chapter 10 potentials and fields
DESCRIPTION
10.1 The Potential Formulation 10.1.1 Scalar and vector potentials 10.1.2 Gauge transformation 10.1.3 Coulomb gauge and Lorentz gaugeTRANSCRIPT
Chapter 10 Potentials and Fields
10.1 The Potential Formulation
10.2 Continuous Distributions
10.3 Point Charges
10.1 The Potential Formulation
10.1.1 Scalar and vector potentials
10.1.2 Gauge transformation
10.1.3 Coulomb gauge and Lorentz gauge
10.1.1 Scalar and Vector Potentials
0
E
tBE
tEJB
000
0B
Maxwell’s eqs
),(),(
),(),(
trBtrE
trJtr
),(),(
trAtrV
or
field formulism
potential formulism
MS
ES
0E
VE
B
AB
)( At
E
0At
E )(
VAt
E
AB
tAVE
10.1.1 (2)
0
E
0tAV
][
0
2 At
V
)(
][)( At
Vt
JA000
AAA 2 )()(
JtVA
tAA 0002
2
002
)()(
tEJB 000
Ex.10.1
10.1.1 (3)
0V
ctxfor0
ctxforzxctc4k
A20 ˆ)(
where k is a constant,
?, J
Sol:
yxctc2kyxct
xc4kAB
zxct2k
tAE
020
0
ˆ)(ˆ)(
ˆ)(
for ctx )(, ctxfor0BE
0E
0B
y2kyxct
x2kE 00 ˆ)ˆ)((
z2kzxct
xc2kB 00 ˆ)ˆ)((
21
00c
)(
10.1.1 (4)
z2kc
tE
0 ˆ
y2k
tB
0 ˆ
0E0
0z2kcz
c2k
tEB1J 0
00
0
ˆˆ)(
)(
200 c1
210x
z0x
z EEas0xat EE , 0Ex ,
xK fxy
xy
xy
xy BBBB ˆ11 0000
00
)ˆˆ(ˆ)]()[( xzKy2kt
2kt
f
yzktK f ˆ
10.1.2 Gauge transformation
tAVE
0ttrVVforV )(),(,
0trAAforAABAA
),(,?aboutHow
tAV
ttAV
ttAV
tAVE
)(
)(tk
t
ttk
t
)(
tiontransformaGauge
tVV
AAWhen
tAV
tAVE
AAB
The fields are independent of the gauges.(note: physics is independent of the coordinates.)
10.1.2 (2)
10.1.3 Coulomb gauge and Lorentz gaugePotential formulation
)( At
V2
JtVA
tAA 0002
2
002
)()(
Sources: J
, AV
,AB
tAVE
Coulomb gauge: 0A
( )A VA Jt t
22
0 0 0 0 02
d
Rtr
41trVV
0
2 ),(),( easy to solve
difficult to solve
V
A
10.1.3 (2)Lorentz gauge:
tVA 00
02
2
002
02
2
002
tVV
JtAA
0
2 V
JA 02
2
2
0022
t
inhomogeneous wave eq.
the d’Alembertion
[Note:Since is with ,the potentials with both and are solutions.]
2 2tt t
0f wave equation2
Then, you have a solution and
Gauge transformation AA
tVV
Coulomb gauge : 0A
If you have a and ,A 0A
2AA
Find , A2
0A
A
10.1.3 (3)
10.1.3 (4)
Lorentz gauge :tVA 00
If you have a set of and , andA
Vt
VA 00
2
2
002
0000 ttVA
tVA
Find ,t
VAt 002
2
002
Then ,you have a set if solutions and , andA V
tVA 00
10.2 Continuous Distributions :
With the Lorentz gauge ,
tVA 00
0
2 1V wheret
2
0022
JA 02
,t
AVE
AB
In the static case ,)( 0t
,22
0
2 1V
JA 02
d
Rr
41rV
0
)()(
dRrJ
4rA 0 )()(
10.2 (2)For nonstatic case, the above solutions only valid when for , and due to and , where is the retarded time. Because the message ofthe pensence of and must travel a distance the delay is ; that is ,
),( trV ),( trA ),( rtr ),( rtrJ
rt J
,rrR
cR /
cRtt r (Causality)
cttrr r )(
10.2 (3)The solutions of retarded potentials for nonstatic sources are
d
Rtr
41trV r
0
),(),(
dR
trJ4
trA r0 ),(),(
Proof:
d
RRtrV )1(1)(
41),(
0
rrRRRR
R1
R1
22 ˆ
)(
Rc
Rc1t
tcRttr r
rr
ˆ)(),(
10.2 (4)
d
RR
RR
c41V 2
0
ˆˆ
dRR
RR
RR
RR
c1
41V 22
0
2 )ˆ
()(ˆˆ
)(ˆ
Rc
tt r
r
ˆ
22
2 R1
R1R
RR1
RR
)()
ˆ( )()
ˆ( R4RR 3
2
dR4
Rc1
Rc1
Rc1
41V 3
2220
2 )(
0
rr
02
2
2ttrd
Rtr
41
tc1
),()),((
),( tr1tV
c1
02
2
2
The same procedure is for proving .A
Example 10.2
Solution:
0V0
dzRtIz
4tsA r0
)(
ˆ),(
10.2 (5)
?),(,?),(,)(
tsB0tforItsE0tfor0tI
0
scontributesctzonlycstfor
0tsBtsE0tsAcstfor
22
)(,
),(),(),(,
22 sct
0 2200
zsdz2z
4ItsA )()ˆ(),(
22 sct
0
2200 zzsz2
I
)()ln(ˆ
ssctctz2
I 2200 ln))(ln(ˆ
zs
sctct2
I 2200 ˆ)
)(ln(
)ln( zzsdzd 22
zzs1
22zs
z221
22
22 zs1
zsct2
cItAtsE
2200 ˆ)(
),(
ˆ),(
sAAtsB z
ˆ))(()(
)()(
22
2222200 sctcts
sct2s2
s1
sctcts
2I
10.2 (6)
22
22222
2200
sctsctsctcts
sctct1
s2I
)()()(
)(
ˆ
)(),(
2200
sctct
s2ItsB
Note:
sct1
ssctctD 2
22
)(
1221
D1
scD
scD
t 2lnln
11
sc
11
D1
sc
22
2
222 sct
c
1sct
1sc
)()(
10.2 (7)
Dt
ts
Ds
Ds
lnlnln
Dtc
ss1ct 2 ln)(
22 sctc
st
)(
,t0E
ˆ
s2IB 00
recover the static case
10.2 (8)
10.3 Point Charges10.3.1 Lienard-Wiechert Potentials
10.3.2 The Fields of a Moving Point Charge
10.3.1 Lienard-Wiechert potentials
Consider a point charge q moving on a trajectory )(tW
retarded position )( rtwrR
location of the observer at time t
cRtt r
Two issues•There is at most one point on the trajectory communicating with at any time t.r
),()( 2211 ttcRttcR
Since q can not move at the speed of light, there is only one point at meet.
Suppose there are two points:
cttRRVttcRR
12
211221
)(
10.3.1 (2)
• qdtr r ),( the point chage
cVR1qdtr r /ˆ),(
due to Doppler –shift effect as the point charge is considered as an extended charge.
cx
vLL
cxL
Proof.consider the extended charge has a length L as a train
(a) moving directly to the observer
time for the light to arrive the observer.
E F
cv1LL
/
(b)moving with an angle to the observer
10.3.1 (3)
cosL x L L xc v c
1 cos /LL
v c
The apparent volumec
vR1
ˆ actual volume
cvR1qdtr r /ˆ),(
10.3.1 (4)
)ˆ
(
),(),(
c
vR1R
q4
1dR
tr4
1trV0
r
0
dtr
Rv
4d
Rtvtr
4trA r
0rr0 ),()(),(),(
),()(
),( trVcv
vRRcvqc
4trA 2
0
Lienard-Wiechert Potentials for a moving point charge
vRRcqc
41trV
0
),(
10.3.1 (5)Example 10.3
constv ?),(?),(
trAtrV
22
22222222
r
2rr
222r
2r
2
rr
vctcrvcvrtcvrtct
ttt2tctvtvr2r
ttctvrR
tvtw00twlet
))(()()(
)(
)(
)()(
Solution:
1
consider
signchoose
retardedcrt
crtt0v r,
q
10.3.1 (6)
))(()(
)(
)()()
ˆ(
2222222
r
2
r
r
rr
tcrvcvrtcc1
tc
vc
rvttc
ttctvr
cv1ttc
cvR1R
1
))(()()ˆ
(),(
222222200 tcrvcvrtc
qc4
1
cvR1Rc
qc4
1trV
),(),( trVcvtrA 2
))(()( 22222220
tcrvcvrtcvqc
4
002c1
)(ˆ)(
r
rr ttc
tvrRttcR
10.3.2 The Fields of a Moving Point ChargeLienard-Weichert potentials:
)(),(
vRRcqc
41trV
0
),(),( trV
cvtrA 2
tAVE
AB
)(),(),( rrr twvandttcRtwrR
Math., Math., and Math,…. are in the following:
)()(
vRRcvRRc
14
qcV 20
10.3.2 (2)
rtcR
)()()()()( RvvRRvvRvR
)(ˆˆ)()( rr
jr
irjii tRajitv
tRjtvRvR
wvrvRv )()()( va
vjvjzrvrv ijijii ˆˆ)(
)(ˆˆ)()( rr
r
jirjii tvvj
it
tw
vjtwvwv
rrijr
r
jrji taktak
it
tv
ktvv
ˆˆˆ)(
rrr tvtvtwrR )()(
0
)()()()()( rrrr tvvtaRtvvvtRavR
)()( aRttRa rr
=
0
10.3.2 (4)
)()(
vRRcvRRc
14
qcV 20
r2
r2
20tvaRvtc
vRRc1
4qc
)()(
vRRcRaRvcv
vRRcqc
41 22
20
)()(
RaRvcvvRRcvRRc
qc4
1 2230
)()()(
vaRvc
cR
caRvvRRc
vRRcqc
41
tA 22
30
)())(()(
Prob.10.17
10.3.2 (3)
)()()( vvttvvtvv0 rrr
r2
r tvtvv )(
r2 tvaRvvR )()(
)()( RRRR2
1RRRtc r
)()( RRRRR1
vRRcRt r
)( rtRvR
rtv
=
rtvRRR1 )(
)()( rr tvRtRvRR1
=
)()( vRttRv rr
10.3.2 (5)
tAVtrE
),( RaRvcvvRRc
vRRcqc
41 22
30
)()()(
vaRvc
cR
caRvvRRc 22
)())((
])ˆ()ˆ)(()(
avRcRvRcaRvcvRRc
qR4
1 223
0
define vRRcuRvRcu ˆ
)()()()(
),( uRaaRuuvcuR
qR4
1trE 223
0
)()()(
),( auRuvcuR
qR4
1trE 223
0
vRcu ˆ
generalized Coulomb field radiation field or acceleration field dominates at large R
if RRq
41trE0v0a 2
0
ˆ),(,,
Electrostatic field
10.3.2 (6)
)()()( VvvVc1vV
c1AB 22 V
cvA 2
vRRcRatav r
RaRvcvvRRcvRRc
qc4
1V 223
0
)()()(
)()( vRRc
RavRRc
qc4
1c1B
02
RvaRvc
vRRcqc
41 22
30
)()(
10.3.2 (7)
)()()()(
223
0vcvaRvuRaR
uR1
4q
c1
)())(())(()(
ˆ uRaaRvRvcvRuR
qR4
1Rc1 22
30
)()()()(
ˆ uRaaRuvcuuR
qR4
1Rc1 22
30
)()()(
ˆ auRvcuuR
qR4
1Rc1 22
30
),(ˆ trERc1
),(ˆ),( trERc1trB
10.3.2 (8)The force on a test charge Q with velocity due to a moving charge q with velocity is
V
v
)( BVEQF
)()(ˆ)()(
)(auRuvcR
cVauRuvc
uRR
4qQ 2222
30
Where rtatevaluatedallareaandvuR ,,,
10.3.2 (9)Example 10.4 q constv ?),( trE
?),( trB
Solution:0a tvw 0t originatw
uuR
Rvc4
qtrE 3
22
0
)()(),(
)()()( tvrcvttcVtrcvRRcuR rr
21
2222222 tcrvcvrtcvRRcuR ))(()( Ex.10.3
Prob.10.14cv1Rc 22 /sin
3
21
22
22
0cv1Rc
Rcvc4
qtrE
)/sin(
)(),(
tvrR
10.3.2 (10)
22
3222
22
0 RR
cv1
cv14
qtrEˆ
)/sin(
/),(
),(ˆ),( trERc1trB
),(),( trEvc1trB 2
c
vRP
Rvtttvr
RtvrR rr
)()(
)ˆ(4
),(,ˆ4
1),(2
02
0
PvPqtrBR
PqtrE
Coulomb`s law “Biot-savart Law for a point charge.”
ecoincidencbyptontpoiE ˆ
p
,22 cvwhen