Abraham-Lorentz Self -force
S.M.Lea
Abraham and Lorentz (circa 1904) attempted a rigorous derivation fo the radiation
reaction force. The idea is to model a particle as a collection of charges, currents and ®elds.
The basic equation of motion is:
dP~particle
= F~ ext
(1)
dt
~
where Fext is the sum of the external forces. We assume for the moment that all the forces
acting are electromagnetic in origin. Then the systemzs total momentum is the sum of its
(apparently) mechanical momentum and the momentum of the ®elds. Thus:
~ ®elds
dP~m ech
dP
+
=0
dt
dt
The rate of change of ®eld momentum may be expressed in terms of the Lorentz force
density:
¶
Z µ
dP~® elds
~ 1 ~j £ B
~ dV
=¡
½E+
dt
c
(See eg equation 12.120). We decompose the ®elds into the sum of the self ®elds plus the
external ®elds:
~ =E
~s + E
~ ext
E
Thus:
¶
¶
¶
Z µ
Z µ
Z µ
1~ ~
1~
1~
~
~
~
~
~
½E + j £ B dV =
½Eext + j £ Bext dV +
½E s + j £ Bs dV
c
c
c
¶
Z µ
~ s + 1 ~j £ B
~ s dV
= F~ ext +
½E
c
Thus we have:
¶
Z µ
dP~ mech
dP~®elds
~ s + 1 ~j £ B
~ s dV
=¡
= F~ ext +
½E
(2)
dt
dt
c
where the second term on the right hand side is the self force.
To calculate this integral we assume:
² The particle is instantaneously at rest
² the charge distribution is rigid and spherically symmetric
1
Then the expression for the self-force simpli®es, since ~j = ½~v = 0 :
Z
~ self =
~ s dV
F
½E
!
Z Ã
~
1
@
A
~ +
= ¡ ½ r©
dV
c @t
where
©=
~:
and similarly for A
~=
A
Z
½ (~x 0 ; t0 )jret 0
dV
R
¯
Z ~j (~x0 ; t0) ¯¯
1
ret
dV 0
c
R
The retarded time t0 = t ¡ R=c differs from t by a term of order a=c where a is the radius of
the particle. This time is extremely short, and the particle cannot move far during this time
interval. Thus we may expand
R @½
½ (~x 0 ; t0 )jret = ½ (~x 0 ; t) ¡
+¢¢¢
c @t
µ ¶n n
1
X
(¡1)n R
@ ½
=
n!
c
@tn
n=0
and thus:
!
Z Ã
~
1
@
A
~ +
½ r©
dV
c @t
=
=
1
n Z
n
0
n~
0
X
(¡1)
~ n¡1 @ ½ (~x ; t) + ½ (~x;t) @ Rn¡ 1 @ j (~x ; t) d3 x 0 d3 x
½
(~
x
;t)
rR
n!cn
@tn
c2 @t
@tn
n=0
"
#
Z
1
n¡1 ~
0
X
(¡1)n
@n
R
@
j
(~
x
;
t)
0
n¡1
~
½ (~x; t) n ½ (~x ; t) rR
+
d 3 x 0d 3 x
n!cn
@t
c2
@t
n=0
Now we can evaluate this expression term by term.
First the scalar potential part:
n=0 :
Z
Z
0
~ 1 d3 x 0d3 x = ¡ ½ (~x;t) ½ (~x0 ; t) ~x ¡ ~x d3 x 0d3 x = 0
½ (~x;t) ½ (~x 0 ; t) r
R
R3
0
by symmetry. (Interchange ~x and ~x : The integral should not change, I1 = I2 , but the
integrand changes sign, implying I1 = ¡I2 : So I = 0.)
(In detail...For a spherically symmetric distribution, ½ (~x ;t) = ½ (r;t) : Write 1=R as
an expansion in Legendre polynomials. Put the polar axis along ~x while we do the x 0
integration:
Z
Z
X rl
<
~
d3 x½ (r;t) r
½ (r 0;t)
P (¹0) r02 dr0 d¹0 dÁ0
l+1 l
r
>
l
The integral over ¹0 is zero unless l = 0: So we have:
Z
Z
Z
µZ r
Z a
¶
1 02 0
1
1
~
~
d3 x½ (r;t) r
½ (r 0;t)
r dr 4¼ = 4¼ d3 x½ (r;t) r
½ (r 0;t) r02 dr 0 +
½ (r0 ;t) 0 r0 2dr 0
r>
r
r
0
r
Z
~ (r)
= 24¼ d3 x½ (r;t) rf
where q (r) is the charge inside radius r : The right hand side is then:
µ
¶
µ
¶
Z
Z a
Z
@ q (r)
q (r)
4¼ d3 x½ (r;t)^r
+
½ (r 0 ;t) r 0 dr 0 = 4¼ d3 x½ (r;t) ^
r ¡ 2 +
@r
r
r
r
Now ^r = ^z¹ + sin µ (^
x cos Á+^
y sin Á) ;so integrating over Á reduces the x¡ and
y¡components to zero, while integrating over ¹ reduces the ^z component to zero. )
~ = 0;so this term is zero too.
n = 1 : Rn¡ 1 ´ 1 and r1
n > 1 : Relabel by setting m = n ¡ 2
Now we want to combine the remaining (n > 1) terms with the vector potential terms,
so letzs relabel the ½ terms by setting n = m + 2
"
#
Z
1
X
(¡1)n
@n
Rn¡ 1 @~j (~x 0 ; t) 3 0 3
0
n¡1
~
½ (~x; t) n ½ (~x ; t) rR
+
d xd x
n!cn
@t
c2
@t
n=0
1
1
n Z
n Z
X
X
(¡1)
@n
(¡1)
@ n R n¡1 @~j (~x 0; t) 3 0 3
0
~ n¡1 +
=
½
(~
x
;
t)
½
(~
x
;
t)
rR
½ (~x ; t) n
d xd x
n
n
n
n!c
@t
n!c
@t
c2
@t
n=2
n=0
Z
1
1
m
n Z
X
X
(¡1)
@ m+2
(¡1)
@ n Rn¡ 1 @~j (~x 0; t) 3 0 3
0
m+1
~
=
½ (~x; t) m+2 ½ (~x ; t) rR
+
½ (~x ; t) n
d xd x
m+2
n
(m + 2)!c
@t
n!c
@t
c2
@t
m=0
n=0
"
#
1
n Z
n+1
X
~ n+1
(¡1)
@½ (~x0 ; t)
rR
n¡1 @
0
~
=
½ (~x ; t) R
+ j (~x ; t) d3 x 0 d3 x
n!cn+ 2
@tn+1
@t
(n + 2) (n + 1) Rn¡1
n=0
Now use charge continuity to write
0
@½ (~x0 ; t)
~ ¢ ~j (~x0 ; t)
= ¡r
@t
Our expression becomes:
"
#
1
n Z
n+1
n^
X
0
(¡1)
(n
+
1)
R
R
n¡1 @
0
0
~ ¢ ~j (~x ; t)
½ (~x ; t) R
¡r
+ ~j (~x ; t) d3 x0 d3 x
n+ 2
n+1
n¡1
n!c
@t
(n
+
2)
(n
+
1)
R
n=0
"
#
1
n Z
n+1
X
~
0
(¡1)
@
R
~ ¢ ~j (~x 0 ; t)
=
½ (~x ; t) Rn¡1 n+1 ~j (~x0 ; t) ¡ r
d3 x0 d3 x
(3)
n+ 2
n!c
@t
(n
+
2)
n=0
Now integrate the last term over the prime variables. We use the usual bag of tricks:
³
´
³ 0
´
³ 0 ´
³
´ ³
´
~ 0 £ ~j £ RR
~ n¡1 = ~j r
~ ¢ RR
~ n¡1 ¡R n¡1 R
~ r
~ ¢ ~j +Rn¡ 1 R
~ ¢r
~ 0 ~j ¡ ~j ¢ r
~ 0 RR
~ n¡1
r
Thus:
³ 0 ´
³
´ ³ 0
´
³
´ ³
´
~ r
~ ¢ ~j = ¡ r
~ 0£ ~j £ RR
~ n¡1 +~j r
~ ¢ RR
~ n¡ 1 +Rn¡1 R
~ ¢r
~ 0 ~j¡ ~j ¢ r
~ 0 RR
~ n¡1
Rn¡ 1 R
The integral of the curl converts to
Z
³
´
~ n¡ 1 d2 x0 = 0
n
^ 0£ ~j £ RR
S
3
and
³ 0
´
~ ¢ RR
~ n¡ 1
~j r
=
=
=
=
³
´
~ ¢ RR
~ n¡1
¡~j r
³
´
³
´ ³
´
³
´
~ £ RR
~ n¡1 £ ~j ¡ Rn¡ 1 R
~ r
~ ¢ ~j ¡ ~j ¢ r
~ RR
~ n¡1 + R n¡1 R
~ ¢r
~ ~j
r
³
´
~ RR
~ n¡1 + 0
(integrates to zero) ¡ 0 ¡ ~j ¢ r
³
´
~j ¢ r
~ 0 RR
~ n¡1
Also: ³
´
³
´ ³
´
³ 0
´
³ 0
´
0
0
0
n¡1 ~
~ ¢r
~ ~j = r
~ R n¡ 1R
~ ¢ ~j ¡ ~j ¢ r
~ R n¡ 1R¡R
~
~ £ ~j ¡~j£ r
~ £ RR
~ n¡ 1
Rn¡ 1 R
R£ r
The ®rst term converts to a surface integral via a divergence theorem variant (see cover of J),
~0 £R
~ = 0: Since our object is rigid, ~j
and the integral is zero. The last term is zero since r
can either be in a constant direction, or due to rotation of the sphere. Jackson does not seem
to allow for rotation, in which case
0
0
0
~ £ ~j = r
~ £ [½ (r0 ) ~v ] = @½ ^
~ £ ~v = @½ ^
r
r 0 £ ~v + ½r
r 0 £ ~v
0
@r
@r0
Then
Z
Z
µ
¶
³ 0
´
0
~£ r
~ £ ~j d3 x0 = Rn¡1 R
~ £ @½ ^
Rn¡1 R
r
£
~
v
d3 x 0
@r0
If we choose polar axis along ~v ;then the right hand side is:
µ
¶
Z
~ £ @½ Á
^ sin µ d3 x 0 = 0
¡ Rn¡1 R
@r0
Thus:
³
´
³
´
0
0
~ ¢r
~ ~j = ¡ ~j ¢ r
~ Rn¡1 R
~
Rn¡1 R
and thus:
Z
Z h³
³ 0 ´
´
³
´
³
´
i
n¡1 ~
3 0
~
~ 0 RR
~ n¡1 ¡ ~j ¢ r
~ 0 Rn¡1 R
~ ¡ ~j ¢ r
~ 0 RR
~ n¡1 d3 x0
~
~j ¢ r
R
R r ¢j d x =
Z ³
´
~ 0 RR
~ n¡1 d3 x 0
~j ¢ r
= ¡
and ®nally!
¡
Z
R
n¡ 1 ~ 0
r ¢ ~j (~x0 ; t)
~
R
d3 x 0
(n + 2)
Z ³
´ n¡1 R
~
~j (~x 0 ; t) ¢ r
~0 R
=
d3 x 0
(n + 2)
Z
1
^ n¡2 Rd
~ 3 x0
=
Rn¡1~j + (n ¡ 1) ~j ¢ RR
n+2
and so from equation (3) we get
·
1
n Z
´ ¸
n+1
X
(¡1)
n+1
n ¡ 1 ³~ 0
n¡1 @
0
~
^
^ d3 x0 d3 x
½ (~x ; t) R
j (~x ; t)
¡
j (~x ; t) ¢ R R
n+2
n+1
n!c
@t
n
+
2
n
+
2
n=0
4
Now we set ~j = ½~v (rigid assumption):
·
1
n Z
´ ¸
n+ 1
X
(¡1)
n+1
n¡1 ³
n¡1 @
^
^ d3 x0 d3 x
½ (~x ; t) R
½ ~v (t)
¡
~v (t) ¢ R R
n!cn+2
@tn+1
n+2
n+2
n= 0
Next choose polar axis along ~
v; and the integral becomes:
·
¸
Z
n+1
@
n+1
n¡1
n¡1
^ d3 x 0 d3 R
In =
½ (~x ; t) Rn¡1 n+1 ½ ~v (t)
¡
~v cos2 ° ¡
v cos ° ?
@t
n+2
n+2
n+2
Because of the symmetry, the direction of ~v is the only direction singled out, and so the
?¡component must integrate to zero. (It is not trivial to demonstrate this explicitly.) Then:
·
¸
Z
@ n+ 1~v (t) n + 1
n¡1
2
In = ½ (~x) R n¡ 1
½
¡
cos
°
d3 x0 R2 dRd¹
@tn+1
n+2
n+2
Term by term: The easiest term is:
n=1
Z
Z
@2
2
2 @ 2~v (t)
I1 =
½ (~x; t) 2 ½~v (t) d3 x 0 d3 R =
½ (~x ) d3 x½ (~x 0) d3 x0
@t
3
3 @t2
2 @ 2 ~v (t) 2
=
e
3 @t2
and thus the n = 1 term in our series is:
2 e2 @ 2~v (t)
¡ 3
3 c @t2
2
@ 2~v( t)
and this contributes a term Fself = 23 ec 3 @ t2 which is exactly the expression we obtained
before for the radiation damping self-force
n = 0 To do the prime integral, put ~x in the x ¡ z plane. Then Á = 0 :
Z
¢
1 @~v
½ (~x; t) ¡
I0 =
½ 1 + cos 2 ° d3 x0 d3 x
2 @t
R
~ and ~v : Now here Jackson inserts the average value of
where ° is the angle between R
2
cos ° = 1=3; which I think is incorrect (or at least needs better justi®cation) to obtain for
the n = 0 term:
Z
~a 4 1
½ (r) ½ (r0 ) 3 0 3
d xd x
2
c 32
R
~a 4
=
U
c2 3
where U is the electromagnetic energy of the sphere. Then if we interpret U=c2 = m; the
mass of the particle, we get:
F0
=
F0 =
5
4
m~a
3
Thus we have , including only the ®rst 2 terms in the series,
4
2 e2 @ 2~v (t)
m~a ¡
= F~ ext
3
3 c3 @t2
In an alternative approach, we take the Fourier transformof equation (1) or equivalently
(2) which now reads:
1
n
X (¡1)
dP~ mech
= F~ext ¡
dt
n!cn+2
n=0
Z
½ (~x) Rn¡ 1
·
¸
@ n+1~v (t) n + 1
n¡1
2
½
¡
cos
°
d 3 x 0d 3 x
@tn+1
n+2
n+2
Taking the FT we get:
·
¸
1
n
n+1 Z
X
(¡1) (¡i!)
n+1
n¡1
n¡ 1
0
2
¡i!m 0~v (!) = F~ext (!)¡
½
(~
x
)
R
~
v
(!)
½
(~
x
)
¡
cos
°
d3 x 0 d3x
n!cn+ 2
n+2
n+2
n=0
and taking the average of cos 2 ° per JDJ we have
1
n Z
X
2
(i!)
~
¡i!m 0~v (!) = F ext (!) + i!~v (!)
½ (~x ) R n¡1 ½ (~x0 ) d3 x 0 d3 x
n+2
3
n!c
n= 0
We can rewrite this expression as:
¡i!M (!) ~v (!) = F~ext (!)
where
1
n Z
2 X (i!)
½ (~x ) R n¡ 1½ (~x0 ) d3 x 0 d3x
3c2 n=0 n!cn
Z
1
n
X
2
(i!R) ½ (~x 0 ) 3 0 3
= m0 + 2
½ (~x )
d xd x
3c
n!cn
R
n=0
Z
2
½ (~x0 ) i!R=c 3 0 3
M (!) = m0 + 2
½ (~x )
e
d xd x
3c
R
Now the form factor for a particle is de®ned as the Fourier transform of ½ :
Z ³ ´
e
½ (~x ) =
f ~k eik¢~x d3~k
3
(2¼)
So we may write:
Z
Z Z
Z ³ ´
Z ³ ´
i!R=c
½ (~x 0 ) i!R=c 3 0 3
e2
1
ik¢~
x 3~
~
~k0 eik 0 ¢~x0 d3~k0 e
½ (~x )
e
d xd x =
f
k
e
d
k
f
d 3 x 0d 3 x
R
R
(2¼)3
(2¼ )3
Z
Z
³ ´ ³ ´Z Z
2
i!R=c
0 0 e
3~
3~ 0 e
~
~0
=
d k d k
eik¢~x eik ¢~x
d3 x0 d3 x
6f k f k
R
(2¼)
M (!) =
m0 +
6
~ = ~x ¡ ~x0
Change variables to R
Z Z
i!R=c
~
~0 0 e
eik¢~xe ik ¢~x
d 3 x 0d 3 x
R
Z Z
=
=
Now put the polar axis along ~k :
Z
~ ~
eik¢R ei!R=c RdRd-
Z
=
i!R=c
0
~ ~
~0 0 e
eik¢(R+~x ) eik ¢~x
d3 x 0 R2 dRdR
Z
Z
0
0
~
~ ~
e ik¢R ei!R=c RdRd- e i( k+ k )¢~x d3x 0
Z
³
´
~ ~
3
(2¼)
eik¢R ei!R=c RdRd-± ~k + ~k0
=
= 2¼
=
2¼
ik
=
2¼
ik
eikR¹ ei!R=c RdRd¹dÁ
¯ +1
eikR¹ ¯¯
ei!R=c RdR
ikR ¯ ¡1
Z
Z
1
0
µ
ei(k+!=c) R ¡ e i( ¡k+!=c)R dR
¶ ¯1
ei(k+!=c) R
e i(¡ k+!=c)R ¯¯
¡
i (k + !=c)
i (¡k + !=c) ¯0
To make the upper limit vanish, we let ! have a small, positive imaginary part, so that
ei!R=c = e i! r R=c e¡! iR=c ! 0 as R ! 1
Then:
Z
e
~ i!R=c
i~
k¢R
e
RdRd-
2¼
ik
1
1
+
i (k + !=c)
i (¡k + !=c)
1
4¼ 2
k ¡ ! 2 =c2
=
=
Thus
Z Z
~
~0
eik¢~x eik ¢~x
0
¶
³
´
e i!R=c 3 0 3
1
3
~k + ~k0
d x d x = (2¼) 4¼ 2
±
R
k ¡ ! 2 =c2
and hence
Z
½ (~x 0 ) i!R=c 3 0 3
½ (~x )
e
d xd x
R
and ®nally
µ
¡
=
Z
=
e2
2¼ 2
d3~k
Z
Z
³ ´ ³ ´
³
´
1
~ f ~k0 (2¼)3 4¼
~k + ~k0
f
k
±
k2 ¡ ! 2 =c2
(2¼)6
¯ ³ ´¯ 2
¯ ~ ¯
¯f k ¯
d3~k0
d3 ~k
e2
k2 ¡ !2 =c2
¯ ³ ´¯ 2
¯ ~ ¯
Z
¯f k ¯
e2
3~
M (!) = m 0 + 2 2
d k 2
3c ¼
k ¡ !2 =c2
which is J equation 16.32. Letting ! ! 0; we obtain the time-independent }physical} mass
7
of the particle, including the effect of the self ®elds. It is
¯ ³ ´¯ 2
¯ ~ ¯
Z
2
¯f k ¯
e
3~
m = m0 + 2 2
d k
(4)
3c ¼
k2
³ ´
For a point particle, for form factor f ~k = 1 (This gives a delta-function density.). The
mass function is divergent for such a particle. The effective mass
Z
¶
¯ ³ ´¯2 µ
e2
1
1
¯
3~ ¯
~
M (!) = m + 2 2
d k ¯f k ¯
¡ 2
3c ¼
k2 ¡ ! 2 =c2
k
Z
¯ ³ ´¯2
e 2 !2
1
¯
¯
= m+ 4 2
d3 ~k ¯f ~k ¯ 2 2
3c ¼
k (k ¡ ! 2 =c2 )
³ ´
The integral in this expression is convergent even when f ~k = 1:
For a point particle:
Z
e2 ! 2
1
M (!) = m + 4 2
d3~k 2 2
3c ¼
k (k ¡ ! 2 =c2 )
Z 1
e2 ! 2
dk
= m + 4 2 4¼
2 ¡ ! 2 =c2 )
3c ¼
(k
0
Z 1
2e2 ! 2
dk
= m+
3 c4 ¼ ¡ 1 (k2 ¡ ! 2 =c2 )
We do the integral by contour integration. There are poles at k = §!=c: Recall that ! has
a small, positive imaginary part, so these poles are not on the real axis. Closing in the upper
half-plane, we enclose the pole at k = + !c and the result is:
M (!) =
=
2 !2
c
2 e2
m + e 2 4 (2¼i)
= m + i 3!
3 c ¼
2!
3c
m (1 + i!¿ )
Looking back at the transformed equation of motion, the solution is
F~ext (!)
M (!)
Z ~
1
F ext (!) ¡i!t
~a (t) = p
e
d!
M (!)
2¼
and so for a point particle, the velocity is
Z
1
F~ ext (!)
~a (t) = p
e¡i!t d!
m (1 + i!¿ )
2¼
Z
1
F~ ext (!)
=
p
e¡i!t d!
m (! ¡ i=¿ )
i¿ 2¼
~a (!) =
8
(5)
We can invert the transform using the convolution theorem, where
Z
1
1
f (t) = p
e¡i!t d!
(! ¡ i=¿ )
2¼
For t > 0 we close downward and f (t) = 0: But for t < 0 we must close upward, enclosing
the pole at i=¿ : The solution is
1
f (t < 0) = p 2¼iet=¿
2¼
Then
Z +1
1
~a (t) = p
F~ ext (t ¡ u) f (u) du
2¼ ¡1
Z 0
1
1
=
p
F~ ext (t ¡ u) p 2¼ieu =¿ du
im¿ 2¼ ¡ 1
2¼
Z 0 ~
F ext (t ¡ u) u =¿ du
=
e
m
¿
¡1
Z 1 ~
Fext (t + x¿ ) ¡ x
=
e dx
m
0
Once again we get an integral over the future.
9