1. No category

Sample Problems
0
Q.1 Consider the Lorentz transformation x µ = Λµν (v)xν . Which of the following set of equations respect the
0
0
0
0
0
principle of relativity ? Prove your answer by actual calculation[ x = γx−γvt, t = γt−γvx/c2 but ψ (x , t ) = ψ(x, t),
Lorentz scalar ].
∂
¯h2 ∂ 2
ψ(x, t) = −
ψ(x, t)
∂t
2m ∂x2
(1)
∂
~
ψ(x, t) = −i¯hc(~σ · ∇)ψ(x,
t)
∂t
(2)
i¯h
i¯h
0
0
0
0
0
0
Consider A 0 (x , t ) = γA0 (x, t) − γ vc A1 (x, t), A 1 (x , t ) = γA1 (x, t) − γ vc A0 (x, t) and the equation,
∂2 µ
1 ∂2 µ
A
(x,
t)
=
A (x, t)
∂x2
c2 ∂t2
0
(3)
0
A. x = γx − γvt, t = γt − γ vx
c2 . Thus,
0
0
0
0
∂x ∂
∂t ∂
∂
∂x ∂
∂t ∂
∂
=
+
;
=
+
∂x
∂x ∂x0
∂x ∂t0 ∂t
∂t ∂x0
∂t ∂t0
0
0
0
0
∂x
∂t
v
∂x
∂t
=γ ;
= −γ 2 ;
= −γv ;
=γ
∂x
∂x
c
∂t
∂t
Or,
∂
∂
γv ∂
∂
∂
∂
=γ 0 − 2 0 ;
= −γv 0 + γ 0
∂x
∂x
c ∂t
∂t
∂x
∂t
(4)
(a) If the first equation respects relativity then we must assume also that,
i¯h
∂ 0 0 0
¯h2 ∂ 2 0 0 0
ψ (x , t )
0 ψ (x , t ) = −
∂t
2m ∂x0 2
0
0
(5)
0
Substitute Eq.(4) into Eq.(1) to get (it is given that ψ(x, t) = ψ (x , t )),
µ
¶
µ
¶2
0
0
0
0
0
0
∂
γv ∂
∂
¯h2
∂
i¯h −γv 0 + γ 0 ψ (x , t ) = −
γ 0 − 2 0
ψ (x , t )
∂x
∂t
2m
∂x
c ∂t
(6)
∂ ∂
This is not the same equation as Eq.(5) as terms such as ∂x
are present in the transformed version of Eq.(1)
0
∂t0
whereas there should not be such terms if the equation respected the principle of relativity.
0
0
0
(b) Substitute Eq.(4) into Eq.(2) to get (again ψ(x, t) = ψ (x , t ), also assume that ψ does not depend on y or z,
∂
∂
= ∂z
= 0),
thus ∂y
µ
¶
µ
¶
0
0
0
0
0
0
∂
∂
∂
γv ∂
i¯h −γv 0 + γ 0 ψ (x , t ) = −i¯hc σx γ 0 − 2 0 ψ (x , t )
∂x
∂t
∂x
c ∂t
This is consistent for all values of v only if,
µ
¶
µ
¶
0
0
0
0
0
0
∂
∂
i¯h γ 0 ψ (x , t ) = −i¯hc σx γ 0 ψ (x , t )
∂t
∂x
(7)
(8)
2
and,
µ
¶
µ
¶
0
0
0
0
0
0
∂
γv ∂
i¯h −γv 0 ψ (x , t ) = −i¯hc σx − 2 0 ψ (x , t )
∂x
c ∂t
(9)
In the first one we cancel γ from both sides and this is nothing but Eq.(2) in primed coordinates. In the second one
we cancel −γv from both sides and multiply by c σx and use σx2 = 1 to get the same equation.
0
0
(c) Inverting Eq.(4) is the same as choosing −v instead of v and choosing t , x instead of t, x and vice versa.
∂
∂
γv ∂
∂
∂
∂
+ 2
;
+γ
0 = γ
0 = γv
∂x
∂x
c ∂t ∂t
∂x
∂t
(10)
Let us calculate the D’Almbertian.
∂2
1 ∂2
∆ ≡
=
02 −
∂x
c2 ∂t0 2
µ
0
γv ∂
∂
+ 2
γ
∂x
c ∂t
¶2
1
− 2
c
µ
¶2
∂
∂2
∂
1 ∂2
+γ
=
γv
−
≡∆
∂x
∂t
∂x2
c2 ∂t2
Thus we have to show that
0
0
0
0
∆ Aµ (x , t ) = 0
is the same as,
∆Aµ (x, t) = 0
we write,
0
0
0
A µ (x , t ) = Λµν Aν (x, t)
Then,
0
0
0
0
∆ Aµ (x , t ) = Λµν ∆Aν (x, t) = Λµν × 0 = 0
0
0
0
0
Thus, ∆Aν (x, t) = 0 means ∆ Aµ (x , t ) = 0.
Q.2 A mass m is suspended by a string of constant length l from a fixed support. Another string of the same length
is fixed to this mass which holds another mass also m tied to its free end. Write the diagram of this system displaying
the generalized coordinates. Write down the lagrangian of the system. Write down the lagrange equations.
A. 1 : (x1 , y1 ) and 2 : (x2 , y2 ).
x1 = l sin(θ1 ) ; y1 = −l cos(θ1 )
x2 = l (sin(θ1 ) + sin(θ2 ) ; y2 = −l (cos(θ1 ) + cos(θ2 ))
T =
1
1
m(x˙ 21 + y˙ 12 ) + m(x˙ 22 + y˙ 22 )
2
2
x˙ 2 = l (θ˙1 cos(θ1 ) + θ˙2 cos(θ2 )) ; y˙ 2 = l (θ˙1 sin(θ1 ) + θ˙2 sin(θ2 ))
T =
1 2 ˙2 1 2 ˙2 ˙2
ml θ1 + ml (θ1 + θ2 + 2θ˙1 θ˙2 cos(θ1 − θ2 ))
2
2
3
V = mgy1 + mgy2 = −mgl cos(θ1 ) − mgl (cos(θ1 ) + cos(θ2 ))
L=T −V =
1 2 ˙2 ˙2
ml (2θ1 + θ2 + 2θ˙1 θ˙2 cos(θ1 − θ2 )) + mgl (2 cos(θ1 ) + cos(θ2 ))
2
p1 =
∂L
1
= ml2 (4θ˙1 + 2θ˙2 cos(θ1 − θ2 ))
2
∂ θ˙1
p2 =
∂L
= ml2 (θ˙2 + θ˙1 cos(θ1 − θ2 ))
∂ θ˙2
∂L
1
= ml2 (−2θ˙1 θ˙2 sin(θ1 − θ2 )) − mgl (2 sin(θ1 ))
∂θ1
2
∂L
1
= ml2 (2θ˙1 θ˙2 sin(θ1 − θ2 )) + mgl (−sin(θ2 ))
∂θ2
2
d
1
1
1
p1 = ml2 (4θ¨1 +2θ¨2 cos(θ1 −θ2 ))+ ml2 (−2θ˙2 (θ˙1 − θ˙2 ) sin(θ1 −θ2 )) = ml2 (−2θ˙1 θ˙2 sin(θ1 −θ2 ))−mgl (2 sin(θ1 ))
dt
2
2
2
d
1
p2 = ml2 (θ¨2 + θ¨1 cos(θ1 − θ2 )) + ml2 (−θ˙1 sin(θ1 − θ2 ))(θ˙1 − θ˙2 ) = ml2 (2θ˙1 θ˙2 sin(θ1 − θ2 )) + mgl (−sin(θ2 ))
dt
2
thus,
g
(2θ¨1 + θ¨2 cos(θ1 − θ2 )) + θ˙22 sin(θ1 − θ2 ) = −2 sin(θ1 )
l
(θ¨2 + θ¨1 cos(θ1 − θ2 )) − θ˙12 sin(θ1 − θ2 ) = −
g
sin(θ2 )
l
Q.3 A massless rod of length 2l is free to oscillate in the vertical plane by fixed nail in the middle of the rod. At the
two ends of the rod two strings of length l each are attached. At the lower end of each string a mass m is attached.
Write the diagram of this system displaying the generalized coordinates. Write down the lagrangian of the system.
Write down the lagrange equations.
Q.4 A rod of length l and mass m is fixed at one end. The other end makes an angle θ from the vertical. Assume
that a tiny spring is hidden in the fixed end so that if the rod tilts by an angle θ the spring acquires a potential energy
1
2
2 αθ . Write down the lagrangian of the rod taking into account gravity also. Write down the lagrange equations.
Solve them in the small angle approximation.
A. The potential energy is V = 12 αθ2 + M g(y − L/2) where y = L/2 cos(θ). The kinetic energy is,
T =
1
2
Z
s=L
s=0
˙ 2
dm (sθ)
4
but dm =
M
L ds.
Thus,
T =
Therefore, I =
ML
3
2
1
2
Z
s=L
˙ 2=
dm (sθ)
s=0
1
1 M L2 ˙ 2
θ = I θ˙2
2 3
2
. The lagrangian is,
2
˙ = 1 M L θ˙2 − 1 αθ2 − M gL (cos(θ) − 1)
L(θ, θ)
2 3
2
2
pθ =
∂L
M L2 ˙
=
θ
3
∂ θ˙
∂L
M gL
= −αθ +
sin(θ)
∂θ
2
The equations of motion are,
M L2 ¨
M gL
M gL
θ = −αθ +
sin(θ) ≈ −(α −
)θ
3
2
2
If α >
is Ω =
M gL
q2
then the motion is simple harmonic, otherwise it is unstable. For simple harmonic motion the frequency
3
M L2 (α
−
M gL
2 )
and the period is Tp =
2π
Ω
=
q
2π
3
M L2
(α− M2gL )
.
Q.5 Repeat Q.3 if there are two rods (mass m each) – second is attached to the end of the first and there is the same
tiny spring in this connection point. Now the potential energy in the spring in between the two rods is 12 α(θ1 − θ2 )2 .
Generalize to N rods.
A. The potential energy is simple. V = M gy1 + M gy2 where y1 and y2 are the heights of the centers of mass of
the two rods. y1 = L2 cos(θ1 ) and y2 = L cos(θ1 ) + L2 cos(θ2 ).
2
2
From Q.4 the kinetic energy of the first rod is T1 = M6L θ˙1 . For the kinetic energy of the second rod, we imagine
a small piece of the second rod dm of length ds.
Z s=L
Z s=L
0
0
1
T2 =
dT2 =
dm (x˙ 2 + y˙ 2 )
2
s=0
s=0
0
0
0
0
where x = L sin(θ1 ) + s sin(θ2 ) and y = L cos(θ1 ) + s cos(θ2 ) also x˙ = Lθ˙1 cos(θ1 ) + s θ˙2 cos(θ2 ) and y˙ =
−L θ˙1 sin(θ1 ) − s θ˙2 sin(θ2 ) and dm = M
L ds.
Z
Z
s=L
T2 =
s=L
dT2 =
s=0
s=0
Z
1
dm ((Lθ˙1 cos(θ1 ) + s θ˙2 cos(θ2 ))2 + (−L θ˙1 sin(θ1 ) − s θ˙2 sin(θ2 ))2 )
2
s=L
T2 =
s=0
2
2
1 M
ds (L2 θ˙1 + s2 θ˙2 + 2Lθ˙1 s θ˙2 cos(θ1 − θ2 ))
2 L
or,
T2 =
2
1 2
1
M L2 (θ˙1 + θ˙2 + θ˙1 θ˙2 cos(θ1 − θ2 ))
2
3
This may also be written as,
T2 = T2,CM + T2,about−CM
5
T2,about−CM =
where ICM =
1
ICM θ˙22
2
M L2
12 .
1
2
M (x˙ 22,CM + y˙ 2,CM
)
2
T2,CM =
x2,CM = L sin(θ1 ) +
L
L
sin(θ2 ) ; y2,CM = L cos(θ1 ) + cos(θ2 )
2
2
L
L
x˙ 2,CM = Lθ˙1 cos(θ1 ) + θ˙2 cos(θ2 ) ; y2,CM = −Lθ˙1 sin(θ1 ) − θ˙2 sin(θ2 )
2
2
or,
2
1
1 2
1
M L2 (θ˙1 + θ˙2 ) + M L2 θ˙1 θ˙2 cos(θ1 − θ2 )
2
4
2
T2,CM =
Thus T2 = T2,CM + T2,about−CM . For N rods we proceed as follows.
0
xk+1 = L
k
X
0
sin(θj ) + s sin(θk+1 ) ; yk+1 = L
j=1
k
X
cos(θj ) + s cos(θk+1 )
j=1
The total kinetic energy is,
T =
N Z
X
k=1
0
x˙ k = L
k−1
X
s=L
s=0
0
0
1
dm (x˙ k2 + y˙ k2 )
2
0
cos(θj ) θ˙j + s θ˙k cos(θk ) ; y˙ k = −L
j=1
0
0
k−1
X
2
θ˙j θ˙l cos(θj − θl ) + s2 θ˙k +
k=1
s=L
s=0
k−1
X
2Ls θ˙k θ˙j cos(θj − θk )
j=1
l,j=1
T =
θ˙j sin(θj ) − s θ˙k sin(θk )
j=1
x˙ k2 + y˙ k2 = L2
N Z
X
k−1
X
k−1
k−1
X
X
2
1
dm (L2
θ˙j θ˙l cos(θj − θl ) + s2 θ˙k +
2Ls θ˙k θ˙j cos(θj − θk ))
2
j=1
l,j=1
In other words,
T =
N
X
1
k=1
2
k−1
X
M L2 (
l,j=1
k−1
1 2 X ˙ ˙
θ˙j θ˙l cos(θj − θl ) + θ˙k +
θk θj cos(θj − θk ))
3
j=1
and
V =
N
X
k=1
M g yk,CM =
N
X
k=1
M g(L
k−1
X
j=1
cos(θj ) +
N −1
L
1 X
cos(θk )) + α
α(θk+1 − θk )2
2
2
The lagrangian is,
˙ =T −V =
L(θ, θ)
k=1
6
N
X
1
k=1
2
M L2 (
k−1
X
l,j=1
k−1
N
k−1
N −1
X
X
L
1 2 X ˙ ˙
1 X
θ˙j θ˙l cos(θj −θl )+ θ˙k +
θk θj cos(θj −θk ))−
M g(L
cos(θj )+ cos(θk ))− α
α(θk+1 −θk )2
3
2
2
j=1
j=1
k=1
k=1
Q.6 A system is described by the following lagrangian.
L=
1
1
m q˙1 q˙2 − mα q1 q2
2
2
Show that q1 → q1 (s) = (1 + s)q1 and q2 → q2 (s) = q2 /(1 + s) is a symmetry of the lagrangian. Using Noether’s
theorem find the conserved quantity associated with this symmetry. Verify that this is a constant explicitly by
differentiating it.
A. Since q1 (s)q2 (s) = q1 q2 and the lagrangian depends only on such products, this is a symmetry. The generalized
momentum and generalized force are,
∂L
1
∂L
1
∂L
1
∂L
1
= mq˙2 ;
= − mα q2 ;
= mq˙1 ;
= − mα q1
∂ q˙1
2
∂q1
2
∂ q˙2
2
∂q2
2
Noether’s constant is,


X dqj ∂L

Q=
ds
∂
q
˙
j
j
1
1
1
= q1 mq˙2 − q2 mq˙1 = m(q1 q˙2 − q2 q˙1 )
2
2
2
s=0
The equations of motion are,
1
1
1
1
m¨
q2 = − mα q2 ; m¨
q1 = − mα q1
2
2
2
2
Hence,
dQ
1
1
1
= m(q1 q¨2 − q2 q¨1 ) = (−q1 mα q2 + q2 mα q1 ) = 0
dt
2
2
2
Q.7 A system is described by the following lagrangian.
L=
(1+s)
1 q˙1 q˙2
1
m
− mα Log[q1 ] Log[q2 ]
2 q1 q2
2
1
and q2 → q2 (s) = q2(1+s) is a symmetry of the lagrangian. Using Noether’s theorem
Show that q1 → q1 (s) = q1
find the conserved quantity associated with this symmetry. Verify that this is a constant explicitly by differentiating
it.
A : To verify that it is a symmetry we proceed as follows. Log[q1 (s)] = (1+s)Log[q1 ] and Log[q2 (s)] =
Hence,
Log[q1 (s)]Log[q2 (s)] = Log[q1 ]Log[q2 ]
and,
(
d
d
d 1
d
Log[q1 (s)])( Log[q1 (s)]) = ( (1 + s)Log[q1 ])(
Log[q1 (s)])
dt
dt
dt
dt 1 + s
1
(1+s) Log[q2 ].
7
Or,
q˙1 (s)q˙2 (s)
q˙1 q˙2
=
q1 (s)q2 (s)
q1 q2
These two put together show that this is a symmetry.
∂L
1
q˙2
∂L
1 q˙1 q˙2
1
1
= m
;
= − m 2 − mα
Log[q2 ]
∂ q˙1
2 q1 q2 ∂q1
2 q1 q2
2
q1
∂L
1
q˙1
∂L
1 q˙1 q˙2
1
1
= m
;
= − m 2 − mα
Log[q1 ]
∂ q˙2
2 q1 q2 ∂q2
2 q2 q1
2
q2
Noether’s constant is,


X dqj ∂L

Q=
ds ∂ q˙j
j
s=0
d
Log[q1 (s)] = Log[q1 ]
ds
But,
dq1 1
d
Log[q1 (s)] =
ds
ds q (1+s)
1
Hence,
dq1
(1+s)
= q1
Log[q1 ]
ds
d
1
Log[q2 (s)] = −
Log[q2 ]
ds
(1 + s)2
But,
1
dq2
q 1+s
= − 2 2 Log[q2 ]
ds
(1 + s)
1
q˙2
1
q˙1
Q = q1 Log[q1 ] m
− q2 Log[q2 ] m
2 q1 q2
2 q1 q2
or,
Q=
µ
¶
1
q˙2
q˙1
m Log[q1 ] − Log[q2 ]
2
q2
q1
Q.8 A system is described by the following lagrangian.
L=
1
1 2
(q˙1 + q˙22 ) − α(q14 − 8q13 q2 + 24q12 q22 − 32q1 q23 + 16q24 )
2
2
Show that q1 → q1 (s) = q1 + 2s and q2 → q2 (s) = q2 + s is a symmetry of the lagrangian. Using Noether’s theorem
find the conserved quantity associated with this symmetry. Verify that this is a constant explicitly by differentiating
it.
8
A. An easy way to do this is if you see that, (q14 −8q13 q2 +24q12 q22 −32q1 q23 +16q24 ) = (q1 −2q2 )4 . Since (q1 (s)−2q2 (s)) =
(q1 − 2q2 ), and q˙1 (s) = q˙1 and q˙2 (s) = q˙2 , this is a symmetry. Noether’s constant is,
¶
µ
dq2 ∂L
dq1 ∂L
Q=
+
= 2q˙1 + q˙2
ds ∂ q˙1 (s)
ds ∂ q˙2 (s) s=0
The equations of motion are,
q¨1 = 4(q1 − 2q3 )3 ; q¨2 = −8(q1 − 2q3 )3
Hence,
d
Q = 2¨
q1 + q¨2 = 8(q1 − 2q3 )3 − 8(q1 − 2q3 )3 = 0
dt
Q.9 Consider N masses subject to a gravitational acceleration g. These masses also experience an internal friction
known as viscosity. This can be modelled by an addition terms show below.
m¨
xk = η(x˙ k+1 + x˙ k−1 − 2x˙ k ) + mg ≡ Fk
(11)
This system corresponds to particles moving under the action of a constant force and also has internal friction or
viscosity. Find the velocity of k-th particle system with periodic boundary condition and initial conditions where all
velocities are zero. In the continuum limit what is the nature of this equation Eq.(11) ?
A. Define vk = x˙ k and G(z, t) =
PN
k=1
vk (t) z k . Therefore,
m
dvk
= η(vk+1 + vk−1 − 2vk ) + mg
dt
(12)
and,
N
N
k=1
k=1
X dvk (t)
X
∂G(z, t)
=m
zk =
(η(vk+1 + vk−1 − 2vk ) + mg) z k
m
∂t
dt
=
N
X
η(vk+1 + vk−1 − 2vk ) z k +
k=1
=
N
X
mg z k
(13)
k=1
ηvk+1 z k +
k=1
N
X
N
X
ηvk−1 z k − 2
k=1
N
X
ηvk z k +
k=1
N
X
mg z k
k=1
For example,
N
X
k
ηvk+1 z = z
k=1
−1
N
X
ηvk+1 z
k+1
=z
−1
k=1
N
+1
X
l
ηvl z = z
−1
l=2
= z −1 η
N
X
N
X
ηvl z l − z −1 ηv1 z + z −1 ηvN +1 z N +1
l=1
vl z l − ηv1 + ηvN +1 z N
l=1
and so on. Thus,
N
X
k=1
η(vk+1 + vk−1 − 2vk ) z k =
N
X
k=1
ηvk+1 z k +
N
X
k=1
ηvk−1 z k − 2
N
X
k=1
ηvk z k
9
= z −1 η
N
X
vl z l − ηv1 + ηvN +1 z N +
l=1
N
X
ηvl z (l+1) − ηvN z (N +1) + ηv0 z − 2η
l=1
N
X
vl z l
l=1
= (z −1 + z − 2)ηG(z, t) − ηv1 + ηvN +1 z N − ηvN z (N +1) + ηv0 z
(14)
and,
N
X
mg z k = mgz
k=1
1 − zN
1−z
(15)
Substituting Eq.(14) and Eq.(15) into Eq.(13) we get,
m
∂G(z, t)
= (z −1 + z − 2) η G(z, t) + m b(z, t)
∂t
(16)
where,
m b(z, t) ≡ −ηv1 (t) + ηvN +1 (t) z N − ηvN (t)z (N +1) + ηv0 (t)z + mgz
1 − zN
1−z
or,
−1
∂ h
G(z, t) e−(z +z−2)
∂t
η
m
t
i
= b(z, t) e−(z
−1
+z−2)
η
m
t
Let us choose periodic boundary v1 (t) = vN +1 (t) and v0 (t) = vN (t) and initial conditions as v1 (0) = v2 (0) = ... =
vN −1 (0) = vN (0) = 0 (nothing is moving at time t = 0). In this case G(z, 0) = 0 and m b(z, t) = −ηv1 (t) (1 − z N ) +
N
ηvN (t)z(1 − z N ) + mgz 1−z
1−z .
Z
G(z, t) = e(z
−1
+z−2)
η
m
t
t
0
0
dt b(z, t ) e−(z
−1
+z−2)
η
m
t
0
0
iθ
Set z = e . Then,
N
X
G(z, t) =
eikθ vk (t) = e−4sin
η
2 θ
(2) m
Z
t
t
0
0
dt b(eiθ , t ) e4sin
η
2 θ
(2) m
t
0
0
k=1
We now multiply both sides by e−inθ and integrate from 0 to 2π.
Z 2π
Z t
0
0
2 θ
2π vn (t) =
dθ e−inθ
dt b(eiθ , t ) e4sin ( 2 )
0
η
m
0
(t −t)
0
Or,
Z
2π
2π vn (t) =
Z
t
dθ e−inθ
0
0
dt (−
0
0
0
η
η
1 − eiN θ 4sin2 ( θ )
2
v1 (t ) (1 − eiN θ ) + vN (t )eiθ (1 − eiN θ ) + geiθ
)e
m
m
1 − eiθ
From this we may write two simultaneous equations for v1 (t) and vN (t).
Z t
Z 2π
0
0
0
η
η
1 − eiN θ 4sin2 ( θ )
2
dt (− v1 (t ) (1 − eiN θ ) + vN (t )eiθ (1 − eiN θ ) + geiθ
2π v1 (t) =
dθ e−iθ
)e
m
m
1 − eiθ
0
0
Z
2π
2π vN (t) =
0
Z
dθ e−iN θ
t
0
dt (−
0
0
η
m
η
m
0
0
η
η
1 − eiN θ 4sin2 ( θ )
2
v1 (t ) (1 − eiN θ ) + vN (t )eiθ (1 − eiN θ ) + geiθ
)e
m
m
1 − eiθ
(t −t)
0
(t −t)
η
m
0
(t −t)
First we have to solve these two equations and substitute the answers into the earlier equation for vn (t). The continuum
limit is easier. Define v(s, t) ≡ vk (t) where s = k² where ² small quantity. Thus v(s ± ², t) = vk±1 (t). Or,
vk+1 + vk−1 − 2vk = v(s + ², t) + v(s − ², t) − 2v(s, t) = ²2
∂ 2 v(s, t)
∂s2
10
Hence the equation for v˙ (Eq.(12)) is,
m
∂v(s, t)
∂ 2 v(s, t)
= η²2
+ mg
∂t
∂s2
This is nothing but the diffusion equation with a driving term. The diffusion constant is D =
η²2
m .
Q.10 Consider the N masses with coordinates x1 , x2 , ...., xN moving in one dimension. Define the density and
current density as follows.
ρ(x, t) = m
N
X
δ(xk (t) − x)
k=1
j(x, t) =
N
X
pk (t) δ(xk (t) − x)
k=1
where pk = mx˙ k . Prove the continuity equation.
∂
∂
ρ(x, t) +
j(x, t) = 0
∂t
∂x
0
0
0
Find the Poisson brackets {ρ(x, t), j(x , t)}, {ρ(x, t), ρ(x , t)} and {j(x, t), j(x , t)}. Write j(x, t) = ρ(x, t)v(x, t) and
0
0
find {v(x, t), ρ(x , t)} and {v(x, t), v(x , t)}.
A. Upon differentiation we get,
N
N
k=1
k=1
X
X
0
0
∂
∂
ρ(x, t) = m
x˙ k (t)δ (xk (t) − x) ;
j(x, t) = −
pk (t) δ (xk (t) − x)
∂t
∂x
Adding these two we get the continuity equation since mx˙ k = pk . For evaluate Poisson brackets, we proceed as
follows.
!
Ã
0
0
N
X
0
∂ρ(x, t) ∂ρ(x , t) ∂ρ(x, t) ∂ρ(x , t)
{ρ(x, t), ρ(x , t)} =
−
∂xk
∂pk
∂pk
∂xk
k=1
0
{ρ(x, t), j(x , t)} =
N
X
Ã
k=1
0
{j(x, t), j(x , t)} =
N
X
k=1
Ã
0
0
0
0
∂ρ(x, t) ∂j(x , t) ∂ρ(x, t) ∂j(x , t)
−
∂xk
∂pk
∂pk
∂xk
∂j(x, t) ∂j(x , t) ∂j(x, t) ∂j(x , t)
−
∂xk
∂pk
∂pk
∂xk
Since ρ does not depend on pk we get,
0
∂ρ(x, t)
∂ρ(x, t)
=0;
= mδ (xk (t) − x)
∂pk
∂xk
and,
0
∂j(x, t)
∂j(x, t)
= δ(xk (t) − x) ;
= pk (t)δ (xk (t) − x)
∂pk
∂xk
Therefore,
0
{ρ(x, t), ρ(x , t)} = 0
!
!
11
0
{ρ(x, t), j(x , t)} =
N
X
0
0
0
0
0
mδ (xk (t) − x)δ(xk (t) − x ) = δ (x − x)ρ(x , t)
k=1
0
0
0
{j(x, t), j(x , t)} = δ (x − x)
N ³
X
´
0
0
0
0
pk (t)δ(xk (t) − x ) + δ(xk (t) − x)pk (t) = δ (x − x)(j(x, t) + j(x , t))
k=1
Since j(x, t) = ρ(x, t)v(x, t), and {A, BC} = B{A, C} + {A, B}C we get,
0
0
0
0
0
0
{ρ(x, t), j(x , t)} = ρ(x , t){ρ(x, t), v(x , t)} = δ (x − x)ρ(x , t)
Or,
0
0
0
{ρ(x, t), v(x , t)} = δ (x − x)
Also,
0
0
0
0
0
0
0
{j(x, t), j(x , t)} = {ρ(x, t)v(x, t), ρ(x , t)v(x , t)} = {ρ(x, t)v(x, t), ρ(x , t)}v(x , t) + ρ(x , t){ρ(x, t)v(x, t), v(x , t)}
0
0
0
0
0
0
0
0
= −δ (x − x )ρ(x, t)v(x , t) + ρ(x , t)δ (x − x)v(x, t) + ρ(x , t)ρ(x, t){v(x, t), v(x , t)}
But,
0
0
0
0
0
0
−δ (x − x )ρ(x, t)v(x , t) + ρ(x , t)δ (x − x)v(x, t) = −[
= −[
0
0
0
0
0
0
d
d
δ(x − x )v(x , t)]ρ(x, t) + ρ(x , t)[ 0 δ(x − x)v(x, t)]
dx
dx
0
0
0
0
d
d
δ(x − x )v(x, t)]ρ(x, t) + ρ(x , t)[ 0 δ(x − x)v(x , t)]
dx
dx
0
0
0
0
0
0
0
0
0
0
= −δ (x − x )v(x, t)ρ(x, t) − δ(x − x )v (x, t)ρ(x, t) + ρ(x , t)δ (x − x)v(x , t) + ρ(x , t)δ(x − x)v (x , t)
0
0
0
= δ (x − x)(j(x, t) + j(x , t))
Hence,
0
{v(x, t), v(x , t)} = 0
Q.11 Consider the same ρ(x, t) and j(x, t) = ρ(x, t)v(x, t) but with the equations of motion as given in Q.9 plus
another given force which is due to pressure fk (x1 , x2 , ...xN ). This corresponds to particles moving under the action
of a constant force like gravity, a given pressure distribution and also has internal friction or viscosity. Derive a closed
set of equations for ρ(x, t) and v(x, t) called the Navier-Stokes equations. You may have to work in the continuum
limit at the outset.
A. One such equation we already know. This is the equation of continuity.
∂ρ(x, t)
∂
= − (ρ(x, t)v(x, t))
∂t
∂x
We now wish to derive a similar formula for
∂v(x,t)
∂t .
(17)
The dynamical variable xk obeys the following set of equations.
m¨
xk = η(x˙ k+1 + x˙ k−1 − 2x˙ k ) + mg + fk (x1 , ..., xN )
(18)
12
Since j(x, t) = ρ(x, t)v(x, t) and j and ρ are the sum of several delta functions, it follows that v is an ordinary function
of x.
j(x, t) =
N
X
pk (t)δ(x − xk (t)) = ρ(x, t)v(x, t) = m
N
X
δ(x − xk (t))v(x, t) = m
k=1
k=1
N
X
δ(x − xk (t))v(xk (t), t)
k=1
Therefore,
x˙ k (t) ≡ v(xk (t), t)
(19)
Thus,
x
¨k (t) =
where vx (x, t) =
d
v(xk (t), t) = x˙ k (t)vx (xk (t), t) + vt (xk (t), t) = vx (xk (t), t)vx (xk (t), t) + vt (xk (t), t)
dt
∂
∂x v(x, t)
and vt (x, t) =
∂v(x,t)
∂t .
(20)
We substitute Eq.(19) and Eq.(20) into Eq.(18) to get,
m (v(xk (t), t)vx (xk (t), t) + vt (xk (t), t)) = η(v(xk+1 (t), t) + v(xk−1 (t), t) − 2v(xk (t), t)) + mg + fk (x1 , ..., xN )
(21)
We multiply this with δ(x − xk (t)) and sum over all k to get,
ÃN
!
N
X
X
m
δ(x − xk (t)) v(xk (t), t)vx (xk (t), t) +
δ(x − xk (t)) vt (xk (t), t)
k=1
= η(
N
X
k=1
δ(x − xk (t)) v(xk+1 (t), t) +
k=1
N
X
δ(x − xk (t)) v(xk−1 (t), t) − 2
k=1
+
N
X
δ(x − xk (t)) mg +
k=1
N
X
δ(x − xk (t)) v(xk (t), t))
k=1
N
X
δ(x − xk (t)) fk (x1 , ..., xN )
(22)
k=1
Because the delta function forces xk (t) = x where ever there is xk we can replace by x. But it is not clear what we
should use when there is xk+1 or xk−1 . Define,
u(x) = (
N
X
δ(x−xk (t)) v(x+xk+1 (t)−x, t)+
k=1
N
X
δ(x−xk (t)) v(x+xk−1 (t)−x, t)−2
k=1
N
X
δ(x−xk (t)) v(x+xk (t)−x, t))
k=1
In the continuum limit,
u(x) =
N
X
δ(x − xk (t)) vx (x, t)(xk+1 (t) + xk−1 (t) − 2x)
k=1
N
N
1X
1X
δ(x − xk (t)) vxx (x, t)(xk+1 (t) − x)2 +
δ(x − xk (t)) vxx (x, t)(xk−1 (t) − x)2
2
2
k=1
k=1
P
2
In the continuum limit, we replace (xk+1 − xk ) ≈ P < (xk+1 − xk )2 >= N1 k (xk+1 − xk )2 = ∆x2 . Therefore
xk+1 + xk−1 − 2xk ≈< xk+1
P+ xk−1 − 2xk >= 0. Also k δ(x − xk (t)) fk is the force density at point x, this is nothing
but the pressure gradient k δ(x−xk (t)) fk = −∇p. Thus we may rewrite Eq.(22) as follows. The parameter η ∆x2 is
0
0
now identified with the viscosity coefficient η. The idea being that η → ∞ while ∆x → 0 such that η ∆x2 = η < ∞.
η
ρ(x, t) (v(x, t)vx (x, t) + vt (x, t)) = ρ(x, t)vxx (x, t) + ρ(x, t) g − ∇p
(23)
m
This is nothing but Navier Stokes equation. In three dimensions we have,
η
ρ(x, t) (v(x, t) · ∇ v(x, t) + vt (x, t)) = ρ(x, t) ∇2 v(x, t) + ρ(x, t) g − ∇p(x, t)
(24)
m
The term below is also called the convective (time) derivative.
+
Dv
= v(x, t) · ∇ v(x, t) + vt (x, t)
Dt