Manner of energy exchange between an accelerating charge and the fields it produce The fields produced by a charged particle in accelerated motion produces an electric and magnetic field given by $${\vec E}(\vec r,t)=\frac{q}{4\pi\varepsilon_0}\left[\frac{(\hat{n}-{\vec \beta})}{\gamma^2(1-{\vec\beta}\cdot\hat{n})^3R^2}+\frac{\hat{n}\times\left[(\hat{n}-\vec\beta)\times{\dot{\vec\beta}}\right]}{c(1-{\vec\beta}\cdot\hat{n})^3R}\right]_{\rm ret}={\vec E}_{v}+{\vec E}_a.\\
{\vec B}_{v}=\hat{n}_{\rm ret}\times{\vec E}_v,~{\vec B}_{a}=\hat{n}_{\rm ret}\times{\vec E}_a$$ where ${\vec E}_{v}, {\vec B}_{v}$ and ${\vec E}_a, {\vec B}_{a}$ are called velocity and acceleration fields, respectively.
Let me quotes a few lineas from Griffiths:

... the velocity felds do carry energy—they just don’t transport it out to infnity. As the particle accelerates and decelerates, energy is exchanged between it and the velocity felds, at the same time as energy is irretrievably radiated away by the acceleration felds.

I want to understand this statement.
Because the acceleration field carries energy off to infinity while the velocity field does not, the energy carried by the acceleration field is irretrievably lost. However, there is something I do not understand.
He seems to be saying that there can be an exchange of energy from the charge to the velocity field and vice-versa. The energy transfer is back-and-forth in this case.  However, there is no such back-and-forth of energy between the charge and the acceleration field. There can only be a one-way transfer of energy from the charge to the acceleration field but not from the acceleration field back to the charge. Why is this so? How can we understand this?
 A: The velocity field is a local field "attached" to the particle. Since it never really leaves, you can think of a generalized "dressed" particle consisting of the particle itself and its local field. The energy within this "dressed" particle can be converted between the velocity field and the kinetic energy of the particle. But the more important thing is that the dressed particle is losing energy to the acceleration field.
The acceleration field is a radiative field that escapes to infinity. This energy is truly lost to the system. Quite literally, the accelerating charge is generating light, and this light carries energy away.
However, this energy is not irreversibly lost -- this only appears to be the case due to the boundary conditions of this particular problem. It is certainly possible for incoming radiation to interact with the charged particle. To give a concrete example, supposed we had two widely separated accelerating charges. Then the acceleration field of particle 1 will cause energy to leave from particle 1. But some of this acceleration field can interact with particle 2 and impart energy to it.
