Physics 505, Classical Electrodynamics Homework …jbourj/jackson/6-2.pdfPhysics 505, Classical...
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Physics 505, Classical Electrodynamics Homework 10 Due Thursday, 18 th November 2004 Jacob Lewis Bourjaily Problem 6.2 Consider the charge and current densities for a single point charge q. Formally, these are given by ρ(x 0 ,t 0 )= qδ [x 0 - r(t 0 )] J(x 0 ,t 0 )= qv(t 0 )δ [x 0 - r(t 0 )] , where r(t 0 ) is the charge’s position at time t 0 and v(t 0 ) is its velocity. While evaluating the expressions involving retarded time, we must put t 0 = t ret = t - R(t 0 )/c, where R = x - r(t 0 ) and R = x - x 0 (t 0 ) inside delta functions. a) We are to show that Z d 3 x 0 δ [x 0 - r(t ret )] = 1 κ , where κ ≡ 1 - v · ˆ R/c, where κ is evaluated at retarded time. Recall the trivial identity of the Dirac δ-function, δ [f (x)] = δ(x - x 0 ) (∂f/∂x)| x=x 0 , where x 0 is a root of f (x). Setting f (x 0 )=[x 0 - r(t ret )] in the expression above, x 0 is such that x 0 = r(t ret ). Inserting this directly, we find δ [x 0 - r(t ret )] = δ(x 0 - x 0 ) ∂ ∂ x 0 (x 0 - r(t ret )) ¶ -1 x 0 =x 0 , = δ(x 0 - x 0 ) 1 - ∂ ∂ x 0 r(t ret ) ¶ -1 x 0 =x 0 , = δ(x 0 - x 0 ) 1 - ∂ r ∂t ∂ ∂ x 0 t - 1 c |x - x 0 (t 0 )| ¶¶ -1 x 0 =x 0 , = δ(x 0 - x 0 ) 1 - ∂ r ∂t 1 c x - x 0 |x - x 0 | ¶ -1 x 0 =x0 , = δ(x 0 - x 0 ) 1 - ∂ r ∂t 1 c x - r(t 0 ) |x - r(t 0 )| ¶ -1 , = δ(x 0 - x 0 ) ‡ 1 - v · ˆ R/c · -1 , = δ(x 0 - x 0 ) κ , ∴ Z d 3 x 0 δ [x 0 - r(t ret )] = 1 κ . ‘ ´ oπ²ρ ’ ´ ²δ²ι δ² ιξαι b) Starting with the Jefimenko generalizations of the Coulomb and Biot-Savart laws, we are to use the expressions for the charge and current densities for a point charge and the result of part a above to obtain the Heaviside-Feynman expressions for the electric and magnetic fields of a point charge. Let us begin with the electric field. The Jefimenko generalization is given by E(x,t)= 1 4π² 0 Z d 3 x 0 ( ˆ R R 2 [ρ(x 0 ,t 0 )] ret + ˆ R cR • ∂ρ(x 0 ,t 0 ) ∂t 0 ‚ ret - 1 c 2 R • ∂ J(x 0 ,t 0 ) ∂t 0 ‚ ret ) . 1
Transcript of Physics 505, Classical Electrodynamics Homework …jbourj/jackson/6-2.pdfPhysics 505, Classical...
Physics 505, Classical ElectrodynamicsHomework 10
Due Thursday, 18th November 2004
Jacob Lewis Bourjaily
Problem 6.2Consider the charge and current densities for a single point charge q. Formally, these are given by
ρ(x′, t′) = qδ [x′ − r(t′)] J(x′, t′) = qv(t′)δ [x′ − r(t′)] ,
where r(t′) is the charge’s position at time t′ and v(t′) is its velocity. While evaluating the expressionsinvolving retarded time, we must put t′ = tret = t − R(t′)/c, where R = x − r(t′) and R = x − x′(t′)inside delta functions.
a) We are to show that ∫d3x′δ [x′ − r(tret)] =
1κ
,
where κ ≡ 1− v · R/c, where κ is evaluated at retarded time.Recall the trivial identity of the Dirac δ-function,
δ [f(x)] =δ(x− x0)(∂f/∂x)|x=x0
,
where x0 is a root of f(x). Setting f(x′) = [x′ − r(tret)] in the expression above, x0
is such that x0 = r(tret). Inserting this directly, we find
δ [x′ − r(tret)] = δ(x′ − x0)(
∂
∂x′(x′ − r(tret))
)−1
x′=x0
,
= δ(x′ − x0)(
1− ∂
∂x′r(tret)
)−1
x′=x0
,
= δ(x′ − x0)(
1− ∂r∂t
∂
∂x′
(t− 1
c|x− x′(t′)|
))−1
x′=x0
,
= δ(x′ − x0)(
1− ∂r∂t
1c
x− x′
|x− x′|)−1
x′=x0
,
= δ(x′ − x0)(
1− ∂r∂t
1c
x− r(t′)|x− r(t′)|
)−1
,
= δ(x′ − x0)(1− v · R/c
)−1
,
=δ(x′ − x0)
κ,
∴∫
d3x′δ [x′ − r(tret)] =1κ
.
‘oπερ ’εδει δε�ιξαι
b) Starting with the Jefimenko generalizations of the Coulomb and Biot-Savart laws, we are to usethe expressions for the charge and current densities for a point charge and the result of parta above to obtain the Heaviside-Feynman expressions for the electric and magnetic fields of apoint charge.Let us begin with the electric field. The Jefimenko generalization is given by
E(x, t) =1
4πε0
∫d3x′
{RR2
[ρ(x′, t′)]ret +RcR
[∂ρ(x′, t′)
∂t′
]
ret
− 1c2R
[∂J(x′, t′)
∂t′
]
ret
}.
1
2 JACOB LEWIS BOURJAILY
Directly inserting our current and charge densities, following out the algebra andusing our result from part a, we have
E(x, t) =1
4πε0
∫d3x′
{RR2
[ρ(x′, t′)]ret +RcR
[∂ρ(x′, t′)
∂t′
]
ret
− 1c2R
[∂J(x′, t′)
∂t′
]
ret
},
=1
4πε0
∫d3x′
{RR2
q [δ (x′ − r(t′))]ret +RcR
q
[∂
∂t′δ (x′ − r(t′))
]
ret
− 1c2R
q
[∂
∂t′v(t′)δ (x′ − r(t′))
]
ret
},
=q
4πε0
∫d3x′
{RR2
[δ (x′ − r(t′))]ret +RcR
∂
∂t[δ (x′ − r(t′))]ret −
1c2R
∂
∂t[v(t′)δ (x′ − r(t′))]ret
},
=q
4πε0
{∫d3x′
[RR2
δ (x′ − r(t′))
]
ret
+∂
∂t
∫d3x′
[RcR
δ (x′ − r(t′))
]
ret
− ∂
∂t
∫d3x′
[1
c2Rv(t′)δ (x′ − r(t′))
]
ret
},
=q
4πε0
{[R
κR2
]
ret
+∂
∂t
[R
cκR
]
ret
− ∂
∂t
[1
c2κRv(t′)
]
ret
},
∴ E(x, t) =q
4πε0
{[R
κR2
]
ret
+∂
c∂t
[RκR
]
ret
− ∂
c2∂t
[1
κRv(t′)
]
ret
}.
‘oπερ ’εδει δε�ιξαι
Notice that in the above, we have used the fact that R does not explicitly dependon t.
Let us now do the analogous calculation for the magnetic field. The Jefimenko general-ization is given by
B(x, t) =µ0
4π
∫d3x′
{[J(x′, t′)]ret ×
RR2
+[∂J(x′, t′)
∂t′
]
ret
× RcR
}.
Directly inserting our current and charge densities and computing directly, we have
B(x, t) =µ0
4π
∫d3x′
{[J(x′, t′)]ret ×
RR2
+[∂J(x′, t′)
∂t′
]
ret
× RcR
},
=qµ0
4π
∫d3x′
{[v(t′)δ (x′ − r(t′))]ret ×
RR2
+[
∂
∂t′v(t′)δ (x′ − r(t′))
]
ret
× RcR
},
=qµ0
4π
{∫d3x′
[v × R
R2δ (x′ − r(t′))
]
ret
+∂
∂t
∫d3x′
[v × R
cRδ (x′ − r(t′))
]
ret
},
=qµ0
4π
{[v × RcR2
]
ret
+∂
∂t
[v × RcκR
]
ret
},
∴ B(x, t) =µ0q
4π
{[v × RcR2
]
ret
+∂
c∂t
[v × R
κR
]
ret
}.
‘oπερ ’εδει δε�ιξαι
