1/21/2015phy 712 spring 2015 -- lecture 31 phy 712 electrodynamics 9-9:50 am mwf olin 103 plan for...
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PHY 712 Spring 2015 -- Lecture 3 11/21/2015
PHY 712 Electrodynamics9-9:50 AM MWF Olin 103
Plan for Lecture 3:
Reading: Chapter 1 in JDJ
1. Review of electrostatics with one-dimensional examples
2. Poisson and Laplace Equations
3. Green’s Theorem and their use in electrostatics
PHY 712 Spring 2015 -- Lecture 3 21/21/2015
PHY 712 Spring 2015 -- Lecture 3 31/21/2015
PHY 712 Spring 2015 -- Lecture 3 41/21/2015
Poisson and Laplace EquationsWe are concerned with finding solutions to the Poisson equation:
and the Laplace equation:
The Laplace equation is the “homogeneous” version of the Poisson equation. The Green's theorem allows us to determine the electrostatic potential from volume and surface integrals:
2
0
( )( )P
r
r
2 ( ) 0L r
3
0
2
1( ) ( ) ( , )
4
1ˆ ( , ) ( ) ( ) ( , ) .
4
V
S
d r G
d r G G
r r r r
r r r r r r r
PHY 712 Spring 2015 -- Lecture 3 51/21/2015
General comments on Green’s theorem
3
0
2
1( ) ( ) ( , )
4
1ˆ ( , ) ( ) ( ) ( , ) .
4
V
S
d r G
d r G G
r r r r
r r r r r r r
This general form can be used in 1, 2, or 3 dimensions. In general, the Green's function must be constructed to satisfy the appropriate (Dirichlet or Neumann) boundary conditions. Alternatively or in addition, boundary conditions can be adjusted using the fact that for any solution to the Poisson equation, other solutions may be generated by use of solutions of the Laplace equation
( )P r
( ) ( ) ( ),for any constant .P LC C r r r
PHY 712 Spring 2015 -- Lecture 3 61/21/2015
“Derivation” of Green’s Theorem
2
0
2 3
( )Poisson equation: ( )
Green's relation: '( , ) 4 '( ).G
rr
r r r r
2
3
3
2Divergence theorm:
Let
ˆ
ˆ
V S
V S
r r
f g g f
r f g g f r f g g
d d
d fd
A A r
A r r r r
r r r r r r r r r
3 2 2 V
r f gd g f r r r r
PHY 712 Spring 2015 -- Lecture 3 71/21/2015
“Derivation” of Green’s Theorem
2
0
2 3
( )Poisson equation: ( )
Green's relation: '( , ) 4 '( ).G
rr
r r r r
2 23 2 ˆ V S
r f g g f r f g g fd d r r r r r r r r r
3
0
2
( )
1( ) ( ) ( , )
4
1ˆ ( , ) ( ) ( ) ( , ) .
, '
4
V
S
g
d r G
d r
f
G G
G
r r
r r r r
r r r r r
r
r r
r r
PHY 712 Spring 2015 -- Lecture 3 81/21/2015
PHY 712 Spring 2015 -- Lecture 3 91/21/2015
PHY 712 Spring 2015 -- Lecture 3 101/21/2015
PHY 712 Spring 2015 -- Lecture 3 111/21/2015
Comment about the example and solution
This particular example is one that is used to model semiconductor junctions where the charge density is controlled by introducing charged impurities nearthe junction.
The solution of the Poisson equation for this case can be determined by piecewise solution within each of the four regions. Alternatively, from Green's theorem in one-dimension, one can use the Green's function
0
1( ) ( , ) ( ) ( , ) 4 where
should be take as the smaller of and '.
4x G x
x x x
x x dx G x x x
PHY 712 Spring 2015 -- Lecture 3 121/21/2015
Notes on the one-dimensional Green’s function
2
The Green's function for the one-dimensional
Poisson equation can be defined as a solution to
the equ ( , ) 4 ( )
Here the factor o
ation:
is not really necessary, bf ut
ensures con
4
sis
G x x x x
tency with your text's treatment of
the 3-dimensional case. The meaning of this expression
is that ' is held fixed while taking the derivative with
respect to .
x
x
PHY 712 Spring 2015 -- Lecture 3 131/21/2015
Construction of a Green’s function in one dimension
2
1 2
( ) 0
where 1 or 2. Let
4( , ) ( )
Co
(
nsider two independent solution
).
This notation means that
s to the h
sho
omogeneous equation
uld be taken as the
smaller of and ' and sho
i x
i
G x x x xW
x
x x x
uld be taken as the larger.
1 22 1
is defined as the "Wronskin":
( ) ( )( ) ( ) .
d x d xW x x
dx dx
W
PHY 712 Spring 2015 -- Lecture 3 141/21/2015
Summary
2
1 2
1 22 1
( , ) 4 ( )
4( , ) ( ) ( )
( ) ( )( ) ( )
( , ) ( , )4
x x x x
G x x x x
G x x x xW
d x d xW x x
dx dx
dG x x dG x x
dx dx
ò ò
t t
PHY 712 Spring 2015 -- Lecture 3 151/21/2015
One dimensional Green’s function in practice
0
0
21
1( ) ( , ') ( ') '
4
1 = ( , ) ( ) ( , ) ( )
4
For the one-dimensional Poisson equation, we can construct
the Green's function by choosing ( ) and ( ) 1; 1:
(
x
x
x G x x x dx
G x x x dx G x x x dx
x x x W
0
1) ( ) ( ) .
x
xx x x dx x x dx
This expression gives the same result as previously obtained for the example r(x) and more generally is appropriate for any neutral charge distribution.
PHY 712 Spring 2015 -- Lecture 3 161/21/2015
PHY 712 Spring 2015 -- Lecture 3 171/21/2015
PHY 712 Spring 2015 -- Lecture 3 181/21/2015
PHY 712 Spring 2015 -- Lecture 3 191/21/2015