Download E8481: Mass on a spring

E8481: Mass on a spring
Submitted by: Michaele Epshtein
The problem:
A balance for measuring weight consists of a sensitive spring which hangs from a xed point. The
spring constant is K . The balance is at temperature T and gravity acceleration is g in x direction.
A small mass m hangs at the end of the spring. There is an option to apply an external force F (t),
to which x is conjugate or apply an external Vector potential A(t).
(a)
Find the partition function Z.
(b)
Find the average hx(t)i and hx2 (t)i.
(c)
Find the uctuation hδx2 i = h(x − hxi)2 i, what is the minimal m which can be meaningfully
measured?
(d)
Write the Langevin equation for x(t) with friction γ and a random force f (t).
(e)
Assuming hf (t)f (0)i = Cδ(t), Find hx̃2 (t)i and the intensity of the random force f (t) that acts
on the mass , from (b) nd the coecient C.
(f )
Describe the external force by a scalar potential and demonstrate FDT.
(g)
Describe the external force by a vector potential and demonstrate FDT.
The solution:
(a) The Hamiltonian of the system is:
H=
p2
1
+ Kx2 − mgx
2m 2
1
Z=
2π~
+∞
2
− βp
2m
e
+∞
dp
−∞
1
Z=
2π~
+∞
r
+∞
2
− βp
2m
e
m
Kπ
e
1
dx =
2π~
r
2mπ
β
−∞
−∞
1
=
~β
2
−( βKx
−mgβx)
2
dp
2
−( βKx
−mgβx)
2
e
q
β
−(y− 2K
mg)2
e
−∞
e
−(y 2 −
q
−(y 2 −
q
2β
mgy)
K
r
βK
x}
2
r
βK
x}
2
dy = {y =
−∞
1
dx =
2π~
−∞
+∞
+∞
r
2mπ
β
+∞
e
−∞
e
βm2 g 2
2K
1
dy =
~β
(b) By denition of hxi
1
r
m βm2 g2
e 2K
K
2β
mgy)
K
dy = {y =
1 1
hx(t)i =
Z 2π~
+∞
e
+∞
2
− βp
2m
−∞
r
=
x e−(
dp
βKx2
−mgβx)
2
dx
−∞
βK − βm2 g2
1 −( βKx2 −mgβx) +∞ mg
2
| +
e 2K {−
e
2π
βK
K
−∞
+∞
e−(
βKx2
−mgβx)
2
dx}
−∞
The rst part of the integral goes to zero and we get
hxi =
mg
k
From the virial theoremhx ∂H
∂x i = kB T
hx(Kx − mg)i = Khx2 i − mghxi = Khx2 i − mg(
hx2 i =
mg
) = kB T
K
kB T
mg 2
+(
)
K
K
(c) hδx2 i = h(x − hxi)2 i = hx2 i − hx2 i =
kB T
K
The Fluctuation of the spring should be much smaller than the tension itself, Therefor
kB T
hxi2 hδx2 i →
K
mg
k
!2
And, nally, we get following condition: m q
KkB T
g2
(d) We use Hamilton equations:
1.
∂H
∂p
=
2.
∂H
∂x
= Kx − mg = −ṗ = −mẍ
p
m
= ẋ
from the two equations, friction γ that interact with velocity and a random force A(t) we nd a
Langevin's equation (the system is in equilibrium)
mẍ + mγ ẋ + Kx − mg = mf (t)
(e) By set x̃(t) = x(t) + hx(t)iinto the Langevin's equation that we got and using Fourier transform
∂n
n
functional relationships ∂t
n x(t) = (iω) X(ω) we get
(−mω 2 + iωmγ + K)X̃(ω) = mf (ω) we rearrange a little to have
2
|X̃(ω)|2
1 |Ṽ (ω)|2
m2
=
=
|f (ω)|2
ω 2 |f (ω)|2
(mω 2 − k)2 + m2 ω 2 γ 2
The intensity of the random force A(t) that acts on the mass
+∞
νA (ω) =
+∞
CA (t) e
iωt
−∞
−∞
From the following relation
hx̃(t)x̃(t + τ )i =
+∞
Cδ(t)eiωt dt = C where CA (t) = hf (t)f (0)i
dt =
2
˜
νx (ω)
|X(ω)|
C
=
→
ν
(ω)
=
x
νA (ω)
|f (ω)|2
(ω 2 − k/m)2 + ω 2 γ 2
νx (ω)e−iωτ dω → hx̃2 (t)i =
−∞
+∞
−∞
νx (ω)dω =
1
2π
dω
(ω 2 −K/m)2 +γ 2 ω 2
=
mπC
2πγK
By comparing this result with (b) we can nd C
hx̃2 (t)i = h(x − hxi)2 i = hx2 i − hxi2 =
hx̃2 (t)i =
kB T
K
kB T
2kB T γ
mπC
=
→ C=
2πγK
K
m
and nelly
νx (ω) =
2kB T γm
(ω 2 − k/m)2 + ω 2 γ 2
νv (ω) =
2kB T γmω 2
(ω 2 − k/m)2 + ω 2 γ 2
(f) Because of the external force that couples to x the system isn't in equilibrium, namely there is
dissipation
H=
p2
1
+ Kx2 − mgx + F (t)x
2m 2
We use Hamilton equations:
1.
∂H
∂p
=
2.
∂H
∂x
= Kx − mg + F = −ṗ = −mẍ
p
m
= ẋ
from the two equations, friction γ that interact with velocity and a random force f (t) like in (d) we
nd a Langevin's equation
mẍ + mγ ẋ + Kx − mg = −F + mf (t)
3
By same way as in (e) we get (mω 2 − iωmγ − K)X̃(ω) = F (ω) we rearrange a little to get the
susceptibility χx that describes the linear response of x to F
hX̃(ω)i
1
=
= χx
2
F (ω)
mω − K − iωmγ
Im χx =
(mω 2
mωγ
− k)2 + m2 ω 2 γ 2
By compering with νx (ω) we show that FDT holds.
νx (ω) =
2kB T
2kB T mγ
Im χx =
2
ω
(mω − k)2 + m2 ω 2 γ 2
(g) Because of the external vector potential A(t) that couples to v the system isn't in equilibrium,
namely there is dissipation
H=
p
p2
1
+ Kx2 − mgx + A(t)
2m 2
m
We use Hamilton equations:
1.
∂H
∂p
=
2.
∂H
∂x
= Kx − mg = −ṗ = −mẍ + Ȧ
p
m
+
A
m
= ẋ
from the two equations, friction γ that interact with velocity and a random force f (t) like in (d) we
nd a Langevin's equation
mẍ + mγ ẋ + Kx − mg = Ȧ + mf (t)
By same way as in (e) we get (−mω 2 − iωmγ + K)X̃(ω) = iωA(ω) we rearrange a little to get
the susceptibility χv that describes the linear response of v to A
hṼ (ω)i
hX̃(ω)i
ω2
= iω
=
= χv
A(ω)
A(ω)
mω 2 − K − iωmγ
Im χv =
mω 3 γ
(mω 2 − k)2 + m2 ω 2 γ 2
By compering with νv (ω) we show that FDT holds.
νx (ω) =
2kB T
2kB T mγω 2
Im χx =
ω
(mω 2 − k)2 + m2 ω 2 γ 2
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