Few question in retarded fields E and B

[yungman]

Given from the book:

$$\vec E_{(\vec r, t)} = \frac q {4\pi\epsilon_0}\frac {\eta}{(\vec{\eta}\cdot\vec u)^3} [(c^2-v^2)\vec u + \vec{\eta} \times (\vec u \times \vec a)]$$ (10.65)

$$\vec B_{(\vec r,t)} =\nabla \times \vec A_{(\vec r,t)} =\frac 1 c \frac q {4\pi\epsilon_0}\frac {\eta}{(\vec{\eta}\cdot\vec u)^3} \vec {\eta} \times [(c^2-v^2)\vec v + (\vec{\eta} \cdot \vec a)\vec v + (\vec{\eta} \cdot \vec u)\vec a]$$ (10.66)

Where $$\vec{\eta} = \vec r -\vec w(t_r)$$ is the distance vector from the source point charge pointed by position vector $$\vec w (t_r)$$ at retarded time, to the field point pointed by position vector $$\vec r$$.

$$\vec u=c\hat{\eta}-\vec v\;,\; \vec v = \frac { d \vec w(t_r)}{dt_r} \;\hbox { is the velocity of point charge and retarded time.$$


Below I copy straight from the book and I will point out my question at the end:


[BOOK]

The quantity in brackets of (10.66) is strikingly similar to the one in (10.55) , which can be written, using BAC-CAB rule as $$[(c^2-v^2)\vec u + (\vec{\eta}\cdot\vec a)\vec u-(\vec{\eta}\cdot\vec u)\vec a]$$; the main difference is that we have $$\vec v$$ instead of $$\vec u$$ in the first two terms. In fact since it's all crossed into $$\vec{\eta}$$ anyway, we can with impunity change these $$\vec v$$ into $$-\vec u$$; the extra term proportional to $\hat {\eta}$ disappears in the crass product.

The first term in $$\vec E$$ in (10.65) (the one involve $$(c^2-v^2)\vec u$$) falls off as the inverse square of the distance from the particle.

The second term in (10.65) (the one involving $$\vec{\eta}\times(\vec u \times \vec a)$$) falls off as the inverse first power of $\eta$ is therefore dominant at large distance.

[END BOOK]


My question is in Blue, Green and Orange. I just don't get these. Can anyone explain these to me.

Thanks

Alan

 

[yungman]

To clarify the three question I copied from the first post in Blue, Lime and Orange:



1) In fact since it's all crossed into $$\vec{\eta}$$ anyway, we can with impunity change these $$\vec v$$ into $$-\vec u$$; the extra term proportional to $\hat {\eta}$ disappears in the crass product.

I don’t understand the above at all.








2) The first term in $$\vec E$$ in (10.65) (the one involve $$(c^2-v^2)\vec u$$) falls off as the inverse square of the distance from the particle.

Is the above imply using the first part of the equation of (10.65) below that it is proportion to $$\frac 1 {\eta^2}$$

$$\vec E_{(\vec r, t)} = \frac q {4\pi\epsilon_0}\frac {\eta}{(\vec{\eta}\cdot\vec u)^3} [(c^2-v^2)\vec u$$








3) The second term in (10.65) (the one involving $$\vec{\eta}\times(\vec u \times \vec a)$$) falls off as the inverse first power of $\eta$ is therefore dominant at large distance.

Does this imply $$\vec E$$ proportion to $$\frac 1 {\eta}$$ as show below?

$$\vec E_{(\vec r, t)} = \frac q {4\pi\epsilon_0}\frac {\eta}{(\vec{\eta}\cdot\vec u)^3} [\vec{\eta} \times (\vec u \times \vec a)]$$


Thanks

Alan