Physics 460
1
Fall 2006
Susan M. Lea
Potentials for Dynamic ¤elds
We have spent the last few weeks discussing the propagation of electromagnetic
waves, so now it is time to think about how the waves are produced. We have
already shown that we can express the ¤elds in terms of the scalar and vector
potentials:
~
B
=
~ £A
~
r
~
E
=
~ ¡
¡rV
(1)
~
@A
@t
(2)
By inserting these expressions into Gaussz Law and Amperezs law, we can show
how the potentials are related to the sources:
~
~ ¢E
~ = ½ = ¡r2 V ¡ r
~ ¢ @A
r
"0
@t
and
³
´
~ £ r
~ £A
~ = ¹ ~j + ¹ "0 @
r
0
0
@t
Ã
~
~ ¡ @A
¡ rV
@t
(3)
!
Expand the term on the left, to get:
³
´
~ r
~ ¢A
~ ¡ r2 A
~ = ¹0~j ¡ ¹0 "0
r
Ã
~
@ ~
@2A
rV +
2
@t
@t
!
We can rearrange a bit, to get
·
¸
~
1 @2A
1 @V
2~
~
~
~
~
¡ r A = ¹0 j ¡ r 2
+r ¢A
c2 @t2
c @t
The left hand side of this equation is nice: it has the wave operator operating
~ and the right hand side has the source ~j of A;
~ but there is an additional
on A;
~ are not unique. Since
term that spoils it. But now we recall that V and A
~ without
the curl of a gradient is zero, we can add the gradient of a scalar to A
~
changing B
~0
A
~0
B
=
=
~ + rÂ
~
A
~ £A
~0 = r
~ £A
~ =B
~
r
~ ¤eld we have:
Then for the new E
~0
~
~
~ 0 = ¡ rV
~ 0 ¡ @ A = ¡rV
~ 0 ¡ @A ¡ @ rÂ
~ =E
~ = ¡ rV
~ ¡ @A
E
@t
@t
@t
@t
1
(4)
~ we need a new V :
So to get the same E
V0=V ¡
@Â
@t
(5)
Equations (4) and (5) guarantee that the ¤elds are the same whether we use
the unprime or prime potentials. They describe the Gauge transformations for
the ¤elds.
Knowing that we have this ¢exibility, we can choose the potentials so that
1 @V
~ ¢A
~=0
+r
c2 @t
(6)
For suppose this is not true. Then we apply a Gauge tranformation, and require
that the new potentials satisfy equation (6). Then
µ
¶
³
´
1 @
@Â
~ ¢ A
~ + rÂ
~
V
¡
+
r
=0
c2 @t
@t
or
1 @2Â
r ¡ 2 2 = ¡
c @t
2
µ
1 @V
~ ¢A
~
+r
c2 @t
¶
So we have a prescription for ¤nding the necessary Â: In fact any two di¤erent
values of  that di¤er by a solution of the homogeneous wave equation are
equally valid. Equation (6) is the Lorentz Gauge condition. Choosing this
~ is a wave equation with source ~j :
gauge, the equation for A
~
1 @2 A
~ = ¹ ~j
¡ r2 A
0
2
2
c @t
(7)
Then putting the gauge condition into the equation (3) for V , we have:
½
"0
or
@ ~ ~
r¢A
@t
µ
¶
@
1 @V
= ¡r2 V ¡
¡ 2
@t
c @t
= ¡r2 V ¡
1 @2V
½
¡ r 2V =
c2 @t2
"0
(8)
~ and the scalar potential V satisfy similar
Thus each Cartesian component of A
~ or
equations: a wave equation with source equal to a component of ~j (for A)
the charge density ½ (for V ).
Because of this nice symmetry (and some other reasons wezll discuss when
we study relativity) the Lorentz gauge is particularly useful for calculating wave
¤elds. But it is not the only choice. We could use the Coulomb gauge
~ ¢A
~=0
r
2
which gives a very simple equation for V : In fact it is the same equation as
when the sources are time independent:
r2 V (~r; t) = ¡
½ (~r ; t)
"0
This is rather odd, however, because it means that as we change ½; the potential
changes instantly everywhere in the universe! This de¤es all of our expectations
~ is also rather ugly:
about causality. The equation for A
~
1 @2 A
~ = ¹ ~j ¡ r
~ 1 @V
¡ r2 A
0
2
2
c @t
c2 @t
There are ways to simplify this (see Physics 704) but we still have to deal with
the acausal nature of V : In classical physics the potentials are merely useful
~ and B;
~ and it turns out tha the physical ¤elds E
~ and
tools for calculating E
~ donzt change everywhere at once, but obey the expected rules that signals
B
travel at c: For now, however, letzs stick with Lorentz gauge.
The wave equations we have found show that electromagnetic signals travel
at the speed of light. If a source 1 km from me changes its magnitude at time
t0 , I wonzt know about it until time t = t0 + 1000 m/(3 £ 10 8 m/s) = t0 + 30
¹s. Another way of saying this is that the charge density that determines V
at my position is the charge density 30 ¹s ago. We can formalize this idea by
noting that we must use the charge density at the retarded time:
tr et = t ¡ R=c
to get the potential at time t: Using this idea, we guess that the correct expression for potential is
Z
1
½ (~r 0; t ¡ R=c) 0
V (~r; t) =
d¿
(9)
4¼"0
R
where, as usual,
R = j~r ¡ ~r0 j
Equation (9) is the retarded potential. Letzs check that it works by stu¢ng
into the wave equation. We have to be careful in taking the space derivatives
because ~r appears in R and R appears in ½ as well as in the denominator. We
use the chain rule to take the derivatives of ½:
µ
¶
@½
@½ @tre t
@½
1 @R
=
=
¡
@x
@tre t @x
@tre t
c @x
@½
@½ @tre t
@½
=
=
@t
@tre t @t
@tre t
3
Thus
r2 V
³
´
~ Z ½ (~r0 ; t ¡ R=c)
~ ¢ rV
~
~ ¢ r
= r
=r
d¿ 0
4¼" 0
R
Z ·
¸
1
1
1~
0
0
~
~
= r¢
½ (~r ; tre t ) r + r½ (~r ; tre t ) d¿ 0
4¼" 0
R R
¸
Z ·
1
1
1 @½ (~r 0 ; tre t ) ~
0
~
~
= r¢
½ (~r ; tre t ) r ¡
rR d¿ 0
4¼" 0
R Rc
@tre t
Z
1
=
(T 1 + T 2 ) d¿ 0
4¼"0
Now we take the next derivative. The ¤rst term in the integrand is
T 1 = ½ (~r 0 ; tre t ) r2
where
0
1
~ 1 ¢ 1 @½ (~r ; tret ) rR
~
¡r
R
R Rc
@tre t
1
= ¡4¼± (~r ¡ ~r0 )
R
~
~ 1 =¡R;
r
(R 6= 0)
R
R2
r2
and
~
~ = R =R
^
rR
R
so
T1
=
=
~
R
1 @½ (~r 0 ; tre t ) ^
¢
R
2
R Rc
@tre t
1 @½ (~r 0 ; tre t )
¡4¼½ (~r0 ; tre t ) ± (~r ¡ ~r 0) +
cR2
@tre t
¡4¼½ (~r0 ; tre t ) ± (~r ¡ ~r 0) +
The second term is
( ·
)
¸
0
~
1 ~ 1 @½ (~r 0; tre t )
1
@½
(~
r
;
t
)
R
r
et
^+
~ ¢
T2 = ¡
r
¢R
r
c
R
@tret
R
@tre t
R
(
Ã
!)
µ
¶
~
^¢R
~
1 @½ (~r0 ; tr et ) ~ 1 ^
@
1 @½
rR
1 @½ (~r0 ; tre t )
R
3
^
= ¡
r ¢R+
¡
¢R +
¡ 2 +
c
@tre t
R
@tret
c @tre t
R
R
@tre t
R
R
(
Ã
!
)
2
0
~
1 @½ (~r0 ; tr et )
R
^ ¡ 1 @ ½ 1 + 2 @½ (~r ; tret )
= ¡
¡ 2 ¢R
2
2
c
@tre t
R
c @tre t R R
@tre t
1 @2½
1 @½ (~r 0 ; tre t )
2 ¡
2
Rc @tre t
R2c
@tre t
=
Putting it all together, the ¤rst order time derivatives cancel:
¸
Z ·
1
1 @½ (~r 0 ; tre t )
1 @2½
1 @½ (~r 0 ; tre t )
2
0
0
r V =
¡4¼½ (~r ; tre t ) ± (~r ¡ ~r ) +
+
¡ 2
d¿ 0
4¼"0
cR2
@tre t
Rc2 @t2re t
R c
@tre t
=
½ (~r; t)
1 @2
+ 2 2V
"0
c @t
4
Thus we have retrieved equation (8).
~ is
Similarly, the solution for A
~ (~r ; t) = ¹0
A
4¼
Z
~j (~r0 ; tre t ) 0
d¿
R
(10)
It remains to show that these potentials also satisfy the Lorentz gauge condition.
See problem 10.8
The retarded potentials (9) and (10) are often easier to write down than they
are to use. But we can calculate the potentials in some simple cases. Imagine
that we could set the current up everywhere in an in¤nitely long wire at time
t = 0: (I donzt know how wezd do this, but letzs pretend.)
I
=
0
=
I0
t<0
t>0
~ ¢~j = 0; so if ½ is zero at any
Then V ´ 0 since ½ ´ 0 everywhere, (notice that r
time it stays zero), and, with z¡axis along the wire:
Z +1
I (tr et ) 0
~ (~r; t) = ¹0 z^
A
dz
4¼ ¡ 1
R
The current function is zero unless tre t = t ¡ R=c > 0; so we need
q
2
R = s2 + (z ¡ z 0 ) < ct
This condition restricts the range of integration:
2
<
jz ¡ z0 j
<
(z ¡ z 0 )
5
c2t2 ¡ s 2
p
c2 t2 ¡ s2
or
z¡
p
c2 t2 ¡ s 2 < z 0 < z +
p
c2 t2 ¡ s 2
Note that s < ct is also required, so that the square root is real.
~ (~r; t) = ¹0 z^
A
4¼
Z
z+
p
c 2 t2 ¡ s 2
p
z¡ c 2t2¡ s2
I0
q
s 2 + (z ¡
z 0 )2
dz 0
Let z 0 ¡ z = s tan µ: Then
Z µ2
s sec2 µ
~ (~r; t) = ¹0 I0 z^
A
dµ
4¼
s sec µ
µ1
¹0 I 0
=
z^ ln (sec µ + tan µ)jµµ21
4¼
0s
1¯ z+p c 2t2¡s2
¯
µ
¶2
0
0
¹0 I 0
z¡ z
z ¡ z ¯¯
@
A
=
z^ ln
1+
+
¯
4¼
s
s
¯ p
=
=
=
Ãr
p
c2 t2 ¡ s2
c2 t2 ¡ s 2
1+
+
2
s
s
Ã
!
p
¹0 I 0
ct + c2 t2 ¡ s2
p
z^ ln
for s < ct
4¼
ct ¡ c2 t2 ¡ s2
¹0 I 0
z^ ln
4¼
z¡ c 2t2¡s2
!
¡ ln
Ãr
c2 t2 ¡ s2
1+
¡
s2
p
c2 t2 ¡ s2
s
!
0 otherwise
Note that the result is independent of z; as we would expect1 .
Now we can ¤nd the ¤elds:
~
E
=
=
=
=
=
~
@A
@t (Ã
! Ã
!)
p
p
¹0 I 0
c + c2 t= c2t2 ¡ s 2
c ¡ c2 t= c2 t2 ¡ s2
z^
p
¡
p
S (ct ¡ s)
4¼
ct + c2 t2 ¡ s 2
ct ¡ c2 t2 ¡ s2
Ã
!
p
³
¹0 I 0
ct + c2 t2 ¡ s2
s´
¡
z^ ln
p
± t¡
4¼
c
ct ¡ c2 t2 ¡ s2
à ¡p
p
p
¢¡
¢ ¡p
¢¡
¢!
c2t2 ¡ s 2 + ct ct ¡ c2 t2 ¡ s 2 ¡
c2 t2 ¡ s2 ¡ ct ct + c2 t2 ¡ s2
¹0 I 0 c
S (ct ¡ s)
^z
p
2
2
2
2
2
4¼
c t ¡ (c t ¡ s )
c2 t2 ¡ s2
¡
¹0 I 0 c
2
^z p
S (ct ¡ s)
2
2
4¼
c t ¡ s2
I0
z^
p
S (ct ¡ s)
2¼"0 c c2 t2 ¡ s2
1 This is not quite the same as Gri¢thsz result. The two di¤er by a gauge transformation.
Try to ¤x up Gri¢thzs sloppy derivation by putting in the nec essary sentences!
6
In the second form it is easier to check that the result is dimensionally correct.
The term with the ±¡function is zero because when s = ct the log is ln (1) = 0:
~ :
Finally we get B
~
B
=
=
=
=
=
@Az
@Az ^
~ £ A^
r
z=
s^ ¡
Á
@Á
@s
Ã
!
p
¹0 I 0 ^ @
ct + c2 t2 ¡ s2
¡
Á
ln
p
S (ct ¡ s)
4¼ @s
ct ¡ c2 t2 ¡ s2
(Ã
!
Ã
!
)
p
p
p
¹0 I 0 ^
¡s= c2 t2 ¡ s2
s= c2 t2 ¡ s2
ct + c2 t2 ¡ s2
¡
Á
p
¡
p
S (ct ¡ s) ¡ ln
p
± (s ¡ ct)
4¼
ct + c2 t2 ¡ s2
ct ¡ c2 t2 ¡ s2
ct ¡ c2 t2 ¡ s2
Ã
!
¹0 I0 ^
s
2ct
Áp
S (ct ¡ s)
4¼
c2 t2 ¡ s2 (ct)2 ¡ c2 t2 + s2
¹0 I0 ^
ct
Á p
S (ct ¡ s)
2¼ s c2t2 ¡ s 2
As t ! 1 we get back the static limit:
~
E
~
B
2
2.1
!
0
¹0 I 0 ^
Á
2¼s
!
Fields due to a moving point charge
Potentials
One of the important applications of the retarded potentials is to calculate the
¤elds due to a moving point charge. The charge is located at a point with
position vector r0 (t) ; where its velocity is
~v =
d~r0
dt
~a =
d~v
dt
and its acceleration is
The charge density is then
½ (~r; t) = q± (~r ¡ ~r0 (t))
and the current density is
~j = ½~v = q~v± (~r ¡ ~r0 (t))
The potential at a point P with position ~rP at time t is due to the charge at
position ~r0 (tre t ) : Thus tre t is determined by the equation
R (tre t ) =
j~rP ¡ ~r0 (tre t )j
=
7
c (t ¡ tret )
c (t ¡ tret )
(11)
To get the potential we back up one step and write:
Z
1
½ (~r 0; t0 ) ± (t0 ¡ tr et) 0 0
V (~r; t) =
d¿ dt
4¼"0
R
The potential is then
V (~r; t) =
1
4¼"0
Z
q± (~r ¡ ~r0 (t)) ± (t0 ¡ tret ) 0 0
d¿ dt
R
We do the integral over the volume (in primed variables) ¤rst, to get
Z
q
± (t0 ¡ tre t (~r0 )) 0
V (~r; t) =
dt
4¼"0
R
We can do the integral if we can get the delta function into the form
± (t0 ¡ something independent of t0) : It is not in that form yet, because tre t
contains ~r0 (t0) : We have ± [f (t0 )] where
µ
¶
j~r ¡ ~r0 (t0)j
0
0
f (t ) = t ¡ t ¡
c
Those of you taking Physics 485/785 know that
± [f (x)] =
± (x ¡ x0 )
jf 0 (x0 )j
where f (x 0 ) = 0: So here the root is tr et and
f 0 (t0 ) =
1¡
=
1¡
1 @
1 (~r ¡ ~r0 )
j~r ¡ ~r0 (t0 )j = 1 ¡ ~v ¢ 2
c @t0
2 c j(~r ¡ ~r0 )j
^
~v ¢ R
c
Thus
V (~r; t) =
q
³
´
^
4¼"0 j~r ¡ ~r0 (tre t )j 1 ¡ ~v ¢ R=c
(12)
Thus the potential looks exactly like the potential due to a static point charge
except that ~r0 now changes with time, and must be evaluated at tre t , and there
^ in the denominator. When the speed
is an extra geometric factor 1 ¡ ~v ¢ R=c
of the charge approches the speed of light, this factor makes the potential very
^ » 1; that is, where R
^ is parallel to ~v :
large at points where ~v ¢ R=c
To understand why this factor appears, consider the observed volume of a
train travelling at speed v; as shown in the diagram.
8
Light from the front of the train leaves the train at time t0 and reaches the
observer at time T where T ¡ t0 = R=c: If light from the rear is to reach the
observed at the same time, it has to start earlier, at time t1 , but then the rear
of the train was father back along the tracks, where L0 ¡ L = v (t0 ¡ t1 ) and
T ¡ t1 =
or
R + L0 cos µ
c
R
R + L0 cos µ
+ t0 ¡ t1 =
c
c
So
L0 ¡ L
L0
=
cos µ
v
c
or
³
´
v
L0 1 ¡ cos µ = L
c
Then the observed volume of the train is
V 0 = L 0A = ¡
LA
V
¢ =¡
¢
v
1 ¡ c cos µ
1 ¡ vc cos µ
The geometrical factor in the denominator here is exactly the same as the factor
in the potenial (12).
~ is the same as for V ; so together we have
The integral we need to do to get A
the Lienard Wiechert potentials:
¯
¯
¯
q
³
´ ¯¯
V =
^
¯
4¼"0 j~r ¡ ~r0 j 1 ¡ ~v ¢ R=c
t
~
A
=
¯ r et
¯
¯
q~v
³
´ ¯¯
^
¯
4¼"0 j~r ¡ ~r0 j 1 ¡ ~v ¢ R=c
(13)
tr e t
where the whole expression is evaluated at the retarded time. As v ! 0 we get
back the usual result for a stationary charge.
9
2.2
Fields due to a moving charge.
To get the ¤elds from the potentials we use equations (1) and (2), This is not
an easy task because tr et depends on the spatial coordinates.
2.2.1
Non-relativistic limit:
In Lorentz Gauge, the ¤elds are found using
~
~ = ¡r©
~ ¡ @A
E
@t
and
~ =r
~ £A
~
B
But our expressions for the potentials are in terms of ~x and tre t ; not ~x and t;
so we have to be very careful in taking the partial derivatives. We can put
the origin at the instantaneous position of the charge to simplify things. Then
R = r: Our potential may be written:
© (~x ; t) ´ ª (~x ;tre t )
Then
¯
~ ¯¯
d© = r©
c on st t
¢ d~x +
¯
@©
@ª
~ ¯¯
dt ´ rª
¢ d~x +
dtre t
@t
c on s t t ret
@tre t
But dtr et = dt ¡ dr=c; so
¯
¯
@©
~ ¯¯
~ ¯¯
r©
¢ d~x +
dt ´ rª
co n st t
@t
c on st
t ret
¢ d~x ¡
@ª dr
@ª
+
dt
@tre t c
@tre t
~ must be modi¤ed:
Thus the r¡component of r©
¯
¯
@© ¯¯
@ª ¯¯
1 @ª
=
¡
@r ¯c on st t
@r ¯ co ns t t ret c @tr et
(14)
Now we can calculate the ¤elds:
~ =
r©
1 ~ q
1 q ~
r =¡
rrv
4¼"0 rv
4¼"0 rv2
and
~ v= @
rr
@r
µ
¶
µ
¶
µ
¶
^
^
~r ¢ ~v
µ @
~r ¢ ~v
Á
@
~r ¢ ~v
r¡
r^ +
r¡
+
r¡
c
r @µ
c
r sin µ @Á
c
We can choose our axes with polar axis along the instantaneous direction of ~v :
Then ~r ¢ ~v = rv cos µ; and
³
´
´
^³
~ v = 1 ¡ v cos µ r^ + µ r v sin µ
rr
c
r
c
10
In the non-relativistic limit, v=c ¿ 1; to zeroth order in v=c; this is
~ v = ^r
rr
We are also going to need
@rv
~r ¢ ~a
=¡
@t
c
Then we have
~
E
~
1 ~ ¯¯
1 @©
@A
r©¯
+
r^ ¡
4¼"0
co ns t t ret
4¼" 0c @t
@t
µ
¶
1 q
1
q
~r ¢ ~a
¹ @ ~v
^r¡
¡
r ¡ 0q
^
4¼"0 r2v
4¼"0 c r2v
c
4¼ @t rv
µ
¶
µ
¶
1 q
~r ¢ ~a
¹0 ~a
¹0 q~v
~r ¢ ~a
^
r
1
+
¡
q
+
r
¡
4¼"0 r2v
c2
4¼ rv
4¼ rv2
c
= ¡
=
=
and again taking the non-relativistic limit, this becomes:
·
µ
¶
¸
1
q
~r ¢ ~a
q ~a
~
E =
r 1+ 2
^
¡ 2
4¼"0 r 2
c
c r
The ¤rst term is the usual Coulomb ¤eld. The other two terms depend on ~a:
these are the radiation ¤eld.
~ rad
E
=
=
1
q
[(^
r ¢ ~a) r^¡~a]
2
4¼"0 c r
¹0 q
[r^ £ (^r £ ~a)]
4¼ r
(15)
Next letzs calculate the magnetic ¤eld:
¯
~ = r
~ £A
~ ¯¯
B
µ ¯ c on st t
¶
1 @
¹ q~v
~ ¯¯
=
r
¡ ^r
£ 0
c on st tr e t
c @t
4¼ rv
· µ ¶
µ
¶¸
¹0
1
1 ~a
~v @rv
~
=
qr
£ ~v ¡ q^
r£
¡ 2
4¼
rv
c rv
rv @t
·
µ
¶¸
¹
q
q
~a
~v ~r ¢ ~a
= ¡ 0
r
^
£
v
~
+
r
^
£
+
4¼ r2
c
rv
r2v c
·
¸
¹0 q
q r^ £ ~a
'
~v £ r^¡
4¼ r 2
r c
where again we have negelected terms of order v=c compared with those retained.
The ¤rst term is the usual Biot-Savart law result. The second term is the
radiation ¤eld:
~ ra d = ¹0 q ~a £ r^
B
(16)
4¼ rc
11
Notice that
as expected for a plane wave.
The Poynting ¢ux is:
~
S
~ r ad = cB
~ rad £ r^
E
=
1 ~
~ = 1 E
~ £
E£B
¹0
¹0
=
1 2
E r^
¹0 c
Ã
~
^r £ E
c
!
where from equation 25:
E=
¹0 q
1
q
a sin µ =
a sin µ
4¼ r
4¼"0 rc2
and µ is the angle between ~a and ^r: Thus
µ
¶2
1
1 q
q2
a2
S=
a
sin
µ
=
sin2 µ
¹0 c 4¼"0 rc2
(4¼) 2 "0 c3 r2
and the power radiated per unit solid angle is:
dP
q 2 a2
= r2S =
sin2 µ
d(4¼)2 "0 c3
(17)
Finally the total power radiated is
Z
Z 2¼ Z +1
¡
¢
dP
q 2 a2
P =
d- =
1 ¡ ¹2 dÁd¹
2
3
d(4¼) "0 c 0
¡1
where as usual ¹ = cos µ: Thus
P
=
q 2 a2
(4¼"0 ) 2c3
=
2 q 2 a2
3 4¼"0 c3
µ
¶ ¯+1
¹3 ¯¯
¹¡
3 ¯¡1
(18)
This result is called the Larmor formula.
2.2.2
Relativistic "brute force" caculation, as per Gri¢ths
Ths is tough. However, we did some of the work already in ¤nding the potentials. First letzs look at the gradient:
4¼ "0 ~
rV
q
=
=
~
r
³
1
~ ³
´ =r
^
R 1 ¡ ~v ¢ R=c
³
´
~ R ¡ ~v ¢ R=c
~
¡r
³
´2
~
R ¡ ~v ¢ R=c
12
1
~
R ¡ ~v ¢ R=c
´
where
~ = ~r ¡ ~r0 (tr et )
R
and
tr et = t ¡ R=c
Thus
~ = rc
~ (t ¡ tre t ) = ¡crt
~ r et
rR
while
h
i ³
´
³
´
³
´
³
´
~ R
~ ¢ ~v (tret ) = R
~ ¢r
~ ~v + ~v ¢ r
~ R
~+R
~ £ r
~ £ ~v + ~v £ r
~ £R
~
r
To evaluate these derivatives, again we use the chain rule. So, for example,
@~v
d~v @tre t
@tr et
=
= ~a
@x
dtre t @x
@x
Thus
h
i
³
´
³
´
³
´
h
i
~ R
~ ¢ ~v (tre t ) = ~a R
~ ¢r
~ tret + ~v ¢ r
~ (~r ¡ ~r0 )¡R£
~
~ re t +~v £ r
~ £ (~r ¡ ~r0 )
r
~a £ rt
~ £ ~r = 0;
But r
³
´
~ r~ = ~v ; and
~v ¢ r
@~r0
d~r0 @tret
@tret
=
= ~v
@x
dtre t @x
@x
(19)
So, using (19), we get
h
i
³
´
³
´
³
´
~ R
~ ¢ ~v (tre t )
~ ¢r
~ tre t + ~v ¡ ~v ~v ¢ rt
~ re t ¡ ~a R
~ ¢r
~ tret
r
= ~a R
³
´
h
i
~ ¢ ~a rt
~ ret + ~v £ ~v £ rt
~ re t
+ R
³
´
³
´
³
´
~ ¢ ~a rt
~ re t ¡ ~v ~v ¢ rt
~ ret + ~v ~v ¢ rt
~ re t ¡ v 2 rt
~ r et
= ~v + R
h
i
~ ¢ ~a ¡ v 2 rt
~ re t
= ~v + R
Thus
4¼"0 ~
rV
q
¡1
=
³
´2
~
R ¡ ~v ¢ R=c
=
³
´2
~
R ¡ ~v ¢ R=c
1
½
·
h³
´
i
~ re t ¡ ~v ¡ 1 R
~ ¢ ~a ¡ v 2 rt
~ ret
¡crt
c
c
i
~v
1 h³ ~ ´
~ re t
+
R ¢ ~a + c2 ¡ v 2 rt
c
c
~ re t :
Now what about rt
~
rR
=
=
p
³
´
~ R
~ ¢R
~
r
1
~ re t = r
~ R
~ ¢R
~ =
¡crt
2
R
³
´i
1 h³ ~ ~ ´ ~
~£ r
~ £R
~
R ¢r R +R
R
13
¸
¾
(20)
~ = B
~ = R:
~
where we used product rule (4) from the front cover of G with A
Thus, again using (19), we have
~ re t
¡crt
³
´
³
´i
1 h~
~ ¢r
~ tret + R
~ £ ~v £ rt
~ re t
R ¡ ~v R
R
³
´
³
´
³
´
i
1 h~
~ ¢r
~ tret + ~v R
~ ¢r
~ tr et ¡ R
~ ¢ ~v rt
~ re t
=
R ¡ ~v R
R ³
´
^¡ R
^ ¢ ~v rt
~ ret
= R
=
Rearranging, we get
~ re t =
rt
^
R
^ ¢ ~v ¡ c
R
=¡
^
R=c
^ ¢ ~v =c
1¡R
(21)
Then putting this result into (20), we get
(
)
h³
´
i
^
q
1
v
~
1
R
~ =
~ ¢ ~a + c2 ¡ v 2
rV
¡
R
³
´2
^ ¢ ~v =c
4¼" 0 2
c c2
1¡R
^
R 1 ¡ ~v ¢ R=c
Next
~
4¼"0 @ A
@
~v
³
´
=
q @t
@t c2R 1 ¡ ~v ¢ R=c
^
So here we need
@R
@R @tre t
=
@t
@tre t @t
µ
¶
@tre t
@
R
1 @R
=
t¡
=1¡
@t
@t
c
c @t
@R
@tre t
=
=
~ @
@ p~ ~
R¢
R¢R=
(~r ¡ ~r0 (tre t ))
@tre t
R @tre t
~
~v ¢ R
¡
R
Thus
@R
=
@t
Ã
!
^ ¢ ~v
@R
R
1¡
=
@t
c
@R
@t
=
14
µ
¶
^ ¢ ~v 1 ¡ 1 @R
¡R
c @t
^ ¢ ~v
¡R
³
^ ¢ ~v
¡R
´
^ ¢ v~ =c
1¡R
(22)
and
^ ¢ ~v=c
@tre t
¡R
1
´ = ³
´
= 1¡ ³
@t
^
^
1 ¡ R ¢ ~v =c
1 ¡ R ¢ ~v=c
Thus
@~v (tre t )
~a
´
= ³
@t
^ ¢ ~v =c
1¡ R
and
~
4¼"0 @ A
q @t
(23)
=
=
=
=
@
~v
³
´
@t c2 R 1 ¡ ~v ¢ R=c
^
2
3
1 @ 4
~v
1
³
´5 ³
´
c2 @tre t R 1 ¡ ~v ¢ R=c
^
^ ¢ ~v =c
1¡ R
2
0
13
^ ¢ ~v ¡ ~a ¢ R=c
~ + ~v ¢ ~v =c C7
1
1
~a
B ¡R
³
´6
¡ ~v @
4
A5
³
´2
2
~
c 1¡ R
^ ¢ ~v =c
R ¡ ~v ¢ R=c
~
R ¡ ~v ¢ R=c
2
0
13
2
^
~
1
1
~a
B R ¢ ~v + ~a ¢ R=c ¡ v =c C7
³
´6
+ ~v @
4
A5
³
´2
~
c2 1 ¡ R
^ ¢ ~v =c
R ¡ ~v ¢ R=c
~
R ¡ ~v ¢ R=c
15
So, putting the pieces together, we get
8
9
<
h³
´
i
=
^
4¼"0 ~
¡1
~v
1
R
~ ¢ ~a + c2 ¡ v 2 ³
´
E =
¡ 2 R
³
´2
q
:c
c
^ ¢ ~v =c ;
^
1¡ R
R 2 1 ¡ ~v ¢ R=c
2
0
13
^ ¢ ~v + ~a ¢ R=c
~ ¡ v 2 =c C7
1
1
~a
6
BR
´4
¡ 2³
+ ~v @
A5
³
´2
~
c 1¡R
^ ¢ ~v =c
R ¡ ~v ¢ R=c
~
R ¡ ~v ¢ R=c
2 0
1
3
µ
¶
^ v
R¢~
v2
2
^
¡
¡1
~
v
v
R
c
c2
´A ¡ 1 ¡ 2 ³
´5
=
³
´2 4 @1 + ³
c
c
^
^
2
^
1
¡
R
¢
v
~
=c
1
¡
R
¢
~
v
=c
R 1 ¡ ~v ¢ R=c
´
³
´
8 ³
9
^ ¢ ~a R
^
^ =c =
<
R
~v ~a ¢ R
1
´ ¡ ~a ¡ ³
´
+
³
´2 ³
;
^
^
^ ¢ ~v =c : 1 ¡ ~v ¢ R=c
1 ¡ ~v ¢ R=c
c2R 1 ¡ R
2
3
^ ¡ ~v =c
1 ¡ v 2 =c2
R
´5
=
³
´2 4 ³
^
2
^
1 ¡ R ¢ ~v =c
R 1 ¡ ~v ¢ R=c
´
³
´
³
´ 9
8³
^ ¢ ~a R
^ ¡ ~a + ~a ~v ¢ R
^ =c ¡ ~v ~a ¢ R
^ =c =
< R
³
´
+
³
´2
;
^
^ ¢ ~v =c :
1 ¡ ~v ¢ R=c
c2R 1 ¡ R
1
Thus the electric ¤eld has two terms. The ¤rst goes as 1=R 2; and in the limit
v ¿ c becomes the usual Coulomb ¤eld
2
3
2 2
^ ¡ ~v =c
^
q
1
¡
v
=c
R
q R
~ C ou l =
´5 !
E
for v ¿ c (24)
³
´2 4 ³
4¼"0 2
4¼"0 R2
^
^
1
¡
R
¢
v
~
=c
R 1 ¡ ~v ¢ R=c
The second term is
² proportional to the acceleration
² proportional to 1=R not 1=R2
This term is the radiation ¤eld.
8³
´³
´
³
´9
>
^ ¢ ~a R
^ ¡ ~v =c ¡ ~aR
^¢ R
^ ¡ ~v =c >
< R
=
q
~ rad =
E
³
´3
>
4¼"0 R >
:
;
^
1 ¡ ~v ¢ R=c
c2
h³
´
i
^
^
q R £ R ¡ ~v =c £ ~a
=
³
´3
4¼"0 R
^
1 ¡ ~v ¢ R=c
c2
16
(25)
~
Before we comment further on these terms, letzs work on B:
~
B
=
=
~ £A
~=r
~ £ (V ~v )
r
~ £ ~v + rV
~ £ ~v
Vr
So, making use of our previous results, we get
Ã
!
^
4¼"0 ~
1
R=c
³
´ ~a £
B =
^ ¢ ~v =c
q
^
1¡R
R 1 ¡ ~v ¢ R=c
(
)
i
^
1
~v
1 h³ ~ ´
R
2
2
+
¡ 2 R ¢ ~a + c ¡ v
£ ~v
³
´2
^ ¢ ~v =c
c
c
1¡R
^
R2 1 ¡ ~v ¢ R=c
=
³
^
~a £ R=c
^
R 1 ¡ ~v ¢ R=c
´2 ¡
1
³
^
R2 1 ¡ ~v ¢ R=c
´2
i R
^ £ ~v
1 h³ ~ ´
2
2
R
¢
~
a
+
c
¡
v
^ ¢ ~v =c
c2
1¡R
Again we get a term in 1=R 2 : This is the Biot-Savart law ¤eld.
¡
¢
2 2
~
~ B -S = q~v £ R ³ 1 ¡ v =c ´
B
3
4¼"0 R2
^
1 ¡ ~v ¢ R=c
(26)
The radiation ¤eld is again proportional to a and to 1=R :
³
´ ³
´³
´
^
^
~ ¢ ~a R
^ £ ~v =c2
~
a
£
R=c
1
¡
~
v
¢
R=c
¡
R
q
~ ra d =
B
³
´3
4¼"0 R
^
1 ¡ ~v ¢ R=c
h³
´
³
´i
^£ R
~ ¢ ~a ~v =c + ~a 1 ¡ ~v ¢ R=c
^
¡
R
q
=
³
´3
4¼"0 cR
^
1 ¡ ~v ¢ R=c
n ³
´³
´
h
³
´io
^£ ¡ R
~ ¢ ~a R
^ ¡ ~v =c + ~a R
^¢ R
^ ¡ ~v =c
¡
R
q
=
³
´3
4¼"0 cR
^
1 ¡ ~v ¢ R=c
^£R
^ , which is identically
In the last step we added a term proportional to R
zero. Thus
n
h³
´
io
^£ R
^£ R
^ ¡ ~v =c £ ~a
¡
R
q
~ r ad =
B
³
´3
4¼ "0 cR
^
1 ¡ ~v ¢ R=c
=
1^
~ ra d
R£E
c
(27)
Relation (27) is exactly what we expect for an EM wave.
17
2.2.3
Properties of the Coulomb ¤elds
Itzs interesting to see what happens to the Coulomb ¤elds as v approaches c:
~ (equation 24).
Letzs look at E
2
3
2
2
^ ¡ ~v =c
q
1
¡
v
=c
R
~ C ou l =
´5
E
³
´2 4 ³
4¼"0 2
^
^
1
¡
R
¢
v
~
=c
R 1 ¡ ~v ¢ R=c
2
3
^
q (1 + v=c) (1 ¡ v=c) 4 R ¡ ~v =c 5
´
=
³
´2 ³
4¼"0 2
^
^
1
¡
R
¢
v
~
=c
R 1 ¡ ~v ¢ R=c
³
´
^ ¡ ~v =c
(1
¡
v=c)
R
q
³
´3 as v ! c
2¼"0
^
R 2 1 ¡ ~v ¢ R=c
!
^ v : We get
First look in the direction Rk~
³
´
~ C ou l R
^ = ^v !
E
^
q
R
2¼"0 R2 (1 ¡ v=c)
^ v ; we get
which becomes very large as v ! c: In the direction with R?~
³
´
^ ¡ ~v =c
³
´
(1 ¡ v=c) R
~ Co ul R?^
^ v ! q
E
!0
2¼"0
R2
But we must be careful in interpreting these results, because
~ = ~r ¡ ~r0 (tr et )
R
If the charge has a constant velocity (~a = 0); then
~r0 = ~v tre t
and
~ ¡ R ~v
R
c
=
~r ¡ ~v tre t ¡ R
=
~r ¡ ~v t ´ ~s
R
2
v~
~v
= ~r ¡ ~v tret ¡ c (t ¡ tret )
c
c
=
·
µ
=
(~s + ~v R=c)
=
s2 + 2~s ¢ ~v R=c + R2 v 2 =c2
R
~r ¡ ~v t ¡
c
2
¶¸2
where ~s = ~r ¡ ~v t; and so we can solve for R as a function of ~s :
µ
¶
v2
~s ¢ ~v
2
R 1 ¡ 2 ¡ 2R
¡ s2 = 0
c
c
18
R
=
=
q
¡
2
4s 2 vc2 cos2 µ + 4s2 1 ¡
¡
2¢
2 1 ¡ vc 2
q
~s ¢ ~v =c § s 1 ¡ v 2 sin 2 µ=c2
v
2 ~s¢~
§
c
v2
c2
¢
1 ¡ v 2 =c2
Since R is positive, we need the plus sign. (Check the limit v ! 0:) Then
µ
¶
R
2
~
cR ¡ ~v ¢ R = cR ¡ ~v ¢ ~r + v t ¡
c
µ
¶
v2
= cR 1 ¡ 2 ¡ ~v ¢ ~r + v 2t
c
p
= ~s ¢ ~v + s c2 ¡ v 2 sin2 µ ¡ ~v ¢ ~s
p
= s c2 ¡ v 2 sin2 µ
Thus
~ Co ul =
E
Thus
2
2
2
3
q
1 ¡ v =c
~s
5
¡
¢4 q
4¼"0 s 2 1 ¡ v 2 sin2 µ=c2
2
s 1 ¡ v sin2 µ=c2
~ C ou l =
E
¡
¢
q
1 ¡ v 2 =c2 ^s
4¼"0 s2 ¡1 ¡ v2 sin2 µ=c2 ¢3=2
In the direction µ ' 0; (~s k~v ); we get
~ C ou l =
E
¡
¢
q^
s
s^ 1 ¡ v 2 =c2 ! 0 as v ! c
2
4¼" 0 s
whereas for µ ' ¼=2; (~s?~v ); we get
~ c ou l =
E
q
^s
p
! 1 as v ! c
2
4¼"0 s
1 ¡ v 2 =c2
Thus the ¤eld is concentrated in the direction ~s?~v ; and becomes very large as v
increases. A stationary observer sees a pulse of ¤eld as the charge moves past.
2.2.4
Properties of the radiation ¤elds
The radiation ¤eld decreases more slowly with distance than the Coulomb ¤eld,
so at large distances, the radiation ¤eld dominates. The Poynting vector describes the energy radiated:
~ rad
S
=
=
=
1 ~
~ rad
E rad £ B
¹0
µ
¶
1 ~
1^
~
E rad £
R £ E rad
¹0
c
³
´
1 ^ 2
~ rad R
^¢E
~ r ad
RE rad ¡ E
¹0 c
19
^¢E
~ rad = 0; and then
From equation (25), we see that R
³
h³
´
i´2
µ
¶2 R
^£ R
^ ¡ ~v =c £ ~a
q
~ = 1 R
^
S
³
´6
¹0 c
4¼"0 R
^
1 ¡ ~v ¢ R=c
c4
In the non-relativistic case, v ¿ c;
~
S
=
=
³
h
i´2
^£ R
^ £ ~a
³
´
R
2
1 ^
q
R
¹0" 20 c
4¼R
c4
³
´
1 ^
q 2 2
R
a sin2 µ
3
"0 c
4¼R
^ and ~a: Notice that S
~ / 1=R2 : This is the usual
where µ is the angle between R
inverse square law for light. We can compute the power radiated per unit solid
angle:
dP
1 ³ q ´2 2
= R2S =
a sin2 µ
(28)
d"0 c3 4¼
This is the Larmor formula for power radiated. The power is proportional to the
square of the particlezs acceleration. Notice that no power is radiated along the
line of the acceleration (µ = 0) and power radiated is maximum perpendicular
to the acceleration. See LB page 763.
The total power radiated is
Z
Z
dP
1 ³ q ´2 2 ¼
P =
d- =
a
sin2 µ2¼ sin µdµ
d" 0 c3 4¼
0
Let ¹ = cos µ: then
P
=
=
q2
a2
8¼"0 c3
Z
+1
¡1
¡
1¡¹
2
¢
q2
d¹ =
a2
8¼" 0 c3
q2
4
q2
a2 =
a2
3
8¼"0 c
3
6¼"0 c3
µ
¶¯+1
¹3 ¯¯
¹¡
3 ¯¡1
(29)
There are some interesting and important corrections when the motion is
relativistic. Most importantly, the total power radiated is increased signi¤cantly, and it is also strongly beamed in the direction of the particlezs velocity
~v : With the relativistic factor ° de¤ned as
°=p
1
1 ¡ v 2 =c2
so that ° ' 1 corresponds to non-relativistic motion, and ° ! 1 as v ! c; we
¤nd that
Ã
!
2
2
q
j~
v
£
~
a
j
P = °6
a2 ¡
6¼"0 c3
c2
and the radiation is emitted within a cone of opening angle of about 1=° around
the direction of ~v : The derivation of these results is not easy, and we shall omit
it.
20