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I'm working on a revision of the absorber theory of radiation proposed by Wheeler and Feynman on their paper "Interaction with the Absorber as the Mechanism of Radiation". On page 161 they say the retarded field of the source reduces to \begin{equation} -\left(\frac{e\mathfrak{U}}{r_k c^2}\right) \sin(\mathfrak{U},r_k) \end{equation} together with a term of electrostatic origin. My problem is that I don't completely understand where this terms comes from.

Using the Liénard-Wiechert potentials, I get that the field for a point charge with arbitrary motion is

\begin{equation} \mathbf{E} = e \left[\left(\frac{\hat{\mathbf{n}}-\boldsymbol{\beta}}{\gamma^2 R^2 (1-\hat{\mathbf{n}} \cdot \boldsymbol{\beta})^3} \right) + \frac{1}{c} \frac{\hat{\mathbf{n}}\times\left((\hat{\mathbf{n}}-\boldsymbol{\beta})\times\dot{\boldsymbol{\beta}}\right)}{R(1-\hat{\mathbf{n}} \cdot \boldsymbol{\beta})^3 } \right] \end{equation}

We're interested in the limit $\beta\ll 1$, so we are left with

\begin{equation} \mathbf{E} = e \left[\frac{\hat{\mathbf{n}}}{\gamma^2 R^2 } + \frac{1}{c} \frac{\hat{\mathbf{n}}\times\left(\hat{\mathbf{n}}\times\dot{\boldsymbol{\beta}}\right)}{R } \right] = \frac{e \hat{\mathbf{n}}}{\gamma^2 R^2 } + \frac{e\hat{\mathbf{n}}\times\left(\hat{\mathbf{n}}\times\mathfrak{U}\right)}{R c^2} \end{equation}

Does this mean that $\hat{\mathbf{n}}\times\left(\hat{\mathbf{n}}\times\mathfrak{U}\right)$ is equal to $-\mathfrak{U}\sin(\mathfrak{U},r_k)$ or am I doing something wrong?

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$-\hat n \times (\hat n \times \vec{\mathfrak{U}})$ is the component of $\vec{\mathfrak{U}}$ perpendicular to $\hat n$ and has a length $\mathfrak{U}\sin(\hat n \cdot \vec{\mathfrak{U}})$, so it appears they are just giving the value of this component.

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