Conservation of Momentum Assignment

Solution 1
a) As be definitions
L=KE−PE
L= 1
2 m ̇x1
2 + 1
2 m ̇x2
2− 1
2 K ( x1
2+ x2
2
)
Now momentum is
p1= ∂ L
∂ ̇x1
=m ̇x1
p2= ∂ L
∂ ̇x2
=m ̇x2
As
L= 1
2 m ̇xI ̇x J−e (φ− ̇xJ A j)
So
pi= ∂ L
∂ ̇xi
= ∂
∂ ̇xi ( 1
2 m ̇xI ̇xJ−e ( φ− ̇xJ A j ) )
pi= 1
2 m ̇¿ 2 xI −e (0−1∗A j)
pi=m ̇xi+ e A j ¿
b) The conservation is
The second reads
pi=m ̇xi+ e Ai
d
dt pi= d
dt ( m ̇xi +e Ai ) =0
d
dt ( m ̇xi )=0
That x component of momentum is conserved.

c)
H= ̇x j
dL
d ̇x j
−L
L= 1
2 m ̇x j ̇x j−e (∅− ̇x j A j)
So in
H= ̇x j
dL
d ̇x j
−L
H= ̇x j
d
d ̇x j ( 1
2 m ̇x j ̇x j−e ( ∅− ̇x j A j ) )−( 1
2 m ̇x j ̇x j−e ( ∅− ̇x j A j ) )
H= ̇x j ( m ̇x j− d
d ̇x j
e ( ∅− ̇x j A j ) )−( 1
2 m ̇x j ̇x j−e ( ∅− ̇x j A j ) )
H= ( p j− d
d ̇x j
e ( ∅− ̇x j A j ) )− (1
2 p j−e ( ∅− ̇x j A j ))
H= ( 1
2 pj − d
d ̇x j
e ( ∅− ̇x j A j ) )+ ( e ( ∅− ̇x j A j ) )
H= ( 1
2 m ( p j−e A j ) 2+ e ∅ )
d)
H= ( 1
2 m ( p j−e A j ) 2+ e ∅ )
Now
H= ̇x j
dL
d ̇x j
−L
Also
p1= ∂ L
∂ ̇x1
=m ̇x1

H= ̇x j pi−mx
Hence
̇x j pi−mx=( 1
2 m ( p j−e A j ) 2 +e ∅ )
̇x j pi−mx=( 1
2 m ( p j
2 +e2 A j
2−2 p j e A j )+e ∅ )
̇x j pi−mx=( 1
2 m ( p j
2 +e2 A j
2¿ )+e ∅−−p j e A j
m )
̇x j pi−mx=( 1
2 m ( p j
2 +e2 A j
2¿ )+e ∅−−p j e A j
m )
So on comparison
̇xi= 1
m ( pi −e Ai )
̇pi= e
m ( p j−e A j ) ∂ A j
∂ x j
−e ∂ ∅
∂ xi
e) As we can see that there should be balancing momentum for pz component of the system.
The subtraction of kinetic and potential energy is not conservative in z direction. Mainly
because of no balancing component available and hence this is not conservative.
f)
From
̇xi= 1
m ( pi −e Ai )
̇pi= e
m ( p j−e A j ) ∂ A j
∂ x j
−e ∂ ∅
∂ xi
Here
Bey=e Ai
So now

̇x= 1
m ( px+ Bey )
̇px=−Ee
̇py=−Be ̇x
̇y= py
m
g) Magic velocity
By law of conservation
1
2 B v2= 1
2 Ev
Bv=E
v= E
B
h) In case if E>cB
In this part we have higher component of em wave than velocity of light. This comes under
uncertainty.
This is physically not possible to have this condition.
Solution 2
a) Given
h ( t )= ∑
n=−∞
∞
f ( x +n )
now as we know that series converges to the point were summation and avering is given.
Using standard fourier formula, we can see period is 1.

b)
∂u
∂ t =2 ∂2 u
∂ x2
u ( 0 , t ) =u ( 5 ,t ) =0
u ( x , 0 )=f ( x )=−4 sin ( πx )+ 3sin ( 2 πx ) 0 ≤ x ≤ 5
Now
Now assuming
u( x , t)= X (x)T ( t)
Now
X ( x ) T ' ( t ) =2 X' ' ( x ) T ( t )
X' ' (x )
X ( x ) =1
2
T ' ( t )
T ( t )
Now assuming the independency of both the sides we have below condition for any given
constant say K.
X' ' ( x )
X ( x ) =K
1
2
T ' ( t )
T ( t ) =K
So solution for above two variables given non zero K

X ( x )=l1 e√ K x+l2 e− √K x
T ( t ) =l3 e2 Kt
Hence finally we have
u ( x , t ) =X ( x ) T ( t )
u ( x , t )=(l1 e √K x+ l2 e− √ K x )l3 e2 Kt
c) Heat equation
Now for above equation when we apply boundary conditions
u ( x , t )= An sin ( nπ
L x )e−2 π 2 n2 t
L2
Here
An =−2i l1 l3
√ K= nπ
L i
Now heat equation satisfies the below equation
u ( x , t )=∑
n=1
∞
An sin ( nπ
L x )e−2 π 2 n2 t
L2
Considering