Physics Help Forum Questions about retarded field E and B

 May 13th 2011, 09:38 AM #1 Member   Join Date: Nov 2010 Posts: 40 Questions about retarded field E and B I have a few question from the book. I am going to split into multiple posts due to the limitation of the number of images per post. These are the formulas from the book which I understand: $\vec E_{(\vec r, t)} = \frac q {4\pi\epsilon_0}\frac {\eta}{(\vec{\eta}\cdot\vec u)^3} [(c^2-v^2)\vec u + \vec{\eta} \times (\vec u \times \vec a)]$ (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 $\vec{\eta} = \vec r -\vec w(t_r) \;\hbox { is the distance vector from the source point to field point}, \vec w (t_r) }\hbox { points to the source point},\vec r \;\hbox { points to the field point.}$ $\vec u=c\hat{\eta}-\vec v\;,\; \vec v = \frac { d \vec w(t_r)}{dt_r} \;\hbox { is the velocity of point charge at retarded time.$ Last edited by yungman; May 13th 2011 at 09:51 AM.
 May 13th 2011, 09:48 AM #2 Member   Join Date: Nov 2010 Posts: 40 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 (10.65), 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 v instead of u in the first two terms. 1) In fact since it's all crossed into $\vec {\eta}$ anyway, we can with impunity change these v into -u; the extra term proportional to $\hat {\eta}$ disappears in the crass product. 2) The first term in E in (10.65) ( the one involve (c^2-v^2)u) falls off as the inverse square of the distance from the particle. 1) The second term( the one involving $\vec{\eta}\times(\vec u \times \vec a)\;\hbox { falls off as the inverse first power of }\;\eta$ is therefore dominant at large distance. [END BOOK] Last edited by topsquark; May 13th 2011 at 12:38 PM. Reason: Removed the color from line 2.
 May 13th 2011, 09:49 AM #3 Member   Join Date: Nov 2010 Posts: 40 These are my questions, the numbers correspond to the number of the last post: 1) I have no ideas what the book said. 2) Is this imply using the first part of the equation of (10.65) shown 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) Does this imply E proportion to $\frac 1 {\eta}$ as shown 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
May 13th 2011, 12:50 PM   #4

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 Originally Posted by yungman These are my questions, the numbers correspond to the number of the last post: 1) I have no ideas what the book said.
$\begin{array}{lclcl} \vec{u} = c \hat{\eta} - \vec{v} \\ . \\ \vec{\eta} \times \vec{u} = \vec{\eta} \times c \hat{\eta} - \vec{\eta} \times \vec{v} \\ . \\\vec{\eta} \times \vec{u} = -\vec{\eta} \times \vec{v} \end{array}$

so we might as well just sub in -v for u.

 Originally Posted by yungman 2) Is this imply using the first part of the equation of (10.65) shown 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) Does this imply E proportion to 1/(eta) as shown below? $\vec E_{(\vec r, t)} = \frac q {4\pi\epsilon_0}\frac {\eta}{(\vec{\eta}\cdot\vec u)^3} [\vec{\eta} \times (\vec u \times \vec a)]$
Yes.

-Dan
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 May 13th 2011, 02:40 PM #5 Member   Join Date: Nov 2010 Posts: 40 Thanks, I really appreciate your help. Last edited by yungman; May 13th 2011 at 02:42 PM.
May 19th 2011, 06:12 AM   #6
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 Originally Posted by topsquark $\begin{array}{lclcl} \vec{u} = c \hat{\eta} - \vec{v} \\ . \\ \vec{\eta} \times \vec{u} = \vec{\eta} \times c \hat{\eta} - \vec{\eta} \times \vec{v} \\ . \\\vec{\eta} \times \vec{u} = -\vec{\eta} \times \vec{v} \end{array}$ -Dan

What is the physical meaning of u?

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