Question in Electric field of a moving point charge

[yungman]

The electric field at a point P pointed by the position vector $\vec r \;\hbox {, from a moving point charge pointed by a position vector }\;\vec w(t_r)\;$ at the retarded time is given by:

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

$$\hbox{Where }\; \vec {\eta}=\vec r -\vec w(t_r) \;,\;\; \vec u = c\hat{\eta}-\vec v\;,\;\; \vec v =\frac{d \vec w(t_r)}{dt} \;\hbox { is the velocity vector of point charge } \;,\; \vec a = \frac{d \vec v}{dt}$$

My question is what is the physical meaning of $$\vec u = c\hat{\eta}-\vec v$$

 

[Bill_K]

There may be some misprints in the above, the units don't seem to be consistent. Anyway, η is supposed to be the position vector of the field point relative to the retarded position of the particle, and u (up to an added factor of c) is supposed to be the position of the field point relative to the current particle position. 'Current position' meaning that you blithely ignore the fact that the particle is accelerating, and extrapolate to where you anticipate it would be at the present time if it had just kept moving with constant velocity.

 

[Meir Achuz]

The formula looks right to me, but it is not in a familiar notation.
I don't think u has any physical significance, but is just a short hand way of writing the formula instead of keeping \eta and v separately.

 

[yungman]

Thank for the answer. I have more questions regarding to an example.

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:

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





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.

 

[Born2bwire]

Thank for the answer. I have more questions regarding to an example.

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:

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





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]

Just plot it out. It's obvious that at zero velocity the field is uniformly distributed. As the charge increases in velocity (ignoring the effects of acceleration) then we see that the field distribution compresses along the direction of propagation and increases in the plane normal to the direction of propagation. The other two are simply the limit conditions.

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


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

No.

 

[yungman]

Just plot it out. It's obvious that at zero velocity the field is uniformly distributed. As the charge increases in velocity (ignoring the effects of acceleration) then we see that the field distribution compresses along the direction of propagation and increases in the plane normal to the direction of propagation. The other two are simply the limit conditions.




No.

Thanks for your reply,


Can you explain a little bit more in detail about why in higth rate of speed, the electric field is perpendicular to the direction of the charge particle? I understand in stationary situation, the field is a central field that radiates out to all direction...like a ball. Aas speed foes up, the field at the front get compressed. But I cannot see how the field become like a circular disk with the normal being the direction of the moving charge.


Regarding to the magnetic field, what do you mean by "No"? I had a typo and I corrected it in my original post, you can see there is a difference of t in the two method.

 

[yungman]

Anyone please? Something is very wrong with this site. The Latex is so slow!

 

[Drakkith]

Anyone please? Something is very wrong with this site. The Latex is so slow!

Huh?

 

[yungman]

Huh?

I resolve the problem. I have to get Firefox 4 and it is a lot better now.

So anyone can help my questions? Please!

Alan

 

[Born2bwire]

Like I said, just plot it out. Plot the function,

$$f(\theta) = \frac{1-\alpha}{1-\alpha\sin\theta}$$

As a polar plot and adjust \alpha from 0 to 1. But we can already see that for alpha that is non-zero that the function approaches unity at \theta = \pi/2. So at \theta = 0, the function is 1-\alpha and at \theta = \pi/2, the function is 1. So the more relativistic you travel, the smaller the function becomes at \theta = 0. So assuming that this is a simple smooth curve, we can infer that the function increases to unity as we approach \pi/2. So as we go faster, the field becomes compressed along the direction of propagation (\theta = 0).