Q.1 Set up an expression for Electromagnetic Lagrengian Density
Or Prove
1 
1
d F  j
4
c
Lagrangian
Euler Lagrangian Equation given by
 dL 
dL
 d 
0
 d  d  Av  
dAv


Lagrangian Density =
1
4 0
1
A J 
C
ic
ys
R
dL
  v J 
dA
oy
Now First term in Euler-Lagrange Equation
Ph
1 v
1
F Fv  A J 
4
c
al
Total Lagrangian (L) =
is
t
Interacting Lagrangian=
F v Fv
We can write
1
Fv F v  F F     o  F Fp
dF
d  d  Av 


 v    v
dL
  v J 
dA
d  F F  
d  d  Av 
T   
k
dL
d   k  g  L
d  d  k 
dL
d  d A 
is
ic
ys
Ph
al
oy
R
For Free Electromagnetic Field
1
L=Fv F v
4
Such That,
t
Fv  d  Av  d v A
dL
T   
d  A  g  L

k d  d A 
v
dL
1 d  Fv F 
4  d  d A 
d  d A 
d  F v  
1  v d  Fv 
F

= Fv


4 
d
d
A
d
d
A
 
   

Using Relation
Fv  d  Av  d v A
F
v
d  Fv 
d  d A 
F
=F
v
v
d  d  Av  d v A 
d  d  A 
d  g  d Av  g v d A 
d  d A 
=F v  g    v  g v    
2
=F v  g  F  g v Fv   2 g  F
Proceeding Similarly we get same for 2nd Term
d  F F  
d  d  Av 
o
 

 dF
 dFp  



Fp  F 
 d  d  Av   
 d  d  Av 


Since
F  d A  d  A
dF
d  d  Av 
   v    v
On Combining

t
is
ic
d  d  Av 

   p   v   v  Fp    pv    pv  F 
ys
d  F F  
al
Ph
=   p v   vp   Fp     v    v   F
R
=4Fv
oy
=Fuv  Fv  Fuv  Fv
=-4Fv
Substituting in Euler-Lagrange Equation , we get
d  F v  J v
Q. Define Canonical Stress Tensor for the free Electromagnetic field and show that the total Energy
and momentum of the fields are given by time time component of that Tensor.
Canonical Stress Tensor
Covariant generalization of Hamiltonian density is canonical stress Tensor
T   
k
dL
d   k  g  L
d  d  k 
For Electromagnetic Field ,
3
 k  A
d   k  d  A
T   
k
dL
d  A  g  L

d  d A 
For Free Electromagnetic Field
1
L=Fv F v
4
Such That,
Fv  d  Av  d v A
F
d  d  Av  d v A 
d  d A 
al
d  d A 
v
oy
d  Fv 
R
F
v
Ph
ys
ic
is
d  F v  
1  v d  Fv 
= Fv
F



4 
d
d
A
d
d
A
 
   

Using Relation
t
v
dL
1 d  Fv F 
4  d  d A 
d  d A 
=F
v
d  g  d Av  g v d A 
d  d A 
=F v  g    v  g v    
=F v  g  F  g v Fv   2 g  F
Proceeding Similarly we get same for 2nd Term
dL
1
1 




4
g
F


g F



4

d  d A 
Hence Canonical Stress Tensor
1
T    g  F d  A  g  L

4
In Terms of Electric and Magnetic Fields
1
L=Fv F v
4
 1  E2
1   E2
2
2
=L= 2  2  B   
 2 B 
4   c
  2  c

Now,
T 00  
1
 g  0 F d 0 A   g 00 L

 g ij  1 if i=j 


 =0 if i  j 




1

 g  0 F d 0 A   g 00 L

=
1
 g 00 F d 0 A   g 00 L

t
dAy
 dAx
dAz  1  E 2
2
 Ey
 Ez
 2 B 
 Ex

dt
dt
dt  2  c


dA
1
1  E2
2
=  2 Ei i 
 2 B 
c
dt 2  c

But E= - 
dA
dt

R
oy
al
Ph
ys
ic
is
1
= 2
c
dA
 E  
dt
E  0
1
1  E2
1  2 B2 
2
T  2 E  E    
 B    E 

    E  
c
2  c 2
2




Now,
1
T 0i    g  0 F d i A   g 0i L

00

1
 g 00 F0  d i A 

dA3 
dA2
1  dA1
E

E

E
1
2
3
 c  dt
dt
dt 
1
d

Ej
A
 c dxi j

5
Further
E  B   E     A  


d
d
 E j
Aj  E j
Ai 
dx j 
 dx j
d
d
Ej
A j   E  B i  E j
Ai
dx j
dx j
  E  B  i  E A
  E  B i    EA 
On Substituting
1
1
T 0 i    E  B i 
  EA 
c
c
and  12
**Drawbacks are Discussed at End
is
11
ic
evaluate 
t
Q. What are the Drawbacks of Canonical Stress Tensor? Deduce Symmetric Stress Tesnor and Hence
Ph
ys
F  d  A  d  A
Canonical Stress Tensor is
1
T    g  F d  A  g  L
R
oy
al
 d  A  d  A  F

T   
=
=
=
1

1

1

g  F  F  d  A   g  L
g  F F 
 1

g  F d  A  g   
Fv F v 

 4

1
g  F F  g 
1
1
Fv F v  g  F d  A
4

1  
 1

 1
Fv F v   g  F d  A
 g F F  g

4
 
6
TD  
1

g  F d  A 
1

F  d  A
1
1
 F  d  A  A d  F    d   F  A 


1
d TD  d d   F  A   0

=
Since d d  is Symmetric while F is antisymmetric
1  


 1
g
F
F

g
Fv F v 



4

This is Stress Tensor
T  TD 

is
1  0
0
00 1
v 
g
F
F

g
F
F




v

 
4
ic
 00 
ys
1  

 1
v 
g
F
F

g
F
F




v

 
4
Ph
  
t
1. Time Time Component     0 

R
oy
al
1  0

0
00 1
g
F
F

g
2  E 2  B 2  





4

1
1

  g 00 F0  F 0    E 2  B 2  

2



1 
1 2
00
10
20
30
00
2 


F
F

F
F

F
F

F
F

g

E

B

 
00
01
02
03

  
2


1 
1 2
2
2
2
00
2 


0

E

E

E

g

E

B

 
x
y
z


  
2

1  1 2
E 2  B2
2 
   E  B    
  2
2

2. Time-Space Component   0,   i 
 oi 
1  0

0i
0i 1
g
F
F

g
Fv F v 





4

7
1  00
0i
  1   F00 F0i  F01F1i  F02 F2i  F03 F3i    1   F01F1i  F02 F2i  F03 F3i  
g
F
F


0

  

  

=
Case if i =1
1
1
1
 E y Bz  Ez By  
 o1    F01F11  F02 F2i  F03 F3i   
 E  B x

4
4
Similarly
1
 o1 
 E  By
4
1
 o2 
 E  B z
4
3. Space Space Component (  =1,   j )
  
1  
1

g F F   g  Fv F v 



4

1  j
j
ij 1
v 
g
F
F

g
F
F




v

 
4
1
 ij   g  j F F j  

al
Ph
ys
ic
is
t
 ij 
R
oy
let i =1 , j = 2
1
1
1
 12   g 1 F F 2     g 11 F1 F 2     g 11 F1 F 2   g11  1



=
1
1
 F10 F02  F11F12  F12 F22  F13 F32       F10 F02  F13 F32  




1
 12     Ex E y  Bx By  

 ij  
1
  Ei E j  Bi B j  

 
i  j 
let i =j
 ij 
1   1
1   1
1
1
11 1
v  
1
v  
g
F
F

g
F
F

g
F
F

F
F


v


v






 
4
4
   

8
=
1  11
1
1
g
F
F

2  E 2  B 2 
1



 
4

 
1 
1
01
11
21
31


F
F

F
F

F
F

F
F

2  E 2  B 2 
10
11
12
13




 
4
1 
1

=-   Ex2  Bz2  By2    E 2  B 2   
 
2

1 
1

=-   Ex2  Bz2  By2  Bx2  Bx2    E 2  B 2   
 
2


=-
1  2
1 2
22
2 



  
E

B

E

B
x 
   x
2

1  2
1 2
22
2 



  
E

B

E

B
i
i

 2
  

When i  j There is absence of
1 2
E  B2 

2
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Comparing  ii and  ij
1
When i = j There is  E 2  B 2 
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ys
 ii  -
is
1  2
1 2
2
2
2 
  Ex  Bx  B    E  B   

 
2

ic
=-
 
t
=-
In Combine Form
 ij  
 ij 2
1 
2 
 Ei E j  Bi B j   E  B   Where  ij is Krneckar Delta
4 
2

Drawbacks:
1. T 00 and T 01 differ from the ussual expression of energy and momentum density
due to the presence of the added Divergence term
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2. It Doesnot have Symmetry under the interacting of i and j
i.e T ij  T ji
3.It is not traceless as required for zero mass Photon
i.e T 00  T11  T 22  T 33  0
4.It involves the potential explicitely, So it is not gauge invariant
Q.4 Introduce Quantum Electrodynamcis. Discuss about the laws of Quantum Electrodynamcis
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OR What is Quantum Electrodynamics? Discuss in Brief the Fermi’s method about the
interaction of light with matter and hence state the laws of QED
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Selection Rules
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