ELECTRO ELECTROMAG MAGNETI NETIC C FIELD FIELD OF A MOVING MOVING POINT CHARGE
1. Maxwell’s Equations and source densities Maxwell’s Maxwell’s Equations Equations for the electric electric potential potential V and and the magnetic (vector) potential A in vacuum are the following: ρ ε0 A = − = − µ0 J
(1)
V = −
(2)
where ρ is the electric charge density, J the electric current density, ε0 the vacuum electric permettivity, µ0 the vacuum vacuum magnetic permeabilit permeability y. We are using the Lorenz Gauge, meaning that the following condition is to be imposed on the electromagnetic potentials for equations (1) and (2) to be valid ( c is the speed of light in vacuum, c = √ ε1 µ ): 0
0
(3)
∇·A+
1 ∂V c2 ∂t
=0
The charg chargee and curren currentt densit densities ies for a single single mo movin vingg point point charg chargee q , whose whose position vector with respect to some frame of reference is r S (t) and whose velocity is v S (t), can be written as: (4)
ρ(r, t) = q δ (r − rS (t))
(5)
J(r, t) = q vS (t) δ (r − rS (t))
δ (r) is the Dirac Delta distribution evaluated at a generic point in space with position vector r . enardena rd-Wieche Wiechert rt Potentials Potenti als 2. Li´
The equations for the electromagnetic potentials, derived solving Maxwell’s Equations (1) and (2) with Green’s Lemma, are: V (r, t) =
A(r, t) =
Where t r is the retarded time:
1
4πε 0 µ0 4π
R3
R3
tr = t −
ρ(r , tr ) dr |r − r|
J(r , tr ) dr |r − r|
| r − r|
1
c
2
ELECTROMAGNETI C FIELD OF A MOVI NG POINT CHARGE
Substituting equations (4) and (5) we have: (6) (7)
q V (r, t) = 4πε 0 µ0 q A(r, t) = 4π
δ (r − rS (tr )) dr |r − r| R vS (tr ) δ (r − rS (tr )) dr |r − r| R
3
3
Let us consider the electric potential first. In order to evaluate the integral, which is not trivial due to the retarded time’s dependence on r , we rewrite it as a double integral using a Dirac Delta with the time as argument, then exchange the order of integration to obtain: q V (r, t) = 4πε 0
+
∞
δ (r − rS (τ )) δ (τ − tr ) dr dτ |r − r|
−∞
R3
The integral over r simply picks out r = r S (τ ): (8)
q V (r, t) = 4πε 0
(9)
+
∞ δ (τ − τ r ) dτ −∞ |rS (τ ) − r|
τ r = t −
| rS (τ ) − r| c
Now, to compute the integral over τ we use the general rule for the evaluation of an integral with a Dirac Delta distribution with a function as argument: n
+
∞
f (x) δ (g (x)) dx =
−∞
k=1
f (xk ) |g (xk )|
where the xk are the simple zeroes of g (x) (i.e. with g (xk ) = 0, ∀k ∈ [1, n]), assuming g (x) has no multiple zeroes. In our case, the argument of the Dirac Delta is τ − τ r , which is zero only when τ = τ r . Taking the derivative of both sides of equation (9) with respect to τ we have: vS (τ ) rS (τ ) − r ∂τ r = − · ∂τ c |rS (τ ) − r|
(10)
v S (τ ) rS (τ ) − r ∂ (τ − τ r ) = 1 + · |rS (τ ) − r| ∂τ c
The right-hand side of equation (10) is certainly positive and lower than or equal to r 1, since | vS | < c and | |rr ((τ τ ))− −r| | = 1. Therefore, evaluating the integral in equation (8) results in: S S
(11)
(12)
V (r, t) =
q 1 4πε 0 |rS (tr ) − r| 1 +
tr = t −
1 vS (tr )
| rS (tr ) − r| c
c
· |rr
S S
(t ) (t ) r
r
−r −r |
ELECTROMAGNETI C FIELD OF A MOVI NG POINT CHARGE
3
which is the explicit expression of the electric potential generated by a moving point charge. Introducing the quantities: R(tr ) = | rS (tr ) − r|
ˆ(tr ) = e
(13)
rS (tr ) − r |rS (tr ) − r|
R(tr ) = R (tr )ˆ e(tr )
and since: ˆ drS dR dR de ˆe + R = = dt dt dt dt ˆ 1 d 1 d de ˆ = e·e ˆ) = ·e (ˆ (1) = 0 2 dt 2 dt dt vS =
equation (11) can be rewritten as: (14)
V (r, t) =
1 q 1 4πε 0 R(tr ) 1 + c dR ( tr ) dt
Proceeding in the same way for the magnetic potential A we have (from equation (7)): µ0 q A(r, t) = 4π
= Therefore: (15)
µ0 q 4π
+
∞
A(r, t) =
3
vS (τ ) δ (r − rS (τ )) δ (τ − tr ) dr dτ |r − r|
−∞ R +∞ vS (τ ) δ (τ − τ r ) |rS − r(τ )|
−∞
µ0 q vS (tr ) 4π |rS (tr ) − r| 1 +
dτ
1 vS (tr )
Or: (16)
c
· |rr
S S
(t ) (t ) r
r
−r −r |
µ0 q dR 1 (tr ) 1 4πR (tr ) dt 1 + c dR ( tr ) dt
A(r, t) =
3. Electric field The electric field generated by the point charge is given by: (17)
E = −∇ V −
∂ A ∂t
In order to calculate the potentials’ derivatives with respect to the spatial coordinates x, y and z and to the time t , it is convenient to first compute the gradient and the time derivative of the retarded time tr ; from here on, primed variables will be intended as evaluated at the retarded time (so rS ≡ r S (tr )). Taking the gradient of both sides of equation (12) we have: ∇tr = − (18)
1 c
∇tr =
vS ·
rS − r rS − r t ∇ − r |rS − r| |rS − r|
1 c 1+
vS
c
1 rS − r r · |rr − −r| |rS − r| S S
4
ELECTROMAGNETI C FIELD OF A MOVI NG POINT CHARGE
While derivating equation (12) with respect to t brings to: v S rS − r ∂t r ∂t r =1− · ∂t c |rS − r| ∂t
∂t r = ∂t 1+
(19)
vS
c
1 r · |rr − −r| S S
Rearranging equation (11) as: V (r, t) =
q q 1 ≡ 4πε 0 |rS − r| + vcS · (rS − r) 4πε 0 ξ
and taking the gradient of both sides we find: ∇V = −
q ∇ξ ; 4πε 0 ξ 2
so in order to determine the gradient of the electric potential we simply need to calculate ∇ ξ . We have: vS · (rS − r) ξ = | rS − r| +
(20)
c
∂ξ ∂ |rS − r| ∂ vS = + · (rS − r) = ∂x ∂x ∂x c 1 ∂ vS ∂ (rS − r) rS − r − r) + v S · ∂ (rS − r) = r = · + · ( S |rS − r| c ∂x ∂x c ∂x
2
rS − r ∂t r aS xS − x − r) ∂t r + | vS | ∂t r − v Sx = r − + · ( S c ∂x c ∂x c |rS − r| ∂x |rS − r|
= v S ·
∂t r = vS · ∂x
rS − r v S + |rS − r| c
Therefore (using equation (18)):
a xS − x v Sx + S · (rS − r) − − |rS − r| c c
(21) ∇ξ = ∇ tr
vS ·
rS − r v S + c |rS − r|
1 r 1 + c · |rr − −r| Finally, we have: =
vS
S S
(22)
a rS − r v S + S · (rS − r) − = − c c |rS − r|
rS − r |vS |2 −1+ |rS − r| c2
q 1 ∇V = 4πε0 |rS − r|2 1+
c
c2
· (rS − r)
vS c
1 vS
aS
−
v S
c
1+
v S rS − r 1+ · c |rS − r|
v S
vS q q vS A(r, t) = ≡ 4πε 0 c2 |rS − r| + vcS · (rS − r) 4πε0 c2 ξ
c
+
3 r ·| − −r| 2 rS − r |vS | a S − − 1 + · (rS − r) c2 c2 |rS − r| rS rS
From equation (15), the magnetic potential can be written as: (23)
rS − r · |rS − r|
ELECTROMAGNETI C FIELD OF A MOVI NG POINT CHARGE
Therefore: ∂ A q = ∂t 4πε 0 c2
∂ vS 1 v S ∂ξ − 2 ∂t ξ ξ ∂t
5
The charged particle’s velocity depends on the time t only through the retarded time’s dependence on t ; so, using equations (19) and (20): |rS − r| ∂ vS ∂t r aS = a S = ∂t ∂t ξ Likewise: ∂ξ ∂ |rS − r| ∂ vS = + · (rS − r) = ∂t ∂t ∂t c
1 ∂ vS vS ∂ (rS − r) ∂ (rS − r) rS − r = · + · (rS − r) + · = |rS − r| c ∂t ∂t c ∂t =
|rS − r| ξ
vS ·
rS − r v S + c |rS − r|
a S + · (rS − r) c
The time derivative of the magnetic potential is then: (24) 1 ∂ A q = 4πε 0 |rS − r|2 ∂t 1+
1
−r 3 · c | −r| 2 v S |vS |
vS
−
rS rS
c
c2
aS − r| 1 + v S · rS − r | r S c2 c |rS − r|
+
v S rS − r a S · + · (rS − r) c c2 |rS − r|
We finally find the expression for the electric field by putting together equations (17), (22) and (24): (25) q 1 E(r, t) = − 4πε 0 |rS − r|2 1+
−
1 vS
c
r · |rr − −r| S S
aS − r| 1 + v S · rS − r r | S c2 c |rS − r|
3
vS rS − r + |rS − r| c
2
|vS | c2
a S − 1 + 2 · (rS − r) c
4. Magnetic field The magnetic field produced by the moving point charge is given by the curl of the magnetic potential: (26)
B = ∇ × A
Using equation (23), we have: (27)
µ0 q ∇ × A = ∇× 4π
vS ξ
µ0 q ξ ∇ × vS − ∇ξ ∧ vS = 4π ξ 2
6
ELECTROMAGNETI C FIELD OF A MOVI NG POINT CHARGE
Let us first compute the curl of the velocity vS . Its dependence on the spatial coordinates x, y , and z is only due to being evaluated at the retarded time tr , therefore: ∇ × vS = −aS ∧ ∇tr = ∇tr ∧ aS =
1+
vS
c
1 rS − r a S ∧ r c · |rr − −r| |rS − r| S S
As for the second term in equation (27), using equation (21): ∇ξ ∧ vS =
1+
vS
c
1 r · |rr − −r| S S
|vS |2 c2
a S rS − r − 1 + 2 · (rS − r) ∧ vS c |rS − r|
where we have used the fact that the cross product of two parallel vectors (in this case v S and itself) is 0. Substituting in equation (27) we obtain: (28) 1 µ0 q B(r, t) = 4π |rS − r|2
1
1+
vS
c
r ·| − −r| rS rS
3
(rS −r)∧
aS vS rS − r |rS −r| 1+ · c c |rS − r|
|vS |2 a S − vS − 1 + 2 · (rS − r) 2
c
c
Which, as it can be easily shown, is equivalent to: (29)
B(r, t) =
1 rS − r ∧ E(r, t) c |rS − r|
5. Heaviside-Feynman formula The expressions we have determined for the electric and magnetic fields generated by the moving point charge can be rewritten in terms of the variables introduced with equations (13). We have: ˆ drS dR dR de ˆe + R ˆ + R ˙e ˆ = = ≡ R˙ e dt dt dt dt dvS ¨ˆ ¨e ˆ + 2 R˙ ˙e ˆ + R e aS = =R dt vS rS − r R˙ · = c c |rS − r| vS =
2 ˙ )2 + (R )2 ˙e ˆ · ˙e ˆ |vS | = (R ¨ˆ ) · R e ¨ˆ = ¨ e ¨ + ( R )2 e ˆ + 2 R˙ ˙e ˆ + R e ˆ = R R ˆ · e aS · (rS − r) = (R
¨ + ( R )2 = R R
d ˙ ¨ − (R )2 ˙e ˆ ) − ˙e ˆ · ˙e ˆ = R R ˆ · e ˆ˙ (ˆ e · e dt
Substituting in equation (25), after some calculations, results in (from now, the on quantities evaluated at the retarded time will be omitted for simplicity: in the
ELECTROMAGNETI C FIELD OF A MOVI NG POINT CHARGE
7
following formulae, R , ˆe and their derivatives are always intended as calculated at t = t r ): (30)
1 1 q E = − 4πε 0 R2 1 +
˙ R c
e ˆ−
1 R˙
1
R2 c 1 +
+
˙ R c
1 c2
e ˆ+
Now: d dtr d 1 = = dt dt dtr 1+
ˆ(tr ) d2 e d = 2 dt dt R d c dt
ˆ e R2
=
R c
1 1+ 1
˙ (t ) R
˙ R c
ˆ˙ (tr ) = e
c
1
R2 1 +
˙ R c
1
Rc 1 +
1 2
1+
˙ R c
˙ R c
e ˆ˙
¨ ¨ˆ − R e c
1
1+
˙ R c
3
ˆ˙ e
d dtr
r
1
e ˆ˙ − 2
1 2
1+
1 R˙ 3 R 1+
˙ R c
˙ R
¨ R c
¨ˆ − e
c
e ˆ =
1
1
1+
1
Rc 1 +
˙ R c
˙ R
3
ˆ˙ e
c
e ˆ˙ − 2
1 R˙ 2 R c 1+
˙ R c
e ˆ
Thus, substituting in equation (30) and simplifying: (31)
e ˆ q R d E = − + 2 c dt 4πε 0 R
e ˆ R2
+
ˆ 1 d2 e c2 dt2
This is the Heaviside-Feynman formula for the electric field of a moving point charge. An analogous formula can be found for the magnetic field simply by substitution of the above into equation (29): (32)
ˆ ˆ 1 1 q de d2 e B = ˆe ∧ E = − ˆe ∧ ˆ∧ 2 + 2e 4πε0 c Rc c dt c dt 1
In each of the above two equations, R and ˆe are meant calculated at the retarded time t r .
