1、非齐次(有源)波动方程的球面波解
电磁场的矢势和标势中给出的标量势的波动方程为:
∇
2
ϕ
(
r
⃗
,
t
)
−
1
c
2
∂
2
∂
t
2
ϕ
(
r
⃗
,
t
)
=
−
ρ
(
r
⃗
,
t
)
ϵ
0
(1)
\nabla^2\phi(\vec{r},t)-\frac{1}{c^2}\frac{\partial^2}{\partial t^2}\phi(\vec{r},t)=-\frac{\rho(\vec{r},t)}{\epsilon_0} \tag{1}
∇2ϕ(r
,t)−c21∂t2∂2ϕ(r
,t)=−ϵ0ρ(r
,t)(1)
首先考虑点源的贡献。假设点源位于
r
⃗
′
\vec{r}'
r
′处,它在场点
r
⃗
\vec{r}
r
处产生的电势为
U
U
U,它满足的方程为:
∇
r
⃗
2
U
(
r
⃗
,
r
⃗
′
,
t
)
−
1
c
2
∂
2
∂
t
2
U
(
r
⃗
,
r
⃗
′
,
t
)
=
−
1
ϵ
0
δ
(
r
⃗
−
r
⃗
′
)
ρ
(
r
⃗
′
,
t
)
(2)
\nabla^2_{\vec{r}}U(\vec{r},\vec{r}',t)-\frac{1}{c^2}\frac{\partial^2}{\partial t^2}U(\vec{r},\vec{r}',t)=-\frac{1}{\epsilon_0}\delta(\vec{r}-\vec{r}')\rho(\vec{r}',t) \tag{2}
∇r
2U(r
,r
′,t)−c21∂t2∂2U(r
,r
′,t)=−ϵ01δ(r
−r
′)ρ(r
′,t)(2)
那么总的电势
ϕ
\phi
ϕ为:
ϕ
(
r
⃗
,
t
)
=
∫
d
τ
′
U
(
r
⃗
,
r
⃗
′
,
t
)
(3)
\phi(\vec{r},t)=\int d\tau' U(\vec{r},\vec{r}',t) \tag{3}
ϕ(r
,t)=∫dτ′U(r
,r
′,t)(3)
这里的
U
U
U就相当与静电场中的格林函数。点源产生的势是关于点源呈球对称分布的,令:
U
(
r
⃗
,
r
⃗
′
,
t
)
=
U
(
R
,
r
⃗
′
,
t
)
(4)
U(\vec{r},\vec{r}',t)=U(R,\vec{r}',t) \tag{4}
U(r
,r
′,t)=U(R,r
′,t)(4)
其中
R
R
R为点源到场点的距离:
R
⃗
=
r
⃗
−
r
⃗
′
=
∑
i
=
1
3
X
i
e
⃗
i
(5)
\vec{R}=\vec{r}-\vec{r}' = \sum_{i=1}^3 X_i\vec{e}_i \tag{5}
R
=r
−r
′=i=1∑3Xie
i(5)
将式(4)代入到方程(2),对等号左侧第一项:
∇
r
⃗
2
U
(
R
,
r
⃗
′
,
t
)
=
∑
i
=
1
3
∂
i
∂
i
U
(
R
,
r
⃗
′
,
t
)
=
∑
i
=
1
3
∂
i
(
∂
U
∂
R
∂
i
R
)
=
∑
i
=
1
3
∂
i
(
∂
U
∂
R
X
i
R
)
=
∑
i
=
1
3
(
X
i
R
∂
2
U
∂
R
2
∂
i
R
+
1
R
∂
U
∂
R
+
∂
U
∂
R
X
i
∂
i
1
R
)
=
∑
i
=
1
3
(
X
i
X
i
R
2
∂
2
U
∂
R
2
+
1
R
∂
U
∂
R
−
X
i
X
i
R
3
∂
U
∂
R
)
=
∂
2
U
∂
R
2
+
2
R
∂
U
∂
R
=
1
R
(
∂
2
U
∂
R
2
R
+
2
∂
U
∂
R
)
=
1
R
∂
2
(
R
U
)
∂
R
2
(6)
\begin{aligned} \nabla^2_{\vec{r}}U(R,\vec{r}',t) & = \sum_{i=1}^3 \partial_i\partial_iU(R,\vec{r}',t) \\ & = \sum_{i=1}^3\partial_i(\frac{\partial U}{\partial R}\partial_iR) \\ & = \sum_{i=1}^3\partial_i(\frac{\partial U}{\partial R}\frac{X_i}{R}) \\ & = \sum_{i=1}^3(\frac{X_i}{R}\frac{\partial^2 U}{\partial R^2}\partial_i{R} + \frac{1}{R}\frac{\partial U}{\partial R} + \frac{\partial U}{\partial R} X_i\partial_i\frac{1}{R}) \\ & = \sum_{i=1}^3(\frac{X_iX_i}{R^2}\frac{\partial^2 U}{\partial R^2}+ \frac{1}{R}\frac{\partial U}{\partial R} - \frac{X_iX_i}{R^3}\frac{\partial U}{\partial R}) \\ & = \frac{\partial^2U}{\partial R^2} + \frac{2}{R}\frac{\partial U}{\partial R} \\ & = \frac{1}{R}(\frac{\partial^2U}{\partial R^2}R + 2\frac{\partial U}{\partial R}) \\ & = \frac{1}{R}\frac{\partial^2(RU)}{\partial R^2} \end{aligned} \tag{6}
∇r
2U(R,r
′,t)=i=1∑3∂i∂iU(R,r
′,t)=i=1∑3∂i(∂R∂U∂iR)=i=1∑3∂i(∂R∂URXi)=i=1∑3(RXi∂R2∂2U∂iR+R1∂R∂U+∂R∂UXi∂iR1)=i=1∑3(R2XiXi∂R2∂2U+R1∂R∂U−R3XiXi∂R∂U)=∂R2∂2U+R2∂R∂U=R1(∂R2∂2UR+2∂R∂U)=R1∂R2∂2(RU)(6)
即:
∇
r
⃗
2
U
(
R
,
r
⃗
′
,
t
)
=
1
R
∂
2
(
R
U
)
∂
R
2
(7)
\nabla^2_{\vec{r}}U(R,\vec{r}',t) = \frac{1}{R}\frac{\partial^2(RU)}{\partial R^2} \tag{7}
∇r
2U(R,r
′,t)=R1∂R2∂2(RU)(7)
将式(7)代入到式(2),得到:
1
R
∂
2
(
R
U
)
∂
R
2
−
1
c
2
∂
2
∂
t
2
U
=
−
1
ϵ
0
δ
(
R
⃗
)
ρ
(
r
⃗
′
,
t
)
(8)
\frac{1}{R}\frac{\partial^2(RU)}{\partial R^2} - \frac{1}{c^2}\frac{\partial^2}{\partial t^2}U=-\frac{1}{\epsilon_0}\delta(\vec{R})\rho(\vec{r}',t) \tag{8}
R1∂R2∂2(RU)−c21∂t2∂2U=−ϵ01δ(R
)ρ(r
′,t)(8)
方程(8)两侧同时乘以
R
R
R,得到:
∂
2
(
U
R
)
∂
R
2
−
1
c
2
∂
2
(
U
R
)
∂
t
2
=
0
(9)
\frac{\partial^2(UR)}{\partial R^2} - \frac{1}{c^2}\frac{\partial^2(UR)}{\partial t^2} = 0 \tag{9}
∂R2∂2(UR)−c21∂t2∂2(UR)=0(9)
上式中等号左侧第二项,
R
R
R与时间无关,将
U
U
U和
R
R
R放一起。等号右侧,只有当
R
=
0
R=0
R=0时,
δ
\delta
δ函数才有贡献,但此时
R
=
0
R=0
R=0,该项仍然为零。将
U
R
UR
UR看成是要求解函数函数,那么方程(9)就变成了一维的波动方程,波速是光速
c
c
c。对一维波动方程,其解可以为:
R
U
(
R
,
r
⃗
′
,
t
)
=
f
(
t
−
R
c
,
r
⃗
′
)
+
g
(
t
+
R
c
,
r
⃗
′
)
(10)
RU(R, \vec{r}', t) = f(t-\frac{R}{c},\vec{r}') + g(t+\frac{R}{c},\vec{r}') \tag{10}
RU(R,r
′,t)=f(t−cR,r
′)+g(t+cR,r
′)(10)
其中
f
f
f和
g
g
g可以为任意函数。定义
U
1
≡
1
R
f
(
t
−
R
c
,
r
⃗
′
)
(11)
U_1 \equiv \frac{1}{R}f(t-\frac{R}{c},\vec{r}') \tag{11}
U1≡R1f(t−cR,r
′)(11)
U
1
U_1
U1称为推迟解(推迟势)。
U
2
≡
1
R
g
(
t
+
R
c
,
r
⃗
′
)
(12)
U_2 \equiv \frac{1}{R}g(t+\frac{R}{c},\vec{r}') \tag{12}
U2≡R1g(t+cR,r
′)(12)
U
2
U_2
U2称为超前解(超前势)。式(10),式(11)和式(12)是方程(9)的解,还不是方程(8)的解。需要将这组解放回到方程(8),当
R
→
0
R\to 0
R→0时,使方程(8)等号两侧的奇异行为是一致的。式(12)相当于是将式(11)中的
c
c
c换成
−
c
-c
−c,只考虑式(11)的情况。将式(11)代入方程(9),并利用式(7),得到:
−
1
ϵ
0
δ
(
R
⃗
)
ρ
(
r
⃗
′
,
t
)
=
(
∇
r
⃗
2
−
1
c
2
∂
2
∂
t
2
)
[
1
R
f
(
t
−
R
c
,
r
⃗
′
)
]
=
(
∇
r
⃗
2
1
R
)
f
(
t
−
R
c
,
r
⃗
′
)
+
2
∇
r
⃗
1
R
⋅
∇
r
⃗
f
(
t
−
R
c
,
r
⃗
′
)
+
1
R
(
∇
r
⃗
2
−
1
c
2
∂
2
∂
t
2
)
f
(
t
−
R
c
,
r
⃗
′
)
=
−
4
π
δ
(
R
⃗
)
f
(
t
−
R
c
,
r
⃗
′
)
+
2
c
R
⃗
R
3
⋅
R
⃗
R
∂
∂
t
f
(
t
−
R
c
,
r
⃗
′
)
+
1
R
(
∂
2
∂
R
2
+
2
R
∂
∂
R
−
1
c
2
∂
2
∂
t
2
)
f
(
t
−
R
c
,
r
⃗
′
)
=
−
4
π
δ
(
R
⃗
)
f
(
t
−
R
c
,
r
⃗
′
)
=
−
4
π
δ
(
R
⃗
)
f
(
t
,
r
⃗
′
)
(13)
\begin{aligned} -\frac{1}{\epsilon_0}\delta(\vec{R})\rho(\vec{r}',t) & = (\nabla^2_{\vec{r}} - \frac{1}{c^2}\frac{\partial^2}{\partial t^2})[\frac{1}{R}f(t-\frac{R}{c},\vec{r}')] \\ & = (\nabla^2_{\vec{r}}\frac{1}{R})f(t-\frac{R}{c},\vec{r}') + 2\nabla_{\vec{r}}\frac{1}{R}\cdot \nabla_{\vec{r}}f(t-\frac{R}{c},\vec{r}') + \frac{1}{R}(\nabla^2_{\vec{r}} - \frac{1}{c^2}\frac{\partial^2}{\partial t^2})f(t-\frac{R}{c},\vec{r}') \\ & = -4\pi \delta(\vec{R}) f(t-\frac{R}{c},\vec{r}') + \frac{2}{c}\frac{\vec{R}}{R^3}\cdot \frac{\vec{R}}{R}\frac{\partial}{\partial t}f(t-\frac{R}{c},\vec{r}') + \frac{1}{R}(\frac{\partial^2}{\partial R^2} + \frac{2}{R}\frac{\partial}{\partial R} - \frac{1}{c^2}\frac{\partial^2}{\partial t^2})f(t-\frac{R}{c},\vec{r}') \\ & = -4\pi \delta(\vec{R}) f(t-\frac{R}{c},\vec{r}') \\ & = -4\pi \delta(\vec{R})f(t,\vec{r}') \end{aligned} \tag{13}
−ϵ01δ(R
)ρ(r
′,t)=(∇r
2−c21∂t2∂2)[R1f(t−cR,r
′)]=(∇r
2R1)f(t−cR,r
′)+2∇r
R1⋅∇r
f(t−cR,r
′)+R1(∇r
2−c21∂t2∂2)f(t−cR,r
′)=−4πδ(R
)f(t−cR,r
′)+c2R3R
⋅RR
∂t∂f(t−cR,r
′)+R1(∂R2∂2+R2∂R∂−c21∂t2∂2)f(t−cR,r
′)=−4πδ(R
)f(t−cR,r
′)=−4πδ(R
)f(t,r
′)(13)
上式中第四个等号利用了波动方程(9)的结果,第五个等号直接取
R
=
0
R=0
R=0得到的结果。如果只考虑
R
→
0
R\to 0
R→0的奇异,在第二个等号中只保留第一项即可。比较式(13)得到的结果,函数
f
f
f的表达式为:
f
(
t
,
r
⃗
′
)
=
ρ
(
r
⃗
′
,
t
)
4
π
ϵ
0
(14)
f(t,\vec{r}') = \frac{\rho(\vec{r}',t)}{4\pi\epsilon_0} \tag{14}
f(t,r
′)=4πϵ0ρ(r
′,t)(14)
将式(14)代入到式(11)和式(12),得到推迟解和超前解的表达式为:
{
U
1
(
R
,
r
⃗
′
,
t
)
=
ρ
(
r
⃗
′
,
t
−
R
c
)
4
π
ϵ
0
R
U
2
(
R
,
r
⃗
′
,
t
)
=
ρ
(
r
⃗
′
,
t
+
R
c
)
4
π
ϵ
0
R
(15)
\left \{ \begin{aligned} & U_1(R, \vec{r}', t) = \frac{\rho(\vec{r}',t-\frac{R}{c})}{4\pi\epsilon_0R} \\ & U_2(R, \vec{r}', t) = \frac{\rho(\vec{r}',t+\frac{R}{c})}{4\pi\epsilon_0R} \end{aligned} \right. \tag{15}
⎩⎪⎪⎪⎨⎪⎪⎪⎧U1(R,r
′,t)=4πϵ0Rρ(r
′,t−cR)U2(R,r
′,t)=4πϵ0Rρ(r
′,t+cR)(15)
- 对给定时刻 t t t,等值面为以 r ⃗ \vec{r} r 为球心的球面, t − R c = 常 数 → R = ∣ r ⃗ − r ⃗ ′ ∣ = 常 数 t-\frac{R}{c}=常数\to R=|\vec{r}-\vec{r}'|=常数 t−cR=常数→R=∣r −r ′∣=常数。
- 等值面以速率 c c c运动,等值面的值的大小按 1 R \frac{1}{R} R1变化。如果 R R R是增加的,等值面扩张,等值面的值缩小,即推迟势(波由点源向外传播);如果 R R R是减小的,等值面收缩,等值面的值增大,即超前势(波向点源传播,通常它是非物理的解)。
2、洛伦兹规范,库伦规范
在洛伦兹规范下,导出的达朗伯方程中,标量势和矢量势具有相同的形式再根据式(15),得到的标量势和矢量势的表达式为:
{
ϕ
(
r
⃗
,
t
)
=
∫
d
τ
′
ρ
(
r
⃗
′
,
t
−
R
c
)
4
π
ϵ
0
R
A
⃗
(
r
⃗
,
t
)
=
∫
d
τ
′
μ
0
j
⃗
(
r
⃗
′
,
t
−
R
c
)
4
π
R
(16)
\left \{ \begin{aligned} & \phi(\vec{r}, t) = \int d\tau' \frac{\rho(\vec{r}', t-\frac{R}{c})}{4\pi\epsilon_0R} \\ & \vec{A}(\vec{r}, t) = \int d\tau' \frac{\mu_0\vec{j}(\vec{r}', t-\frac{R}{c})}{4\pi R} \end{aligned} \right. \tag{16}
⎩⎪⎪⎪⎨⎪⎪⎪⎧ϕ(r
,t)=∫dτ′4πϵ0Rρ(r
′,t−cR)A
(r
,t)=∫dτ′4πRμ0j
(r
′,t−cR)(16)
与静电静磁的情况(见真空中电磁相互作用的场方程中式(3)和式(8))比较可以发现,式(16)就多了
t
−
R
c
t-\frac{R}{c}
t−cR这一项,但是静电静磁情况下得到的标量势和矢量势的表达式不是麦克斯韦方程组的解(考虑时变电磁场),而式(16)是麦克斯韦方程组的解。在静电和静磁情况下时,标量势和矢量势在无穷远处按
1
r
\frac{1}{r}
r1衰减,对应的电场强度和磁感应强度按
1
r
2
\frac{1}{r^2}
r21衰减。考虑电磁场随时间变化后,式(16)中给出的标量势和矢量势在无穷远处仍然按
1
r
\frac{1}{r}
r1衰减,但是对应的电场强度和磁感应强度在无穷远处不全按
1
r
2
\frac{1}{r^2}
r21衰减(
t
−
R
c
t-\frac{R}{c}
t−cR有贡献),产生按
1
R
\frac{1}{R}
R1衰减的部分,即推迟效应。式(16)是在洛伦兹规范下得到的波动方程推导出来的,接下来证明式(16)满足洛伦兹规范。令:
t
∗
=
t
−
R
c
(17)
t^* = t-\frac{R}{c} \tag{17}
t∗=t−cR(17)
对矢量势做散度:
∇
⋅
A
⃗
=
μ
0
4
π
∇
⋅
∫
d
τ
′
j
⃗
(
r
⃗
′
,
t
∗
)
R
=
μ
0
4
π
∫
d
τ
′
[
(
∇
1
R
)
⋅
j
⃗
(
r
⃗
′
,
t
∗
)
+
1
R
∇
⋅
j
⃗
(
r
⃗
′
,
t
∗
)
]
=
μ
0
4
π
∫
d
τ
′
[
−
(
∇
′
1
R
)
⋅
j
⃗
(
r
⃗
′
,
t
∗
)
+
1
R
∂
j
⃗
(
r
⃗
′
,
t
∗
)
∂
t
∗
⋅
∇
t
∗
]
=
μ
0
4
π
∫
d
τ
′
{
−
∇
′
⋅
[
j
⃗
(
r
⃗
′
,
t
∗
)
R
]
+
1
R
∇
′
⋅
j
⃗
(
r
⃗
′
,
t
∗
)
+
1
R
∂
j
⃗
(
r
⃗
′
,
t
∗
)
∂
t
∗
⋅
∇
t
∗
}
=
μ
0
4
π
∫
d
τ
′
[
1
R
∇
′
⋅
j
⃗
(
r
⃗
′
,
t
∗
)
∣
t
∗
+
1
R
∂
j
⃗
(
r
⃗
′
,
t
∗
)
∂
t
∗
⋅
(
∇
t
∗
+
∇
′
t
∗
)
]
=
μ
0
4
π
∫
d
τ
′
(
−
1
R
∂
ρ
(
r
⃗
′
,
t
∗
)
∂
t
∗
)
=
μ
0
4
π
∫
d
τ
′
(
−
1
R
∂
ρ
(
r
⃗
′
,
t
∗
)
∂
t
)
=
−
∂
∂
t
μ
0
4
π
∫
d
τ
′
ρ
(
r
⃗
′
,
t
∗
)
R
=
−
μ
0
ϵ
0
∂
ϕ
∂
t
(18)
\begin{aligned} \nabla \cdot \vec{A} & = \frac{\mu_0}{4\pi} \nabla \cdot \int d\tau' \frac{\vec{j}(\vec{r}', t^*)}{R} \\ & = \frac{\mu_0}{4\pi} \int d\tau' [(\nabla\frac{1}{R})\cdot\vec{j}(\vec{r}', t^*)+\frac{1}{R}\nabla\cdot\vec{j}(\vec{r}', t^*)] \\ & = \frac{\mu_0}{4\pi} \int d\tau' [-(\nabla'\frac{1}{R})\cdot\vec{j}(\vec{r}', t^*) + \frac{1}{R}\frac{\partial \vec{j}(\vec{r}', t^*)}{\partial t^*} \cdot\nabla t^*] \\ & = \frac{\mu_0}{4\pi} \int d\tau' \{ -\nabla' \cdot [\frac{\vec{j}(\vec{r}', t^*)}{R}] + \frac{1}{R}\nabla'\cdot \vec{j}(\vec{r}', t^*) + \frac{1}{R}\frac{\partial \vec{j}(\vec{r}', t^*)}{\partial t^*} \cdot\nabla t^* \} \\ & = \frac{\mu_0}{4\pi} \int d\tau' [ \frac{1}{R}\nabla'\cdot \vec{j}(\vec{r}', t^*)|_{t^*} + \frac{1}{R}\frac{\partial \vec{j}(\vec{r}', t^*)}{\partial t^*} \cdot (\nabla t^* + \nabla' t^* )] \\ & = \frac{\mu_0}{4\pi} \int d\tau' (-\frac{1}{R}\frac{\partial \rho(\vec{r}', t^*)}{\partial t^*}) \\ & = \frac{\mu_0}{4\pi} \int d\tau' (-\frac{1}{R}\frac{\partial \rho(\vec{r}', t^*)}{\partial t}) \\ & = -\frac{\partial}{\partial t}\frac{\mu_0}{4\pi} \int d\tau'\frac{\rho(\vec{r}', t^*)}{R} \\ & = -\mu_0\epsilon_0\frac{\partial\phi}{\partial t} \end{aligned} \tag{18}
∇⋅A
=4πμ0∇⋅∫dτ′Rj
(r
′,t∗)=4πμ0∫dτ′[(∇R1)⋅j
(r
′,t∗)+R1∇⋅j
(r
′,t∗)]=4πμ0∫dτ′[−(∇′R1)⋅j
(r
′,t∗)+R1∂t∗∂j
(r
′,t∗)⋅∇t∗]=4πμ0∫dτ′{−∇′⋅[Rj
(r
′,t∗)]+R1∇′⋅j
(r
′,t∗)+R1∂t∗∂j
(r
′,t∗)⋅∇t∗}=4πμ0∫dτ′[R1∇′⋅j
(r
′,t∗)∣t∗+R1∂t∗∂j
(r
′,t∗)⋅(∇t∗+∇′t∗)]=4πμ0∫dτ′(−R1∂t∗∂ρ(r
′,t∗))=4πμ0∫dτ′(−R1∂t∂ρ(r
′,t∗))=−∂t∂4πμ0∫dτ′Rρ(r
′,t∗)=−μ0ϵ0∂t∂ϕ(18)
即式(16)给出的解满足洛伦兹规范。上式中,第四个等号中的第一项可以换成面积分,在无穷远处电流为零,其积分值为零;第四个等号中的第二项,
∇
′
\nabla'
∇′算符作用于
r
⃗
′
\vec{r}'
r
′,得到第五个等号中的第一项,作用于
t
∗
t^*
t∗,得到第五个等号中的第二项中的
∇
′
t
∗
\nabla't^*
∇′t∗项;第五个等号中的第一项利用了电荷守恒定律。对超前势,同样满足洛伦兹规范固定条件。对库仑规范,标量势满足拉普拉斯方程,所以标量势的表达式为:
ϕ
(
r
⃗
,
t
)
=
∫
d
τ
′
ρ
(
r
⃗
′
,
t
)
4
π
ϵ
0
R
(19)
\phi(\vec{r},t)=\int d\tau' \frac{\rho(\vec{r}', t)}{4\pi\epsilon_0 R} \tag{19}
ϕ(r
,t)=∫dτ′4πϵ0Rρ(r
′,t)(19)
相比于库仑定律给出的电势的表达式,式(19)仅仅多了对时间
t
t
t的依赖。矢量势满足满足方程,其源项为:
j
⃗
∗
(
r
⃗
,
t
)
=
j
⃗
(
r
⃗
,
t
)
−
ϵ
0
∇
∂
∂
t
ϕ
(
r
⃗
,
t
)
(20)
\vec{j}^*(\vec{r},t)=\vec{j}(\vec{r},t)-\epsilon_0 \nabla\frac{\partial }{\partial t}\phi(\vec{r},t) \tag{20}
j
∗(r
,t)=j
(r
,t)−ϵ0∇∂t∂ϕ(r
,t)(20)
将式(16)中的
j
⃗
\vec{j}
j
替换为
j
⃗
∗
\vec{j}^*
j
∗,得到矢量势的解:
A
⃗
(
r
⃗
,
t
)
=
∫
d
τ
′
μ
0
j
⃗
∗
(
r
⃗
′
,
t
−
R
c
)
4
π
R
(21)
\vec{A}(\vec{r}, t) = \int d\tau' \frac{\mu_0\vec{j}^*(\vec{r}', t-\frac{R}{c})}{4\pi R} \tag{21}
A
(r
,t)=∫dτ′4πRμ0j
∗(r
′,t−cR)(21)
- 对一个纯电流场, ρ ( r ⃗ , t ) = 0 → ϕ ( r ⃗ , t ) = 0 \rho(\vec{r},t)=0\to\phi(\vec{r},t)=0 ρ(r ,t)=0→ϕ(r ,t)=0,即不存在净电荷密度分布时,库仑规范与洛伦兹规范相同
- 对标量势不随时间变化的情形,库仑规范与洛伦兹规范相同
- 洛伦兹规范下,矢量势和标量势的表达式类似,可借鉴标量势的计算结果
3、光子质量对平方反比率的修正
如果光子有非零静止质量
m
光
子
m_{光子}
m光子,那么方程(1)应当加上光子质量的贡献(见理想绝缘介质中的波动方程中的式(24)):
(
∇
2
−
1
c
2
∂
2
∂
t
2
−
m
光
子
2
c
2
ℏ
2
)
ϕ
~
(
r
⃗
,
t
)
=
−
ρ
(
r
⃗
,
t
)
ϵ
0
(22)
(\nabla^2 - \frac{1}{c^2}\frac{\partial^2}{\partial t^2}-\frac{m^2_{光子}c^2}{\hbar^2})\tilde{\phi}(\vec{r},t)=-\frac{\rho(\vec{r},t)}{\epsilon_0} \tag{22}
(∇2−c21∂t2∂2−ℏ2m光子2c2)ϕ~(r
,t)=−ϵ0ρ(r
,t)(22)
对点源满足的方程为:
(
∇
r
⃗
2
−
1
c
2
∂
2
∂
t
2
−
m
光
子
2
c
2
ℏ
2
)
U
~
(
r
⃗
,
r
⃗
′
,
t
)
=
−
1
ϵ
0
δ
(
R
⃗
)
ρ
(
r
⃗
,
t
)
(23)
(\nabla^2_{\vec{r}} - \frac{1}{c^2}\frac{\partial^2}{\partial t^2}-\frac{m^2_{光子}c^2}{\hbar^2})\tilde{U}(\vec{r},\vec{r}',t)=-\frac{1}{\epsilon_0}\delta(\vec{R})\rho(\vec{r},t) \tag{23}
(∇r
2−c21∂t2∂2−ℏ2m光子2c2)U~(r
,r
′,t)=−ϵ01δ(R
)ρ(r
,t)(23)
与式(3)和式(4)类似:
{
ϕ
~
(
r
⃗
,
t
)
=
∫
d
τ
′
U
~
(
r
⃗
,
r
⃗
′
,
t
)
U
~
(
r
⃗
,
r
⃗
′
,
t
)
=
U
~
(
R
,
r
⃗
′
,
t
)
(24)
\left \{ \begin{aligned} & \tilde{\phi}(\vec{r},t)=\int d\tau' \tilde{U}(\vec{r},\vec{r}',t) \\ & \tilde{U}(\vec{r},\vec{r}',t)=\tilde{U}(R,\vec{r}',t) \end{aligned} \right. \tag{24}
⎩⎪⎨⎪⎧ϕ~(r
,t)=∫dτ′U~(r
,r
′,t)U~(r
,r
′,t)=U~(R,r
′,t)(24)
当光子质量等于零时,即得到式(15)的结果:
U
~
(
R
,
r
⃗
′
,
t
)
⟹
m
光
子
2
=
0
ρ
(
r
⃗
′
,
t
∓
R
c
)
4
π
ϵ
0
R
(25)
\tilde{U}(R,\vec{r}',t) \stackrel{m_{光子}^2=0}{\Longrightarrow}\frac{\rho(\vec{r}',t\mp\frac{R}{c})}{4\pi\epsilon_0R} \tag{25}
U~(R,r
′,t)⟹m光子2=04πϵ0Rρ(r
′,t∓cR)(25)
无质量粒子一定会导致平方反比率!光子质量对平方反比率的修正来自于
U
~
(
R
,
r
⃗
′
,
t
)
\tilde{U}(R,\vec{r}',t)
U~(R,r
′,t)与
ρ
(
r
⃗
′
,
t
∓
R
c
)
4
π
ϵ
0
R
\frac{\rho(\vec{r}',t\mp\frac{R}{c})}{4\pi\epsilon_0R}
4πϵ0Rρ(r
′,t∓cR)的差别。当光子有非零的静止质量时,对上式乘以一个修正函数
h
(
R
)
h(R)
h(R):
U
~
(
R
,
r
⃗
′
,
t
)
=
ρ
(
r
⃗
′
,
t
∓
R
c
)
4
π
ϵ
0
R
h
(
R
)
(26)
\tilde{U}(R,\vec{r}',t) = \frac{\rho(\vec{r}',t\mp\frac{R}{c})}{4\pi\epsilon_0R}h(R) \tag{26}
U~(R,r
′,t)=4πϵ0Rρ(r
′,t∓cR)h(R)(26)
那么式(26)要满足方程(23):
−
1
ϵ
0
δ
(
R
⃗
)
ρ
(
r
⃗
,
t
)
=
(
∇
r
⃗
2
−
1
c
2
∂
2
∂
t
2
−
m
光
子
2
c
2
ℏ
2
)
U
~
(
r
⃗
,
r
⃗
′
,
t
)
=
h
(
R
)
(
∇
r
⃗
2
−
1
c
2
∂
2
∂
t
2
−
m
光
子
2
c
2
ℏ
2
)
ρ
(
r
⃗
′
,
t
∓
R
c
)
4
π
ϵ
0
R
+
ρ
(
r
⃗
′
,
t
∓
R
c
)
4
π
ϵ
0
R
∇
r
⃗
2
h
(
R
)
+
2
[
∇
r
⃗
ρ
(
r
⃗
′
,
t
∓
R
c
)
4
π
ϵ
0
R
]
⋅
∇
r
⃗
h
(
R
)
=
−
1
ϵ
0
δ
(
R
)
ρ
(
r
⃗
′
,
t
)
h
(
R
)
+
ρ
(
r
⃗
′
,
t
∓
R
c
)
4
π
ϵ
0
R
(
∂
2
∂
R
2
+
2
R
∂
R
−
m
光
子
2
c
2
ℏ
2
)
h
(
R
)
+
2
[
∂
∂
R
ρ
(
r
⃗
′
,
t
∓
R
c
)
4
π
ϵ
0
R
]
∂
∂
R
h
(
R
)
=
−
h
(
0
)
ϵ
0
δ
(
R
⃗
)
ρ
(
r
⃗
′
,
t
)
+
ρ
(
r
⃗
′
,
t
∓
R
c
)
4
π
ϵ
0
R
(
∂
2
∂
R
2
−
m
光
子
2
c
2
ℏ
2
)
h
(
R
)
+
2
R
[
∂
∂
R
ρ
(
r
⃗
′
,
t
∓
R
c
)
4
π
ϵ
0
R
]
∂
∂
R
h
(
R
)
\begin{aligned} -\frac{1}{\epsilon_0}\delta(\vec{R})\rho(\vec{r},t) & = (\nabla^2_{\vec{r}} - \frac{1}{c^2}\frac{\partial^2}{\partial t^2}-\frac{m^2_{光子}c^2}{\hbar^2})\tilde{U}(\vec{r},\vec{r}',t) \\ & = h(R)(\nabla^2_{\vec{r}} - \frac{1}{c^2}\frac{\partial^2}{\partial t^2}-\frac{m^2_{光子}c^2}{\hbar^2}) \frac{\rho(\vec{r}',t\mp\frac{R}{c})}{4\pi\epsilon_0R} + \frac{\rho(\vec{r}',t\mp\frac{R}{c})}{4\pi\epsilon_0R} \nabla^2_{\vec{r}} h(R) + 2[\nabla_{\vec{r}} \frac{\rho(\vec{r}',t\mp\frac{R}{c})}{4\pi\epsilon_0R}]\cdot \nabla_{\vec{r}}h(R) \\ & = -\frac{1}{\epsilon_0} \delta(R)\rho(\vec{r}', t)h(R) + \frac{\rho(\vec{r}',t\mp\frac{R}{c})}{4\pi\epsilon_0R}(\frac{\partial ^2}{\partial R^2} + \frac{2}{R}\frac{\partial }{R}-\frac{m_{光子}^2c^2}{\hbar^2})h(R) + 2[\frac{\partial}{\partial R} \frac{\rho(\vec{r}',t\mp\frac{R}{c})}{4\pi\epsilon_0R}]\frac{\partial}{\partial R}h(R) \\ & = -\frac{h(0)}{\epsilon_0}\delta(\vec{R})\rho(\vec{r}',t)+ \frac{\rho(\vec{r}',t\mp\frac{R}{c})}{4\pi\epsilon_0R}(\frac{\partial^2}{\partial R^2}-\frac{m_{光子}^2c^2}{\hbar^2})h(R)+\frac{2}{R}[\frac{\partial }{\partial R} \frac{\rho(\vec{r}',t\mp\frac{R}{c})}{4\pi\epsilon_0R}]\frac{\partial}{\partial R}h(R) \end{aligned}
−ϵ01δ(R
)ρ(r
,t)=(∇r
2−c21∂t2∂2−ℏ2m光子2c2)U~(r
,r
′,t)=h(R)(∇r
2−c21∂t2∂2−ℏ2m光子2c2)4πϵ0Rρ(r
′,t∓cR)+4πϵ0Rρ(r
′,t∓cR)∇r
2h(R)+2[∇r
4πϵ0Rρ(r
′,t∓cR)]⋅∇r
h(R)=−ϵ01δ(R)ρ(r
′,t)h(R)+4πϵ0Rρ(r
′,t∓cR)(∂R2∂2+R2R∂−ℏ2m光子2c2)h(R)+2[∂R∂4πϵ0Rρ(r
′,t∓cR)]∂R∂h(R)=−ϵ0h(0)δ(R
)ρ(r
′,t)+4πϵ0Rρ(r
′,t∓cR)(∂R2∂2−ℏ2m光子2c2)h(R)+R2[∂R∂4πϵ0Rρ(r
′,t∓cR)]∂R∂h(R)
即:
−
1
ϵ
0
δ
(
R
⃗
)
ρ
(
r
⃗
,
t
)
=
−
h
(
0
)
ϵ
0
δ
(
R
⃗
)
ρ
(
r
⃗
′
,
t
)
+
ρ
(
r
⃗
′
,
t
∓
R
c
)
4
π
ϵ
0
R
(
∂
2
∂
R
2
−
m
光
子
2
c
2
ℏ
2
)
h
(
R
)
+
2
R
[
∂
∂
R
ρ
(
r
⃗
′
,
t
∓
R
c
)
4
π
ϵ
0
R
]
∂
∂
R
h
(
R
)
(27)
-\frac{1}{\epsilon_0}\delta(\vec{R})\rho(\vec{r},t) =-\frac{h(0)}{\epsilon_0}\delta(\vec{R})\rho(\vec{r}',t)+ \frac{\rho(\vec{r}',t\mp\frac{R}{c})}{4\pi\epsilon_0R}(\frac{\partial^2}{\partial R^2}-\frac{m_{光子}^2c^2}{\hbar^2})h(R)+\frac{2}{R}[\frac{\partial }{\partial R} \frac{\rho(\vec{r}',t\mp\frac{R}{c})}{4\pi\epsilon_0R}]\frac{\partial}{\partial R}h(R) \tag{27}
−ϵ01δ(R
)ρ(r
,t)=−ϵ0h(0)δ(R
)ρ(r
′,t)+4πϵ0Rρ(r
′,t∓cR)(∂R2∂2−ℏ2m光子2c2)h(R)+R2[∂R∂4πϵ0Rρ(r
′,t∓cR)]∂R∂h(R)(27)
对比上式中等号的两侧,等号要成立,上式中等号右侧除了第一项,剩余项之和要等于零:
{
∂
2
∂
R
2
−
m
光
子
2
c
2
ℏ
2
+
2
[
∂
∂
R
ln
ρ
(
r
⃗
′
,
t
∓
R
c
)
]
∂
∂
R
}
h
(
R
)
=
0
(28)
\{ \frac{\partial ^2}{\partial R^2} - \frac{m_{光子}^2c^2}{\hbar^2} + 2[\frac{\partial}{\partial R}\ln\rho(\vec{r}',t\mp\frac{R}{c})] \frac{\partial}{\partial R} \} h(R) = 0 \tag{28}
{∂R2∂2−ℏ2m光子2c2+2[∂R∂lnρ(r
′,t∓cR)]∂R∂}h(R)=0(28)
当
R
=
0
R=0
R=0时,式(27)中第一项要求:
h
(
0
)
=
1
(29)
h(0) = 1 \tag{29}
h(0)=1(29)
在这里没有考虑
h
(
R
)
h(R)
h(R)与时间的关系,所以只有电荷密度不随时间变化时才自恰:
h
(
R
)
=
e
±
c
ℏ
m
光
子
R
(30)
h(R) = e^{\pm \frac{c}{\hbar}m_{光子}R} \tag{30}
h(R)=e±ℏcm光子R(30)
考虑无穷远边界条件:
U
~
(
R
,
r
⃗
′
,
t
)
=
ρ
(
r
⃗
′
,
t
∓
R
c
)
4
π
ϵ
0
R
e
−
c
ℏ
m
光
子
R
(31)
\tilde{U}(R,\vec{r}',t) = \frac{\rho(\vec{r}',t\mp\frac{R}{c})}{4\pi\epsilon_0R}e^{- \frac{c}{\hbar}m_{光子}R} \tag{31}
U~(R,r
′,t)=4πϵ0Rρ(r
′,t∓cR)e−ℏcm光子R(31)
此时在无穷远处,势不再是按
1
r
\frac{1}{r}
r1衰减,而是按指数衰减。光子质量对平方反比率的修正:
1
R
2
⇒
1
R
2
e
−
c
ℏ
m
光
子
R
(32)
\frac{1}{R^2}\Rightarrow \frac{1}{R^2}e^{- \frac{c}{\hbar}m_{光子}R} \tag{32}
R21⇒R21e−ℏcm光子R(32)
光子静止质量使使长程作用力变成短程作用力。
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光子非零静止质量的理论估计 m 光 子 m_{光子} m光子:
m 光 子 c 2 Δ t ⟹ 宇 宙 年 龄 : Δ t ∼ 1 0 10 年 m 光 子 ≈ 3.7 × 1 0 − 66 克 m_{光子}c^2\Delta t \stackrel{宇宙年龄:\Delta t\sim 10^{10}年}{\Longrightarrow} m_{光子}\approx 3.7\times 10^{-66}克 m光子c2Δt⟹宇宙年龄:Δt∼1010年m光子≈3.7×10−66克 -
平方反比率修正指数因子里系数的理论估计:
e − c ℏ m 光 子 R : c ℏ m 光 子 ∼ 1 c Δ t = 1.1 × 1 0 − 16 米 − 1 e^{- \frac{c}{\hbar}m_{光子}R}:\frac{c}{\hbar}m_{光子} \sim \frac{1}{c\Delta t}=1.1 \times 10^{-16} 米^{-1} e−ℏcm光子R:ℏcm光子∼cΔt1=1.1×10−16米−1 -
最新实验结果(华中科技大学罗俊等,2010):
m 光 子 < 1 0 − 51 克 m_{光子}<10^{-51}克 m光子<10−51克