1
Chaire Européenne du College de France (2004/2005)
Sandro Stringari
Lecture 6
21 Mar 05
MOMENT OF INERTIA
AND SUPERFLUIDITY
Previous lecture: BEC in low dimensions
- Theorems on long range order. Algebraic decay in low D.
- Mean field and beyond mean field.
- Collective oscillations in 1D gas.
This lecture .
Irrotational vs rotatational flow.
Moment of inertia and scissors mode.
Expansion of rotating BEC.
2
Rotating Bose-Einstein condensates
(low angular velocities)
Superfluids rotate very differently from classical fluids
In classical fluid, due to viscosity, the velocity field of steady
rotation is given by the rigid value v( r ) = Ω × r
and is characterized by uniform vorticity ∇ × v(r ) = 2Ω
Superfluids are characterized by irrotationality constraint (lecture 2)
consequence of the phase of order parameter ( Ψ = n0 exp iS ),
yielding irrotational value v ( r ) = (h / m)∇S
vorticity is hence vanishing ( ∇ × v(r ) = 0 ) except along lines of
singularity (vorticites, see lecture 7).
Difference in velocity field shows up in observable quantities
3
Moment of inertia
When the container (bucket in helium, magnetic or optical trap
In atomic gases) is put in rotation with angular velocity Ω the
system acquires angular momentum
Lz (Ω) = m ∫ drn(r )r × v
∂Lz
Moment of inertia is defined by relationship Θ =
∂Ω
If the system rotates classically Lz = ΩΘ rig
( v = Ω × r ) one finds rigid
Θ rig = N < r⊥2 >
value of moment of inertia
rigid value
Superfluid exhibits different behaviour:
- at low angular velocity has Lz = 0
- at higher Ω exhibits jumps in
angular momentum due to vortices
Angular momentum vs angular velocity
measured in helium (Hess and Fairbank, 1967)
4
Moment of inertia of a trapped gas
Questions:
- can we evaluate Θ in a trapped gas ?
- what is the role of the “roughness” of the trap
(needed to transfer angular momentum)
- can we measure Θ and probe superfluidity ?
Some answers:
- calculation of Θ in
- ideal gas (role of statistics, temperature)
- hydrodynamic theory of superfluids at T=0
- experimental information from
- scissors mode,
- expansion of a rotating gas,
5
Moment of inertia of harmonically trapped quantum gas
Moment of inertia can be regarded as linear response
to static perturbation H pert = −ΩLz .
Perturbation theory then provides result
n | Lz | m
∂L
2
Θ = z = ∑n , m e − β Em
∂Ω Q
En − Em
with
2
Q = ∑n e − βEn
n , En eigenstates and eigenvalues of unperturbed Hamiltonian H
- Response function can be calculated if
H is not axi-symmetric ( [ H , Lz ] ≠ 0 ).
- Deformation of harmonic trap provides natural
symmetry breaking ( “mesosocpic roughening” ) :
Vho =
(
m 2 2
ω x x + ω y2 y 2 + ω z2 z 2
2
)
ω x ≠ ω y ⇒ [ H , Lz ] = im(ω y2 − ω x2 )∑i xi yi ≠ 0
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Moment of inertia is obtained by solution of
[ H , X ] = Lz
Θ=
2
− βEm
e
∑
Q n ,m
n | Lz | m
En − Em
yielding
2
=
1
− βEm
( m | Lz | n n | X | m − m | X | n n | Lz | m
e
∑
n ,m
Q
Θ = [ Lz , X ]
completeness relation
Algebraic solution for X available with Hamiltonian
X =−
)
[
i
2
2
x y
(
ω
+
ω
)
x
y
+
2
p
pi / m
∑
x
y
i
i
i
2
2
i
h (ω x − ω y )
(
]
(ω x2 + ω y2 ) y 2 − x 2 + 2 ω y2 y 2 − ω x2 x 2
Θ
=
Θ rig
(ω x2 − ω y2 ) y 2 + x 2
H
sp
p2
=
+ Vho
2m
yielding
)
holds for Bose and
Fermi statistics. Any T
Result for moment of inertia admits well defined limit when
ωx → ω y
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Moment of inertia of ideal Bose gas
For a Bose gas above critical temperature or for Fermi gas one
can use semiclassical result for square radii: x 2 ∝ 1 / ω x2 , y 2 ∝ 1 / ω y2
Moment of inertia takes rigid value Θ = Θ rig
For a Bose gas at T=0 one instead has x 2 ∝ 1 / ω x , y 2 ∝ 1 / ω y
2
Moment of inertia takes irrotational value Θ = δ Θ rig
δ=
x2 − y2
x + y
2
2
where
is deformation of the condensate. Vanishes
for axi-symmetric configuration (superfluidity)
Calculation is easily extended to finite T (Stringari, 1996)
condensate
Θ
=
Θ rig
thermal component
δ 2 r⊥2 0 N 0 + r⊥2
r⊥2 N 0 + r⊥2
0
T
T
NT
NT
8
Moment of inertia in thermodynamic limit
For fixed value of T / Tc one has
as N → ∞ (see Lecture 4)
< r⊥2 > 0 TC 1
≈
→0
2
1/ 3
< r⊥ >T T N
Hence Θ → Θ rig also below critical temperature
When N → ∞ , moment of inertia of ideal gas
is determined by thermal component even if fraction
of atoms in the condensate is finite
Interactions stabilize the
ratio < r⊥2 > 0 TC
3 1/ 6
< r⊥2 >T
≈
T
N = 5× 107
N = 5× 10 4
ideal gas
( n ( 0) a )
and hence provide finite
reduction of moment of
inertia for any T below
critical temperature
thermodynamic limit
with interaction
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Irrotational hydrodynamics
and T=0 value of moment of inertia
Role of interactions in a superfluid at T=0 can be investigated using
equations of irrotational hydrodynamics (Lecture 2)
It is convenient to write HD equations in rotating frame where
rotating trap is at rest and one can look for steady solutions.
Equations are derived using Hamiltonian H − ΩLz = H − mΩ ⋅ r × v
∂
n + ∇[n(vS − Ω × r )] = 0
∂t
∂
1 2
m vS + ∇( mvS + µ (n) + Vext − vS ⋅ Ω × r ) = 0
∂t
2
where vS =
h
∇S
m
is superfluid velocity (irrotational) in lab frame
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For harmonic trapping the hydrodynamic equations
admit stationary solutions of the form
vS = α∇xy
Equation of continuity yields relationship
δ=
x2 − y2
x + y
2
2
α = −δΩ
where
is deformation of the condensate.
Angular momentum is given by
Lz (Ω) = m ∫ drn(r )r × vS = δ 2 Θ rig Ω
Irrotationality of moment of inertia
follows from irrotationality of superfluid motion
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Scissors mode
Direct measurement of moment of inertia is difficult because images
of atomic cloud probe density distribution (not velocity distribution)
In deformed traps rotation is however coupled to density oscillations.
Exact relation, holding also in the presence of 2-body forces:
[ H , Lz ] = im(ω y2 − ω x2 )∑i xi yi
angular momentum
quadrupole operator
Response to transverse probe measurable
thorugh density response function !!
Example of coupling is provided by scissor mode.
If confining (deformed) trap is suddenly rotated by
angle θ the gas is no longer in equilibrium.
Behaviour of resulting oscillation depends crucially
on value of moment of inertia (irrotational vs rigid)
12
Qualitative estimate of
scissors frequency
(role of moment of inertia)
ω=
K
Θ
deformation of harmonic trap
Restoring force
K
Mass parameter
is proportional to ε (no energy cost for symmetric trap)
2
Θ is given by
2
δ
Θ rig in superfluid phase ( δ ∝ ε )
- irrotational value
- rigid value
Θ rig in non superfluid phase
As ε → 0
scissors frequency
- approaches finite value in superlfuid
- vanishes in non superfluid phase
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Scissors frequencies (Guery-Odelin and Stringari, 1999)
Superfluid (T=0)
With the irrotational ansatz v S = α ( t ) ∇ xy
one finds exact solution
of HD equations. If trap is deformed ( ω x ≠ ω y ) the solution
corresponds to rotation of the gas around the principal axis in x,y plane
ω = ω x2 + ω y2
Result is independent of equation of state (surface mode)
Normal gas (above TC ).
Gas is dilute and interactions can be ignored (collisionless regime).
Excitations are provided by ideal gas Hamiltonian. Two frequencies:
ω± = ω x ± ω y
Differently from superfluid the normal gas exhibits low frequency
mode ω x − ω y ∝ ε (crucial to ensure rigid value of moment of inertia)
14
Scissors measured at Oxford
(Marago’et al, PRL 84, 2056 (2000))
Above
TC
2 modes:
(normal)
ω± = ω x ± ω y
Below TC (superfluid) :
single mode:
ω = ω x2 + ω y2
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Scissors and superfluidity
Is the measurement of the scissors mode at hydrodynamic
frequency
2
2 a proof of superfluidity ?
ω = ωx + ω y
- If normal gas is in collisional regime its dynamics is governed
by same hydrodynamic equations as in the superfluid.
- In this case study of scissors mode does not permit to distinguish
between superfluid and non superfluid regimes.
- Question relevant in Fermi gases near Feshbach resonance (Lecture 8)
In general to exploit superfluidity one should study collective
oscillations in the presence of rotating trap !
In this case superfluid and normal gas behave differently
even if normal gas is in collisional regime.
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Irrotational vs rotational flow
In the presence of rotating trap the stationary velocity field behaves
differently depending on whether the system is superfluid or normal.
In a superfluid the
velocity field is subject
to the constraint of
Irrotationality :
v = α ∇ xy
A normal gas, in steady
configuration, instead
rotates in rigid way
v = Ω×r
Rotating Anisotropic Trap
Vext =
(
)
[
M 2 2
M 2
ωx x + ω2y y 2 + ω2z z 2 =
ω⊥ (1 + ε )x 2 + ω2⊥ (1 − ε )y 2 + ω2z z 2
2
2
Rigid rotation v0 = Ω ∧ r :
y
]
Irrotational flow v0 = α∇(xy ) :
y
x
x
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Time needed to achieve rigid
rotation in non superflid phase
- In the absence of viscosity the rotating trap will never be
able to transfer angular momentum to the normal gas
and to generate rigid rotation.
- Time needed to spin up the normal gas can be calculated by
solving Boltzmann equations.
- In collisional regime ω⊥τ < 1 one finds (Guery-Odelin, 2000)
τ up = 1 /(ε ω τ )
2
where
τ
2
⊥
is average collisional time.
- Time τ up becomes large if
- deformation of trap is small
- system is too deeply in hydrodynamic regime ( ω⊥τ << 1 )
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Excitation of scissors mode with rotating trap
What happens to the cloud if we suddenly stop the rotation Ω
of the confining trap? System will be no longer in equilibrium
and will start oscillate (scissors mode).
If the gas is superfluid we excite the scissors mode
according to the predictions
irrotational hydrodynamics
ω = ω x2 + ω y2
If the gas is normal and collisional, the scissors mode
will be described by the equations of rotational hydrodynamics
(in laboratory frame)
∂
∇P
1 2
= mv × ∇ × v
m v + ∇( mv + Vext ) +
∂t
n
2
term absent in superfluid HD
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Scissors mode (after stopping rotation of the trap)
Superfluid (T=0)
ω = ω x2 + ω y2
Normal (collisional)
beating between
ω x2 + ω y2 ± Ω
Normal (collisionless)
beating between
ωx ± ω y
Ω = 0.2ω⊥ , ε = 0.2
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Expansion of a rotating superfluid gas :
consequences of irrotationality
In the absence of rotation the expansion
of a cigar condensate is faster in the radial
direction (Lecture 2). After time t c such
that R⊥ω⊥tc = Z the shape of the system
becomes spherical (aspect ratio = 1)
For longer times the density profile
takes a pancake form.
What happens if the gas is rotating?
At t=0 the gas carries irrotational angular momentum Lz = δ Θ rig Ω
2
A superfluid cannot appraoch spherical shape during
the expansion because the moment of inertia would vanish
and angular momentum would not be conserved.
The gas starts rotating fast when t approaches tc ,
but deformation remains finite (aspect ratio ≠ 1 ).
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Skater increases
angular velocity
by reducing radial size
Superfluid gas increases
angular velocity during the
expansion.
It cannot reach symmetric
configuration (aspect ratio =1)
because of angular momentum
conservation.
Theory: Edwards et al., 2002
Exp: Hechenblaickner et al. 2002
Ω / 2π = 28 Hz
Ω / 2π = 20 Hz
Ω / 2π = 0
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What happens at higher angular velocities?
By increasing Ω vortex lines become energetically favourable
(see lecture 7). System has then two main possibilities:
A) System keeps irrotationality and is still described by HD
(metastability, angular velocity should be increased adiabatically)
B) Lines of singular vorticity are created if system is allowed to jump into
lowest energy configuration (lecture 7)
Hypothesis A) can be explored by finding stationary solutions of
irrotational HD equations as a function of angular velocity Ω
∂
n + ∇[n(vS − Ω × r )] = 0
∂t
∂
1
m vS + ∇( mvS2 + µ (n) + Vext − vS ⋅ Ω × r ) = 0
∂t
2
equations in
rotating frame
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For a dilute Bose gas (
with irrotational velocity
µ = gn ) one finds stationary solutions
α = −δΩ
v = α ∇ xy
and parabolic density profile
n( r ) =
1
g
 ~ m ~2 2 ~2 2
2 2 
µ
ω
x
ω
y
ω
−
(
+
+
x
y
z z )

2

The new distribution is characterized by the renormalized oscillator
frequencies:
ω~x2 = ω x2 + α 2 − 2αΩ
ω~ 2 = ω 2 + α 2 + 2αΩ
y
y
Hydrodynamic equations yield cubic equation
α 3 + α (ω⊥2 − 2Ω 2 ) + Ωεω⊥2 = 0
(Recati et al. 2001)
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Isotropic trapping ( ε = 0 )
- If Ω > ω⊥ / 2 one finds 3 solutions for α = −δΩ
- solutions with α ≠ 0 have lowest energy.
(for α = 0 m=2 quadrupole oscillation becomes
energetically unstable ( δE − mΩ = 2ω⊥ − 2Ω < 0 ).
Spontaneous breaking of rotational symmetry
(similar to bifurcation phenomena in rotating classical fluids)
ε =0
1/ 2
Ω / ω⊥
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Role of trap deformation
Even small trap deformations can produce sizable effects
For ε ≠ 0 one identifies two different branches:
Main branch starting from Ω = 0 . This branch can be followed
adiabatically by increasing slowly the angular velocity up to
some critical angular velocity where the system exhibits
dynamic instability (Sinha, Castin, 2001).
α
Second branch extends up to
angular velocities larger than
trapping oscillator frequency
(overcritical branch).
ideal gas
main branch
x
overcritical
dynamic instability
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Rotating configurations can be
realized experimentally by
ramping up adiabatically the
angular velcocity (black circles).
For larger angular velocities the
system nucleates vortices due
to dynamic instability of
collective frequencies
Overcritical branch can be
also followed experimentally
(white circles).
Madison et al., 2001
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Energetic vs dynamic instability
Systems out of equilibrium can develop instabilities
- Energetic instability. Corresponds to occurrence of oscillations
with negative excitation energy. Energetic instability is effective
only in the presence of thermalization effects or mechanical
activation of the unstable mode.
- Dynamic instability. Corresponds to oscillations with
complex frequency. In this case the perturbation growth up
spontaneously, also without thermal activation.
BEC gases in rotating harmonic traps
exploit both cases of instability.
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This Lecture . Moment of inertia and superfluidity
Irrotational vs rotatational flow.
Moment of inertia and scissors mode.
Expansion of rotating BEC.
Next Lecture. Quantized vortices.
Quantization of circulation. Nucleation of vortices.
Measurement of angular momentum.
Vortex lattice. Collective oscillations.