- #1
yungman
- 5,742
- 290
Given from the book:
[tex]\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)] [/tex] (10.65)
[tex]\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] [/tex] (10.66)
Where [tex]\vec{\eta} = \vec r -\vec w(t_r)[/tex] is the distance vector from the source point charge pointed by position vector [tex]\vec w (t_r)[/tex] at retarded time, to the field point pointed by position vector [tex]\vec r[/tex].
[tex]\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.[/tex]
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 [tex][(c^2-v^2)\vec u + (\vec{\eta}\cdot\vec a)\vec u-(\vec{\eta}\cdot\vec u)\vec a][/tex]; the main difference is that we have [tex]\vec v[/tex] instead of [tex]\vec u[/tex] in the first two terms. In fact since it's all crossed into [tex]\vec{\eta}[/tex] anyway, we can with impunity change these [tex]\vec v[/tex] into [tex]-\vec u[/tex]; the extra term proportional to [itex]\hat {\eta} [/itex] disappears in the crass product.
The first term in [tex]\vec E[/tex] in (10.65) (the one involve [tex](c^2-v^2)\vec u [/tex]) falls off as the inverse square of the distance from the particle.
The second term in (10.65) (the one involving [tex]\vec{\eta}\times(\vec u \times \vec a)[/tex]) falls off as the inverse first power of [itex]\eta[/itex] 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
[tex]\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)] [/tex] (10.65)
[tex]\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] [/tex] (10.66)
Where [tex]\vec{\eta} = \vec r -\vec w(t_r)[/tex] is the distance vector from the source point charge pointed by position vector [tex]\vec w (t_r)[/tex] at retarded time, to the field point pointed by position vector [tex]\vec r[/tex].
[tex]\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.[/tex]
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 [tex][(c^2-v^2)\vec u + (\vec{\eta}\cdot\vec a)\vec u-(\vec{\eta}\cdot\vec u)\vec a][/tex]; the main difference is that we have [tex]\vec v[/tex] instead of [tex]\vec u[/tex] in the first two terms. In fact since it's all crossed into [tex]\vec{\eta}[/tex] anyway, we can with impunity change these [tex]\vec v[/tex] into [tex]-\vec u[/tex]; the extra term proportional to [itex]\hat {\eta} [/itex] disappears in the crass product.
The first term in [tex]\vec E[/tex] in (10.65) (the one involve [tex](c^2-v^2)\vec u [/tex]) falls off as the inverse square of the distance from the particle.
The second term in (10.65) (the one involving [tex]\vec{\eta}\times(\vec u \times \vec a)[/tex]) falls off as the inverse first power of [itex]\eta[/itex] 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
Last edited: