Molecular Dynamics And Simulation

1. ASSIGNMENT 1 SOLUTIONS (FM: 25)
The marking scheme is indicated.
1-3: [4]
Given,
𝑉(𝑥) = − ∫ 𝐹(𝜉)𝑑𝜉
𝑥
0
=
1
2
𝑘𝑥2
Where,
𝐹(𝜉) = −𝑘𝜉 -(1)
And 𝜉 = 𝑥 − 𝑥0
Thus, 𝑑𝜉 = 𝑑𝑥
So, − ∫ −𝑘(𝑥 − 𝑥0)𝑑𝑥 =
1
2
𝑘𝑥2
𝑥
0
Or
𝑘𝑥2
2
− 𝑘𝑥𝑥0 + 0 =
1
2
𝑘𝑥2
[0.5]
Or 𝑘𝑥𝑥0 = 0
Since, 𝑘, 𝑥 ≠ 0; therefore, 𝑥0 = 0
Also,
𝐹 = 𝑚𝑎 = 𝑚
𝑑2𝑥
𝑑𝑡2
= 𝑚
𝑑2(𝜉+𝑥0)
𝑑𝑡2
= 𝑚
𝑑2𝜉
𝑑𝑡2
-(2)
From eq. (1) and (2):
−𝑘𝜉 = 𝑚
𝑑2𝜉
𝑑𝑡2
Or
𝑑2𝜉
𝑑𝑡2
+
𝑘
𝑚
𝜉 = 0
On solving,
𝜉(𝑡) = 𝑎𝑠𝑖𝑛𝜔𝑡 + 𝑏𝑐𝑜𝑠𝜔𝑡 where 𝜔 = √
𝑘
𝑚
[0.5]
Or 𝜉(𝑡) = 𝐴𝑠𝑖𝑛(𝜔𝑡 + 𝜙)
Also, 𝑥(𝑡) = 𝑥0 + 𝜉(𝑡)
We can write,
𝑥(𝑡) = 𝐴𝑠𝑖𝑛(𝜔𝑡 + 𝜙) [0.5]
Kinetic energy, 𝐾(𝑥) =
1
2
𝑚𝑣2
=
1
2
𝑚 (
𝑑𝑥
𝑑𝑡
)
2
𝑣(𝑡) =
𝑑𝑥
𝑑𝑡
= 𝐴𝜔cos (𝜔𝑡 + 𝜙) [0.5]
Total energy = 𝐾 + 𝑉

2. =
1
2
𝑚𝐴2
𝜔2
cos2
(𝜔𝑡 + 𝜙) +
1
2
𝑘𝐴2
sin2
(𝜔𝑡 + 𝜙)
=
1
2
𝑚𝐴2
(
𝑘
𝑚
) cos2
(𝜔𝑡 + 𝜙) +
1
2
𝑘𝐴2
sin2
(𝜔𝑡 + 𝜙)
=
1
2
𝑘𝐴2
Hence, Total energy is constant and is not a function of time. [1]
Whereas, kinetic and potential energies behave as a function of time as:
𝐾(𝑡) =
1
2
𝑘𝐴2
cos2
(𝜔𝑡 + 𝜙)
𝑉(𝑡) =
1
2
𝑘𝐴2
sin2
(𝜔𝑡 + 𝜙)
[1]
1-10: [5]
Lagrange equation for particle moving in two dimensions under a central potential 𝑢(𝑟).
Converting cartesian coordinates to polar coordinates,
𝑥 = 𝑟𝑐𝑜𝑠𝜃 and 𝑦 = 𝑟𝑠𝑖𝑛𝜃
So,
𝑥̇ = −𝑟𝑠𝑖𝑛𝜃 𝜃̇ + 𝑟̇ cos 𝜃
𝑦̇ = 𝑟𝑐𝑜𝑠𝜃 𝜃̇ + 𝑟̇ 𝑠𝑖𝑛𝜃 [1]
𝑥̇2
= 𝑟2
sin2
𝜃 𝜃̇2
+ 𝑟̇2
cos2
𝜃 − 2𝑟𝑟̇𝜃̇𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃
𝑦̇2
= 𝑟2
cos2
𝜃 𝜃̇2
+ 𝑟̇2
sin2
𝜃 + 2𝑟𝑟̇𝜃̇𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃
𝑥̇2
+ 𝑦̇2
= 𝑟2
𝜃̇2
+ 𝑟̇2
𝐾 =
𝑚
2
(𝑥̇2
+ 𝑦̇2) =
𝑚
2
(𝑟2
𝜃̇2
+ 𝑟̇2
) [1]
And 𝑈 = 𝑢(𝑟) = −𝑘𝑟
𝐿 = 𝐾 − 𝑈 =
𝑚
2
(𝑟2
𝜃̇2
+ 𝑟̇2
) + 𝑘𝑟
Lagrange’s equations:
𝑑
𝑑𝑡
(
𝜕𝐿
𝜕𝑟̇
) =
𝜕𝐿
𝜕𝑟
-(1)
𝑑
𝑑𝑡
(
𝜕𝐿
𝜕𝜃̇ ) =
𝜕𝐿
𝜕𝜃
-(2)

3. Solving eqn. (1),
𝑑
𝑑𝑡
(
𝜕𝐿
𝜕𝑟̇
) =
𝑑
𝑑𝑡
(
𝑚
2
(2𝑟̇)) = 𝑚𝑟̈
𝜕𝐿
𝜕𝑟
=
𝑚
2
(2𝑟𝜃̇2
) + 𝑘𝑟̇
Therefore,
𝑚𝑟̈ = 𝑚𝑟𝜃̇2
+ 𝑘𝑟̇ -(3)
Solving eqn. (2),
𝑑
𝑑𝑡
(
𝜕𝐿
𝜕𝜃̇
) =
𝑑
𝑑𝑡
(
𝑚
2
(𝑟2
2𝜃̇)) =
𝑑
𝑑𝑡
(𝑚𝑟2
𝜃̇) = 𝑚. 2𝑟𝑟̇𝜃̇ + 𝑚𝑟2
𝜃̈
𝜕𝐿
𝜕𝜃
= 0
Therefore,
𝑚. 2𝑟𝑟̇𝜃̇ + 𝑚𝑟2
𝜃̈ = 0 -(4)
[2]
Eqn. (4) can also be written as
𝑑
𝑑𝑡
(𝑚𝑟2
𝜃̇) = 0
Which is the law of conservation of angular momentum. [0.5]
Angular momentum remains conserved even if the potential depends on 𝜃 as well, since it is
given that the particle is moving under a central potential. [0.5]
1-11: [6]
Let x, y, z be the cartesian coordinates of point. Then the relationship between cartesian and
spherical coordinates is given as
𝑥 = 𝑟 sin 𝜃 cos 𝜙 ; 𝑦 = 𝑟 sin 𝜃 sin 𝜙 ; 𝑧 = 𝑟 cos 𝜃 [0.5]
𝑣𝑥 =
𝑑𝑥
𝑑𝑡
=
𝑑
𝑑𝑡
( 𝑟 sin 𝜃 cos 𝜙)
𝑣𝑥 = 𝑥̇ = 𝑟̇ sin 𝜃 cos 𝜙 + 𝑟𝜃̇ cos 𝜃 cos 𝜙 − 𝑟𝜙̇ sin 𝜃 sin 𝜙
Similarly,
𝑣𝑦 = 𝑦̇ = 𝑟̇ sin 𝜃 sin 𝜙 + 𝑟𝜃̇ cos 𝜃 sin 𝜙 − 𝑟𝜙̇ sin 𝜃 cos 𝜙

4. 𝑣𝑧 = 𝑧̇ = 𝑟̇ cos 𝜃 − 𝑟𝜃̇ sin 𝜃 [0.5]
(𝑣𝑥)2
= (𝑥̇)2
= ( 𝑟̇ sin 𝜃 cos 𝜙 + 𝑟𝜃̇ cos 𝜃 cos 𝜙 − 𝑟𝜙̇ sin 𝜃 sin 𝜙)2
(𝑣𝑦)2
= (𝑦̇)2
= ( 𝑟̇ sin 𝜃 cos 𝜙 + 𝑟𝜃̇ cos 𝜃 cos 𝜙 − 𝑟𝜙̇ sin 𝜃 sin 𝜙)2
(𝑣𝑧)2
= (𝑧̇)2
= (𝑟̇ cos 𝜃 − 𝑟𝜃̇ sin 𝜃)2
K =
𝑚
2
(𝑣𝜘
2
+ 𝑣𝑦
2
+ 𝑣𝑧
2
) =
𝑚
2
(( 𝑟̇ sin 𝜃 cos 𝜙 + 𝑟𝜃̇ cos 𝜃 cos 𝜙 − 𝑟𝜙̇ sin 𝜃 sin 𝜙)2
+
( 𝑟̇ sin 𝜃 cos 𝜙 + 𝑟𝜃̇ cos 𝜃 cos 𝜙 − 𝑟𝜙̇ sin 𝜃 sin 𝜙)2
+
(𝑟̇ cos 𝜃 − 𝑟𝜃̇ sin 𝜃)2
)
Upon simplification we get
K =
𝑚
2
{ (𝑟̇2
) + (𝑟2
𝜃̇2
) + (𝑟2
sin2
𝜃 𝜙̇2
)} [1]
L = K - U
L =
𝑚
2
{ (𝑟̇2
) + (𝑟2
𝜃̇2
) + (𝑟2
sin2
𝜃 𝜙̇2
)} − 𝑈(𝑟) [0.5]
Lagrangian Equations of motion are:
1.
𝑑
𝑑𝑡
(
𝜕𝐿
𝜕𝑟̇
) −
𝜕𝐿
𝜕𝑟
= 0
2.
𝑑
𝑑𝑡
(
𝜕𝐿
𝜕𝜃̇ ) −
𝜕𝐿
𝜕𝜃
= 0
3.
𝑑
𝑑𝑡
(
𝜕𝐿
𝜕∅̇ ) −
𝜕𝐿
𝜕∅
= 0
Solving,
1.
𝑑
𝑑𝑡
[
𝜕𝐿
𝜕𝑟̇
] =
𝜕𝐿
𝜕𝑟
Substituting values we get,
𝑑
𝑑𝑡
[
1
2
𝑚2𝑟̇] =
1
2
𝑚(2𝑟𝜃̇2
+ 2𝑟𝑠𝑖𝑛2
𝜃𝜙̇2
−
𝜕
𝜕𝑟
𝑈(𝑟)
Upon simplification we get
𝒎𝒓̈ = 𝒎𝒓(𝜽̇ 𝟐
+ 𝐬𝐢𝐧𝟐
𝜽 𝝓
̇ 𝟐
) −
𝝏
𝝏𝒓
𝑼(𝒓)
2.
𝑑
𝑑𝑡
[
𝜕𝐿
𝜕𝜃̇ ] =
𝜕𝐿
𝜕𝜃
𝑑
𝑑𝑡
[
1
2
𝑚𝑟2
2𝜃̇] = 𝑚𝑟2
𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃𝜙̇2
Upon simplification we get
𝒎𝒓𝟐
𝜽̈ + 𝟐𝒎𝒓𝒓̇𝜽̇ = 𝒎𝒓𝟐
𝝓
̇ 𝟐
𝐬𝐢𝐧 𝜽 𝒄𝒐𝒔𝜽
3.
𝑑
𝑑𝑡
(
𝜕𝐿
𝜕∅̇ ) =
𝜕𝐿
𝜕∅

5. 𝑑
𝑑𝑡
[
1
2
𝑚𝑟2
sin2
𝜃 2𝜙̇] = 0
Upon simplification we get
𝒎𝒓𝟐
𝝓
̈ 𝐬𝐢𝐧𝟐
𝜽 + 𝟐𝒎𝒓𝒓̇𝝓
̇ 𝐬𝐢𝐧𝟐
𝜽 + 𝟐𝒎𝒓𝟐
𝛉̇ 𝛟
̇ 𝐬𝐢𝐧 𝜽 𝐜𝐨𝐬 𝜽 = 𝟎
Hence, Lagrangian equation of motions are:
𝒎𝒓̈ − 𝒎𝒓(𝜽̇ 𝟐
+ 𝐬𝐢𝐧𝟐
𝜽 𝝓
̇ 𝟐
) +
𝝏
𝝏𝒓
{𝑼(𝒓)} = 𝟎
𝒎𝜽̈ 𝒓𝟐
+ 𝟐𝒎𝒓𝒓̇𝜽̇ − 𝒎𝒓𝟐
𝝓
̇ 𝟐
𝐬𝐢𝐧 𝜽 𝒄𝒐𝒔𝜽 = 𝟎
𝒎𝒓𝟐
𝝓
̈ 𝐬𝐢𝐧𝟐
𝜽 + 𝟐𝒎𝒓𝒓̇𝝓
̇ 𝐬𝐢𝐧𝟐
𝜽 + 𝟐𝒎𝒓𝟐
𝛉̇ 𝛟
̇ 𝐬𝐢𝐧 𝜽 𝐜𝐨𝐬 𝜽 = 𝟎
[1.5]
Given,
𝐻 = ∑𝑝𝑗𝑞𝑗
̇ − 𝐿
𝑝
𝑗=
𝜕𝐿
𝜕𝑞𝑗
̇
Similarly,
𝑝𝑟 =
𝜕𝐿
𝜕𝑟̇
=
1
2
2𝑚𝑟̇ = 𝑚𝑟̇
𝑝𝜃 =
𝜕𝐿
𝜕𝜃̇
=
1
2
2𝑚𝑟2
𝜃̇ = 𝑚𝑟2
𝜃̇
𝑝𝜙 =
𝜕𝐿
𝜕𝜙̇
=
1
2
2𝑚𝑟2
sin2
𝜃 𝜙̇ = 𝑚𝑟2
sin2
𝜃 𝜙̇
[1]
𝐻 = ∑𝑝𝑗𝑞𝑗
̇ − 𝐿 = 𝑝𝑟𝑟̇ + 𝑝𝜃𝜃̇ + 𝑝𝜙𝜙̇ − 𝐿
= 𝑚𝑟̇ + 𝑚𝑟2
𝜃̇ + 𝑚𝑟2
sin2
𝜃 𝜙̇ −
𝑚
2
{ (𝑟̇2
) + (𝑟2
𝜃̇2
) + (𝑟2
sin2
𝜃 𝜙̇2
)} + 𝑈(𝑟)
=
𝑚
2
{ (𝑟̇2
) + (𝑟2
𝜃̇2
) + (𝑟2
sin2
𝜃 𝜙̇2
)} + 𝑈(𝑟)
= 𝐾 + 𝑈
Hence proved. [1]

6. 1-12: [5]
Consider a system of two masses 𝑚1and 𝑚2 connected by a harmonic spring with force
constant k. The spring is stretched 𝑥1and 𝑥2 in opposite directions as shown:
[0.5]
Solving this by Lagrangian Approach.
Lagrangian Equation of motion given as:
𝑑
𝑑𝑡
(
𝜕𝐿
𝜕𝑞̇
) =
𝜕𝐿
𝜕𝑞
-(1)
for L = K – U
𝐾 =
1
2
𝑚1𝑥1
̇ 2
+
1
2
𝑚2𝑥2
̇ 2
and 𝑈 =
1
2
𝑘𝑥2
[0.5]
Where 𝑥 = 𝑥1 + 𝑥2 = total extension of the string
So, 𝑈 =
1
2
𝑘(𝑥1 + 𝑥2)2
𝐿 =
1
2
𝑚1𝑥1
̇ 2
+
1
2
𝑚2𝑥2
̇ 2
−
1
2
𝑘(𝑥1 + 𝑥2)2
[1]
Now putting 𝑞 = 𝑥1𝑎𝑛𝑑 𝑥2 in eqn. (1)
𝑑
𝑑𝑡
(
𝜕𝐿
𝜕𝑥1
̇
) =
𝜕𝐿
𝜕𝑥1
Or
𝑑
𝑑𝑡
(
1
2
𝑚12𝑥1
̇ ) = −
1
2
2𝑘(𝑥1 + 𝑥2)
Or 𝑚1𝑥1
̈ + 𝑘(𝑥1 + 𝑥2) = 0 -(2)
Since, centre of mass is at rest, 𝑥𝐶𝑀=0, we can write:
𝑚1𝑥1 − 𝑚2𝑥2 = (𝑚1 + 𝑚2)𝑥𝐶𝑀 = 0
Or 𝑚1𝑥1 = 𝑚2𝑥2
Or 𝑥2 =
𝑚1𝑥1
𝑚2
and 𝑥1 =
𝑚2𝑥2
𝑚1
[1]
Putting this is eqn. (2) we can write:
𝑚1𝑥1
̈ + 𝑘 (1 +
𝑚1
𝑚2
) 𝑥1 = 0
𝑘 (1 +
𝑚1
𝑚2
) = 𝑘′
is constant Hence, on solving we can write:
𝑥1 = 𝑎𝑠𝑖𝑛𝑤𝑡 + 𝑏𝑐𝑜𝑠𝑤𝑡 = 𝐴𝑠𝑖𝑛(𝑤𝑡 + 𝜙)

7. Where 𝑤 = √
𝑘′
𝑚1
= √
𝑘(𝑚2+𝑚1)
𝑚1𝑚2
[1]
Similarly we can solve for 𝑥2:
𝑑
𝑑𝑡
(
𝜕𝐿
𝜕𝑥2
̇
) =
𝜕𝐿
𝜕𝑥2
𝑑
𝑑𝑡
(
1
2
𝑚22𝑥2
̇ ) = −
1
2
2𝑘(𝑥1 + 𝑥2)
𝑚2𝑥2
̈ + 𝑘(𝑥1 + 𝑥2) = 0 -(3)
𝑚2𝑥2
̈ + 𝑘 (
𝑚2
𝑚1
+ 1) 𝑥2 = 0
On solving, we can write:
𝑥2 = 𝑎′𝑠𝑖𝑛𝑤𝑡 + 𝑏′𝑐𝑜𝑠𝑤𝑡 = 𝐴′𝑠𝑖𝑛(𝑤𝑡 + 𝜙) -(4)
Where 𝑤 = √
𝑘′′
𝑚2
= √
𝑘(𝑚1+𝑚2)
𝑚1𝑚2
[1]
OR
This can also be solved by Newtonian Approach. Considering Forces 𝐹1and 𝐹2 and total
extension of 𝑥 = 𝑥1 + 𝑥2.
[0.5]
𝐹1 = 𝑚1𝑥1
̈ = −𝑘(𝑥1 + 𝑥2)
𝐹2 = 𝑚2𝑥2
̈ = −𝑘(𝑥1 + 𝑥2) [1]
These ODEs can be solved to get the same results by taking centre of mass at rest.
𝑚1𝑥1 = 𝑚2𝑥2 [1]
𝑥1 = 𝑎𝑠𝑖𝑛𝑤𝑡 + 𝑏𝑐𝑜𝑠𝑤𝑡 = 𝐴𝑠𝑖𝑛(𝑤𝑡 + 𝜙)
𝑥2 = 𝑎′𝑠𝑖𝑛𝑤𝑡 + 𝑏′𝑐𝑜𝑠𝑤𝑡 = 𝐴′𝑠𝑖𝑛(𝑤𝑡 + 𝜙) [2]
Where, 𝑤 = √
𝑘(𝑚1+𝑚2)
𝑚1𝑚2
[0.5]

8. 1-14: [5]
Given,
𝐻 = ∑𝑝𝑗𝑞𝑗
̇ − 𝐿
So, 𝑑𝐻 = ∑𝑞𝑗
̇ 𝑑𝑝𝑗 + ∑𝑝𝑗𝑑𝑞𝑗
̇ − ∑
𝜕𝐿
𝜕𝑞𝑗
̇
𝑑𝑞̇ − ∑
𝜕𝐿
𝜕𝑞𝑗
𝑑𝑞𝑗 [0.5]
We know that,
𝑝𝑗 =
𝜕𝐿
𝜕𝑞𝑗
̇
-(1)
𝑑
𝑑𝑡
(
𝜕𝐿
𝜕𝑞𝑗
̇
) =
𝜕𝐿
𝜕𝑞𝑗
Or
𝑑
𝑑𝑡
𝑝𝑗 =
𝜕𝐿
𝜕𝑞𝑗
{
𝑑
𝑑𝑡
𝑝𝑗 = 𝑝𝑗
̇ } -(2)
𝑑𝐻 = ∑𝑞𝑗
̇ 𝑑𝑝𝑗 + ∑𝑝𝑗𝑑𝑞𝑗
̇ − ∑𝑝𝑗𝑑𝑞𝑗
̇ − ∑𝑝𝑗
̇ 𝑑𝑞𝑗
𝑑𝐻 = ∑𝑞𝑗
̇ 𝑑𝑝𝑗 − ∑𝑝𝑗
̇ 𝑑𝑞𝑗 -(3)
[1]
Given, H has no explicit dependence on time, thus, total derivative of H can be written as-
𝑑𝐻 = ∑ (
𝜕𝐻
𝜕𝑝𝑗
) 𝑑𝑝𝑗 + ∑ (
𝜕𝐻
𝜕𝑞𝑗
) 𝑑𝑞𝑗 -(4)
[1]
Comparing equations (3) and (4):
𝜕𝐻
𝜕𝑝𝑗
= 𝑞𝑗
̇ and
𝜕𝐻
𝜕𝑞𝑗
= −𝑝𝑗
̇ [0.5]
Now taking derivative of H with respect to time from eqn.(4) we have:
𝑑𝐻
𝑑𝑡
= ∑
𝜕𝐻
𝜕𝑝𝑗
𝑝𝑗
̇ + ∑
𝜕𝐻
𝜕𝑞𝑗
𝑞𝑗
̇
= ∑𝑞𝑗
̇ 𝑝𝑗
̇ + ∑(−𝑝𝑗
̇ )𝑞𝑗
̇ = 0
𝑑𝐻
𝑑𝑡
= 0
[1]
This means physically that when Hamiltonian does not depend on time, hence it remains
conserved. Since Hamiltonian is close to total energy, this states that total energy remains
conserved. [0.5]
It is not true if H explicitly depends upon time; then Hamiltonian does not remain conserved.
[0.5]