8.07 Lecture 36: December 10, 2012
POTENTIALS AND FIELDS
Maxwell's Equations with Sources:
1
~
~
(i) r E = 0
~ B
~ =0
(ii) r
~ E
~ =
(iii) r
~
@B
@t
;
~
1
@E
~ B
~ = 0 ~
(iv) r
J+ 2
c @t
(1)
;
Question: If we are given the sources (~r ; t) and ~J (~r ; t), can we
nd E~ and B~ ? If we accept the proposition that all integrals
are in principle doable (at least numerically), then the answer
is YES .
Alan Guth
Massachusetts Institute of Technology
8.07 Lecture 36, December 10, 2012
{1{
Electromagnetic Potentials
~ E
~ =
If B~ depends on time, then r
6 ~0, so we cannot write E~ =
r~ V . BUT: we can still write
~
B
~ A
~ :
=r
(2)
Then notice that
r~ E~ =
@ ~
(
@t
r A~ ) =) r~ ~
E
+
~
@A
!
= 0 : (3)
@t
so we can write
~
E
+
~
@A
@t
~V
= r
=)
~
E
~V
= r
~
@A
@t
:
(4)
Alan Guth
Massachusetts Institute of Technology
8.07 Lecture 36, December 10, 2012
{2{
With
~
B
~ A
~ ;
=r
~
E
~V
= r
~
@A
@t
(5)
;
the source-free Maxwell equations (ii) and (iii),
1
~
~
(i) r E = 0
~ B
~ =0
(ii) r
~ E
~ =
(iii) r
~
@B
@t
;
~
1
@E
~ B
~ = 0 ~
(iv) r
J+ 2
c @t
;
are automatically satised. We must therefore deal with the two
other Maxwell equations, (i) and (iv).
Alan Guth
Massachusetts Institute of Technology
8.07 Lecture 36, December 10, 2012
{3{
Maxwell's Other Equations:
1
@ ~ ~
1
2
(i) r = =) r V + @t (r A) = :
~
~
E
0
0
~
1
@E
~ B
~ = 0 ~
(iv) r
J+ 2
c @t
1r
~
~ (r
~ A
~ ) = 0 ~
=) r
J
=)
rA
where we used
2~
(6)
~
1 @ 2A
c2
@t2
!
c2
r r
~
~
@V
@t
~ +
A
1
c2
1 @ 2A~
c2 @t2
@V
@t
r~ (r~ A~ ) = r~ (r~ A~ ) r A~ :
2
=
0 ~
J
(7)
(8)
Alan Guth
Massachusetts Institute of Technology
8.07 Lecture 36, December 10, 2012
{4{
Gauge Transformations
We have already discussed gauge transformations for statics. But
it easily generalizes to the full theory of electrodynamics.
Let (~r ; t) be an arbitrary scalar function. Then, if we are
~ (~r ; t), we can dene new potentials by a
given V (~r ; t) and A
gauge transformation:
~0
A
=
r
~ + ~ ;
A
V0 =V
@
@t
:
(9)
Then
~0
B
~ A
~0 = r
~ A
~ +r
~ r
~=r
~ A
~ =B
~ :
=r
(10)
0
~
~
@A
@A
@
@ ~
0
0
~
~
~
~
~ :
E = rV
=
r
V
+
r
r
=
E
@t
@t
@t
@t
(11)
{5{
Choice of Gauge:
~ +r
~ , to make r
~ A
~ whatever we
Can use gauge freedom, A~ 0 = A
want.
~ A
~ =0:
Coulomb Gauge: r
(12)
rV+
2
@ ~
(
@t
r
~)
A
1
1
2
= =) r V = :
0
0
(13)
is easy to nd, but A~ is hard. V responds instantaneously to changes in , but V is not measurable. E~ and
~ receive information only at the speed of light.
B
1
@V
~
~
:
Lorentz Gauge: r A =
(14)
V
c2 @t
=) r
2
V
1 @ 2V = 1 :
c2 @t2
0
(15)
Alan Guth
Massachusetts Institute of Technology
8.07 Lecture 36, December 10, 2012
{6{
Dene
2
r
2
1
@2
c2
@t2
Then, in Lorentz gauge,
2
V
= D'Alembertian :
1
= :
(17)
0
In general, A~ obeys Eq. (7):
!
2~
1
@ A
1
@V
2~
~
~
~
rA
r rA+
=
c2 @t2
(16)
c2 @t
J :
0 ~
In Lorentz gauge,
2~
A
=
0 ~
J :
(18)
Alan Guth
Massachusetts Institute of Technology
8.07 Lecture 36, December 10, 2012
{7{
2V
Solution to
=
1
0
Method: Guess a solution and then show that it works.
We know that
Z
0
1
1
(
~
r
; t)
3 0
2
r V = =) V (~r ; t) = 4 d x j~r ~r 0j
0
0
We try the guess
2
V
1
= =)
0
where
tr
=t
1
V (~r ; t) =
40
j~r
~r 0
c
Z
:
(19)
0
(
~
r
; tr )
3 0
d x ~r ~r 0 ;
(20)
j
j = retarded time:
j
(21)
{8{
Testing the trial solution:
~
r
= xi e^i
@i (~
r ; tr )
0
j
; @i ~
r
=
1
c
xi
0
xi
j~r
xi0
j = j~r
_(~
r ; tr )
@i
Then
~
r
0
j
xi
j~r
x0i
~
r0
~
r0 3
xi0
~
r
0
j
j
; tr
;
= 4Æ3 (~r
=t
where _ j~r
~
r0
c
j;
@(~
r 0 ; tr )
@tr
(22)
;
~
r 0) :
0
1
(
~
r
; tr )
3 0
V (~r ; t) =
d
x
:
0
40
j~r ~r j
1
_
3 0
0
c
d x j~r ~r 0j2 (xi xi ) j~r ~r 0j3 (xi
Z
Z
1
@i V =
xi0 )
40
Z
1
1
_

1
2 2
3
0
3 0
c
c
@i V =
d
x
4
Æ (~r ~r ) +
+
0
2
40
j~r ~r j j~r ~r 0j
2
3
_
_
c
c
+ j~r ~r 0j2 j~r ~r 0j2
:
{9{
Testing the trial solution:
~
r
= xi e^i
@i (~
r ; tr )
0
j
; @i ~
r
=
1
c
xi
0
xi
j~r
xi0
j = j~r
_(~
r ; tr )
@i
Then
~
r
0
j
xi
j~r
x0i
~
r0
~
r0 3
xi0
~
r
0
j
j
; tr
;
= 4Æ3 (~r
=t
where _ j~r
~
r0
c
j;
@(~
r 0 ; tr )
@tr
(22)
;
~
r 0) :
0
1
(
~
r
; tr )
3 0
V (~r ; t) =
d
x
:
0
40
j~r ~r j
1
_
3 0
0
c
d x j~r ~r 0j2 (xi xi ) j~r ~r 0j3 (xi
Z
Z
1
@i V =
x0i )
40
Z
1
1
_

1
2 2
3 0
3
0
c
c
@i V =
d
x
4
Æ (~r ~r ) +
+
0
2
40
j~r ~r j j~r ~r 0j
2
3
_
_
c
c
+ j~r ~r 0j2 j~r ~r 0j2
:
{9{
Testing the trial solution:
~
r
= xi e^i
@i (~
r ; tr )
0
j
; @i ~
r
=
1
c
xi
0
xi
j~r
x0i
j = j~r
_(~
r ; tr )
@i
Then
~
r
0
j
xi
j~r
x0i
~
r0
~
r0 3
xi0
~
r
0
j
j
; tr
;
= 4Æ3 (~r
=t
where _ j~r
~
r0
c
j;
@(~
r 0 ; tr )
@tr
(22)
;
~
r 0) :
0
1
(
~
r
; tr )
3 0
V (~r ; t) =
d
x
:
0
40
j~r ~r j
1
_
3 0
0
c
d x j~r ~r 0j2 (xi xi ) j~r ~r 0j3 (xi
Z
Z
1
@i V =
xi0 )
40
Z
1
1
_

1
2 2
3 0
3
0
c
c
@i V =
d
x
4
Æ (~r ~r ) +
+
0
2
40
j~r ~r j j~r ~r 0j
2
3
_
_
c
c
+ j~r ~r 0j2 j~r ~r 0j2
:
{9{
Testing the trial solution:
~
r
= xi e^i
@i (~
r ; tr )
0
j
; @i ~
r
=
1
c
xi
0
xi
j~r
x0i
j = j~r
_(~
r ; tr )
@i
Then
~
r
0
j
xi
j~r
x0i
~
r0
~
r0 3
xi0
~
r
0
j
j
; tr
;
= 4Æ3 (~r
=t
where _ j~r
~
r0
c
j;
@(~
r 0 ; tr )
@tr
(22)
;
~
r 0) :
0
1
(
~
r
; tr )
3 0
V (~r ; t) =
d
x
:
0
40
j~r ~r j
1
_
3 0
0
c
d x j~r ~r 0j2 (xi xi ) j~r ~r 0j3 (xi
Z
Z
1
@i V =
xi0 )
40
Z
1
1
_

1
2 2
3 0
3
0
c
c
@i V =
d
x
4
Æ (~r ~r ) +
+
0
2
40
j~r ~r j j~r ~r 0j
2
3
_
_
c
c
+ j~r ~r 0j2 j~r ~r 0j2
:
{9{
Testing the trial solution:
~
r
= xi e^i
@i (~
r ; tr )
0
j
; @i ~
r
=
1
c
xi
0
xi
j~r
xi0
j = j~r
_(~
r ; tr )
@i
Then
~
r
0
j
xi
j~r
x0i
~
r0
~
r0 3
xi0
~
r
0
j
j
; tr
;
= 4Æ3 (~r
=t
where _ j~r
~
r0
c
j;
@(~
r 0 ; tr )
@tr
(22)
;
~
r 0) :
0
1
(
~
r
; tr )
3 0
V (~r ; t) =
d
x
:
0
40
j~r ~r j
1
_
3 0
0
c
d x j~r ~r 0j2 (xi xi ) j~r ~r 0j3 (xi
Z
Z
1
@i V =
xi0 )
40
Z
1
1

_
1
2 2
3 0
3
0
c
c
@i V =
d
x
4
Æ (~r ~r ) +
+
0
2
40
j~r ~r j j~r ~r 0j
2
3
_
_
c
c
+ j~r ~r 0j2 j~r ~r 0j2
:
{9{
2
@i V
1
= 4
0
Z
+ j~r
=
=
d x0
2
(~r ; t)
0
_
c
~r 0
3
3
j
j~r
2
1
+ 4
0
1
+ 4
1
1
@ 2V
= + c2 @t2
0
~r 0 ) +
4Æ (~r
3
(~r ; t)
0
_
c
~r 0
Z
j
d
3
c2
2
0c
@
2
@t2
Z
c _
~r ~r 0
1
j
+
j j~r
2
1

c2 ~r 0
j
2
@ 2 (~r 0 ;tr )
0 @t2r
x
j~r ~r 0j
0
(
~
r
; tr )
3 0
d x j~r ~r 0j
tr =t
j~r ~r 0 j c
Y ES !
Alan Guth
Massachusetts Institute of Technology
8.07 Lecture 36, December 10, 2012
{10{
2
@i V
1
= 4
0
Z
+ j~r
=
=
d x0
2
(~r ; t)
0
_
c
~r 0
3
3
j
j~r
2
1
+ 4
0
1
+ 4
1
1
@ 2V
= + c2 @t2
0
~r 0 ) +
4Æ (~r
3
(~r ; t)
0
_
c
~r 0
Z
j
d
3
c2
2
0c
@
2
@t2
Z
c _
~r ~r 0
1
j
+
j j~r
2
1

c2 ~r 0
j
2
@ 2 (~r 0 ;tr )
0 @t2r
x
j~r ~r 0j
0
(
~
r
; tr )
3 0
d x j~r ~r 0j
tr =t
j~r ~r 0 j c
Y ES !
Alan Guth
Massachusetts Institute of Technology
8.07 Lecture 36, December 10, 2012
{10{
Retarded Time Solutions
0
1
(
~
r
; tr )
3 0
V (~r ; t) =
d
x
40
j~r ~r 0j
Z
0
~
1
J
(
~
r
; tr )
3 0
~
A(~r ; t) =
d
x
40
j~r ~r 0j
Z
where
tr
=t
j~r
~r 0
c
j:
(23)
;
(24)
Alan Guth
Massachusetts Institute of Technology
8.07 Lecture 36, December 10, 2012
{11{
Advanced Time Solutions??
0
1
(
~
r
; ta )
3 0
V (~r ; t) =
d
x
40
j~r ~r 0j
Z
0
~
1
J
(
~
r
; ta )
3 0
~
A(~r ; t) =
d
x
40
j~r ~r 0j
Z
where
ta
j
~r
=t+
~r 0
c
j:
(25)
;
(26)
Maxwell's equations, and the laws of physics as we know them,
make no distinction between the future and the past.
Alan Guth
Massachusetts Institute of Technology
8.07 Lecture 36, December 10, 2012
{12{
Mystery of the Arrow of Time
Real Events:
Arrow of time
Laws of Physics:
Time symmetric
Alan Guth
Massachusetts Institute of Technology
8.07 Lecture 36, December 10, 2012
{13{
Mystery of the Arrow of Time
Real Events:
Standard YouTube License. This content is excluded from our Creative Commons
license. For more information, see http://ocw.mit.edu/fairuse.
Arrow of time
Laws of Physics:
Time symmetric
Alan Guth
Massachusetts Institute of Technology
8.07 Lecture 36, December 10, 2012
{13{
Mystery of the Arrow of Time
Real Events:
Standard YouTube License. This content is excluded from our Creative Commons
license. For more information, see http://ocw.mit.edu/fairuse.
Arrow of time
Laws of Physics:
Time symmetric
Alan Guth
Massachusetts Institute of Technology
8.07 Lecture 36, December 10, 2012
{13{
Mystery of the Arrow of Time
Real Events:
Standard YouTube License. This content is excluded from our Creative Commons
license. For more information, see http://ocw.mit.edu/fairuse.
Arrow of time
Laws of Physics:
Time symmetric
Alan Guth
Massachusetts Institute of Technology
8.07 Lecture 36, December 10, 2012
{13{
Mystery of the Arrow of Time
Real Events:
Standard YouTube License. This content is excluded from our Creative Commons
license. For more information, see http://ocw.mit.edu/fairuse.
Arrow of time
Laws of Physics:
Time symmetric
Alan Guth
Massachusetts Institute of Technology
8.07 Lecture 36, December 10, 2012
{13{
Mystery of the Arrow of Time
Real Events:
Standard YouTube License. This content is excluded from our Creative Commons
license. For more information, see http://ocw.mit.edu/fairuse.
Arrow of time
Laws of Physics:
Time symmetric
Alan Guth
Massachusetts Institute of Technology
8.07 Lecture 36, December 10, 2012
{13{
Mystery of the Arrow of Time
Real Events:
Standard YouTube License. This content is excluded from our Creative Commons
license. For more information, see http://ocw.mit.edu/fairuse.
Laws of Physics:
Arrow of time
Time symmetric
Alan Guth
Massachusetts Institute of Technology
8.07 Lecture 36, December 10, 2012
{13{
Mystery of the Arrow of Time
Real Events:
Standard YouTube License. This content is excluded from our Creative Commons
license. For more information, see http://ocw.mit.edu/fairuse.
Arrow of time
Laws of Physics:
Time symmetric
Alan Guth
Massachusetts Institute of Technology
8.07 Lecture 36, December 10, 2012
{13{
Mystery of the Arrow of Time
Real Events:
Standard YouTube License. This content is excluded from our Creative Commons
license. For more information, see http://ocw.mit.edu/fairuse.
Arrow of time
Laws of Physics:
Time symmetric
Alan Guth
Massachusetts Institute of Technology
8.07 Lecture 36, December 10, 2012
{13{
Most often heard explanation:
The universe started out in a low entropy (i.e. highly ordered)
state, so the entropy (disorder) has been increasing ever since.
Why did the universe start in a low entropy state?
Who knows?
My preferred explanation:
It is possible that there is no upper limit to the entropy of
the universe, so any state in which it may have started is lowentropy compared to what it can be.
Alan Guth
Massachusetts Institute of Technology
8.07 Lecture 36, December 10, 2012
{14{
Most often heard explanation:
The universe started out in a low entropy (i.e. highly ordered)
state, so the entropy (disorder) has been increasing ever since.
Why did the universe start in a low entropy state?
Who knows?
My preferred explanation:
It is possible that there is no upper limit to the entropy of
the universe, so any state in which it may have started is lowentropy compared to what it can be.
Alan Guth
Massachusetts Institute of Technology
8.07 Lecture 36, December 10, 2012
{14{
Most often heard explanation:
The universe started out in a low entropy (i.e. highly ordered)
state, so the entropy (disorder) has been increasing ever since.
Why did the universe start in a low entropy state?
Who knows?
My preferred explanation:
It is possible that there is no upper limit to the entropy of
the universe, so any state in which it may have started is lowentropy compared to what it can be.
Alan Guth
Massachusetts Institute of Technology
8.07 Lecture 36, December 10, 2012
{14{
Most often heard explanation:
The universe started out in a low entropy (i.e. highly ordered)
state, so the entropy (disorder) has been increasing ever since.
Why did the universe start in a low entropy state?
Who knows?
My preferred explanation:
It is possible that there is no upper limit to the entropy of
the universe, so any state in which it may have started is lowentropy compared to what it can be.
Alan Guth
Massachusetts Institute of Technology
8.07 Lecture 36, December 10, 2012
{14{
Most often heard explanation:
The universe started out in a low entropy (i.e. highly ordered)
state, so the entropy (disorder) has been increasing ever since.
Why did the universe start in a low entropy state?
Who knows?
My preferred explanation:
It is possible that there is no upper limit to the entropy of
the universe, so any state in which it may have started is lowentropy compared to what it can be.
Alan Guth
Massachusetts Institute of Technology
8.07 Lecture 36, December 10, 2012
{14{
Most often heard explanation:
The universe started out in a low entropy (i.e. highly ordered)
state, so the entropy (disorder) has been increasing ever since.
Why did the universe start in a low entropy state?
Who knows?
My preferred explanation:
It is possible that there is no upper limit to the entropy of
the universe, so any state in which it may have started is lowentropy compared to what it can be.
Alan Guth
Massachusetts Institute of Technology
8.07 Lecture 36, December 10, 2012
{14{
The Fields of a Point Charge
From Eq. (23),
0
1
(
~
r
; tr )
3 0
V (~r ; t) =
d
x
:
0
40
j~r ~r j
For a point charge q moving on a trajectory ~r p(t),
(~r ; t) = qÆ 3 ~r ~r p (t) ;
so
3
0 ~r p(tr )
Z
Æ
~
r
q
3 0
V (~r ; t) =
d
x
40
j~r ~r 0 j
= 4 j~r
0
Z
q
Z
j
~r p (tr )
d3x0 Æ3 ~r 0
~r p (tr )
(27)
(28)
:
Alan Guth
Massachusetts Institute of Technology
8.07 Lecture 36, December 10, 2012
{15{
But, perhaps surprisingly,
Z
Z
d3 x0 Æ3 ~r 0 ~r p(tr ) =
(29)
6 1;
where I am calling the integral Z for future reference. Remember,
X Æ (x x )
i
Æ g (x) =
; where g (xi ) = 0 ;
(30)
0
jg (x )j
and
i
i
j~r
~r 0
j:
=t
c
To make things simple, suppose that the particle velocity at tr
points inZthe x-direction. Then
Z=
d3 x0Æ x0 xp(tr ) Æ y0 yp(tr ) Æ z 0 zp(tr )
(31)
Z
= dx0 Æ x0 xp(tr ) ;
where the integrals over y0 and z 0 were simple, since yp0 (tr ) =
zp0 (tr ) = 0.
tr
{16{
So we need to evaluate
Z
=
Z
dx0 Æ x0
xp (tr )
;
So, to use our formula,
g (x0 ) = x0
where tr = t
xp
t
and then
j~r
d
xp dtr
1
d
xp d
0
0
g (x ) = 1
=
1
+
j
~r
0
0
dt dx
c dt dx
and Z = 1=g0(x0). Generalizing,
Z
= 1
~v
j~r
~r
c j~r
c
0
~r j
~r 0
c
j:
;
0
1
d
x
x
p
~r 0 j = 1 +
c dt j~r
0
~r
~r 0 j
1
:
(32)
(33)
x
~r 0 j
;
(34)
(35)
Alan Guth
Massachusetts Institute of Technology
8.07 Lecture 36, December 10, 2012
{17{
The Lienard-Wiechert Potentials
Finally,
1
V (~r ; t) =
40 j~r
~r p
j 1
q
~vp
c
~r ~r p
j~r ~r p j
;
(36)
where ~r p and ~vp are the position and velocity of the particle at tr .
Similarly, starting with
~
J (~r ; t) = q~v Æ 3 ~r ~r p (t)
(37)
for a point particle, we nd
~ (~r ; t)
A
0
= 4
j~r
~r p
j 1
q~vp
~vp
c
~r ~r p
j~r ~r p j
~vp
= c2 V (~r ; t) :
(38)
{18{
The Fields of a Point Charge
Dierentiating the Lienard-Wiechert potentials, after several pages, one nds
~ (~
E
r ; t)
j~r
=
40 (~u (~r
q
~
rp
j
~
r p ))
3
2
(c
2
vp )~
u + (~
r
~
r p)
(~u ~a
p
)
;
(39)
where
~
u
=c
~
r
j~r
~
rp
~
rp
j
~
vp :
(40)
And
~ (~
B
r ; t)
=
1
~
r
~
rp
c ~
r
~
rp
j
~ (~
E
r ; t) :
j
(41)
{19{
Here ~r p, ~vp, and ~ap are the position, velocity, and acceleration,
respectively, of the particle at the retarded time.
If the particle is moving at constant velocity, then the acceleration
term in Eq. (39) is absent, and the electric eld points along
~
u. Note that ~
u can also be written as
~
u
= j~r
c
~r p
j
h
~r
~r p + ~vp (t
i
tr )
:
(42)
In this form one can see that, for the case of constant velocity,
~
u points outward from the current position of the particle,
which is ~r p + ~vp(t tr ).
Alan Guth
Massachusetts Institute of Technology
8.07 Lecture 36, December 10, 2012
{20{
MIT OpenCourseWare
http://ocw.mit.edu
8.07 Electromagnetism II
Fall 2012
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