Help in explaining what the book said in fields of a moving point charge.

[yungman]

My questions are what the book said after working out the solution that I have no issue of finding. The original question is to find E and B at a field point pointed by r due to a moving point charge pointed by w(t_r) at retard time moving at a constant velocity v.


For constant velocity:

\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 ]

Where

\vec{\eta} = \vec r -\vec w(t_r), \; \vec u=c\hat{\eta}-\vec v\;,\; \vec v = \frac { d \vec w(t_r)}{dt_r}

I have no problem finding the answer:

\vec E_{(\vec r, t)} = \frac q {4\pi\epsilon_0} \frac {1-\frac{v^2}{c^2}} {\left (1-\frac{v^2}{c^2} sin^2\theta \right )^{\frac 3 2}} \frac {\hat R}{R^2} \;\hbox { where }\; \vec R = \vec r –t\vec v





Below are the three statements directly quoted by the book and I have no idea what they mean:

[BOOK]

1) Because of the sin^2\theta in the denominator, the field of a fast-moving charge is flattened out like a pancake in the direction perpendicular to the motion.

2) In the forward and backward directions, E is reduced by a factor (1 - \frac {v^2}{c^2} ) relative to the field of a charge at rest;

3) In the perpendicular direction it is enhanced by a factor \frac 1 {\sqrt { \left (1 - \frac {v^2}{c^2}\right )}}

[END BOOK]







Also in finding B

\vec B =\frac 1 c \hat{\eta}\times \vec E = \frac 1 c \hat{\eta} \times \frac {q\left ( 1-\frac {v^2}{c^2}\right )}{4\pi\epsilon_0\left ( 1-\frac {v^2}{c^2} sin^2\theta \right )^{\frac 3 2}} \frac {\vec R}{R^3}

In two different ways, I get different answer.


1)
\hat {\eta} \times \vec R = \frac {1}{\eta} [(\vec r -t_r \vec v)\times(c\frac{\vec{\eta}}{\eta}-t\vec v)]=\frac {1}{\eta}[-\frac {ct_r}{\eta} \vec r\times \vec v -t(\vec r \times \vec v) -\frac{ct_r}{\eta}(\vec v\times \vec r)]=-\frac {t} {\eta} (\vec r \times \vec v)

\Rightarrow\;\vec B = \frac 1 c \frac {q\left ( 1-\frac {v^2}{c^2} \right )}{4\pi\epsilon_0\eta \left ( 1-\frac {v^2}{c^2} sin^2\theta\right )^{\frac 3 2}R^3}\;[ t (\vec v \times \vec r)]



2)
\hat{\eta}=\frac{\vec r-t_r\vec v}{\eta}=\frac {(\vec-t\vec v)+(t-t_r)\vec v}{\eta}=\frac {\vec R}{\eta} + \frac {\vec v}{c} \Rightarrow\; \vec B = \frac 1 c \frac {q\left ( 1-\frac {v^2}{c^2} \right )}{4\pi\epsilon_0\eta \left ( 1-\frac {v^2}{c^2} sin^2\theta\right )^{\frac 3 2}R^3} \;(\vec v \times \vec r)


Notice a difference of t between the two method? I cannot resolve this.

 

[Bill_K]

Why are you posting this twice? This is identical to your post of a week ago, https://www.physicsforums.com/showthread.php?t=500092

 

[yungman]

Anyone please, even if you don't have the answer, give me a link or point me to materials that explain more on this. I have been stuck for like two weeks regarding to the electric field of the moving charge when moving at high rate of speed.

Thanks

Alan