classical and quantum...
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Classical and Quantum Turbulence
Andrew W. Baggaley
School of Mathematics and StatisticsUniversity of Glasgow
August 18, 2014
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Motivation
Da Vinci, circa 1500
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The cliche quote
“ Finally there is a physical problem that is common to manyfields, that is very old, and that has not been solved. It is not theproblem of finding new fundamental particles, but something leftover from a long time ago over a hundred years. Nobody inphysics has really been able to analyze it mathematicallysatisfactorily in spite of its importance to the sister sciences. It isthe analysis of circulating or turbulent fluids.”
Feynman, 1963
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Classical fluids
Water, air, etc. – Viscous fluids
Mathematical description - Navier-Stokes equation
∂v
∂t+ (v · ∇)v = −1
ρ∇p+ ν∇2v
Key parameter Re =UL
νIf Re 1, ‘easy’ – Stokes flow
Typically Re 1, nonlinearity bites
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Classical turbulence
In a 3D classical turbulent flow,large scale eddies break up intosmaller eddies, these intosmaller ones and soon...(Richardson Cascade)
If there is a large inertial rangebetween the forcing anddissipation scale (i.e. high Re)then the flow of energy throughscales is characterised by aconstant energy flux ε.
Dimensional analysis leads to apower-law scaling for the energyspectrum E(k)
E(k) = Cε2/3k−5/3.
∂v
∂t+ (v · ∇)v = −1
ρ∇p+ ν∇2v
In the first part of this talk we will tryto understand where this came from.
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The Navier-Stokes Equation
Reynolds transport theorem, consider the change of somequantity J in a volume Ω,
d
dt
∫ΩJ dV = −
∫∂ΩJv · n dA−
∫ΩQ dV
Q sources and sinks of J in the volume.
After a little algebra (i.e. Divergence theorem):
∂J
∂t+∇ · (Jv) +Q = 0
Apply this to the density (J = ρ), Q = 0:
∂ρ
∂t+∇ · (ρv) = 0
Incompressibility ρ = const.→ ∇ · v = 0
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The Navier-Stokes Equation
Now apply to momentum J = ρv,
∂
∂t(ρv) +∇ · (ρvv) + Q = 0
Expand and tidy up (using continuity equation derived above):
ρ
(∂v
∂t+ v · ∇v
)= b
Q = −b, body forces which represent sources (and sinks) ofmomentum.
To understand the form of b consider the forces acting on asmall volume of the fluid, can reason b is composed of thedivergence of the Cauchy stress tensor (denoted σ) and otherbody forces (f), i.e. gravity, Lorentz force etc.
ρ
(∂v
∂t+ v · ∇v
)= ∇ · σ + f
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The Navier-Stokes Equation
Decompose σ into normal (diagonal) and shear (off-diagonal)stresses:
σij =
σxx τxy τxzτyx σyy τyzτzx τzy σzz
= −
P 0 00 P 00 0 P
+
σxx + P τxy τxzτyx σyy + P τyzτzx τzy σzz + P
σ = −P I + T, pressure P = −1
3(σxx + σyy + σzz) , i.e.
minus the mean of the normal stresses.
T (Deviatoric stress), basically what’s left after subtractingthe hydrostatic stress, note T = 0 if the fluid is at rest.
Newtonian fluid, assume shear stress is linearly proportional tothe strain rate, if incompressible viscosity µ is constant
Tij = µ
(∂vi∂xj
+∂vj∂xi
)Hence total stress tensor σij = −Pδij + µ
(∂vi∂xj
+∂vj∂xi
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The Navier-Stokes Equation
Plugging this in leaves the Navier-Stokes we know and love
ρ
(∂v
∂t+ v · ∇v
)= −∇p+ µ∇2v + f
Note if we cannot assume incompressibility then thecompressible Navier-Stokes must be considered
ρ
(∂v
∂t+ v · ∇v
)= −∇p+ µ∇2v +
(ζ +
µ
3
)∇(∇ · v) + f
where ζ is the bulk viscosity
More convenient to nondimensionalise and write the equationas
∂tv + v · ∇v = −∇p+ Re−1∇2v
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Reynolds number
For incompressible flow Buckingham π theorem Re = UL/ν(ν = µ/ρ), for a given geometrical shape of the boundaries,the Reynolds number Re is the only control parameter.
The Navier-Stokes equations probably contains all of Classicalturbulence.
Now following Van Dyke’s album of fluid motion lets considerthe flow past a cylinder at increasing Re
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Re = 0.16
Flow appears to possess the following symmetries
left - right (x-reversal)
up - down (y-reversal)
time translation (t-invariance)
space translation (z-invariance)
v = (u, v, w)
left - right: (x, y, z)→ (−x, y, z), (u, v, w)→ (u,−v,−w)Only true if v · ∇v = 0 (Stoke’s flow, Re = 0).
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Re = 1.54
Noticeable left-right asymmetry, at Re ≈ 5, flow separation begins,recirculating eddies are found.
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Re = 26
Increasing Re we start to see these eddies being shed leading tothe Karmaan vortex street.
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Re = 140
There is a critical Reynolds number (not accurately known andgeometry dependent) above which the flow becomes chaotic,further increasing Re and only a few distinct eddies are foundbefore merging into a turbulent wake.
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Re = 1800
At very large Re, a sequence of instabilities produceturbulence as a superposition of motions of different scales.
The resulting flow is irregular both spatially and temporally sowe need to describe it statistically.
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The Cascade
The most revealing insight into the nature of turbulence givesa cascade picture.
Pump in energy at large scales L, inertial effects dominateover viscous effects UL/ν 1.
At smaller scales, ` energy can be dissipated if u(`)`/ν ≈ 1.
The nonlinearity (v · ∇v) gives us an idea of how large scaleenergy can be transmitted to small scales
v = sin(x)⇒ v∂v
∂x∼ sin(2x)
For a truly heroic early effort to understand the cascade ofenergy consult Taylor, G. I. and Green, A. E., Mechanism ofthe Production of Small Eddies from Large Ones, Proc. R.Soc. Lond. A, 158, 499521 (1937)
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The Cascade
Consider a fan, with blades rotating at a constant rate. Thisgenerate some flow, whose magnitude tends to somestatistically steady value after an initial transient.
The input of energy from the blades is balanced by viscousdissipation.
The energy dissipation rate ε remains finite as Re→∞.
We shall derive this result from an empirical observation madein the 1920’s.
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The Cascade
Richardson(1926) found experimentally that the relativeseparation of particles suspended in a turbulent flow grows asa cubic law of time (super diffusively),
15 years before the theoretical work by Kolmogorov andObukhov.
R(t) is the separation of two particles, Richardson dispersion< R2 >∝ t3,
the constant of proportionality has units cm2s−3, i.e. theenergy dissipation rate ε
< R2 >≈ εt3
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The Cascade
Richardsons law is often interpreted as the increase of thetypical velocity difference δv(R) with separation R.
Consider, there are vortices of different scales in a turbulentflow, the velocity difference at a given distance is due tovortices with comparable scales and smaller.
As R increases, more (and larger) vortices contribute to therelative velocity, which makes separation faster than diffusive(where the velocity is independent of the distance).
Now R = ε1/2t3/2 is a solution of the equationdR
dt= (εR)1/3,
as dR/dt = δv(R)
δv(R) ≈ (εR)1/3,(δv)3
R≈ ε
This last result sets the scene for the idea of the turbulentcascade
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The Cascade
(δv)3/R ≈ ε
We pump energy in at large scales L, δv(L) ∼ U , energy isdissipated at the viscous scale, l
The energy flux through a given scale is the energy (δv)2,divided by the time R/δv.
Within the inertial range L R l, there is neither forcingnor dissipation, and so the energy fluxε(R) =< (δv)3(R) > /R is expected to be R independent.
As ν → 0 (Re→∞) the cascade gets longer, but the energyflux and the rate of dissipation stay the same (for the sameenergy input).
We interpret the finiteness of ε as ν → 0 as the locality ofenergy transfer in Fourier space.
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The Cascade
Kolmogorov proposed the first statistical theory of turbulence,based on the notion of the energy cascade and the concept ofself-similarity.
Here we simply state Kolmogorov’s hypothesis and providetwo of his most famous results.
Kolmogorov assumed at very high, but not infinite, Reynoldsnumber, all of the small-scale statistical properties areuniquely and universally determined by the length scale `, themean dissipation rate (per unit mass) ε and the viscosity ν.
Further assumptions must also be made, principallyhomogeneity and isotropy.
Kolmogorov originally defined the following scaling lawsdirectly from the Navier-Stokes equation, we shall simply usedimensional analysis
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K41
As above denote size of smallest eddies `, assume
` ∼ ναεβ
dimensional analysis gives
` ∼ (ν3/ε)1/4.
We proceed in a similar way for the largest scale in the flow,where we have length L and velocity U .Energy supplied to the large scales is the same as the energytransferred in the cascade, ε, as energy is only dissipated atthe smallest scales. Assume
ε ∼ UγLδ,leaving ε ∼ U3/LCombining these two results we find an important relation
`
L=
1
L
(ν3
ε
)1/4
=1
L
(ν3L
U3
)1/4
=
(UL
ν
)−3/4
≡ Re−3/4.
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K41
The energy spectrum E(K), is defined as∫ ∞0
E(k)dk =1
2〈u2〉,
E(k) is the amount of energy in structures with a size of2π/k, per unit logarithmic interval of k.
Following Kolmogorov’s assumptions one argues
E(k) ∼ εαkβ,
and arrives atE(k) = CKε
2/3k−5/3. (1)
Where CK is a dimensionless (O(1)) constant, known as theKolmogorov constant.
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Experimental verification
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Structure Functions
Whilst the Kolmogorov spectrum is the most famous result inthe theory of turbulence, the most important result isarguably the four-fifths law
We introduce the pth order longitudinal structure function:
Sp(r) ≡⟨
[v(x + r)− v(x)] · rr
p⟩p ∈ Z+, r denotes a spatial vector increment.
K41 theory (self-similarity) predicts
Sp(r) = Cpεp/3rp/3
However the 3rd order structure function, can be shown(exactly from the Karman–Howarth equation) without anyassumption of self-similarity to satisfy
S3(r) = −4
5εr
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Experimental verification
Energy cascade and the four-fifths law in superfluid turbulence,Salor et al. , EPL, (2012)
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Intermittency
But what about the grander claims based on self-similarity?
Sp(r) = Cpεp/3rp/3
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Intermittency
Is this a surprise? Perhaps not, many systems exhibithierarchical clustering.The fact that small-scale activity in high-Reynold numberturbulence becomes increasingly clumpy and thatself-similarity is broken is referred to as intermittencyIntermittency is often be quantified by measuring the flatnessof velocity increments
F (r) ≡ S4(r)/S22(r),
this should be independent of r in K41, in reality it grows as rdecreases.
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A short discourse into rotating turbulence
Alexander Selkirk Island in the southern Pacific Ocean, NASA
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Rotating turbulence
Cape Verde southern Atlantic Ocean, NASA
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The cause?
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But island height ∼ 1000mcloud height ∼ 5000m.
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Equations of motion
Momentum equation:
∂u
∂t+ (u · ∇)u︸ ︷︷ ︸interia
+ Ω× (Ω× r)︸ ︷︷ ︸centrifugal
= −∇Pρ︸ ︷︷ ︸
pressure gradients
+ ν∇2u︸ ︷︷ ︸viscous
− 2Ω× u︸ ︷︷ ︸coriolis
− g︸︷︷︸gravity
ρ – fluid density
ν – kinematic viscosity
P – pressure
Ω = (0, 0, ω) – angularvelocity
g = (0, 0,−g) – externalgravity field
u = (u, v, w) – velocityfield
r – position vector
Missing compressible effects,
(ν
3+ ζ)∇(∇ · u)
Continuity Equation:
∂u
∂t+∇ · (ρu) = 0
If ρ is constant ∇ · u = 0,incompressible flow
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Equations of motion
Momentum equation:
∂u
∂t+ (u · ∇)u︸ ︷︷ ︸interia
+ Ω× (Ω× r)︸ ︷︷ ︸centrifugal
= −∇Pρ︸ ︷︷ ︸
pressure gradients
+ ν∇2u︸ ︷︷ ︸viscous
− 2Ω× u︸ ︷︷ ︸coriolis
− g︸︷︷︸gravity
ρ – fluid density
ν – kinematic viscosity
P – pressure
Ω = (0, 0, ω) – angularvelocity
g = (0, 0,−g) – externalgravity field
u = (u, v, w) – velocityfield
r – position vector
Missing compressible effects,
(ν
3+ ζ)∇(∇ · u)
Continuity Equation:
∂u
∂t+∇ · (ρu) = 0
If ρ is constant ∇ · u = 0,incompressible flow
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Combine gravity (g = ∇φ) and the centrifugal force into a singleterm ∇Φ ≡ ∇(φ+ 1/2(Ω× r)2)
∂u
∂t+ (u · ∇)u = −∇P
ρ+ ν∇2u− 2Ω× u +∇Φ
Steady flow,∂u
∂t= 0
Dimensionless numbers: Re=UL
ν 1, Ro=
U
ωL 1
Can set ν∇2u = 0, (u · ∇)u = 0.
Geostrophic balance,
−∇Pρ
= 2Ω× u−∇Φ.
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Combine gravity (g = ∇φ) and the centrifugal force into a singleterm ∇Φ ≡ ∇(φ+ 1/2(Ω× r)2)
∂u
∂t+ (u · ∇)u = −∇P
ρ+ ν∇2u− 2Ω× u +∇Φ
Steady flow,∂u
∂t= 0
Dimensionless numbers: Re=UL
ν 1, Ro=
U
ωL 1
Can set ν∇2u = 0, (u · ∇)u = 0.
Geostrophic balance,
−∇Pρ
= 2Ω× u−∇Φ.
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Combine gravity (g = ∇φ) and the centrifugal force into a singleterm ∇Φ ≡ ∇(φ+ 1/2(Ω× r)2)
∂u
∂t+ (u · ∇)u = −∇P
ρ+ ν∇2u− 2Ω× u +∇Φ
Steady flow,∂u
∂t= 0
Dimensionless numbers: Re=UL
ν 1, Ro=
U
ωL 1
Can set ν∇2u = 0, (u · ∇)u = 0.
Geostrophic balance,
−∇Pρ
= 2Ω× u−∇Φ.
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Combine gravity (g = ∇φ) and the centrifugal force into a singleterm ∇Φ ≡ ∇(φ+ 1/2(Ω× r)2)
∂u
∂t+ (u · ∇)u = −∇P
ρ+ ν∇2u− 2Ω× u +∇Φ
Steady flow,∂u
∂t= 0
Dimensionless numbers: Re=UL
ν 1, Ro=
U
ωL 1
Can set ν∇2u = 0, (u · ∇)u = 0.
Geostrophic balance,
−∇Pρ
= 2Ω× u−∇Φ.
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Combine gravity (g = ∇φ) and the centrifugal force into a singleterm ∇Φ ≡ ∇(φ+ 1/2(Ω× r)2)
∂u
∂t+ (u · ∇)u = −∇P
ρ+ ν∇2u− 2Ω× u +∇Φ
Steady flow,∂u
∂t= 0
Dimensionless numbers: Re=UL
ν 1, Ro=
U
ωL 1
Can set ν∇2u = 0, (u · ∇)u = 0.
Geostrophic balance,
−∇Pρ
= 2Ω× u−∇Φ.
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−∇Pρ
= 2Ω× u−∇Φ.
Take curl of both sides,
−∇× ∇Pρ
= ∇× (2Ω× u−∇Φ).
For constant ρ – left-hand side vanishes and ∇ · u = 0.
For constant Ω – right-hand side reduces to2(Ω∇ · u− (Ω · ∇)u).
Hence,Ω · ∇u = 0,
and if Ω = ωz,∂u
∂z= 0,
fluid velocity is independent of z.36 / 55
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−∇Pρ
= 2Ω× u−∇Φ.
Take curl of both sides,
−∇× ∇Pρ
= ∇× (2Ω× u−∇Φ).
For constant ρ – left-hand side vanishes and ∇ · u = 0.
For constant Ω – right-hand side reduces to2(Ω∇ · u− (Ω · ∇)u).
Hence,Ω · ∇u = 0,
and if Ω = ωz,∂u
∂z= 0,
fluid velocity is independent of z.36 / 55
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−∇Pρ
= 2Ω× u−∇Φ.
Take curl of both sides,
−∇× ∇Pρ
= ∇× (2Ω× u−∇Φ).
For constant ρ – left-hand side vanishes and ∇ · u = 0.
For constant Ω – right-hand side reduces to2(Ω∇ · u− (Ω · ∇)u).
Hence,Ω · ∇u = 0,
and if Ω = ωz,∂u
∂z= 0,
fluid velocity is independent of z.36 / 55
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−∇Pρ
= 2Ω× u−∇Φ.
Take curl of both sides,
−∇× ∇Pρ
= ∇× (2Ω× u−∇Φ).
For constant ρ – left-hand side vanishes and ∇ · u = 0.
For constant Ω – right-hand side reduces to2(Ω∇ · u− (Ω · ∇)u).
Hence,Ω · ∇u = 0,
and if Ω = ωz,∂u
∂z= 0,
fluid velocity is independent of z.36 / 55
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−∇Pρ
= 2Ω× u−∇Φ.
Take curl of both sides,
−∇× ∇Pρ
= ∇× (2Ω× u−∇Φ).
For constant ρ – left-hand side vanishes and ∇ · u = 0.
For constant Ω – right-hand side reduces to2(Ω∇ · u− (Ω · ∇)u).
Hence,Ω · ∇u = 0,
and if Ω = ωz,∂u
∂z= 0,
fluid velocity is independent of z.36 / 55
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−∇Pρ
= 2Ω× u−∇Φ.
Take curl of both sides,
−∇× ∇Pρ
= ∇× (2Ω× u−∇Φ).
For constant ρ – left-hand side vanishes and ∇ · u = 0.
For constant Ω – right-hand side reduces to2(Ω∇ · u− (Ω · ∇)u).
Hence,Ω · ∇u = 0,
and if Ω = ωz,∂u
∂z= 0,
fluid velocity is independent of z.36 / 55
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Consequences
Flow organises into (Taylor) columns,
Rotating (DNS) turbulence, Turbulence numerics teams, NCAR37 / 55
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What about turbulence in quantum fluids?
Classical fluids:vortex reconnections(Navier-Stokes)approach to possiblesingularities (Euler)
Quantum fluids:turbulence in superfluid 4He,3He, ultra-cold atomic gasesand neutron stars
ξ
v
zero viscosity
quantization of circulation∮v · dr = κ
fixed vortex core radius ξ
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What about turbulence in quantum fluids?
I will focus on turbulence in superfluid helium.
At least four regimes of turbulence:
Finite temperature turbulence 1K . T . 2K:Quasi-classical two fluid turbulence (mechanically stirred)Counterflow turbulence (thermally driven)
Zero temperature (pure) quantum turbulence Kolmogorovtangle (local polarisation)Random Vinen tangle
Note the picture is a little more complex in reality as between0.8K . T . 1K the normal fluid does not ’drive’ turbulencein the superfluid,
however the dissipation mechanism is different here to the 0Kregime.
Here the principal focus will be the zero temperature regime.
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Energy spectrum
At scale larger than intervortex spacing - classical(Kolmogorov) regime. Experimentally (Maurer and Tabeling[1998], Salort et al. 2012) and numerically (Nore et al. [1997],Araki et al. [2002]) verified.
As discussed earlier Kolmogorov’s 4/5 law also verified atT = 1.56K
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Energy spectrum
At small (< `) scales quantum regime - Kelvin wave cascade.Quantised vortices reconnect at the crossover scale (k`,` = 1/
√(L), L = Λ/V ), creating pronounced cusps, which
these propagate as Kelvin waves.Nonlinear (3 wave) interactions lead to the creation of smallerscales.Nonlocal theory of L’vov and Nazarenko numerically verified.At high k energy dissipated as phonon (sound) emission,Caroli-Matricon in 3He-B.
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Energy spectrum
At crossover scale arguments for, L’vov et al. [2007] (left),and against, Kozik & Svistunov [2008] (right), a bottleneck.
This is probably the regime where most work needs to bedone.
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Energy spectrum - Summary
Vinen, [2006]43 / 55
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How is K41 possible in QT?
L’vov et al. , [207] A Kolmogorov spectrum needs polarisationof the tangle, i.e. structures which consist of bundles ofvortices.
Why? Bundles of quantised vortices can exhibit stretching,which an individual quantised vortex cannot.
Vortex stretching is important in understanding the turbulenceenergy flux.
For incompressible flow, stretching a vortex implies a’squeezing’ of the fluid elements in the directionsperpendicular to the stretching direction,
hence smaller scale motion.
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Vortex Bundles
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Vortex filament model
Helium experiments: average distance between lines (` ≈ 10−1 to10−4 cm) is much bigger than core-size 10−8 cm ∴ Model vortexlines as reconnecting space curves s(ξ, t)
Biot-Savart law:
ds
dt= − Γ
4π
∮(s− r)× dr
|s− r|3
LIA:
ds
dt≈ βs′ × s′′
N = number of discretization pointsBiot-Savart is slow: CPU ∼ N2
Tree algorithm is faster: CPU ∼ N logN - (AWB & Barenghi,JLTP [2012])
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Identifying structures
How to identify structures in a discrete tangle?
We convolve the discrete vortex filaments with a Gaussiankernel, and define a smoothed vorticity field ω, on a regularCartesian mesh
ω(r, t) = κ
N∑i=1
s′i(2πσ2)3/2
exp(−|si − r|2/2σ2)∆ξ ,
s′i = dsi/dξ is the unit vector along a vortex at si = si(ξ, t)
N is the number of discretization points.
We choose a smoothing length σ which is of the order of` = 1/
√L, the typical separation of vortices.
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Identifying structures
ω(r, t) = κ
N∑i=1
s′i(2πσ2)3/2
exp(−|si − r|2/2σ2)∆ξ ,
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The punchline: Bundles are important!
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Random tangle
Note the picture is less clear in the random Vinen tangle at0K.
Arguments that loop emission could be important, adissipation non-local in physical space, very different toclassical turbulence.
(left) Kondaurova et al. [2012], (right) Kursa et al. [2011]
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Conclusions
Classical turbulence, established and tested theories,Kolmogorov spectrum, 4/5-law.
Failure of self-similar theory, intermittency.
Quantum turbulence - quasi classical tangle exhibits k−5/3,4/5-law. No work yet on intermittency, interesting theoreticalpredictions at finite temperature.
Crossover regime between Kolmogorov and Kelvin wavecascades is poorly understood.
Wave turbulence theory appears to explain low amplitudeKelvin wave dynamics.
Large scale flow (k−5/3) appears to depend on vortex bundles,which mimic large scale eddies and allow vortex stretching.
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Thank you for your attention!
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