Radiacion: campos retardados

Potenciales de Liénard-Wiechert

Los potenciales de Liénard-Wiechert son los potenciales escalar y vectorial producidos por una carga puntual en movimiento. El potencial escalar esta dado por \[ \phi\left(\mathbf{r},t\right)=\frac{1}{\varepsilon R\left(t_{r}\right)}\frac{e}{\left[1-\boldsymbol{\beta}\left(t_{r}\right)\cdot\mathbf{\hat{\mathbf{n}}}\left(t_{r}\right)\right]}.\label{eq: phi lenard n} \]

Mientras que el potencial vectorial, puesto que $\mathbf{J}=\rho c\boldsymbol{\beta}$, es

\[ \mathbf{A}\left(\mathbf{r},t\right)=\frac{\mu c}{R\left(t_{r}\right)}\frac{e\boldsymbol{\beta}\left(t_{r}\right)}{\left[1-\boldsymbol{\beta}\left(t_{r}\right)\cdot\mathbf{\hat{\mathbf{n}}}\left(t_{r}\right)\right]}.\label{eq: A lenard n} \] donde \[ \hat{\mathbf{n}}\equiv\frac{\mathbf{R}}{\left|\mathbf{R}\right|},\qquad R\equiv\left|\mathbf{R}\right|, \]

si además se define \[ \boldsymbol{\beta}\equiv\frac{1}{c}\frac{d\mathbf{r}_{e}\left(t_{r}\right)}{dt_{r}}=-\frac{1}{c}\frac{\partial\mathbf{R}}{\partial t_{r}},\qquad\mathbf{R}\equiv\mathbf{r}\left(t\right)-\mathbf{r}_{e}\left(t_{r}\right).\label{eq: vel en c} \]

Derivadas temporales

Derivada de la magnitud de R

Puesto que $R=c\left(t-t_{r}\right)$, su derivada con respecto a $t$ es \[ \frac{\partial R}{\partial t}=c\left(1-\frac{\partial t_{r}}{\partial t}\right), \] por otro lado, usando la regla de la cadena \[ \frac{\partial R}{\partial t}=\frac{\partial R}{\partial t_{r}}\frac{\partial t_{r}}{\partial t}; \] al igualar éstas dos expresiones se obtiene

\[ c\left(1-\frac{\partial t_{r}}{\partial t}\right)=\frac{\partial R}{\partial t_{r}}\frac{\partial t_{r}}{\partial t}\quad\Rightarrow\quad\frac{\partial t_{r}}{\partial t}=\frac{1}{1+\frac{1}{c}\frac{\partial R}{\partial t_{r}}} \]

puesto que $R^{2}=\mathbf{\mathbf{R}\cdot R}$ y la derivada de ésta ecuación con respecto a $t_{r}$ es \[ 2R\frac{\partial R}{\partial t_{r}}=2\frac{\partial\mathbf{R}}{\partial t_{r}}\cdot\mathbf{R}=-2c\boldsymbol{\beta}\cdot\mathbf{R}\quad\Rightarrow \] \[ \frac{\partial R}{\partial t_{r}}=-c\boldsymbol{\beta}\cdot\hat{\mathbf{n}}.\label{eq: der R en tr} \] de manera que

\[ \frac{\partial t_{r}}{\partial t}=\frac{1}{1-\boldsymbol{\beta}\cdot\hat{\mathbf{n}}}\label{eq: der t ret en t} \]

y la derivada de $R$ con respecto a $t$ es

\[ \frac{\partial R}{\partial t}=\frac{-c\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}}{1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}}.\label{eq: der R en t} \]

Derivada del vector R

La derivada del vector $\mathbf{R}$ con respecto a $t$ es \[ \frac{\partial\mathbf{R}}{\partial t}=\frac{\partial\mathbf{R}}{\partial t_{r}}\frac{\partial t_{r}}{\partial t}=-c\boldsymbol{\beta}\frac{1}{1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}}.\label{eq: der R vec en t} \]

Derivada del vector unitario

La derivada temporal del vector que une a carga y observador es \[ \frac{\partial\hat{\mathbf{n}}}{\partial t}=\frac{\partial}{\partial t}\left(\frac{\mathbf{R}}{R}\right)=-\frac{\mathbf{R}}{R^{2}}\frac{\partial R}{\partial t}+\frac{1}{R}\frac{\partial\mathbf{R}}{\partial t}=-\frac{\hat{\mathbf{n}}}{R}\left(\frac{-c\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}}{1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}}\right)-\frac{1}{R}\frac{c\boldsymbol{\beta}}{1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}} \] que puede escribirse como \[ \frac{\partial\hat{\mathbf{n}}}{\partial t}=\frac{c}{R\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)}\left[\left(\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)\hat{\mathbf{n}}-\boldsymbol{\beta}\right]\label{eq: der n en t} \]

Derivada de la velocidad

Finalmente, la derivada temporal de la velocidad de la carga es \[ \frac{\partial\boldsymbol{\beta}}{\partial t}=\frac{\partial\boldsymbol{\beta}}{\partial t_{r}}\frac{\partial t_{r}}{\partial t}=\frac{\dot{\boldsymbol{\beta}}}{1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}}\label{eq: der beta} \]

Derivadas espaciales

Gradiente de la magnitud R

Calculemos el gradiente de $R$ \[ \nabla R\left(t_{r}\right)=-c\nabla t_{r}\label{eq: grad r grat tr} \]

\[ \nabla R=\frac{\nabla\left[\left(x-x_{e}\right)^{2}+\left(y-y_{e}\right)^{2}+\left(z-z_{e}\right)^{2}\right]}{2\sqrt{\left(x-x_{e}\right)^{2}+\left(y-y_{e}\right)^{2}+\left(z-z_{e}\right)^{2}}} \]

\[ \nabla R=\frac{\left(x-x_{e}\right)\nabla\left(x-x_{e}\right)+\left(y-y_{e}\right)\nabla\left(y-y_{e}\right)+\left(z-z_{e}\right)\nabla\left(z-z_{e}\right)}{R} \] \[ \nabla R=\frac{\mathbf{R}}{R}+\frac{\left(x-x_{e}\right)\nabla\left(-x_{e}\right)+\left(y-y_{e}\right)\nabla\left(-y_{e}\right)+\left(z-z_{e}\right)\nabla\left(-z_{e}\right)}{R} \] pero $\frac{\partial x_{e}}{\partial x}=\frac{\partial x_{e}}{\partial t_{r}}\frac{\partial t_{r}}{\partial x}=c\beta_{x}\frac{\partial t_{r}}{\partial x}$, $\frac{\partial x_{e}}{\partial y}=\frac{\partial x_{e}}{\partial t_{r}}\frac{\partial t_{r}}{\partial y}=c\beta_{x}\frac{\partial t_{r}}{\partial y}$, etcétera, de manera que \[ \nabla R=\frac{\mathbf{R}}{R}+\frac{-\left(x-x_{e}\right)c\beta_{x}\nabla t_{r}-\left(y-y_{e}\right)c\beta_{y}\nabla t_{r}-\left(z-z_{e}\right)c\beta_{z}\nabla t_{r}}{R} \] \[ \nabla R=\frac{\mathbf{R}}{R}+\frac{c\mathbf{R}\cdot\boldsymbol{\beta}\nabla t_{r}}{R}=\hat{\mathbf{n}}-c\hat{\mathbf{n}}\cdot\boldsymbol{\beta}\nabla t_{r}=\hat{\mathbf{n}}+c\hat{\mathbf{n}}\cdot\boldsymbol{\beta}\frac{\nabla R}{c} \] de manera que se obtiene \[ \nabla R=\frac{\hat{\mathbf{n}}}{1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}}.\label{eq: grad R} \]

Gradiente de $\boldsymbol{\beta}\cdot\hat{\mathbf{n}}$

El gradiente del producto es \[ \nabla\left(\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)=\left(\boldsymbol{\beta}\cdot\nabla\right)\mathbf{\hat{\mathbf{n}}}+\left(\mathbf{\hat{\mathbf{n}}}\cdot\nabla\right)\boldsymbol{\beta}+\mathbf{\hat{\mathbf{n}}}\times\nabla\times\boldsymbol{\beta}+\boldsymbol{\beta}\times\nabla\times\hat{\mathbf{n}} \] o de plano calcular por componentes \[ \nabla\left(\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)=\nabla\left(\boldsymbol{\beta}\cdot\frac{\mathbf{R}}{R}\right)=\left(\boldsymbol{\beta}\cdot\mathbf{R}\right)\nabla\left(\frac{1}{R}\right)+\frac{1}{R}\nabla\left(\boldsymbol{\beta}\cdot\mathbf{R}\right) \] y del cálculo por componentes \[ \nabla\left(\boldsymbol{\beta}\cdot\mathbf{R}\right)=\sum_{j}\frac{\partial}{\partial r_{j}}\hat{e}_{j}\sum_{k}\left[\beta_{k}\left(r_{k}-r_{ek}\right)\right] \] \[ \nabla\left(\boldsymbol{\beta}\cdot\mathbf{R}\right)=\sum_{j}\sum_{k}\frac{\partial\beta_{k}}{\partial r_{j}}\left(r_{k}-r_{ek}\right)\hat{e}_{j}+\sum_{j}\sum_{k}\beta_{k}\left(\frac{\partial r_{k}}{\partial r_{j}}-\frac{\partial r_{ek}}{\partial r_{j}}\right)\hat{e}_{j} \] \[ \nabla\left(\boldsymbol{\beta}\cdot\mathbf{R}\right)=\sum_{j}\sum_{k}\frac{\partial\beta_{k}}{\partial t_{r}}\frac{\partial t_{r}}{\partial r_{j}}\left(r_{k}-r_{ek}\right)\hat{e}_{j}+\sum_{j}\sum_{k}\beta_{k}\frac{\partial r_{k}}{\partial r_{j}}\hat{e}_{j}-\sum_{j}\sum_{k}\beta_{k}\frac{\partial r_{ek}}{\partial t_{r}}\frac{\partial t_{r}}{\partial r_{j}}\hat{e}_{j} \] \[ \nabla\left(\boldsymbol{\beta}\cdot\mathbf{R}\right)=\sum_{j}\left[\sum_{k}\frac{\partial\beta_{k}}{\partial t_{r}}\left(r_{k}-r_{ek}\right)\right]\frac{\partial t_{r}}{\partial r_{j}}\hat{e}_{j}+\sum_{j}\sum_{j}\beta_{j}\frac{\partial r_{j}}{\partial r_{j}}\hat{e}_{j}-\sum_{j}\left[\sum_{k}\beta_{k}\frac{\partial r_{ek}}{\partial t_{r}}\right]\frac{\partial t_{r}}{\partial r_{j}}\hat{e}_{j} \] pero $\frac{\partial\beta_{k}}{\partial t_{r}}=\dot{\beta}_{k}$, $\frac{\partial r_{ek}}{\partial t_{r}}=c\beta_{k}$ \[ \nabla\left(\boldsymbol{\beta}\cdot\mathbf{R}\right)=\mathbf{\dot{\boldsymbol{\beta}}}\cdot\mathbf{R}\sum_{j}\frac{\partial t_{r}}{\partial r_{j}}\hat{e}_{j}+\boldsymbol{\beta}-c\beta^{2}\sum_{j}\frac{\partial t_{r}}{\partial r_{j}}\hat{e}_{j} \] y recordando que $\sum_{j}\frac{\partial t_{r}}{\partial r_{j}}\hat{e}_{j}=\nabla t_{r}=\frac{-\nabla R}{c}$

\[ \nabla\left(\boldsymbol{\beta}\cdot\mathbf{R}\right)=\left(\mathbf{\dot{\boldsymbol{\beta}}}\cdot\mathbf{R}-c\beta^{2}\right)\left(\frac{-\nabla R}{c}\right)+\boldsymbol{\beta} \] de manera que \[ \nabla\left(\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)=\left(\boldsymbol{\beta}\cdot\mathbf{R}\right)\left(\frac{-1}{R^{2}}\right)\nabla R+\frac{1}{R}\left[\left(\mathbf{\dot{\boldsymbol{\beta}}}\cdot\mathbf{R}-c\beta^{2}\right)\left(\frac{-\nabla R}{c}\right)+\boldsymbol{\beta}\right] \] \[ \nabla\left(\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)=\left(\boldsymbol{\beta}\cdot\hat{\mathbf{n}}\right)\left(\frac{-1}{R}\right)\frac{\hat{\mathbf{n}}}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)}+\frac{1}{R}\left[\left(\mathbf{\dot{\boldsymbol{\beta}}}\cdot\mathbf{R}-c\beta^{2}\right)\left(\frac{-\hat{\mathbf{n}}}{c\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)}\right)+\boldsymbol{\beta}\right] \] \[ \nabla\left(\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)=\frac{1}{R\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)}\left\{ -\hat{\mathbf{n}}\left(\boldsymbol{\beta}\cdot\hat{\mathbf{n}}\right)+\left(R\left(\mathbf{\dot{\boldsymbol{\beta}}}\cdot\hat{\mathbf{n}}\right)-c\beta^{2}\right)\left(\frac{-\hat{\mathbf{n}}}{c}\right)+\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)\boldsymbol{\beta}\right\} \] \[ \nabla\left(\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)=\frac{1}{R\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)}\left\{ -\left(\boldsymbol{\beta}+\hat{\mathbf{n}}\right)\left(\boldsymbol{\beta}\cdot\hat{\mathbf{n}}\right)+\boldsymbol{\beta}+\beta^{2}\hat{\mathbf{n}}-\left(\frac{R\left(\mathbf{\dot{\boldsymbol{\beta}}}\cdot\hat{\mathbf{n}}\right)\hat{\mathbf{n}}}{c}\right)\right\} \]

Campos retardados

El campo eléctrico en términos de los potenciales es \[ \mathbf{E}=-\nabla\phi-\frac{\partial\mathbf{\mathbf{A}}}{\partial t}=-\nabla\left(\frac{1}{\varepsilon R}\frac{e}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)}\right)-\frac{d}{dt}\left(\frac{\mu c}{R}\frac{e\boldsymbol{\beta}}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)}\right). \] donde los operadores gradiente y derivada temporal se evalúan en el punto de observación puesto que los campos se desean conocer en ese punto.

Evaluemos los distintos términos,

Gradiente

El término gradiente es entonces \[ \nabla\left(\frac{1}{\varepsilon R}\frac{e}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)}\right)=\frac{e}{\varepsilon}\frac{1}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)}\nabla\left(\frac{1}{R}\right)+\frac{e}{\varepsilon R}\frac{1}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)^{2}}\nabla\left(\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right). \] el primer término \[ \frac{e}{\varepsilon}\frac{1}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)}\nabla\left(\frac{1}{R}\right)=\frac{e}{\varepsilon}\frac{1}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)}\left(\frac{-1}{R^{2}}\right)\frac{\hat{\mathbf{n}}}{1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}}=-\frac{e}{\varepsilon R^{2}}\frac{\hat{\mathbf{n}}}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)^{2}} \]

\[ \frac{e}{\varepsilon}\frac{1}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)}\nabla\left(\frac{1}{R}\right)=-\frac{e}{\varepsilon R^{2}}\frac{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)\hat{\mathbf{n}}}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)^{3}}\label{eq: pri ter grad phi} \] mientras que el segundo término es \[ \frac{e}{\varepsilon R}\frac{\nabla\left(\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)^{2}}=\frac{e}{\varepsilon R^{2}}\frac{-\left(\boldsymbol{\beta}+\hat{\mathbf{n}}\right)\left(\boldsymbol{\beta}\cdot\hat{\mathbf{n}}\right)+\boldsymbol{\beta}+\beta^{2}\hat{\mathbf{n}}-\frac{1}{c}R\left(\mathbf{\dot{\boldsymbol{\beta}}}\cdot\hat{\mathbf{n}}\right)\hat{\mathbf{n}}}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)^{3}}.\label{eq: seg ter grad phi} \] Agrupando términos de (\ref{eq: pri ter grad phi}) y (\ref{eq: seg ter grad phi}) \[ \left.\nabla\phi\right|_{vel}=\frac{e}{\varepsilon R^{2}}\frac{1}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)^{3}}\left\{ \left(-1+\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)\hat{\mathbf{n}}-\left(\boldsymbol{\beta}+\hat{\mathbf{n}}\right)\left(\boldsymbol{\beta}\cdot\hat{\mathbf{n}}\right)+\boldsymbol{\beta}+\beta^{2}\hat{\mathbf{n}}\right\} \]

\[ \left.\nabla\phi\right|_{vel}=\frac{e}{\varepsilon R^{2}}\frac{1}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)^{3}}\left\{ \left(-1+\beta^{2}\right)\hat{\mathbf{n}}+\left(-\boldsymbol{\beta}\right)\left(\boldsymbol{\beta}\cdot\hat{\mathbf{n}}\right)+\boldsymbol{\beta}\right\} \]

\[\left.\nabla\phi\right|_{vel}=\frac{e}{\varepsilon R^{2}}\frac{-1}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)^{3}}\left\{ \left(1-\beta^{2}\right)\hat{\mathbf{n}}-\left[1-\left(\boldsymbol{\beta}\cdot\hat{\mathbf{n}}\right)\right]\boldsymbol{\beta}\right\} \label{eq: grad phi vel} \]

\[ \left.\nabla\phi\right|_{acel}=-\frac{e}{\varepsilon cR}\frac{\left(\mathbf{\dot{\boldsymbol{\beta}}}\cdot\hat{\mathbf{n}}\right)\hat{\mathbf{n}}}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)^{3}}\label{eq: grad phi acel} \]

Derivada temporal

La derivada temporal del potencial vectorial es

\[ \frac{d}{dt}\left(\frac{\mu c}{R}\frac{e\boldsymbol{\beta}}{1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}}\right)=\left(-\frac{\mu c}{R^{2}}\frac{e\boldsymbol{\beta}}{1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}}\right)\frac{\partial R}{\partial t}+\left(\frac{\mu c}{R}\frac{e}{1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}}\right)\frac{\partial\boldsymbol{\beta}}{\partial t}+\left(\frac{\mu c}{R}\frac{e\boldsymbol{\beta}}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)^{2}}\right)\frac{\partial\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}}{\partial t} \] el primer término con (\ref{eq: der R en t}) es \[ \left(-\frac{\mu c}{R^{2}}\frac{e\boldsymbol{\beta}}{1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}}\right)\frac{\partial R}{\partial t}=\left(-\frac{\mu c}{R^{2}}\frac{e\boldsymbol{\beta}}{1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}}\right)\frac{-c\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}}{1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}} \]

\[ \left(-\frac{\mu c}{R^{2}}\frac{e\boldsymbol{\beta}}{1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}}\right)\frac{\partial R}{\partial t}=\frac{\mu c^{2}e}{R^{2}}\frac{\left(\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)\boldsymbol{\beta}}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)^{2}}\label{eq: primer der t} \] mientras que el segundo término utilizando (\ref{eq: der beta}) es

\[ \left(\frac{\mu c}{R}\frac{e}{1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}}\right)\frac{\partial\boldsymbol{\beta}}{\partial t}=\left(\frac{\mu c}{R}\frac{e}{1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}}\right)\frac{\dot{\boldsymbol{\beta}}}{1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}}=\frac{\mu ce}{R}\frac{\dot{\boldsymbol{\beta}}}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)^{2}}\label{eq: segundo der t} \]

la derivada temporal del producto punto en el último término con (\ref{eq: der n en t}) y (\ref{eq: der beta}) es \[ \frac{\partial\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}}{\partial t}=\frac{\partial\boldsymbol{\beta}}{\partial t}\cdot\mathbf{\hat{\mathbf{n}}}+\boldsymbol{\beta}\cdot\frac{\partial\mathbf{\hat{\mathbf{n}}}}{\partial t}=\frac{\dot{\boldsymbol{\beta}}\cdot\mathbf{\hat{\mathbf{n}}}}{1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}}+\frac{c\left[\left(\hat{\mathbf{n}}\cdot\boldsymbol{\beta}\right)\left(\boldsymbol{\beta}\cdot\hat{\mathbf{n}}\right)-\boldsymbol{\beta}\cdot\boldsymbol{\beta}\right]}{R\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)} \] de manera que el último término es \[ \left(\frac{\mu c}{R}\frac{e\boldsymbol{\beta}}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)^{2}}\right)\left\{ \frac{\dot{\boldsymbol{\beta}}\cdot\mathbf{\hat{\mathbf{n}}}}{1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}}+\frac{c\left[\left(\hat{\mathbf{n}}\cdot\boldsymbol{\beta}\right)^{2}-\beta^{2}\right]}{R\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)}\right\} \] que puede reescribirse como

\[ \left(\frac{\mu ce}{R^{2}}\frac{1}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)^{3}}\right)\left\{ R\left(\dot{\boldsymbol{\beta}}\cdot\mathbf{\hat{\mathbf{n}}}\right)\boldsymbol{\beta}+c\left[\left(\boldsymbol{\beta}\cdot\hat{\mathbf{n}}\right)^{2}-\beta^{2}\right]\boldsymbol{\beta}\right\} .\label{eq: tercer der t} \]

La derivada temporal de (\ref{eq: primer der t}), (\ref{eq: segundo der t}) y (\ref{eq: tercer der t}) es \[ \frac{\partial\mathbf{A}}{\partial t}=\frac{\mu c^{2}e}{R^{2}}\frac{\left(\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)\boldsymbol{\beta}}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)^{2}}+\frac{\mu ce}{R}\frac{\dot{\boldsymbol{\beta}}}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)^{2}}+\left(\frac{\mu ec}{R^{2}}\frac{R\left(\dot{\boldsymbol{\beta}}\cdot\mathbf{\hat{\mathbf{n}}}\right)\boldsymbol{\beta}+c\left[\left(\boldsymbol{\beta}\cdot\hat{\mathbf{n}}\right)^{2}-\beta^{2}\right]\boldsymbol{\beta}}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)^{3}}\right) \]

\[ =\frac{\mu ce\left[c\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)\left(\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)\boldsymbol{\beta}+R\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)\dot{\boldsymbol{\beta}}+R\left(\dot{\boldsymbol{\beta}}\cdot\mathbf{\hat{\mathbf{n}}}\right)\boldsymbol{\beta}+c\left[\left(\boldsymbol{\beta}\cdot\hat{\mathbf{n}}\right)^{2}-\beta^{2}\right]\boldsymbol{\beta}\right]}{R^{2}\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)^{3}} \]

\[ \frac{\partial\mathbf{A}}{\partial t}=\frac{\mu ce\left[c\left(\left(\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)-\beta^{2}\right)\boldsymbol{\beta}+R\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)\dot{\boldsymbol{\beta}}+R\left(\dot{\boldsymbol{\beta}}\cdot\mathbf{\hat{\mathbf{n}}}\right)\boldsymbol{\beta}\right]}{R^{2}\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)^{3}} \] que podemos agrupar en dos partes vinculadas con velocidades y aceleraciones

\[ \left.\frac{\partial\mathbf{A}}{\partial t}\right|_{vel}=\frac{\mu c^{2}e}{R^{2}\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)^{3}}\left[\left(\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)-\beta^{2}\right]\boldsymbol{\beta}\label{eq: der A en t vel} \]

\[ \left.\frac{\partial\mathbf{A}}{\partial t}\right|_{acel}=\frac{\mu ce}{R\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)^{3}}\left[\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)\dot{\boldsymbol{\beta}}+\left(\dot{\boldsymbol{\beta}}\cdot\mathbf{\hat{\mathbf{n}}}\right)\boldsymbol{\beta}\right]\label{eq: der A en t acel} \]

Términos en E de velocidad

Los potenciales involucrando velocidades son

\[ \left.\mathbf{E}\right|_{vel}=-\left.\nabla\phi\right|_{vel}-\left.\frac{\partial\mathbf{A}}{\partial t}\right|_{vel}, \]

de las expresiones (\ref{eq: grad phi vel}) y (\ref{eq: der A en t vel}), se obtiene

\[ \left.\mathbf{E}\right|_{vel}=\frac{e}{\varepsilon R^{2}}\frac{1}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)^{3}}\left\{ \left(1-\beta^{2}\right)\hat{\mathbf{n}}-\left[1-\left(\boldsymbol{\beta}\cdot\hat{\mathbf{n}}\right)\right]\boldsymbol{\beta}\right\} -\frac{\mu c^{2}e\left[\left(\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)-\beta^{2}\right]\boldsymbol{\beta}}{R^{2}\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)^{3}} \]

y puesto que $1/\varepsilon=\mu c^{2}$ se puede factorizar

\[ \left.\mathbf{E}\right|_{vel}=\frac{e}{\varepsilon R^{2}}\frac{1}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)^{3}}\left\{ \left(1-\beta^{2}\right)\hat{\mathbf{n}}-\left[1-\left(\boldsymbol{\beta}\cdot\hat{\mathbf{n}}\right)\right]\boldsymbol{\beta}-\left[\left(\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)-\beta^{2}\right]\boldsymbol{\beta}\right\} \]

\[ \left.\mathbf{E}\right|_{vel}=\frac{e}{\varepsilon R^{2}}\frac{1}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)^{3}}\left\{ \left(1-\beta^{2}\right)\hat{\mathbf{n}}-\boldsymbol{\beta}+\left(\boldsymbol{\beta}\cdot\hat{\mathbf{n}}\right)\boldsymbol{\beta}-\left(\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)\boldsymbol{\beta}+\beta^{2}\boldsymbol{\beta}\right\} \]

\[ \left.\mathbf{E}\right|_{vel}=\frac{e}{\varepsilon R^{2}}\frac{1}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)^{3}}\left\{ \left(1-\beta^{2}\right)\left(\hat{\mathbf{n}}-\boldsymbol{\beta}\right)\right\} \label{eq: E vel} \]

Términos en E de aceleración

La aceleración \[ \left.\mathbf{E}\right|_{acel}=-\left.\nabla\phi\right|_{acel}-\left.\frac{\partial\mathbf{A}}{\partial t}\right|_{acel} \]

\[ \left.\mathbf{E}\right|_{acel}=\frac{e}{\varepsilon cR}\frac{\left(\mathbf{\dot{\boldsymbol{\beta}}}\cdot\hat{\mathbf{n}}\right)\hat{\mathbf{n}}}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)^{3}}-\frac{\mu ce}{R\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)^{3}}\left[\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)\dot{\boldsymbol{\beta}}+\left(\dot{\boldsymbol{\beta}}\cdot\mathbf{\hat{\mathbf{n}}}\right)\boldsymbol{\beta}\right] \]

si $1/c\varepsilon=\mu c$ que es correcto pues implica $1/\mu\varepsilon=c^{2}$

\[ \left.\mathbf{E}\right|_{acel}=\frac{e}{\varepsilon cR}\frac{1}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)^{3}}\left\{ \left(\mathbf{\dot{\boldsymbol{\beta}}}\cdot\hat{\mathbf{n}}\right)\hat{\mathbf{n}}-\left[\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)\dot{\boldsymbol{\beta}}+\left(\dot{\boldsymbol{\beta}}\cdot\mathbf{\hat{\mathbf{n}}}\right)\boldsymbol{\beta}\right]\right\} \]

\[ \left.\mathbf{E}\right|_{acel}=\frac{e}{\varepsilon cR}\frac{1}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)^{3}}\left\{ \left(\mathbf{\dot{\boldsymbol{\beta}}}\cdot\hat{\mathbf{n}}\right)\left(\hat{\mathbf{n}}-\boldsymbol{\beta}\right)-\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)\dot{\boldsymbol{\beta}}\right\} \label{eq: E acel} \]

El término entre corchetes se puede escribir como \[ \left(\mathbf{\hat{\mathbf{n}}}\cdot\dot{\boldsymbol{\beta}}\right)\left(\mathbf{\hat{\mathbf{n}}}-\boldsymbol{\beta}\right)-\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)\dot{\boldsymbol{\beta}}=\mathbf{\hat{\mathbf{n}}}\times\left\{ \left(\mathbf{\hat{\mathbf{n}}}-\boldsymbol{\beta}\right)\times\dot{\boldsymbol{\beta}}\right\} , \]

pues $\mathbf{\hat{\mathbf{n}}}\times\left\{ \left(\mathbf{\hat{\mathbf{n}}}-\boldsymbol{\beta}\right)\times\dot{\boldsymbol{\beta}}\right\} =\mathbf{\hat{\mathbf{n}}}\times\left(\mathbf{\hat{\mathbf{n}}}\times\dot{\boldsymbol{\beta}}\right)-\mathbf{\hat{\mathbf{n}}}\times\left(\boldsymbol{\beta}\times\dot{\boldsymbol{\beta}}\right)$, y cada triple producto puede desarrollarse como $\mathbf{\hat{\mathbf{n}}}\times\left(\mathbf{\hat{\mathbf{n}}}\times\dot{\boldsymbol{\beta}}\right)=\left(\mathbf{\hat{\mathbf{n}}}\cdot\dot{\boldsymbol{\beta}}\right)\mathbf{\hat{\mathbf{n}}}-\dot{\boldsymbol{\beta}}$ y $\mathbf{\hat{\mathbf{n}}}\times\left(\boldsymbol{\beta}\times\dot{\boldsymbol{\beta}}\right)=\left(\mathbf{\hat{\mathbf{n}}}\cdot\dot{\boldsymbol{\beta}}\right)\boldsymbol{\beta}-\left(\mathbf{\hat{\mathbf{n}}}\cdot\boldsymbol{\beta}\right)\dot{\boldsymbol{\beta}}$.

El término acelerado se escribe como

\[ \mathbf{E}_{acel}=\frac{e}{\varepsilon c}\left[\frac{\mathbf{\hat{\mathbf{n}}}\times\left\{ \left(\mathbf{\hat{\mathbf{n}}}-\boldsymbol{\beta}\right)\times\dot{\boldsymbol{\beta}}\right\} }{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)^{3}R}\right]_{ret}.\label{eq: E lenard acel} \]

Campo eléctrico completo

De las ecuaciones (\ref{eq: E vel}) y (\ref{eq: E lenard acel}) se obtiene

\[ \mathbf{E}=\frac{e}{\varepsilon}\left[\frac{\left(\mathbf{\hat{\mathbf{n}}}-\boldsymbol{\beta}\right)\left(1-\beta^{2}\right)}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)^{3}R^{2}}\right]_{ret}+\frac{e}{\varepsilon c}\left[\frac{\mathbf{\hat{\mathbf{n}}}\times\left\{ \left(\mathbf{\hat{\mathbf{n}}}-\boldsymbol{\beta}\right)\times\dot{\boldsymbol{\beta}}\right\} }{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)^{3}R}\right]_{ret}.\label{eq: E lenard} \]

Campo magnético

El campo magnético está dado por

\[ \mathbf{B}=\nabla\times\mathbf{A}=\nabla\times\left\{ \frac{\mu c}{R}\frac{e\boldsymbol{\beta}}{\left[1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right]}\right\} \]

y

\[ \nabla\times\left\{ \frac{\mu c}{R}\frac{e\boldsymbol{\beta}}{\left[1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right]}\right\} =\nabla\left\{ \frac{\mu c}{R}\frac{e}{\left[1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right]}\right\} \times\boldsymbol{\beta}+\left\{ \frac{\mu c}{R}\frac{e}{\left[1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right]}\right\} \nabla\times\boldsymbol{\beta} \]

El primer término es del gradiente de (\ref{eq: pri ter grad phi}) es

\[ \frac{\mu ce}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)}\nabla\left(\frac{1}{R}\right)=-\frac{\mu ce}{R^{2}}\frac{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)\hat{\mathbf{n}}}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)^{3}}\label{eq: pri ter grad A} \]

mientras que el segundo de (\ref{eq: seg ter grad phi}) es

\[ \frac{\mu ce}{R}\frac{\nabla\left(\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)^{2}}=\frac{\mu ce}{R^{2}}\frac{-\left(\boldsymbol{\beta}+\hat{\mathbf{n}}\right)\left(\boldsymbol{\beta}\cdot\hat{\mathbf{n}}\right)+\boldsymbol{\beta}+\beta^{2}\hat{\mathbf{n}}-\frac{1}{c}R\left(\mathbf{\dot{\boldsymbol{\beta}}}\cdot\hat{\mathbf{n}}\right)\hat{\mathbf{n}}}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)^{3}}.\label{eq: seg ter grad A} \]

de manera que el término gradiente es

\[ \nabla\left\{ \frac{\mu c}{R}\frac{e}{\left[1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right]}\right\} =\frac{\mu ce}{R^{2}}\frac{-\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)\hat{\mathbf{n}}-\left(\boldsymbol{\beta}+\hat{\mathbf{n}}\right)\left(\boldsymbol{\beta}\cdot\hat{\mathbf{n}}\right)+\boldsymbol{\beta}+\beta^{2}\hat{\mathbf{n}}-\frac{1}{c}R\left(\mathbf{\dot{\boldsymbol{\beta}}}\cdot\hat{\mathbf{n}}\right)\hat{\mathbf{n}}}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)^{3}} \]

que se puede simplificar a

\[ \nabla\left\{ \frac{\mu c}{R}\frac{e}{\left[1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right]}\right\} =\frac{\mu ce}{R^{2}}\frac{-\hat{\mathbf{n}}-\left(\boldsymbol{\beta}\cdot\hat{\mathbf{n}}\right)\boldsymbol{\beta}+\boldsymbol{\beta}+\beta^{2}\hat{\mathbf{n}}-\frac{1}{c}R\left(\mathbf{\dot{\boldsymbol{\beta}}}\cdot\hat{\mathbf{n}}\right)\hat{\mathbf{n}}}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)^{3}} \]

y al evaluar el producto cruz

\[ \nabla\left\{ \frac{\mu c}{R}\frac{e}{\left[1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right]}\right\} \times\boldsymbol{\beta}=\frac{\mu ce}{R^{2}}\frac{-\left(\hat{\mathbf{n}}\times\boldsymbol{\beta}\right)+\beta^{2}\left(\hat{\mathbf{n}}\times\boldsymbol{\beta}\right)-\frac{1}{c}R\left(\mathbf{\dot{\boldsymbol{\beta}}}\cdot\hat{\mathbf{n}}\right)\hat{\mathbf{n}}\times\boldsymbol{\beta}}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)^{3}} \]

que puede escribirse como

\[ \nabla\left\{ \frac{\mu c}{R}\frac{e}{\left[1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right]}\right\} \times\boldsymbol{\beta}=\frac{\mu ce}{R^{2}}\frac{-\left(1-\beta^{2}\right)\left(\hat{\mathbf{n}}\times\boldsymbol{\beta}\right)-\frac{1}{c}R\left(\mathbf{\dot{\boldsymbol{\beta}}}\cdot\hat{\mathbf{n}}\right)\hat{\mathbf{n}}\times\boldsymbol{\beta}}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)^{3}}\label{eq: ter grad A} \]

Por otro lado, el segundo término que involucra al rotacional de la velocidad

\[ \nabla\times\boldsymbol{\beta}=\left(\frac{\partial\beta_{z}}{\partial y}-\frac{\partial\beta_{y}}{\partial z}\right)\hat{i}+\left(\frac{\partial\beta_{x}}{\partial z}-\frac{\partial\beta_{z}}{\partial x}\right)\hat{j}+\left(\frac{\partial\beta_{y}}{\partial x}-\frac{\partial\beta_{x}}{\partial y}\right)\hat{k} \]

\[ \nabla\times\boldsymbol{\beta}=\left(\dot{\beta}_{z}\frac{\partial t_{r}}{\partial y}-\dot{\beta}_{y}\frac{\partial t_{r}}{\partial z}\right)\hat{i}+\left(\dot{\beta}_{x}\frac{\partial t_{r}}{\partial z}-\dot{\beta}_{z}\frac{\partial t_{r}}{\partial x}\right)\hat{j}+\left(\dot{\beta}_{y}\frac{\partial t_{r}}{\partial x}-\dot{\beta}_{x}\frac{\partial t_{r}}{\partial y}\right)\hat{k} \]

pero este resultado no es sino el producto cruz de la derivada de $\beta$ por el gradiente de $t_{r}$ \[ \nabla\times\boldsymbol{\beta}=-\dot{\boldsymbol{\beta}}\times\nabla t_{r} \]

Se sustituye el valor de $\nabla t_{r}$

\[ \nabla t_{r}=\frac{-\nabla R}{c}=\frac{-\hat{\mathbf{n}}}{c\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)} \]

de manera que el rotacional de la velocidad es

\[ \nabla\times\boldsymbol{\beta}=\frac{\dot{\boldsymbol{\beta}}\times\hat{\mathbf{n}}}{c\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)} \]

El segundo término es entonces

\[ \left\{ \frac{\mu c}{R}\frac{e}{\left[1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right]}\right\} \nabla\times\boldsymbol{\beta}=\left\{ \frac{\mu c}{R}\frac{e}{\left[1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right]}\right\} \frac{\dot{\boldsymbol{\beta}}\times\hat{\mathbf{n}}}{c\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)} \]

que puede reescribirse como

\[ \left\{ \frac{\mu c}{R}\frac{e}{\left[1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right]}\right\} \nabla\times\boldsymbol{\beta}=\frac{\mu e}{R}\frac{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)\dot{\boldsymbol{\beta}}\times\hat{\mathbf{n}}}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)^{3}}\label{eq: ter cruz Beta} \]

Recopilando los términos (\ref{eq: ter grad A}) y (\ref{eq: ter cruz Beta}) obtenemos el campo magnético

\[ \frac{\mu ce}{R^{2}}\frac{-\left(1-\beta^{2}\right)\left(\hat{\mathbf{n}}\times\boldsymbol{\beta}\right)-\frac{1}{c}R\left(\mathbf{\dot{\boldsymbol{\beta}}}\cdot\hat{\mathbf{n}}\right)\hat{\mathbf{n}}\times\boldsymbol{\beta}}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)^{3}}+\frac{\mu e}{R}\frac{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)\dot{\boldsymbol{\beta}}\times\hat{\mathbf{n}}}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)^{3}} \]

términos en B de velocidad

Si se agrupan los términos de velocidad

\[ \left.\mathbf{B}\right|_{vel}=\frac{\mu ce}{R^{2}}\frac{\left(1-\beta^{2}\right)\left(\boldsymbol{\beta}\times\hat{\mathbf{n}}\right)}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)^{3}}\label{eq: B vel lenard} \]

términos en B de aceleración

Mientras que los términos de aceleración son

\[ \left.\mathbf{B}\right|_{acel}=\frac{\mu e}{R}\frac{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)\dot{\boldsymbol{\beta}}\times\hat{\mathbf{n}}-\left(\mathbf{\dot{\boldsymbol{\beta}}}\cdot\hat{\mathbf{n}}\right)\hat{\mathbf{n}}\times\boldsymbol{\beta}}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)^{3}} \]

que puede reescribirse como

\[ \left.\mathbf{B}\right|_{acel}=\frac{\mu e}{R}\frac{-\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)\hat{\mathbf{n}}\times\dot{\boldsymbol{\beta}}-\left(\mathbf{\dot{\boldsymbol{\beta}}}\cdot\hat{\mathbf{n}}\right)\hat{\mathbf{n}}\times\boldsymbol{\beta}}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)^{3}} \]

\[ \left.\mathbf{B}\right|_{acel}=\frac{\mu e}{R}\frac{\hat{\mathbf{n}}\times\left[-\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)\dot{\boldsymbol{\beta}}-\left(\mathbf{\dot{\boldsymbol{\beta}}}\cdot\hat{\mathbf{n}}\right)\boldsymbol{\beta}\right]}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)^{3}}\label{eq: B acel lenard} \]

añado un término $\left(\mathbf{\dot{\boldsymbol{\beta}}}\cdot\hat{\mathbf{n}}\right)\hat{\mathbf{n}}$ que por el producto cruz es cero

\[ \left.\mathbf{B}\right|_{acel}=\frac{\mu e}{R}\frac{\hat{\mathbf{n}}\times\left[-\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)\dot{\boldsymbol{\beta}}+\left(\mathbf{\dot{\boldsymbol{\beta}}}\cdot\hat{\mathbf{n}}\right)\left(\hat{\mathbf{n}}-\boldsymbol{\beta}\right)\right]}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)^{3}} \]

y el término entre paréntesis cuadrados es igual al triple producto cruz

\[ \left.\mathbf{B}\right|_{acel}=\frac{\mu e}{R}\frac{\hat{\mathbf{n}}\times\left[\mathbf{\hat{\mathbf{n}}}\times\left\{ \left(\mathbf{\hat{\mathbf{n}}}-\boldsymbol{\beta}\right)\times\dot{\boldsymbol{\beta}}\right\} \right]}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)^{3}} \]

Entonces la relación entre el campo magnético y eléctrico es

\[ \mathbf{B}=\frac{1}{c}\left[\mathbf{\hat{\mathbf{n}}}\times\mathbf{E}\right]_{ret} \]

puesto que $\mu c=1/\varepsilon c$ y $\mu=1/\varepsilon c^{2}$. Los campos son entonces siempre ortogonales. Al agrupar los términos del campo magnético se obtiene

\[ \mathbf{B}=\mu ce\left[\frac{\left(\boldsymbol{\beta}\times\mathbf{\hat{\mathbf{n}}}\right)\left(1-\beta^{2}\right)}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)^{3}R^{2}}\right]_{ret}+\mu e\left[\frac{\mathbf{\hat{\mathbf{n}}}\times\left(\mathbf{\hat{\mathbf{n}}}\times\left\{ \left(\mathbf{\hat{\mathbf{n}}}-\boldsymbol{\beta}\right)\times\dot{\boldsymbol{\beta}}\right\} \right)}{\left(1-\boldsymbol{\beta}\cdot\mathbf{\hat{\mathbf{n}}}\right)^{3}R}\right]_{ret}.\label{eq: B lenard} \]