推迟势辐射场总结

问题核心：

为什么加速会产生1/R的辐射项，而匀速运动只会产生1/R^2项

\(\phi(\mathbf{r}, t) = \left. \frac{q}{4\pi \epsilon_0} \frac{1}{(1 - \mathbf{n} \cdot \boldsymbol{\beta}) R} \right|_{t_{\text{ret}}}\)

\(\mathbf{A}(\mathbf{r}, t) = \left. \frac{\mu_0 q c}{4\pi} \frac{\boldsymbol{\beta}}{(1 - \mathbf{n} \cdot \boldsymbol{\beta}) R} \right|_{t_{\text{ret}}}\)

\(\mathbf{E} = -\nabla \phi - \frac{\partial \mathbf{A}}{\partial t}\)

\(\mathbf{B} = \nabla \times \mathbf{A}\)

\(\mathbf{E} = \frac{q}{4\pi \epsilon_0} \left[ \frac{ (\mathbf{n} - \boldsymbol{\beta}) (1 - \beta^2) }{ (1 - \mathbf{n} \cdot \boldsymbol{\beta})^3 R^2 }+\frac{ \mathbf{n} \times \left[ (\mathbf{n} - \boldsymbol{\beta}) \times \dot{\boldsymbol{\beta}} \right] }{ c (1 - \mathbf{n} \cdot \boldsymbol{\beta})^3 R } \right]_{\text{ret}}\)

\(\mathbf{B} = \frac{\mu_0 q}{4\pi} \left[ \frac{\mathbf{n} \times \boldsymbol{\beta}}{(1 - \mathbf{n} \cdot \boldsymbol{\beta})^2 R^2} + \frac{\mathbf{n} \times \dot{\boldsymbol{\beta}}}{c (1 - \mathbf{n} \cdot \boldsymbol{\beta}) R} \right]_{\text{ret}}\)

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令\(D = 1 - \boldsymbol{\beta} \cdot \mathbf{n}\),

\(\nabla \phi = \frac{q}{4\pi \epsilon_0} \nabla \left( \frac{1}{D R} \right)\) \( = \frac{q}{4\pi \epsilon_0}\left[-\frac{1}{(D R)^2} \nabla (D R)\right]\)

\( = \frac{q}{4\pi \epsilon_0} \left[ -\frac{D \nabla R + R \nabla D}{(D R)^2} \right]\) \( = \frac{q}{4\pi \epsilon_0} \left[ -\frac{D \mathbf{n} + R \nabla D}{(D R)^2} \right]\)

其中 \(\nabla R = \mathbf{n}\)

\(\nabla D = \frac{\partial D}{\partial t’} \nabla t’ + \frac{\partial D}{\partial R} \nabla R\)

\(= -\frac{\partial D}{\partial t’} \frac{\mathbf{n}}{c} - \frac{\partial D}{\partial R} \mathbf{n}\)

其中 \(\nabla t’ = -\nabla (R/c) = -\mathbf{n}/c\)， \(\nabla R = \mathbf{n}\)

\(\frac{\partial D}{\partial t’} = -\frac{\partial}{\partial t’} (\boldsymbol{\beta} \cdot \mathbf{n}) = -(\dot{\boldsymbol{\beta}} \cdot \mathbf{n} + \boldsymbol{\beta} \cdot \dot{\mathbf{n}})\)

\(\frac{\partial D}{\partial R} = -\boldsymbol{\beta} \cdot \frac{\partial \mathbf{n}}{\partial R} = -\boldsymbol{\beta} \frac{\partial}{\partial R} \left( \frac{\mathbf{r} - \mathbf{r}’}{R} \right) = \frac{\boldsymbol{\beta} \cdot \mathbf{n}}{R}\)

合并 \(\nabla D\)：

\(\nabla D = \left[ -(\dot{\boldsymbol{\beta}} \cdot \mathbf{n} + \boldsymbol{\beta} \cdot \dot{\mathbf{n}}) \right] \left( -\frac{\mathbf{n}}{c} \right) + \left( \frac{\boldsymbol{\beta} \cdot \mathbf{n}}{R} \right) \mathbf{n} = \frac{(\dot{\boldsymbol{\beta}} \cdot \mathbf{n} + \boldsymbol{\beta} \cdot \dot{\mathbf{n}})}{c} \mathbf{n} + \frac{(\boldsymbol{\beta} \cdot \mathbf{n})}{R} \mathbf{n}\)

将 \(\nabla D\) 代入原式：

\(R \nabla D + D \mathbf{n} = R \left( \frac{(\dot{\boldsymbol{\beta}} \cdot \mathbf{n} + \boldsymbol{\beta} \cdot \dot{\mathbf{n}})}{c} \mathbf{n} + \frac{(\boldsymbol{\beta} \cdot \mathbf{n})}{R} \mathbf{n} \right) + D \mathbf{n}\)

\( = \frac{R}{c} (\dot{\boldsymbol{\beta}} \cdot \mathbf{n} + \boldsymbol{\beta} \cdot \dot{\mathbf{n}}) \mathbf{n} + \mathbf{n}\)

最终： \(\nabla \phi = -\frac{q}{4\pi \epsilon_0 D^2 R^2} \left[ \mathbf{n} + \frac{R}{c} (\dot{\boldsymbol{\beta}} \cdot \mathbf{n} + \boldsymbol{\beta} \cdot \dot{\mathbf{n}}) \mathbf{n} \right]\)

辐射部分为：

\((\nabla \phi)_{\text{rad}} = -\frac{q}{4\pi \epsilon_0 c D^2 R} (\dot{\boldsymbol{\beta}} \cdot \mathbf{n} + \boldsymbol{\beta} \cdot \dot{\mathbf{n}}) \mathbf{n}\)

其中仅 \(\dot{\boldsymbol{\beta}} \cdot \mathbf{n}\) 与加速度相关。

但Jackson的书中推导的结果为：

\(-\nabla \phi = \frac{q}{4\pi\epsilon_0} \left[ \frac{ \mathbf{n} - \boldsymbol{\beta} }{ D^2 R^2 } + \frac{ \boldsymbol{\beta} (\mathbf{n} \cdot \dot{\boldsymbol{\beta}}) }{ c D^3 R } - \frac{ (\mathbf{n} \cdot \dot{\boldsymbol{\beta}}) \mathbf{n} }{ c D^3 R } \right]\)

== 再求\(\frac{\partial \mathbf{A}}{\partial t}\)

\(\mathbf{A}=\frac{\mu_0 q c}{4\pi} \frac{\boldsymbol{\beta}}{(1 - \mathbf{n} \cdot \boldsymbol{\beta}) R}\)

\(\frac{\partial \mathbf{A}}{\partial t} = \frac{\partial \mathbf{A}}{\partial t’} \frac{\partial t’}{\partial t}\)

\(\frac{\partial t’}{\partial t} = \frac{1}{1 - \boldsymbol{\beta} \cdot \mathbf{n}}\)

\(\frac{\partial \mathbf{A}}{\partial t’} = \frac{\mu_0 qc}{4\pi} \left[ \frac{\dot{\boldsymbol{\beta}}}{D R} + \boldsymbol{\beta} \frac{\partial}{\partial t’} \left( \frac{1}{D R} \right) \right]\)

\(\frac{\partial}{\partial t’} \left( \frac{1}{D R} \right) = -\frac{1}{D^2 R} \frac{\partial D}{\partial t’} - \frac{1}{D R^2} \frac{\partial R}{\partial t’} = -\frac{1}{D^2 R} \left( -\dot{\boldsymbol{\beta}} \cdot \mathbf{n} - \boldsymbol{\beta} \cdot \dot{\mathbf{n}} \right) - \frac{1}{D R^2} (-\boldsymbol{\beta} \cdot \mathbf{n} c)\)

（因 \(\frac{\partial R}{\partial t’} = -c \boldsymbol{\beta} \cdot \mathbf{n}\)）

代入并保留辐射项：

\(\frac{\partial \mathbf{A}}{\partial t’} = \frac{\mu_0 qc}{4\pi} \left[ \frac{\dot{\boldsymbol{\beta}}}{D R} + \frac{\boldsymbol{\beta} (\dot{\boldsymbol{\beta}} \cdot \mathbf{n})}{D^2 R} + \frac{\boldsymbol{\beta} (\boldsymbol{\beta} \cdot \dot{\mathbf{n}})}{D^2 R} + \frac{\boldsymbol{\beta} (\boldsymbol{\beta} \cdot \mathbf{n}) c}{D R^2} \right]\)

于是：

\(\frac{\partial \mathbf{A}}{\partial t} = \frac{\mu_0 qc}{4\pi} \frac{1}{D} \left[ \frac{\dot{\boldsymbol{\beta}}}{D R} + \frac{\boldsymbol{\beta} (\dot{\boldsymbol{\beta}} \cdot \mathbf{n})}{D^2 R} + \cdots \right]\)

辐射部分（\(\propto \dot{\boldsymbol{\beta}}/R\)）：

\(\left( \frac{\partial \mathbf{A}}{\partial t} \right)_{\text{rad}} = \frac{\mu_0 qc}{4\pi} \frac{1}{D} \left[ \frac{\dot{\boldsymbol{\beta}}}{D R} + \frac{\boldsymbol{\beta} (\dot{\boldsymbol{\beta}} \cdot \mathbf{n})}{D^2 R} \right] = \frac{\mu_0 qc}{4\pi} \left[ \frac{\dot{\boldsymbol{\beta}}}{D^2 R} + \frac{\boldsymbol{\beta} (\dot{\boldsymbol{\beta}} \cdot \mathbf{n})}{D^3 R} \right]\)

\( = \frac{q}{4\pi\epsilon_0} \frac{1}{c} \left[ \frac{\dot{\boldsymbol{\beta}}}{(1 - \boldsymbol{\beta} \cdot \mathbf{n})^2 R} + \frac{\boldsymbol{\beta} (\dot{\boldsymbol{\beta}} \cdot \mathbf{n})}{(1 - \boldsymbol{\beta} \cdot \mathbf{n})^3 R} \right]\)

总的辐射场：

\(\mathbf{E}_{\text{rad}} = - (\nabla \phi)_{\text{rad}} - \left( \frac{\partial \mathbf{A}}{\partial t} \right)_{\text{rad}}\)

\( = -\left[ \frac{q}{4\pi\epsilon_0} \left( -\frac{(\dot{\boldsymbol{\beta}} \cdot \mathbf{n}) \mathbf{n} }{c (1 - \boldsymbol{\beta} \cdot \mathbf{n})^2 R} \right) \right] - \left[ \frac{q}{4\pi\epsilon_0} \frac{1}{c} \left( \frac{\dot{\boldsymbol{\beta}}}{(1 - \boldsymbol{\beta} \cdot \mathbf{n})^2 R} + \frac{\boldsymbol{\beta} (\dot{\boldsymbol{\beta}} \cdot \mathbf{n})}{(1 - \boldsymbol{\beta} \cdot \mathbf{n})^3 R} \right) \right]\)

\( = \frac{q}{4\pi\epsilon_0} \left[ \frac{(\dot{\boldsymbol{\beta}} \cdot \mathbf{n}) \mathbf{n} }{c (1 - \boldsymbol{\beta} \cdot \mathbf{n})^2 R} - \frac{\dot{\boldsymbol{\beta}}}{c (1 - \boldsymbol{\beta} \cdot \mathbf{n})^2 R} - \frac{\boldsymbol{\beta} (\dot{\boldsymbol{\beta}} \cdot \mathbf{n})}{c (1 - \boldsymbol{\beta} \cdot \mathbf{n})^3 R} \right]\)

\( = \frac{q}{4\pi\epsilon_0} \frac{\mathbf{n} \times \left( (\mathbf{n} - \boldsymbol{\beta}) \times \dot{\boldsymbol{\beta}} \right)}{c (1 - \boldsymbol{\beta} \cdot \mathbf{n})^3 R}\)

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\(-\nabla \phi\)中的辐射项(沿x轴匀加速)：

\(\frac{q}{4\pi\epsilon_0} \left[ \frac{ (\dot{\boldsymbol{\beta}} \cdot \mathbf{n} )\mathbf{n} }{c (1 - \boldsymbol{\beta} \cdot \mathbf{n})^3 R} \right]\)

\( = \frac{q}{4\pi\epsilon_0} \cdot \frac{a}{c^2} \cdot \frac{1}{(1 - \boldsymbol{\beta} \cdot \mathbf{n})^3 R^2} \left( (x - x(t’))^2, (x - x(t’)) y, (x - x(t’)) z \right)\)

\(E_x = \frac{q a (x - x(t’))^2}{4\pi\epsilon_0 c^2 (1 - \boldsymbol{\beta} \cdot \mathbf{n})^3 R^3}\)

\(E_y = \frac{q a (x - x(t’)) y}{4\pi\epsilon_0 c^2 (1 - \boldsymbol{\beta} \cdot \mathbf{n})^3 R^3}\)

\(E_z = \frac{q a (x - x(t’)) z}{4\pi\epsilon_0 c^2 (1 - \boldsymbol{\beta} \cdot \mathbf{n})^3 R^3}\)

\(-\frac{\partial A}{\partial t}\)中的辐射项(沿x轴匀加速)：

第一项：\(-\frac{q}{4\pi\epsilon_0} \frac{ \dot{\boldsymbol{\beta}} }{c (1 - \boldsymbol{\beta} \cdot \mathbf{n})^2 R}\)

来自\(\mathbf{A}\)对t求导中的\(\frac{\partial \beta}{\partial t’}\),表明为矢量势在速度方向上的时间变换，即在速度方向上有没有加速度。

\(\mathbf{A}=\frac{\mu_0 q c}{4\pi} \frac{\boldsymbol{\beta}}{(1 - \mathbf{n} \cdot \boldsymbol{\beta}) R}\)

\(E_x = -\frac{q}{4\pi\epsilon_0} \frac{a/c^2}{(1 - \beta n_x)^2 R}\) \( = -\frac{q a}{4\pi\epsilon_0 c^2} \frac{1}{\left[1 - \frac{v(t’) (x-x(t’))}{c R}\right]^2 R}\)

\(E_y=E_z=0\)

第二项：\(\frac{ \boldsymbol{\beta} (\dot{\boldsymbol{\beta}} \cdot \mathbf{n}) }{c (1 - \boldsymbol{\beta} \cdot \mathbf{n})^3 R}\)

\(E_x = -\frac{q}{4\pi\epsilon_0} \frac{\frac{v}{c} \cdot \frac{a}{c}\frac{x-x(t’)}{R}}{c \left(1 - \frac{v}{c}\frac{x-x(t’)}{R}\right)^3 R} \) \( = -\frac{q a v(t’) (x-x(t’))}{4\pi\epsilon_0 c^3 R^2 \left[1 - \frac{v(t’) (x-x(t’))}{c R}\right]^3}\)

\(E_y = E_z =0 \)

\(x(t’) = \frac{c^2}{a}(\sqrt{1 + (a t’/c)^2} - 1)\)

\(v(t’) = \frac{a t’}{\sqrt{1 + (a t’/c)^2}}\)

\(R = \sqrt{(x-x(t’))^2 + y^2 + z^2}\)

\(t’ = t - R/c\)

⚠️ 重要说明：匀加速电荷是否辐射？

这是一个著名的概念性问题！

❗ 经典电动力学中，匀加速电荷（如恒定外力下的直线加速）是否辐射电磁波？

• 数学上，\( \dot{\boldsymbol{\beta}} \ne 0 \)，所以辐射项非零
• 但广义相对论观点：匀加速参考系等效于引力场，电荷静止，不应辐射
• 这是一个长期争论的问题（Paradox of radiation of charged particles in a uniform acceleration）

但在经典李纳-维谢尔场框架下，我们仍然计算出非零辐射场，因为：

• 电荷确实在加速（\( \mathbf{a} \ne 0 \)）
• 场的结构满足 \( 1/r \) 衰减、横向性等辐射特征

所以：在经典理论中，我们接受它会辐射，尽管物理图像有争议。

辐射电场分量： \[ \begin{aligned} E_x(x,y,z,t) &= \frac{q}{4\pi \epsilon_0 c^2 R} \left[ \frac{x (x \ddot{x}_q + y \ddot{y}_q + z \ddot{z}_q)}{R^2} - \ddot{x}_q \right]_{t_{\text{ret}}}, \\ E_y(x,y,z,t) &= \frac{q}{4\pi \epsilon_0 c^2 R} \left[ \frac{y (x \ddot{x}_q + y \ddot{y}_q + z \ddot{z}_q)}{R^2} - \ddot{y}_q \right]_{t_{\text{ret}}}, \\ E_z(x,y,z,t) &= \frac{q}{4\pi \epsilon_0 c^2 R} \left[ \frac{z (x \ddot{x}_q + y \ddot{y}_q + z \ddot{z}_q)}{R^2} - \ddot{z}_q \right]_{t_{\text{ret}}}. \end{aligned} \]

磁场分量

（通过 \(\mathbf{B} = \frac{1}{c} \hat{n} \times \mathbf{E}_{\text{rad}}\)）：

\[ \begin{aligned} B_x &= \frac{1}{c} \left( \frac{y}{R} E_z - \frac{z}{R} E_y \right), \\ B_y &= \frac{1}{c} \left( \frac{z}{R} E_x - \frac{x}{R} E_z \right), \\ B_z &= \frac{1}{c} \left( \frac{x}{R} E_y - \frac{y}{R} E_x \right). \end{aligned} \]

\(\mathbf{B}_{rad}=\frac{1}{c} \mathbf{n} \times \mathbf{E}_{rad}\)