Transcript CARLO_TALK

Dynamics of small tracer particles
in He II
Carlo F. Barenghi
University of Newcastle, UK
Demos Kivotides
Daniel Poole
Yuri Sergeev
Joe Vinen
1. VISUALISATION IN CLASSICAL FLUIDS
Ink, smoke, Kalliroscope, hydrogen bubbles,
Baker’s pH, hot wire, laser Doppler, ultra-sound,
PIV (particle image velocimetry), etc
2. VISUALISATION IN LIQUID HELIUM
second sound, ion trapping, temperature,
pressure, chemical potential gradients
Difficulties:
- only probe averaged properties
- no information about flow patters
and turbulent fluctuations
PIV in liquid helium
-Donnelly, Karpetis, Niemela, Sreenivasan and Vinen
in He I, buoyant hollow glass spheres, 1 - 5 μm size
-VanSciver, Zhang and Celik: in He II, heavier,
hollow glass / polymers / solid neon, 0.8 - 50 μm size
What do the tracer particles actually trace ?
Assuming:
-particles do not disturb fluid
-do not interact with / are trapped in superfluid vortices
-are smaller than vortex spacing and Kolmogorov length
-have small Reynolds number with respect to normal fluid
-neglect Basset history force, Faxen drag correction,
shear-induced lift and Magnus force
1:

du p
dt
Then the equations of motion of a neutrally buoyant
particle of radius ap, position rp, velocity up are:

 
vn  u p





 n vn 
 s vs 
 [
 (vn  )vn ]  [
 (vs  )vs ]
 t
 t
Viscous drag
2:

drp
dt

 up
NF inertia
SF inertia
a p

 10 4 sec
3 n
2
Relaxation time:
typically
Time-dependent, space-independent flows
Assume
then
where


vn  Vn exp( it )
 
vs  Vs exp( it )

0
u p  U p exp( t /  )  U p exp( it )

Up 
Thus, if ωτ>>1
n 
s 
1
[(1  i )Vn  i Vs ]
(1  i )



n  s 
U p  Vn  Vs
particle moves with


mass current
J=ρnVn+ρsVs
that is, Up=0 for second sound,
Up=Vn at high T, and Up=Vs at low T
And, if ωτ<<1, Up=Vn (particles trace normal fluid)

du p
One fluid only:
dt

 
v  up

[

v
t


 (v   )v ]
If fluid has no viscosity, using Euler’s equation,

v


1
 (v  )v   p
t

we get

du p
dt

1

p
hence the force per unit mass acting on a fluid parcel equals
the force on the particle which replaces (and moves with)
that parcel. We expect


up  v
Not quite so !
Consider an ABC flow
V=(Asin2πz+Ccos2πy,
Bsin2πx+Acos2πz,
Csin2πy+Bcos2πx)

dr f
dt

v
Trajectories of inertial particles are
unstable and concentrate in regions
where the magnitude of the rate of
strain tensor Eij=(dVi/dxj+dVj/dxi)/2
is large.
Time scale of instability depends
on intensity of ABC flow and
particle’s relaxation time τ
Particle and vortex
If particle is trapped onto a vortex, helium’s energy is
reduced by the equivalent vortex length lost
s2a p
E 
ln( a p /  )  k BT
4
Particle arrives from
far distance…
… and is trapped
Consider neutrally buoyant particle at distance r0 from straight
vortex in the presence of stationary normal fluid.
For a typical particle size the relaxation time is smaller than
the time to orbit a vortex at the typical intervortex distance
ℓ=1/√L in a tangle, so the orbital motion will be damped.
The radial motion is governed by
du p
dt

up

2

   s  2 / 8 2 
r3
The time for the particle to arrive at distance, say, 2ap
(sufficiently large that the vortex is not much disturbed) is
4
r04  16a p
ta 
1 4

8 
r0




bc=capture cross section
ℓ²/bc=mean free path for capture
Tp=ℓ²/bcvL=mean free time
vL=typical particle velocity with respect to vortices
Assume particle approaches vortex with velocity vL.
Assume capture time from a distance r0 for a particle
initially at rest is is of the order of the previous value
r04
ta 
8
The particle will be probably captured if the time spent
at distance r0 is greater than ta: r0/vL>ta
This yields the cross section and the mean free time:
3
c
b

2  s a 2p 
3n
2 2
Tp 
( / a p ) 2 / 3

1 msec < Tp <10 sec
The dynamics of the close approach may be complicated.
The particle may be trapped or not, and may trigger Kelvin
waves. We need more information using a microscopic model
(eg GP equation used by Berloff & Roberts 2000 and
Winiecki & Adams 2000 to study vortex-ion interaction)
For simplicity, assume that the vortex remains straight.
If the vortex attaches to the particle, it becomes perpendicular
to it and the force will decrease, until it vanishes when the
The position is symmetric.
Crude CUT-OFF model: assume force-free motion if d<ap
Pure superfluid limit

du p

The equations


vs

 ( v s   )v s
reduce to:
dt
t



 
Note that u p (t )  vs (rf (t ), t ) rp (t )  r f (t )

drp
dt

 up
is a formal solution, where rf(t) is a Lagrangian trajectory
of a superfluid particle. One would thus expect that a
small, buoyant, inertial particle, which at t=0 has velocity
equal to the local fluid velocity, will move with velocity up=vs
Unfortunately even in the simplest case of the motion
around a single straight vortex, the particle’s trajectory
is UNSTABLE
Using polar coordinates (r,θ), the particle trajectory obeys
2
d2

2
r

r

p p  
2 p
dt
4 2 rp3
d
d
2 p rp  rp  p  0
dt
dt
where ωp=dθp/dt . Perturb the circular orbit
rp=R, ωp=Ω=Γ/(2π R2) by letting rp=R+r’,ωp=Ω+ω’
with r’<<R, ω’<< Ω. Perturbations obey d3 r’/dr3=0, so
r ' (t ) 


dr ' (0)
[ 2 r ' (0)   ' (0)]t 2 
t  r ' (0)
2R R
dt
ω’ is also quadratic in time. Any mismatch between
initial fluid/particle velocities and any sensitivity on initial
conditions (chaos for sufficient number of vortices, Aref)
will reinforce the instability.
2-dim
motion around 3 vortices
on triangle:
AB= fluid particle
AC=inertial particle
motion within 20 random
vortices (not shown)
assuming cut-off model:
AB=fluid particle
AC=inertial particle
(2-dim)
Particle velocity (dashed) and
fluid velocity (solid) vs time
3-dim
Vortex tangle
Vortex line density vs t
L=18.5 X 10³ cm-2
Intervortex spacing
ℓ ≈1/√L=0.0073 cm
Histogram of fluid particle’s
velocity
vf≈Γ/(2πℓ)=0.02 cm/sec
Inertial particle’s velocity up across
vortex line. The plateau near x=0
arises from cut-off model (force-free
motion at distances less than ap)
500 particles used
x,y projection of the
trajectory of one particle
Histogram of up
Note up>>vf
Velocity saturation
The mean velocity is mainly due to the ballistic parts of
the trajectory. Moving vortices accelerate the particle
until, in the moving frame of reference, the vortices
appear still. At this point the particle speeds up when
going into a potential well and slows down when coming
out, the cut-off model preventing up →∞. During the
interaction, the particle’s velocity changes by an amount
of order of the vortex velocity, which is the same as <vf>.
Saturation takes place when the mean acceleration along
the trajectory, <up2>/ℓ, is of the same order of the mean
acceleration <vf2>/ap of the particle at distance ap from vortex:

 u p 

ap

 vf 
in agreement with 2-dim and 3-dim simulations
CONCLUSIONS
1. Equations of motion
2. Solution for time-dependent, space-independent flows
3. Classical case (one fluid): trajectory of inertial particle
may be unstable and move to regions of low vorticity
and high strain
4. Close particle-vortex interaction needs understanding
5. Pure superfluid:
-even the motion of a particle around a single,
straight vortex is unstable
- 2-dim and 3-dim simulations using cut-off model:
particles’ motion says nothing of fluid motion