λ(Xkµ, X

k) d3a, (2)

Lagrangian density :

Λ = Λ(Xkµ, X

k) = 12 X

k0 X

k0 − ϵ(Xk

l , Xk), (3)

Xkµ = ∂Xk/∂aµ.

Variation :

δΛ =∂Λ

∂XkδXk +

∂Λ

∂Xkµ

δXkµ

=[ ∂Λ∂Xk

− ∂

∂aµ

( ∂Λ

∂Xkµ

)]δXk +

∂

∂aµ

( ∂Λ

∂Xkµ

δXk)

δI =

∫ τ2

τ1

∫Ma

δΛ d4a = 0, d4a = dτ d3a.

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(b) Euler-Lagrange equation

∂

∂aµ

( ∂Λ

∂Xkµ

)− ∂Λ

∂Xk= 0, k = 1, 2, 3; µ = 0, · · · , 3. (4)

Noether’s theorem：

∂

∂aνT νµ = 0

T νµ ≡ Xk

µ

( ∂Λ

∂Xkν

)− Λ δνµ (Energy-momentum tensor)

∂

∂aνT νµ = Xk

µ

[∂ν

( ∂Λ

∂Xkν

)− ∂Λ

∂Xk

]= 0 .

Eckart (1960), Kambe (2008)

(i) Energy equation ( µ = 0):

∂τH + ∂a

[p∂(X, Y, Z)

∂(τ, b, c)

]+ ∂b

[p∂(X,Y, Z)

∂(a, τ, c)

]+ ∂c

[p∂(X,Y, Z)

∂(a, b, τ)

]= 0. (5)

where H = 12 v

2 + ϵ.

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(ii) Equation of motion (µ = 0):

∂τVα + ∂α F = 0, (for α = a, b, c), (6)

Vα ≡ XαXτ + YαYτ + ZαZτ , F = −12 v

2 + h.

(Xτ , Yτ , Zτ) : x-space velocity

Vα : Velocity transformed to the a-space.

From (6), the equation for the acceleration Aα is

Aα(≡ XαXττ + YαYττ + ZαZττ) = −∂αh, (α = a, b, c) (7)

This is the Lagrange’s form of the equation of motion.

Multiplying ∂α/∂x and summing, we obtain

Xττ = −1

ρ∂x p, ∂xp =

∂α

∂x

∂p

∂α. (8)

Euler’s equation of motion.

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3. Transformation between (a, b, c) and (X, Y, Z)

• a-space acceleration,

Aα = XαXττ + YαYττ + ZαZττ = eα · A,

Middle side is a form of an inner product of A and eα (α = a, b, c):

x-space acceleration : A ≡ (Xττ , Yττ , Zττ ),

Direction vector of the α-axis : eα ≡ (Xα, Yα, Zα).

The inner product is invariant with respect to rotation ofthe frame axes (x, y, z) in the x-space.

• a-space velocity:

Vα = XαXτ + YαYτ + ZαZτ = eα · V,

is also an inner product, where two vectors are eα and

x-space velocity : V ≡ (Xτ , Yτ , Zτ ).

• Transformation between (∆a,∆b,∆c) and (∆X,∆Y,∆Z) is given by

∆X = Xa ∆a+Xb∆b+Xc∆c,

∆Y = Ya ∆a+ Yb∆b+ Yc ∆c,

∆Z = Za ∆a+ Zb∆b+ Zc∆c.

There are nine undetermined coefficients: Xa, Xb, Xc, · · · , etc. .But we have six relations of Aα and Vα, not sufficient .

Additional transformation of vorticity must be considered,in order to fix this freedom for uniqueness. Kambe (2008).

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4. Improved Lagrangians

We have three invariances:

∂τ (d3a) = 0, ∂τs = 0, ∂τ Ωa = 0.

Last one ∂τ Ωa = 0 is obtained from invariance of L with respect to particleexchange, where

Ωa = ∇a × V a : vorticity in the a-space,

V a = (Va, Vb, Vc) : velocity in the a-space.

We can define the following improved Lagrangian (Kambe 2008):

L∗ = L − ∂τ

∫ϕ d3a− ∂τ

∫s ψ d3a− ∂τ

∫M

⟨Aa, Ωa ⟩ d3a

= L −∫M

∂τϕ d3a−

∫M

s ∂τψ d3a−∫M

⟨∂τAa, Ωa⟩ d3a.

The action integral is unchanged by added Lagrangians:

I =

∫ τ2

τ1

L∗ dτ =

∫L dτ.

• The Euler-Lagrange equation is unchanged.• However, these play non-trivial role in the Eulerian space.• Particle permutation causes no change,

since material particles are assumed to be uniform.　• Hence, there is arbitrariness in the definition of (a, b, c).• The last term

∫⟨∂τAa,Ωa⟩d3a is analogous to the Chern-Simons

term in the Gauge Theory (where ∂τAa is assumed to be afunction of a only).

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5. Eulerian description

(a) Independent variables are (t, x, y, z).

Time derivative ∂τ is replaced by Dt ≡ ∂t + vk∂k = ∂t + v · ∇ ,

Mass element d3a is replaced by ρ d3x,

Velocity ∂τX is replaced by v = (u, v, w) = Dtx.

(b) Lagrangian in the Eulerian space

L∗ = L −∫M

Dtϕ ρ d3x−

∫M

sDtψ ρ d3x− ∂τ

∫M

⟨Aa, Ωa ⟩ d3a.

The last integral can be given another equivalent forms:

−∂τ∫M

⟨Aa, Ωa ⟩ d3a = −∫M

⟨L∂τA, ω∗ ⟩ d3x =

∫M

⟨A, L∂τω∗ ⟩ d3x+ IntS

The derivative ∂τ is equivalent to the Lie derivative L,which is represented as follows:

Scalar Φ : ∂τΦ = L∂τΦ = DtΦ = ∂tΦ + v · ∇Φ,

Dtak = 0 ,

Cotangent vectorAi : ∂τAk = L∂τAk = ∂tAk + (v · ∇)Ak + Ai∂kvi

⇒ ∂tA+∇(v ·A)− v × (∇×A) ,

Axial vector ω∗ : ∂τω∗ = L∂τω

∗ = ∂tω∗ +∇× (ω∗ × v) + v(∇ · ω∗).

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The action integral is represented as

I =

∫ τ2

τ1

L∗ dτ =

∫Λ(v, ρ, s, ϕ, ψ,A) d4x, d4x = dt d3x.

where the Lagrangian density Λ (excluding surface terms) is defined by

Λ[v, ρ, s, ϕ, ψ,A] ≡ 12 ρ ⟨v,v ⟩ − ρ ϵ(ρ, s)− ρDtϕ− ρsDtψ + ⟨A, L∂τω ⟩.

The action principle is δ I =∫δΛ(v, ρ, s, ϕ, ψ,A) d4x = 0.

In order to derive the Euler-Lagrange equation, we consider infinitesimal trans-formation:

x → x′ = x+ ξ(x, t),

d3x → d3x′ = (1 + ∂kξk)d3x,

Variations :

∆(d3x) = ∂kξk d3x,

∆ρ = −ρ ∂kξk, ∆v = Dtξ, ∆s = 0

∆I =

∫d4x

[ ∂Λ∂v

∆v +∂Λ

∂ρ∆ρ+

∂Λ

∂s∆s+ Λ ∂kξ

k].

Substituting the expressions for ∆ρ, ∆v, ∆s and ∆(d3x),and requiring ∆I vanish for arbitrary variation ξk, we obtain theEuler-Lagrange equation:

∂

∂t

( ∂Λ∂vk

)+

∂

∂xl(vl∂Λ

∂vk)+

∂

∂xk(Λ− ρ

∂Λ

∂ρ

)= 0. (9)

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The Euler-Lagrange equation reduces to the momentum conservation:

∂t(ρv

)+∇ · ρvv +∇ p = 0. (10)

This becomes the Euler’s equation of motion by (12):

∂tv + (v · ∇)v = −1

ρ∇ p. (11)

From the invariance of I with respect to the variations of ϕ, ψ and A,

∆ϕ : ∂tρ+∇ · (ρv) = 0 (Continuity eq.), (12)

∆ψ : ∂t(ρs) +∇ · (ρsv) = 0 (Entropy eq.),

∆A : ∂tω +∇× (ω × v) = 0 (Vorticity eq.).

The energy equation (5) can be transformed to the following equa-tion of energy conservation:

∂t[ρ(12 v

2 + ϵ)]+ ∂k

[ρvk (12 v

2 + h)]= 0. (13)

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Transformation between a and x(a)

• Transformation from the a space to the x space is determinedby nine components of the matrix ∂xk/∂al.

• Remaining three conditions are given by the following two-formequation Ω2 connecting ω in the x-space andΩa in the a-space:

Ω2 = Ωa db ∧ dc+ Ωb dc ∧ da+ Ωc da ∧ db

= ωx dy ∧ dz + ωy dz ∧ dx+ ωz dx ∧ dy,

where Ωa = ∇a × V a = (Ωa,Ωb,Ωc) is the vorticity in the a-space, andω = ∇× v = (ωx, ωy, ωz) is the vorticity in the x-space.For example, Ωa is given by

Ωa = ωx (∂by ∂cz− ∂cy ∂bz) + ωy (∂bz ∂cx− ∂cz ∂bx) + ωz (∂bx ∂cy− ∂cx ∂by),

with Ωb and Ωc being determined cyclically.

• Transformation relations of the three vectors v, A and ω sufficeto determine the nine matrix elements ∂xk/∂al locally.

* Thus, the transformation between the Lagrangian space andEulerian space is determined uniquely.

* In this sense, the equation of vorticity is essential for the unique-ness of the transformation between both spaces.

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Part II. Fluid Maxwell Equations

1. Analogy of equations：Fluid mechanics and Electromagnetism

2. Fluid Maxwell equations

3. Equation of sound wave

4. Equation of motion of a test particle

Colioros force ⇔ Lorentz force.

5. Uniqueness

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1. Analogy of equations of both systems

(a) Fluid mechanics

∂tv + (v · ∇)v +∇h = 0, (14)

∂th+ v · ∇h+ a2∇ · v = 0, (15)

∂tω +∇× (ω × v) = 0 (Vorticity equation). (16)

Euler’s equation of motion : ∂tv + (v · ∇)v = −1

ρ∇ p = −∇h.

Continuity equation : ∂tρ+ (v · ∇)ρ+ ρ∇ · v = 0 .

When the fluid entropy s is uniform, the enthalpy h becomes h = h(ρ):

1

ρ∇p = ∇h, since dh =

1

ρdp+ Tds =

1

ρdp(ρ), (ds = 0),

dρ = (ρ/a2) dh, since dp =(∂p/∂ρ

)s=const

dρ = a2 dρ,

a =√(∂p/∂ρ)s =

√γp/ρ : sound speed,

∂tρ = (ρ/a2)∂th, ∇ρ = (ρ/a2)∇h.

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Linearization:(Continuity eq.) (Eq.of motion)

∂th+ a20 divv = 0, ∂tv +∇h = 0,

(∂ 2t − a2

0∇2) (h, v) = 0, ∂th+ a0 div (a0v) = 0.

(b) Electromagnetism

∇×Eem + ∂tHem = 0, ∇ ·Hem = 0,

∇×Hem − ∂τEem = J , ∇ ·Eem = q.

Eem: electric field,　 Hem: magnetic field, τ = c t (c: light speed),

The equations in vacuum are satisfied by Eem and Hem definedin terms of a vector potential A and a scalar potential ϕ(e):

Eem = −∂τA−∇ϕ(e), Hem = ∇×A,

(∂ 2t − c2∇2) (ϕ(e), A) = 0, ∂tϕ

(e) + c divA = 0.

There is analogy in the wave property.

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2. Fluid Maxwell Equations

Definition: E = −∂tv −∇h, H = ∇× v, (17)

(A) ∇ ·H = 0, (B) ∇×E + ∂tH = 0,

(C) ∇ ·E = q, (D) a 20 ∇×H − ∂tE = J .

q = ∇ · [(v · ∇)v], J = ∂2t v +∇∂th+ a 20∇× (∇× v).

From ∂t(C) + div(D), we have the charge conservation:

∂tq + divJ = 0.

From the equation (14), we obtain

E (≡ −∂tv −∇h) = (v · ∇)v = ω × v +∇(12 v2). (18)

Therefore, the charge density q is given by q = ∇ ·E = div[(v · ∇)v

].

Equation of motion is

∂tv + ω × v +∇(12 v2 + h) = 0. (14)

(A) is deduced from the definition of H = ∇× v.

(B) is just curl [Eq.(14)]. (⇒ Vorticity equation).

(C) is just div [Eq.(18)].

(D) is what ?

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• Eq.(D) can be derived from the continuity equation:

∂th+ (v · ∇)h+ a2∇ · v = 0.

Applying ∂t to E = −∂tv −∇h, and substituting the above,

−∂tE − ∂2t v = ∇∂th = −∇(a2∇ · v + (v · ∇)h

).

This can be rewritten in the form of Eq.(D):

a 20 ∇×H − ∂tE = J ,

where J = ∂2t v + a20∇× (∇× v)−∇(a2∇ · v

)−∇

((v · ∇)h

).

Using the identity,

∇(∇ · v) = ∇× (∇× v) +∇2v,

the vector J can be rewritten as

J = (∂2t −a2∇2)v+(a20−a2)∇× (∇×v)− (∇·v)∇a2−∇(v ·∇h).

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3. Equation of Sound Wave

Sound wave is analogous to the Electromagnetic wave

Assume that a flow is generated in a uniform state at rest,where the pressure is p0, density ρ0 and enthalpy h0.

Differentiating Eq.(D) with respect to t, and eliminating ∂tH by (B), we obtain∂ 2t E + a 2

0∇× (∇×E) = −∂tJ . This reduces to

grad[(a−2

0 ∂ 2t −∇2)h − ∇ ·E − ∂tQ

]= 0, (19)

∇ ·E = div(ω × v) +∇2 12 v

2,

Q = (1− (a/a0)2)∇ · v − a−2

0 (v · ∇)h.

Integrating (19), we obtain the following wave equation:

(a−20 ∂ 2

t −∇2) h = S(x, t), S ≡ ∇ ·E + ∂tQ,

where S(x, t) is a source term of the wave.

• Since ∇ ·E = div(ω × v) +∇2 12 v

2, motion of ω can generate waves.

⇒ Vortex sound.　　• The second term ∂tQ is O(M2).

⇒ Namely, higher order if Mach number M = |v|/a0 is small.

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4. Equation of motion of a Test Particle in a flow field

This is another example of analogy.

Suppose that a test particle of mass m is placed in a flow field v(x, t),which is unsteady, rotational and compressible.

• The particle is assumed to be sufficiently small, so that its influence is regardedas perturbation and the background velocity field is v(x, t) is independentof the position and velocity of the particle.

• Equation of motion of the particle can be written in the form:

d

dtP = mE +mu×H −m∇ϕg, P = (Pi). (E)

Pi = mui +mikuk, E = −∂tv −∇h, H = ∇× v.

• Pi : Total additional momentum due to the particle existence.• u = (ui): Particle velocity relative to the fluid velocity v.• mik : Induced mass tensor of the particle.• 1

2 mikuiuk: Induced kinetic energy due to the particle.

Particle position is expressed by xp(t) = ξ(t) +X(t).Total particle velocity is u+ v, where

u(t) = dξ/dt, v(t,xp(t)) = dX/dt.

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• The equation (E) can be derived from the following Lagrangian：

L(t, ξ,u) = 12 m (u+ v)2 + 1

2 mjkujuk −mϕ, (20)

where mϕ is a force potential. Lagrange’s equation is

d

dt

(∂L∂ξi

)=∂L

∂ξi, where

d

dt= ∂t + u · ∇.

From this, we obtain the equation (E):

d

dt(mui) +

d

dt(mikuk) = −m∂tv +mu× (∇× v)−m∇ϕ. (21)

m∇ϕ = mϕg +mh, Pi = mui +mikuk,

Rewriting, d

dtP = mE + (m/a)u× aH −m∇ϕg. (E)

• The equation of motion of a charged particle in an electromagnetic field is

d

dt(mvp) = eEem + (e/c)vp ×Hem −m∇Φg. (F)

vp : velocity of a charged particle, c : the light velocity.

Hem ⇔ aH .

• Equivalence of both expressions reconfirms validity of the definitions

E = −∂tv −∇h, H = ∇× v.

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5. Uniqueness

We consider here whether the original system of fluid equations (14), (15) and(16) are recovered from the modified fluid Maxwell equations:

(A) ∇ ·H = 0, (B) ∇× (ω × v) + ∂tH = 0,

(C) ∇ ·E = q, (D) a 20 ∇×H − ∂tE = J .

q = ∇ · [(v · ∇)v],

J = ∂2t v −∇[v · ∇h+ a2∇ · v] + a 20∇× (∇× v).

modified : ∇×E is replaced by ∇× (ω × v).

∂tv + (v · ∇)v +∇h = 0, (14)

∂th+ v · ∇h+ a2∇ · v = 0, (15)

∂tω +∇× (ω × v) = 0. (16)

Starting assumption is that the forces acting on a test particle of mass m in flowfield are represented in the same form as those of electromagnetism:

dP /dt = mE +mu×H −m∇ϕg. (22)

We use this as a guiding principle. Comparing this with the equation (21) derivedfrom the Lagrangian (20), we obtain the following definition of E and H :

E = −∂tv −∇h, H = ∇× v. (23)

So that, Eq.(B) is equivalent to the vorticity equation.

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• By the definition H = ∇×v, Eq.(A) is satisfied identically, andEq.(B) is the vorticity equation (16) .

• Next is to consider the remaining equations (C) and (D), and tryto deduce (14) and (15).

Let us consider Eq.(D), from which we try to derive the equation(15) under the condition of no mass source.

In fact, substituting the expressions of E, H , and J ,

∂2t v +∇∂th+ a 20∇× (∇× v) =

∂2t v−∇[v · ∇h+ a2∇ · v] + a 20∇× (∇× v).

From this, we obtain ∇(∂th+ v · ∇h+ a2∇ · v

)= 0. Therefore,

∂th+ v · ∇h+ a2∇ · v = C(t) (a function of time t only).

The term C(t) must vanish in the absence of mass source. Namely,if C(t) = 0, there is a mass source of term (ρ/a2)C. Thus, C(t) = 0,and the equation (15) is recovered:

∂th+ v · ∇h+ a2∇ · v = 0.

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Next, we consider Eq.(C). Substituting q = div[(v · ∇)v] and inte-grating it, we obtain

E = (v · ∇)v +∇× ξ, ξ : arbitrary vector field.

By using the definition of E, this reduces to

∂tv + (v · ∇)v +∇h = −∇× ξ.

Using the identity (v · ∇)v = ω × v +∇(12 v2), this is rewritten as

∂tv + ω × v +∇(h+ 12 v

2) = −∇× ξ. (C∗)

On the other hand, Eq.(B) leads to

∂tv + ω × v = −∇φ, φ : arbitrary scalar field. (B∗)

Comparing (B*) with (C*), it is found that

E = (v · ∇)v, ∇× ξ = 0, φ = h+ 12 v

2.

Thus, we obtain the Euler’s equation (14):

∂tv + (v · ∇)v +∇h = 0.

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Thus, the fluid equations,

∂tv + (v · ∇)v +∇h = 0,

∂th+ v · ∇h+ a2∇ · v = 0,

∂tω +∇× (ω × v) = 0.

are recovered

from modified Maxwell equations:

(A) ∇ ·H = 0, (B) ∇× (ω × v) + ∂tH = 0,

(C) ∇ ·E = q, (D) a 20 ∇×H − ∂tE = J .

q = ∇ · [(v · ∇)v], J = ∂2t v +∇∂th+ a 20∇× (∇× v).

by the definition[E = −∂tv −∇h, H = ∇× v

], and isentropy.

T. Kambe (2010): ”Geometrical Theory of Dynamical Systems and Fluid Flows”,(revised ed., World Scientific, Singapore) Chap.7.

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Summary

Part I

• Lagrangian is defined by

L∗ =

∫Λ(v, ρ, s, ϕ, ψ,A) d3x,

Λ = 12 ρ ⟨v,v ⟩ − ρ ϵ(ρ, s)− ρDtϕ− ρsDtψ + ⟨A, L∂τω ⟩.

• Euler-Lagrange equation and Noether’s theorem are given by

Momentum conservation : ∂t(ρv

)+∇ · ρvv +∇ p = 0.

Energy conservation : ∂t[ρ(12 v

2 + ϵ)]+ ∂k

[ρvk (12 v

2 + h)]= 0.

Action principle with respect to the variations of ϕ, ψ and A:

∂tρ+∇ · (ρv) = 0 (Continuity eq.),

∂t(ρs) +∇ · (ρsv) = 0 (Entropy eq.),

∂tω +∇× (ω × v) = 0 (Vorticity eq.).

• Transformation relations of the three vectors v, A and ω sufficeto determine the nine matrix elements ∂xk/∂al locally.

• Thus, the vorticity equation is essential for the uniqueness of thetransformation between the Lagrangian and Eulerian spaces.

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Part II

• From the fluid equations,

∂tv + (v · ∇)v +∇h = 0,

∂th+ v · ∇h+ a2∇ · v = 0,

∂tω +∇× (ω × v) = 0.

⇓

fluid Maxwell equations are derived:

(A) ∇ ·H = 0, (B) ∇×E + ∂tH = 0,

(C) ∇ ·E = q, (D) a 20 ∇×H − ∂tE = J .

q = ∇ · [(v · ∇)v], J = ∂2t v +∇∂th+ a 20∇× (∇× v).

by the definition[E = −∂tv −∇h, H = ∇× v

], and isentropy.

• From this, equation of sound wave is derived. Hence, the soundwave is analogous to the Electromagnetic wave. Dynamicalmotion of vorticity can generate sound waves.

• Forces acting on a test particle of mass m in a flow field arerepresented in the same form as those of electromagnetism:

dP /dt = mE +mu×H −m∇ϕg.

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Thank you very much for your attention

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Eckart, C., 1960. Variation principles of hydrodynamics, Phys. Fluids, 3,421–427.

Kambe,T. 2008. ”Variational formulation of ideal fluid flows according togauge principle”, Fluid Dyn. Res. 40, 399-426.

Kambe, T. 2010. Geometrical Theory of Dynamical Systems and Fluid Flows(Rev.Ed., World Scientific, Singapore, 2010) Chap.7.

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