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16 Νοε 2013 (πριν από 4 χρόνια και 5 μήνες)

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Lecture 3 Energy density and flow

Again recalling mechanical waves,

the p
ower carried by a wave is force*
velocity.

By the analysis
e,

we might expect the power to be related to the product
E B

.In fact,

since we
expect a vector quantity (since the power flow is dir
ectional), we might begin by
investigating
E B

.

This is also motivated by a dim
ensional argument:

the units of the

Poynting vector
:

S E H
 

A
re W/m^2.

We begin by considering a charge q in an arbitrary electromagnetic field.

The force on the
charge is the
L
orentz force
:

( )
F q E v B
  

If in a time interval
t

,

the c
harge is displaced by an amount
x

,

the work done by the field is:

( )
E F x qE x q v B x
   
      

Now
x v t
 

,

where is orthogonal to
v B

,so the magnetic field does no work,

and:

E qE x
 
 

Therefore the rate at which work is done on the charge is:

( )
x
qE qE v E qv
t t
 
 
     

Now we can generalize to a distribution of charge in a volume
V

.The rate
at which work is
done is just:

arg
| [ ] ( )
V
mechanical
i i
allch es
d
E q v E v V
dt

 
   

Where

is t
he charge density in the volume
V

,

and
v

is the velocity of those charges.

Of course:

J v

i
s just current denisity.

Therefore in any macroscopic volume V,

the total rate of mechanical work
done by the field is:

mechanical
V
d
J EdV
dt

 


This is the power transformed from the field to the charges,

in the form of
mechanical or
thermal energy.

(This should not be a surprise.

Suppose Ohm

s law applies,

J E

then:

2
2
mechanical
V
d U
E dV
dt R

 


(U = voltage)

A
s is familiar from sophomore physics.

By conservation of energy,

the power transformed to the charges must be
accompanied by a
decrease in the energy in the electromagnetic field in the volume V.

In order to express this entirely in forms of the field variables,

we must eliminate
J
.

Ampere+Maxwell say:

D
J H
t

  

S
o

[ ]
V V
D
J EdV E H dV
t

    

 

W
here I

ve been lazy and written
V

as
V

Now a useful vector identity is

( ) ( ) ( )
E H H E E H
       

W
hich is fine for any vector fields
E

and
H
.

Then:

[ ( ) ( ) ]
V V
D
J EdV H E E H dV
t

      

 

[ ( )]
V
B D
H E E H dV
t t
 
      
 

In order to go any further we must make an addtional

assumption that the medium is linear in
its electric and magnetic fields,so that the constitutive relations hold:

D E

and
B H

Then
2
( ) ( )
2 2
D E
E E E E E
t t t t
 

   
     
   

And similarly
2
1
( )
2
B
H B
t t

 
 
 

Thus

2 2
1 1
( ) ( )
2 2
V V V
J EdV E B dV E H dV
t

      

  

From electro
-

and magnetic
-
statics,we
recognize this as

the total energy stored
in the electromagnetic field in V

We can make the extension to dynamic
fields’

as well total e.m. energy density

2 2
1 1
2 2
u E B

 

Now define the
Poynting vector
:

S E H
 

(for real fields)

We have

[ ] 0
V
u
J E S dV
t

   

Since the
integration

volume is
arbitrary, we

may say that the
integrand

must hold at all points in
space, and
:

u
S J E
t

   

T
his is in the form of a continuity
equation, so

if :

u
t

=chan
ge in
energy density in field

A
nd
J E

=change in mechanical energy of particles

T
hen
S
=the flow of electromagnetic energy

It is also useful to go back to
above

and u
se Gauss

s theorem

V S
SdV S ds
  
 

For
shorthand, we

can write:

mech
V
field
V
J EdV
t
udV
t t

 

 

 

and
V S
SdV S ds
  
 
becomes
mech field
S
S ds
t t
 
 
   
 


Thus the total change in energy in the volume V is given by the net flow of energy through the
surface
S

that bound V.

Simplest case: in vaccum:
| |
S cu

Of course,

we are most interested in the power flow that we would measure directly.

Light
detectors are generally

square law detectors

which measure power and not electric field,

for
the simple reason that they are too slow.

The farfest light

detectors we have

picomatrix
,a spin
-
off of UM!)

have a band width of a few hundred GHz,

whereas the optical field
oscillates

at ~10^15Hz,and 10^11Hz<<10^15Hz!

Therefore a
photo detector

measures a time
-
average of the incident pow
er.

E
x.
plane wave

Note:

We have defined the Poynting vector for real fields.If you want to use complex fields,a
slightly different defination is called for.

0 0
cos( ) cos( )
S E H E wt k r H wt k r
       

0 0 0
k
i k E i B H E


      

2
0 0
( )cos ( )
k
S E E wt k r

    

(
k


)

2
2
0
c o s ( )
k
E w t k r
k

 
  
 
 

Note that for our aimple plane wave,the energy propagates in the same direction as the phase
k
).This is generally
true

in isopic media,but turns out not to be

true

in
anisotropic media(such as birefrigent crystals)

Now we want the time average
:

0
2
0
2
0
1
| | | ( )cos ( ) ( )
t T
t
k
I S E wt k r dt I r
T k

    

In general ,T is very long compared to an optical wave,but we can fake
2
T

and

0
0
2 2
1 1
cos cos
2
t T
t
wt wtdt
T

  

So we have:

2
0
1
2
I E

(W/m2)

This is commonly known in the laser community as the
intensity of the beam.

It is worth nothing that this is
not, unfortun
ately
,

a
universal terminology.

In the ra
d
i
om
etry community *which is more careful with its difinitions),

I

is actually known as
the

accordance with common
usage, we

will use these terms interchangeably.

Momentum

Since an e.m. wave carries energy,

it should also carry momentum,

and thus exert a pressure on
anything which reflects or absorbs light.

Consider a wave incident fro
m a vaccum onto a region
confirming charges+

currents
:

Lorentz force on microscopic char
ges:

( )
F q E v B
  

so we have force density
l
F E J B

  
.

Maxwell

s equation may be used to eliminate

and
J
:

0
E

 

0 0 0
E
B J
t
 

  

0 0
0
1
( ) ( )
l
E
F E E B B B
t
 

      

Now consider
( )
E B
E B B E
t t t
  
    
  

Then
0 0 0
0 0
1 1
( ) ( ) ( ) ( ) ( )
l
F E B E E E E E E B B
t
  
 

           

0
B
 

to the right hand side to symmetrize the equation.

The net force exerted on the matter in the volume V is obtained by
integrating

over the
volume,
so
:

0
( ) [..]
tot
V V
F E B dV r h s dV
t

  

 

Now the space force is the time derivative of the momentum,

so we may write:

[..]
mech field
V
dP dP
r l s dV
dt dt
 

W
here:

0
2
1
( )
field
V V
P E B dV SdV
c

  
 

I
s the linear momentum associated with the field.

Often it is said that
2
S
g
c

is the

momentum
density

associated with the electromagnetic wave.

We will not carry out
the calculation,

but it can be shown t
hat it can be written as

V S
TdV T ds
  
 

W
here:

2
0
2
0
0 0
1
( )
2 2
B
T EE BB I E

 
   

I
s called the

Maxwell stress tensor

(2
nd

rank tensor).

Thus * is an expression of a momentum
conservation law for electromagnetic fields.

The r.h.s

is basically the force on the surfaces
bonding the volume V.

We will not further concern ourselves to the most general case,

but consider only simple cases,

such as complete absorption of the wave by the material in volume V.

T
he pressure exerted by
the
ion

on the volume is:

| | | |
P g V
force
t t
P
area A A

 
  

I
n the
Δ
t,

the light propagates a distance c
Δ
t,

so the relevant volume is

V Ac t
 

| |
| |
| |
g Ac t
S
t
P g c
A c

  

O
f co
urse,

a time
average

mu
st be taken :

S I
P
c c
 
  