lecture 12: domains, nucleation and coarsening outline: domain walls nucleation coarsening
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Lecture 12: Domains, nucleation and coarsening
Outline:• domain walls• nucleation• coarsening
Domain walls
Below Tc: m0: uniform solution of
€
m = tanh βJm( )
Domain walls
Below Tc: m0: uniform solution of
€
m = tanh βJm( )
or in Landau model: ϕ0 solves
€
dV
dφ= 0
Domain walls
Below Tc: m0: uniform solution of
€
m = tanh βJm( )
or in Landau model: ϕ0 solves
€
dV
dφ= 0 ⇒ r0φ0 + u0φ0
3 = 0
Domain walls
Below Tc: m0: uniform solution of
€
m = tanh βJm( )
or in Landau model: ϕ0 solves
€
dV
dφ= 0 ⇒ r0φ0 + u0φ0
3 = 0
⇒ φ0 = ±r0
u, r0 < 0
Domain walls
Below Tc: m0: uniform solution of
€
m = tanh βJm( )
or in Landau model: ϕ0 solves
€
dV
dφ= 0 ⇒ r0φ0 + u0φ0
3 = 0
⇒ φ0 = ±r0
u, r0 < 0
Suppose we have boundary conditions
€
φ→r0
u, x → ∞
φ → −r0
u, x → −∞
Non-uniform stationary solution
€
δE
δφ(x)=
δ
δφ(x)dd ′ x ∫ 1
2 r0φ2( ′ x ) + 1
4 u0φ4 ( ′ x ) + 1
2 (∇φ( ′ x ))2[ ]{ } = 0
Non-uniform stationary solution
€
δE
δφ(x)=
δ
δφ(x)dd ′ x ∫ 1
2 r0φ2( ′ x ) + 1
4 u0φ4 ( ′ x ) + 1
2 (∇φ( ′ x ))2[ ]{ } = 0
rewrite this with
€
E[φ] = dd ′ x ∫ 14 u0 φ2( ′ x ) − φ0
2( )
2+ 1
2 (∇φ( ′ x ))2[ ]
Non-uniform stationary solution
€
δE
δφ(x)=
δ
δφ(x)dd ′ x ∫ 1
2 r0φ2( ′ x ) + 1
4 u0φ4 ( ′ x ) + 1
2 (∇φ( ′ x ))2[ ]{ } = 0
rewrite this with
(differs only by an additive constant)
€
E[φ] = dd ′ x ∫ 14 u0 φ2( ′ x ) − φ0
2( )
2+ 1
2 (∇φ( ′ x ))2[ ]
Non-uniform stationary solution
€
δE
δφ(x)=
δ
δφ(x)dd ′ x ∫ 1
2 r0φ2( ′ x ) + 1
4 u0φ4 ( ′ x ) + 1
2 (∇φ( ′ x ))2[ ]{ } = 0
rewrite this with
(differs only by an additive constant)
€
E[φ] = dd ′ x ∫ 14 u0 φ2( ′ x ) − φ0
2( )
2+ 1
2 (∇φ( ′ x ))2[ ]
€
⇒δE
δφ(x)= u0 φ2(x) − φ0
2( )φ(x) −
d2φ
dx 2= 0
Non-uniform stationary solution
€
δE
δφ(x)=
δ
δφ(x)dd ′ x ∫ 1
2 r0φ2( ′ x ) + 1
4 u0φ4 ( ′ x ) + 1
2 (∇φ( ′ x ))2[ ]{ } = 0
rewrite this with
(differs only by an additive constant)
€
E[φ] = dd ′ x ∫ 14 u0 φ2( ′ x ) − φ0
2( )
2+ 1
2 (∇φ( ′ x ))2[ ]
€
⇒δE
δφ(x)= u0 φ2(x) − φ0
2( )φ(x) −
d2φ
dx 2= 0 (d = 1)
Non-uniform stationary solution
€
δE
δφ(x)=
δ
δφ(x)dd ′ x ∫ 1
2 r0φ2( ′ x ) + 1
4 u0φ4 ( ′ x ) + 1
2 (∇φ( ′ x ))2[ ]{ } = 0
rewrite this with
(differs only by an additive constant)
€
E[φ] = dd ′ x ∫ 14 u0 φ2( ′ x ) − φ0
2( )
2+ 1
2 (∇φ( ′ x ))2[ ]
€
φ(x) = φ0 tanh xr0
2
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟solution:
€
⇒δE
δφ(x)= u0 φ2(x) − φ0
2( )φ(x) −
d2φ
dx 2= 0 (d = 1)
Non-uniform stationary solution
€
δE
δφ(x)=
δ
δφ(x)dd ′ x ∫ 1
2 r0φ2( ′ x ) + 1
4 u0φ4 ( ′ x ) + 1
2 (∇φ( ′ x ))2[ ]{ } = 0
rewrite this with
(differs only by an additive constant)
€
E[φ] = dd ′ x ∫ 14 u0 φ2( ′ x ) − φ0
2( )
2+ 1
2 (∇φ( ′ x ))2[ ]
€
φ(x) = φ0 tanh xr0
2
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟solution:
€
⇒δE
δφ(x)= u0 φ2(x) − φ0
2( )φ(x) −
d2φ
dx 2= 0
“domain wall” (“kink”)solution
(d = 1)
Non-uniform stationary solution
€
δE
δφ(x)=
δ
δφ(x)dd ′ x ∫ 1
2 r0φ2( ′ x ) + 1
4 u0φ4 ( ′ x ) + 1
2 (∇φ( ′ x ))2[ ]{ } = 0
rewrite this with
(differs only by an additive constant)
€
E[φ] = dd ′ x ∫ 14 u0 φ2( ′ x ) − φ0
2( )
2+ 1
2 (∇φ( ′ x ))2[ ]
€
φ(x) = φ0 tanh xr0
2
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟solution:
€
⇒δE
δφ(x)= u0 φ2(x) − φ0
2( )φ(x) −
d2φ
dx 2= 0
“domain wall” (“kink”)solutionlocalized: size r0
-½
(d = 1)
Non-uniform stationary solution
€
δE
δφ(x)=
δ
δφ(x)dd ′ x ∫ 1
2 r0φ2( ′ x ) + 1
4 u0φ4 ( ′ x ) + 1
2 (∇φ( ′ x ))2[ ]{ } = 0
rewrite this with
(differs only by an additive constant)
€
E[φ] = dd ′ x ∫ 14 u0 φ2( ′ x ) − φ0
2( )
2+ 1
2 (∇φ( ′ x ))2[ ]
€
φ(x) = φ0 tanh xr0
2
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟solution:
€
⇒δE
δφ(x)= u0 φ2(x) − φ0
2( )φ(x) −
d2φ
dx 2= 0
“domain wall” (“kink”)solutionlocalized: size r0
-½
broad near Tc
(d = 1)
kink energy calculation
€
E[φ] = dx 14 u0 φ2(x) − φ0
2( )
2+ 1
2 (∇φ(x))2[ ]−∞
∞
∫
kink energy calculation
€
E[φ] = dx 14 u0 φ2(x) − φ0
2( )
2+ 1
2 (∇φ(x))2[ ]−∞
∞
∫
= dx 14 u0φ0
4 1− tanh2 xr0
2
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
2
+ 12 φ0
r0
21− tanh2 x
r0
2
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
⎡
⎣
⎢ ⎢
⎤
⎦
⎥ ⎥
2 ⎡
⎣
⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥
−∞
∞
∫
kink energy calculation
€
E[φ] = dx 14 u0 φ2(x) − φ0
2( )
2+ 1
2 (∇φ(x))2[ ]−∞
∞
∫
= dx 14 u0φ0
4 1− tanh2 xr0
2
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
2
+ 12 φ0
r0
21− tanh2 x
r0
2
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
⎡
⎣
⎢ ⎢
⎤
⎦
⎥ ⎥
2 ⎡
⎣
⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥
−∞
∞
∫
= 12 r0 φ0
2 dx 1− tanh2 xr0
2
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
2
−∞
∞
∫
kink energy calculation
€
E[φ] = dx 14 u0 φ2(x) − φ0
2( )
2+ 1
2 (∇φ(x))2[ ]−∞
∞
∫
= dx 14 u0φ0
4 1− tanh2 xr0
2
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
2
+ 12 φ0
r0
21− tanh2 x
r0
2
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
⎡
⎣
⎢ ⎢
⎤
⎦
⎥ ⎥
2 ⎡
⎣
⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥
−∞
∞
∫
= 12 r0 φ0
2 dx 1− tanh2 xr0
2
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
2
−∞
∞
∫ = 12 r0 φ0
2 ⋅2
r0
du 1− tanh2 u( )2
−∞
∞
∫
kink energy calculation
€
E[φ] = dx 14 u0 φ2(x) − φ0
2( )
2+ 1
2 (∇φ(x))2[ ]−∞
∞
∫
= dx 14 u0φ0
4 1− tanh2 xr0
2
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
2
+ 12 φ0
r0
21− tanh2 x
r0
2
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
⎡
⎣
⎢ ⎢
⎤
⎦
⎥ ⎥
2 ⎡
⎣
⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥
−∞
∞
∫
= 12 r0 φ0
2 dx 1− tanh2 xr0
2
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
2
−∞
∞
∫ = 12 r0 φ0
2 ⋅2
r0
du 1− tanh2 u( )2
−∞
∞
∫
= 12 r0 φ0
2 ⋅2
r0
dy
1− y 2−1
1
∫ 1− y 2( )
2
kink energy calculation
€
E[φ] = dx 14 u0 φ2(x) − φ0
2( )
2+ 1
2 (∇φ(x))2[ ]−∞
∞
∫
= dx 14 u0φ0
4 1− tanh2 xr0
2
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
2
+ 12 φ0
r0
21− tanh2 x
r0
2
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
⎡
⎣
⎢ ⎢
⎤
⎦
⎥ ⎥
2 ⎡
⎣
⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥
−∞
∞
∫
= 12 r0 φ0
2 dx 1− tanh2 xr0
2
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
2
−∞
∞
∫ = 12 r0 φ0
2 ⋅2
r0
du 1− tanh2 u( )2
−∞
∞
∫
= 12 r0 φ0
2 ⋅2
r0
dy
1− y 2−1
1
∫ 1− y 2( )
2= 1
2 r0 φ02 ⋅
2
r0
⋅4
3
kink energy calculation
€
E[φ] = dx 14 u0 φ2(x) − φ0
2( )
2+ 1
2 (∇φ(x))2[ ]−∞
∞
∫
= dx 14 u0φ0
4 1− tanh2 xr0
2
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
2
+ 12 φ0
r0
21− tanh2 x
r0
2
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
⎡
⎣
⎢ ⎢
⎤
⎦
⎥ ⎥
2 ⎡
⎣
⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥
−∞
∞
∫
= 12 r0 φ0
2 dx 1− tanh2 xr0
2
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
2
−∞
∞
∫ = 12 r0 φ0
2 ⋅2
r0
du 1− tanh2 u( )2
−∞
∞
∫
= 12 r0 φ0
2 ⋅2
r0
dy
1− y 2−1
1
∫ 1− y 2( )
2= 1
2 r0 φ02 ⋅
2
r0
⋅4
3= 4
3 ε0ξ
kink energy calculation
€
E[φ] = dx 14 u0 φ2(x) − φ0
2( )
2+ 1
2 (∇φ(x))2[ ]−∞
∞
∫
= dx 14 u0φ0
4 1− tanh2 xr0
2
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
2
+ 12 φ0
r0
21− tanh2 x
r0
2
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
⎡
⎣
⎢ ⎢
⎤
⎦
⎥ ⎥
2 ⎡
⎣
⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥
−∞
∞
∫
= 12 r0 φ0
2 dx 1− tanh2 xr0
2
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
2
−∞
∞
∫ = 12 r0 φ0
2 ⋅2
r0
du 1− tanh2 u( )2
−∞
∞
∫
= 12 r0 φ0
2 ⋅2
r0
dy
1− y 2−1
1
∫ 1− y 2( )
2= 1
2 r0 φ02 ⋅
2
r0
⋅4
3= 4
3 ε0ξ________thickness
kink energy calculation
€
E[φ] = dx 14 u0 φ2(x) − φ0
2( )
2+ 1
2 (∇φ(x))2[ ]−∞
∞
∫
= dx 14 u0φ0
4 1− tanh2 xr0
2
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
2
+ 12 φ0
r0
21− tanh2 x
r0
2
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
⎡
⎣
⎢ ⎢
⎤
⎦
⎥ ⎥
2 ⎡
⎣
⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥
−∞
∞
∫
= 12 r0 φ0
2 dx 1− tanh2 xr0
2
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
2
−∞
∞
∫ = 12 r0 φ0
2 ⋅2
r0
du 1− tanh2 u( )2
−∞
∞
∫
= 12 r0 φ0
2 ⋅2
r0
dy
1− y 2−1
1
∫ 1− y 2( )
2= 1
2 r0 φ02 ⋅
2
r0
⋅4
3= 4
3 ε0ξ _____energydensity
________thickness
kink energy calculation
€
E[φ] = dx 14 u0 φ2(x) − φ0
2( )
2+ 1
2 (∇φ(x))2[ ]−∞
∞
∫
= dx 14 u0φ0
4 1− tanh2 xr0
2
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
2
+ 12 φ0
r0
21− tanh2 x
r0
2
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
⎡
⎣
⎢ ⎢
⎤
⎦
⎥ ⎥
2 ⎡
⎣
⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥
−∞
∞
∫
= 12 r0 φ0
2 dx 1− tanh2 xr0
2
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
2
−∞
∞
∫ = 12 r0 φ0
2 ⋅2
r0
du 1− tanh2 u( )2
−∞
∞
∫
= 12 r0 φ0
2 ⋅2
r0
dy
1− y 2−1
1
∫ 1− y 2( )
2= 1
2 r0 φ02 ⋅
2
r0
⋅4
3= 4
3 ε0ξ
=2 2
3
r03 / 2
u0
∝ Tc − T( )3 / 2
_____energydensity
________thickness
Higher dimensions
energy of domain wall of size L is
€
2 2
3
r03 / 2
u0
Ld −1
Higher dimensions
energy of domain wall of size L is
€
2 2
3
r03 / 2
u0
Ld −1
energy of a flipped domain is proportional to itssurface area
Nucleation
€
E[φ] = 12 dd x∫ r0φ
2(x) + 12 u0φ
4 (x) + (∇φ(x))2 − 2hφ(x)[ ]With a field,
Nucleation
Suppose we are below Tc (r0 < 0), ϕ = - ϕ0
€
E[φ] = 12 dd x∫ r0φ
2(x) + 12 u0φ
4 (x) + (∇φ(x))2 − 2hφ(x)[ ]With a field,
Nucleation
Suppose we are below Tc (r0 < 0), ϕ = - ϕ0
Then make h > 0€
E[φ] = 12 dd x∫ r0φ
2(x) + 12 u0φ
4 (x) + (∇φ(x))2 − 2hφ(x)[ ]With a field,
Nucleation
Suppose we are below Tc (r0 < 0), ϕ = - ϕ0
Then make h > 0€
E[φ] = 12 dd x∫ r0φ
2(x) + 12 u0φ
4 (x) + (∇φ(x))2 − 2hφ(x)[ ]With a field,
It is now favorable to change to ϕ = + ϕ0
Nucleation
Suppose we are below Tc (r0 < 0), ϕ = - ϕ0
Then make h > 0€
E[φ] = 12 dd x∫ r0φ
2(x) + 12 u0φ
4 (x) + (∇φ(x))2 − 2hφ(x)[ ]With a field,
It is now favorable to change to ϕ = + ϕ0
but it costs energy to make a local region where ϕ(x) = - ϕ0
Nucleation
Suppose we are below Tc (r0 < 0), ϕ = - ϕ0
Then make h > 0€
E[φ] = 12 dd x∫ r0φ
2(x) + 12 u0φ
4 (x) + (∇φ(x))2 − 2hφ(x)[ ]With a field,
It is now favorable to change to ϕ = + ϕ0
but it costs energy to make a local region where ϕ(x) = - ϕ0
energy of a spherical bubble of radius R of the + phase:
€
E[R] = −2hφ0 ⋅4π
3R3 + 4πR2 ⋅
2 2
3
r03 / 2
u0
Nucleation
Suppose we are below Tc (r0 < 0), ϕ = - ϕ0
Then make h > 0€
E[φ] = 12 dd x∫ r0φ
2(x) + 12 u0φ
4 (x) + (∇φ(x))2 − 2hφ(x)[ ]With a field,
It is now favorable to change to ϕ = + ϕ0
but it costs energy to make a local region where ϕ(x) = - ϕ0
energy of a spherical bubble of radius R of the + phase:
€
E[R] = −2hφ0 ⋅4π
3R3 + 4πR2 ⋅
2 2
3
r03 / 2
u0
=8π
3
r0
u0
−hR3 + 2r0
u0
R2 ⎡
⎣ ⎢
⎤
⎦ ⎥
a Kramers escape problem:
€
E(R) =8π
3
r0
u0
−hR3 + 2r0
u0
R2 ⎡
⎣ ⎢
⎤
⎦ ⎥
a Kramers escape problem:
€
E(R) =8π
3
r0
u0
−hR3 + 2r0
u0
R2 ⎡
⎣ ⎢
⎤
⎦ ⎥
a Kramers escape problem:
€
E(R) =8π
3
r0
u0
−hR3 + 2r0
u0
R2 ⎡
⎣ ⎢
⎤
⎦ ⎥
to find barrier height:maximize E(R)
a Kramers escape problem:
€
E(R) =8π
3
r0
u0
−hR3 + 2r0
u0
R2 ⎡
⎣ ⎢
⎤
⎦ ⎥
to find barrier height:maximize E(R)
€
Eb =64π 2
81⋅
r0
7 / 2
u02h2
a Kramers escape problem:
€
E(R) =8π
3
r0
u0
−hR3 + 2r0
u0
R2 ⎡
⎣ ⎢
⎤
⎦ ⎥
to find barrier height:maximize E(R)
€
Eb =64π 2
81⋅
r0
7 / 2
u02h2
€
τ nucl = exp(Eb ) = (prefactor) ⋅exp const ⋅r0
7 / 2
u02h2
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟ nucleation time
Spinodal decomposition
Ordering by nucleation is a transition from a metastable stateto a stable one.
In spinodal decomposition, one quenches to a temperature belowTc at zero magnetization, an unstable state. Local domains order and grow.
Spinodal decomposition
Ordering by nucleation is a transition from a metastable stateto a stable one.
Spinodal decomposition
Ordering by nucleation is a transition from a metastable stateto a stable one.
In spinodal decomposition, one quenches to a temperature belowTc at zero magnetization
Spinodal decomposition
Ordering by nucleation is a transition from a metastable stateto a stable one.
In spinodal decomposition, one quenches to a temperature belowTc at zero magnetization, an unstable state.
Spinodal decomposition
Ordering by nucleation is a transition from a metastable stateto a stable one.
In spinodal decomposition, one quenches to a temperature belowTc at zero magnetization, an unstable state. Local domains order and grow.
Ising model, T = 0
20 MC sweeps 200 MC sweeps
(from JSethna)
coarsening by shrinkage of small domains
consider a droplet of size R
coarsening by shrinkage of small domains
consider a droplet of size R, use
€
∂φ(x)
∂t= −
δE
δφ(x)
coarsening by shrinkage of small domains
consider a droplet of size R, use
€
∂φ(x)
∂t= −
δE
δφ(x)
spherical symmetry:
€
∂φ∂t
=∂ 2φ
∂r2+
2
r
∂φ
∂r−
∂V (φ)
∂φ
coarsening by shrinkage of small domains
consider a droplet of size R, use
€
∂φ(x)
∂t= −
δE
δφ(x)
spherical symmetry:
€
∂φ∂t
=∂ 2φ
∂r2+
2
r
∂φ
∂r−
∂V (φ)
∂φ
ansatz:
€
φ(r, t) = f r − R(t)( )
coarsening by shrinkage of small domains
consider a droplet of size R, use
€
∂φ(x)
∂t= −
δE
δφ(x)
spherical symmetry:
€
∂φ∂t
=∂ 2φ
∂r2+
2
r
∂φ
∂r−
∂V (φ)
∂φ
ansatz:
€
φ(r, t) = f r − R(t)( )
€
⇒ − ′ f dR
dt= ′ ′ f +
2
r′ f − ′ V ( f )
coarsening by shrinkage of small domains
consider a droplet of size R, use
€
∂φ(x)
∂t= −
δE
δφ(x)
spherical symmetry:
€
∂φ∂t
=∂ 2φ
∂r2+
2
r
∂φ
∂r−
∂V (φ)
∂φ
ansatz:
€
φ(r, t) = f r − R(t)( )
€
⇒ − ′ f dR
dt= ′ ′ f +
2
r′ f − ′ V ( f )
f’(x) is has a localized peak around x = 0 (the domain wall)
coarsening by shrinkage of small domains
consider a droplet of size R, use
€
∂φ(x)
∂t= −
δE
δφ(x)
spherical symmetry:
€
∂φ∂t
=∂ 2φ
∂r2+
2
r
∂φ
∂r−
∂V (φ)
∂φ
ansatz:
€
φ(r, t) = f r − R(t)( )
€
⇒ − ′ f dR
dt= ′ ′ f +
2
r′ f − ′ V ( f )
f’(x) is has a localized peak around x = 0 (the domain wall)Multiply by f’ and integrate through the wall:
coarsening by shrinkage of small domains
consider a droplet of size R, use
€
∂φ(x)
∂t= −
δE
δφ(x)
spherical symmetry:
€
∂φ∂t
=∂ 2φ
∂r2+
2
r
∂φ
∂r−
∂V (φ)
∂φ
ansatz:
€
φ(r, t) = f r − R(t)( )
€
⇒ − ′ f dR
dt= ′ ′ f +
2
r′ f − ′ V ( f )
f’(x) is has a localized peak around x = 0 (the domain wall)Multiply by f’ and integrate through the wall:
€
⇒dR
dt= −
2
R
coarsening by shrinkage of small domains
consider a droplet of size R, use
€
∂φ(x)
∂t= −
δE
δφ(x)
spherical symmetry:
€
∂φ∂t
=∂ 2φ
∂r2+
2
r
∂φ
∂r−
∂V (φ)
∂φ
ansatz:
€
φ(r, t) = f r − R(t)( )
€
⇒ − ′ f dR
dt= ′ ′ f +
2
r′ f − ′ V ( f )
f’(x) is has a localized peak around x = 0 (the domain wall)Multiply by f’ and integrate through the wall:
€
⇒dR
dt= −
2
R
€
⇒ R2(t) = R2(0) − 4t
coarsening by shrinkage of small domains
consider a droplet of size R, use
€
∂φ(x)
∂t= −
δE
δφ(x)
spherical symmetry:
€
∂φ∂t
=∂ 2φ
∂r2+
2
r
∂φ
∂r−
∂V (φ)
∂φ
ansatz:
€
φ(r, t) = f r − R(t)( )
€
⇒ − ′ f dR
dt= ′ ′ f +
2
r′ f − ′ V ( f )
f’(x) is has a localized peak around x = 0 (the domain wall)Multiply by f’ and integrate through the wall:
€
⇒dR
dt= −
2
R
€
⇒ R2(t) = R2(0) − 4t
=> disappearance and coalescence of domainsof size R(0) at time ~ ¼R(0)2
scaling
remaining domains at time t have size ~ t½
scaling
remaining domains at time t have size ~ t½
There is no other (long) length scale for correlations todepend on.
scaling
remaining domains at time t have size ~ t½
suggests scaling of correlations
There is no other (long) length scale for correlations todepend on.
scaling
remaining domains at time t have size ~ t½
suggests scaling of correlations
€
⇒ C(x − ′ x , t) ≡ φ(x, t)φ( ′ x , t) = φ02g
x − ′ x
t1/ 2
⎛
⎝ ⎜
⎞
⎠ ⎟
There is no other (long) length scale for correlations todepend on.
scaling
remaining domains at time t have size ~ t½
suggests scaling of correlations
€
⇒ C(x − ′ x , t) ≡ φ(x, t)φ( ′ x , t) = φ02g
x − ′ x
t1/ 2
⎛
⎝ ⎜
⎞
⎠ ⎟
There is no other (long) length scale for correlations todepend on.
g for Ising model: