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推迟势

作者:互联网

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∂2​ϕ(r ,t)=−ϵ0​ρ(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 2​U(r ,r ′,t)−c21​∂t2∂2​U(r ,r ′,t)=−ϵ0​1​δ(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∑3​Xi​e 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 2​U(R,r ′,t)​=i=1∑3​∂i​∂i​U(R,r ′,t)=i=1∑3​∂i​(∂R∂U​∂i​R)=i=1∑3​∂i​(∂R∂U​RXi​​)=i=1∑3​(RXi​​∂R2∂2U​∂i​R+R1​∂R∂U​+∂R∂U​Xi​∂i​R1​)=i=1∑3​(R2Xi​Xi​​∂R2∂2U​+R1​∂R∂U​−R3Xi​Xi​​∂R∂U​)=∂R2∂2U​+R2​∂R∂U​=R1​(∂R2∂2U​R+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 2​U(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∂2​U=−ϵ0​1​δ(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​≡R1​f(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​≡R1​g(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} −ϵ0​1​δ(R )ρ(r ′,t)​=(∇r 2​−c21​∂t2∂2​)[R1​f(t−cR​,r ′)]=(∇r 2​R1​)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 ′)+c2​R3R ​⋅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 ′)=4πϵ0​ρ(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)=4πϵ0​Rρ(r ′,t−cR​)​U2​(R,r ′,t)=4πϵ0​Rρ(r ′,t+cR​)​​(15)

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′4πϵ0​Rρ(r ′,t−cR​)​A (r ,t)=∫dτ′4πRμ0​j ​(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∗)​=−μ0​ϵ0​∂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′4πϵ0​Rρ(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′4πRμ0​j ​∗(r ′,t−cR​)​(21)

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2​c2​)ϕ~​(r ,t)=−ϵ0​ρ(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2​c2​)U~(r ,r ′,t)=−ϵ0​1​δ(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​=0​4πϵ0​Rρ(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} 4πϵ0​Rρ(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)=4πϵ0​Rρ(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} −ϵ0​1​δ(R )ρ(r ,t)​=(∇r 2​−c21​∂t2∂2​−ℏ2m光子2​c2​)U~(r ,r ′,t)=h(R)(∇r 2​−c21​∂t2∂2​−ℏ2m光子2​c2​)4πϵ0​Rρ(r ′,t∓cR​)​+4πϵ0​Rρ(r ′,t∓cR​)​∇r 2​h(R)+2[∇r ​4πϵ0​Rρ(r ′,t∓cR​)​]⋅∇r ​h(R)=−ϵ0​1​δ(R)ρ(r ′,t)h(R)+4πϵ0​Rρ(r ′,t∓cR​)​(∂R2∂2​+R2​R∂​−ℏ2m光子2​c2​)h(R)+2[∂R∂​4πϵ0​Rρ(r ′,t∓cR​)​]∂R∂​h(R)=−ϵ0​h(0)​δ(R )ρ(r ′,t)+4πϵ0​Rρ(r ′,t∓cR​)​(∂R2∂2​−ℏ2m光子2​c2​)h(R)+R2​[∂R∂​4πϵ0​Rρ(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} −ϵ0​1​δ(R )ρ(r ,t)=−ϵ0​h(0)​δ(R )ρ(r ′,t)+4πϵ0​Rρ(r ′,t∓cR​)​(∂R2∂2​−ℏ2m光子2​c2​)h(R)+R2​[∂R∂​4πϵ0​Rρ(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2​c2​+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±ℏc​m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)=4πϵ0​Rρ(r ′,t∓cR​)​e−ℏc​m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​⇒R21​e−ℏc​m光子​R(32)
光子静止质量使使长程作用力变成短程作用力


在这里插入图片描述

推迟势

标签:frac,推迟势,vec,rho,光子,partial,cR
来源: https://blog.csdn.net/Function_RY/article/details/120555252