Jan 17, 2002

Notes for Lecture Notes for Lecture #1

1  Introduction

  1. Textbook and course structure
  2. Motivation
  3. Chapters I and 1 and Appendix of Jackson

    1. Units - SI vs Gaussian
    2. Laplace and Poisson Equations
    3. Green's Theorm

2  Units - SI vs Gaussian

Coulomb's law has the form:

F = KC q1 q2
r122
.
(1)
Ampere's law has the form:
F = KA i1 i2
r122
  ds1 ×ds2 × ^
r
 
12,
(2)
where the current and charge are related by i1 = dq1/dt for all unit systems. The two constants KC and KA are related so that their ratio KC/KA has the units of (m/s)2 and it is experimentally known that in both the SI and CGS (Gaussian) unit systems, it the value KC/KA = c2, where c is the speed of light.

The choices for these constants in the SI and Gaussian units are given below:

CGS (Gaussian) SI

KC

1 [1/( 4 pe0)]

KA

[1/( c2)] [(m0)/( 4 p)]

Here, [(m0)/( 4 p)] º 10-7 N/A2 and [1/( 4 pe0)] = c2 ·10-7 N/A2 = 8.98755×109 N ·m2/C2.

Below is a table comparing SI and Gaussian unit systems. The fundamental units for each system are so labeled and are used to define the derived units.

Variable SI Gaussian SI/Gaussian
Unit Relation Unit Relation

length

m fundamental cm fundamental 100

mass

kg fundamental gm fundamental 1000

time

s fundamental s fundamental 1

force

N kg ·m2/s dyne gm ·cm2/s 105

current

A fundamental statampere statcoulomb/s [1/ 10 c]

charge

C A ·s statcoulomb Ö{dyne ·cm2} [1/ 10 c]

One advantage of the Gaussian system is that all of the field vectors: E,D,B,H,P,M have the same dimensions, and in vacuum, B = H and E = D and the dielectric and permittivity constants e and m are unitless.

CGS (Gaussian) SI

Ñ·D = 4 pr

Ñ·D = r

Ñ·B = 0

Ñ·B = 0

Ñ×E = - 1/c [(B)/( t)]

Ñ×E = - [(B)/( t)]

Ñ×H = [(4 p)/ c] J + 1/c [(D)/( t)]

Ñ×H = J + [(D)/( t)]

F = q (E + [(v)/ c] ×B

F = q (E + v ×B

u = [1/( 8 p)] (E·D+B·H)

u = 1/2 (E·D+B·H)

S = [c/( 4 p)] (E ×H)

S = (E ×H)

``Proof'' of the identity (Eq. (1.31))

Ñ2 æ
ç
è
1
|r-r¢|
ö
÷
ø
= - 4 pd3(r-r¢).
(3)

Noting that

ó
õ


[b]1insmall sphereabout r¢  
d3r   d3(r-r¢) f(r) = f(r¢),
(4)

we see that we must show that

ó
õ


[b]1insmall sphereaboutr¢  
d3r   Ñ2 æ
ç
è
1
|r-r¢|
ö
÷
ø
f(r) = - 4 pf(r¢).
(5)

We introduce a small radius a such that:

1
|r-r¢|
=
lim
a® 0 
1
Ö
|r-r¢|2 + a2
 
.
(6)

For a fixed value of a,

Ñ2 1
Ö
|r-r¢|2 + a2
 
= -3 a2
(|r-r¢|2 + a2)5/2
.
(7)

If the function f(r) is continuous, we can make a Tayor expansion about the point r = r¢. Jackson's text shows that it is necessary to keep only the leading term. The integral over the small sphere about r¢ can be carried out analytically, by changing to a coordinate system centered at r¢;

u = r-r¢,
(8)

so that

ó
õ


[b]1insmall sphereaboutr¢  
d3r   Ñ2 æ
ç
è
1
|r-r¢|
ö
÷
ø
f(r) » f(r¢)
lim
a ® 0 
ó
õ


u < R 
  d3u -3 a2
(u2 + a2)5/2
.
(9)

We note that

ó
õ


u < R 
  d3u -3 a2
(u2 + a2)5/2
= 4 p ó
õ
R

0 
  du    -3 a2 u2
(u2 + a2)5/2
= 4 p -R3
(R2 + a2)3/2
.
(10)

If the infinitesimal value a is a << R, then (R2 +a2)3/2 » R3 and the right hand side of Eq. 10 is - 4 p. Therefore, Eq. 9 becomes,

ó
õ


[b]1insmall sphereaboutr¢  
d3r   Ñ2 æ
ç
è
1
|r-r¢|
ö
÷
ø
f(r) » f(r¢) (- 4 p),
(11)

which is consistent with Eq. 5.


File translated from TEX by TTH, version 2.20.
On 17 Jan 2002, 14:15.