# Has the Helmholtz decomposition of the $\mathbf{E}$ field from the Liénard–Wiechert potentials been worked out?

If you look at Maxwell's equations for $\mathbf{E}(\mathbf{x},t)$ they split neatly into two categories. They are: \begin{align} \nabla\cdot\mathbf{E}(\mathbf{x},t)&=\frac{\rho(\mathbf{x},t)}{\epsilon_0},\ \mathrm{and} & \nabla\times\mathbf{E}(\mathbf{x},t) & = -\frac{\partial \mathbf{B}(\mathbf{x},t)}{\partial t}. \end{align} Examined in light of Helmholtz decomposition, these equations could be read as: electric charge produces irrotational electric fields, and changing magnetic fields produce solenoidal electric fields.

The electric field from the Liénard–Wiechert potential is given by equation 14.14 from Jackson's "Classical Electrodynamics" (3rd Ed) as $$\mathbf{E}_{LW}(\mathbf{x},t) = \frac{q}{4\pi\epsilon_0}\left[\frac{\mathbf{n}-\boldsymbol{\beta}}{\gamma^2(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3 R^2} \right]_{\mathrm{ret}} + \frac{q}{4\pi}\sqrt{\frac{\mu_0}{\epsilon_0}}\left[\frac{\mathbf{n}\times\left[(\mathbf{n}-\boldsymbol{\beta})\times \dot{\boldsymbol{\beta}}\right]}{(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3 R}\right]_{\mathrm{ret}},$$ where Jackson's emphasis is on the separation of radiation fields (second term) from the near field (first term).

Has someone already done the Helmholtz decomposition of $\mathbf{E}_{\mathrm{LW}}(\mathbf{x},t)$, and if so, what is it?

Equally useful for my purposes would be the Helmholtz decomposition of Jefimenko's equation for the electric field: \begin{align} \mathbf{E}_J(\mathbf{x},t) &= \frac{1}{4\pi\epsilon_0} \int \left[\left(\frac{\rho(\mathbf{x}',t_r)}{\left|\mathbf{x}-\mathbf{x}'\right|^3} + \frac{1}{c\left|\mathbf{x}-\mathbf{x}'\right|^2} \frac{\partial \rho(\mathbf{x}',t_r)}{\partial t}\right)(\mathbf{x}-\mathbf{x}') - \frac{1}{|\mathbf{x}-\mathbf{x}'|c^2} \frac{\partial\mathbf{J}(\mathbf{x}',t_r)}{\partial t}\right]\operatorname{d}^3x' \\ t_r&\equiv t-\frac{|\mathbf{x}-\mathbf{x}'|}{c}. \end{align}

A big part of the reason for being interested in the HD of $\mathbf{E}$ using either of the above formulae for it is because the separation is trivial in the Coulomb gauge because when $\mathbf{A}$ is solenoidal it doesn't contribute to the irrotational part of $\mathbf{E}$. Thus \begin{align} \mathbf{E}_{\mathrm{irrot}}(\mathbf{x},t) & = \frac{1}{4\pi\epsilon_0} \int \frac{\mathbf{x} - \mathbf{x}'}{|\mathbf{x} - \mathbf{x}'|^3} \rho(\mathbf{x}',t)\operatorname{d}^3x', \ \mathrm{and} \\ \mathbf{E}_{\mathrm{sol}}(\mathbf{x},t) & = \frac{1}{4\pi\epsilon_0} \nabla \times \int \left[\int_0^{|\mathbf{x}-\mathbf{x}'|/c} t' \frac{\mathbf{x}-\mathbf{x}'}{\left|\mathbf{x}-\mathbf{x}'\right|^3} \times \dot{\mathbf{J}}(\mathbf{x}',t-t') \operatorname{d}t'\right] \operatorname{d}^3x', \end{align} and I cannot see what property of $\mathbf{E}_{\mathrm{sol}}(\mathbf{x},t)$ allows it to cancel the apparent instantaneous nature of $\mathbf{E}_{\mathrm{irrot}}(\mathbf{x},t)$.

• May be I'am wrong but I think that the HD-problem of $\:\mathbf{E}_{_{\rm LW}}\:$ is equivalent (must be restricted) to the HD-problem of $\:\mathbf{A}_{_{\rm LW}}$, the last being the Liénard–Wiechert vector potential, since if $$\mathbf{A}_{_{\rm LW}}=\boldsymbol{-}\boldsymbol{\nabla}\psi+\boldsymbol{\nabla}\boldsymbol{\times}\boldsymbol{\Psi} \tag{01}$$ – Frobenius Dec 29 '17 at 22:35
• then $$\mathbf{E}_{_{\rm LW}}=\boldsymbol{-}\boldsymbol{\nabla}\phi_{_{\rm LW}}\boldsymbol{-}\dfrac{\partial \mathbf{A}_{_{\rm LW}}}{\partial t}=\boldsymbol{-}\boldsymbol{\nabla}\left(\phi_{_{\rm LW}}-\dfrac{\partial \psi}{\partial t}\right)\boldsymbol{+}\boldsymbol{\nabla}\boldsymbol{\times}\left(\boldsymbol{-}\dfrac{\partial \boldsymbol{\Psi}}{\partial t}\right) \tag{02}$$ .... – Frobenius Dec 29 '17 at 22:45
• May be I'am wrong but I think that the HD-problem of $\:\mathbf{E}_{_{\rm LW}}\:$ could be solved if we could solve the HD-problem of $\:\dfrac{\partial \mathbf{A}_{_{\rm LW}}}{\partial t}\:$ or $\:\mathbf{A}_{_{\rm LW}}\:$ the last being the Liénard–Wiechert vector potential (possibly the two problems are not equivalent, as I think in a previous comment of mine). – Frobenius Dec 29 '17 at 23:13