Electric Field Directly Ahead of or Behind a Moving Charge

[shinobi20]

Then indeed
$$\vec{E}=\frac{q \gamma}{4 \pi} \frac{\vec{x}-\vec{v} t}{[(u \cdot x)^2-x \cdot x]^{3/2}}=\frac{q}{4 \pi \gamma^2 \epsilon^2} \frac{\vec{\beta}}{\beta}.$$
Thus there's an additional factor $1/\gamma^2$ compared to the expression at rest.

The electric field of a charge moving with velocities close to $c$ is sharply peaked in transverse direction (with the longitudinal component suppressed by the said factor $1/\gamma^2=1-\beta^2$.

Just a clarification, there's an additional factor $1/\gamma^2$ compared to the expression at rest with respect to the unprimed frame, i.e.,

$E_x = q\frac{1}{\gamma^2 (x-x_q)^2}~$ compare with $~E_x = q\frac{1}{(x-x_q)^2}~$ (E-field of a charge at rest in the unprimed frame)

if there is a charge at rest in the unprimed frame (although in this case the charge is at rest in the primed frame).

Compared to the expression at rest with respect to the primed frame (which is the case in Ohanian) THEY ARE THE SAME, i.e.,

$E_x = q\frac{1}{\gamma^2 (x-x_q)^2} = q\frac{1}{(x')^2}~$ compare with $~E'_x = q\frac{1}{(x')^2}$

since $x' = \gamma (x - x_q)$.

This is correct right?

 

[TSny]

Here's how Ohanian described it in the first edition of his text:

 

[vanhees71]

I don't understand this. In any textbook you find the important result that the electric field for a relativistically moving point charge is almost transverse. The important point is that you must use all quantities in one frame. In my notation the charge is at rest in the primed frame, and there you have
$$\vec{E}'(t',\vec{x}')=\frac{q}{4 \pi r^{\prime 3}} \vec{e}_{3}'.$$
The field in the original frame, where it moves with velocity $\vec{v}=v \vec{e}_1$ is given by (written in 3D-notation)
$$\vec{E}(t,\vec{x}) = \frac{q}{4 \pi} \frac{\gamma (\vec{x}-v t \vec{e}_1)}{[\gamma^2(x^1-v t)^2+(x^2)^2+(x^3)^2]^{3/2}}.$$
The magnetic field turns out to be
$$\vec{B}=\vec{\beta} \times \vec{E}.$$
This you get either by Lorentz boosting the fields (and writing everything, i.e., the spacetime arguments and the fields with the components in the respetive frame) or just using the Lienard-Wiechert potentials for the uniformly point charge (as given in my SRT FAQ [1]).

Note that the above given result is identical with (3.6.9) in my FAQ since the expression in the denominator is
$$(u \cdot x)^2-x \cdot x=\gamma^2 (ct-v x^1)^2-(c t)^2+\vec{x}^2=\gamma^2(x^1-v t)^2 + (x^2)^2 + (x^3)^2.$$

The text by Ohanian is misleading, because he doesn't take into account that he as to express what he calls "proper distance" in terms of the (in his notation) unprimed coordinates!

[1] https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

 

[pervect]

Just a clarification, there's an additional factor $1/\gamma^2$ compared to the expression at rest with respect to the unprimed frame, i.e.,

$E_x = q\frac{1}{\gamma^2 (x-x_q)^2}~$ compare with $~E_x = q\frac{1}{(x-x_q)^2}~$ (E-field of a charge at rest in the unprimed frame)

if there is a charge at rest in the unprimed frame (although in this case the charge is at rest in the primed frame).

Compared to the expression at rest with respect to the primed frame (which is the case in Ohanian) THEY ARE THE SAME, i.e.,

$E_x = q\frac{1}{\gamma^2 (x-x_q)^2} = q\frac{1}{(x')^2}~$ compare with $~E'_x = q\frac{1}{(x')^2}$

since $x' = \gamma (x - x_q)$.

This is correct right?

Depending on what you are comparing, I believe it's correct. Let us specify exactly what you are comparing. You are comparing the field one unit distance (say 1 meter) away from the charge as measured in the unprimed frame to the field, one meter away, from the charge as measured in the primed frame.

The relationship $E_{\parallel} = E'_{\parallel}$ is also a perfectly correct relationship in the correct context. You don't seem to be understanding correctly the context in which this relationship is correct, because you're more focused on a different context. The context in which $E_{\parallel} = E'_{\parallel}$ is correct is that we are comparing the electric field as the same event as seen by two different observers.

Now we ask, "is that an event 1 unit distance unit (say 1 meter) away from the charge q in the unprimed frame is not the same event as an event 1 meter away from the charge q in the primed frame.". The answer is no, they are different events.

When you suggested that "Lorentz contraction" was the explanation for the factor of gamma discrepancy, way back near the beginning of the post, you were basically on the right track. The distance changes by a factor of gamma due to Lorentz contraction, and the field is inversely proportional to the square of the distance, so there is your factor of gamma^2.

You seem to be trying to convince yourself there is a typo in your textbook, rather than figuring out where you went wrong :(.

Getting the details right and understanding the direction of the correction factor of $\gamma^2$ - is it multiplicative, or does it divide? - takes some more attention , which is what I tried to outline in my post #30, and other posters have outlined other mathematical approaches.

There's a graphical approach based on the "field lines" drawn by Ohanian as quoted by another poster in post #32. With the graphical approach, the field strength is proportional to the density of the "field lines" in the transverse direction. See for instance https://en.wikipedia.org/wiki/Field_line.

An individual field line shows the direction of the vector field but not the magnitude. In order to also depict the magnitude of the field, a selection of field lines can be drawn such that the density of field lines (number of field lines per unit perpendicular area) at any location is proportional to the magnitude of the vector field at that point.

It's hard to get the numbers from the field line approach, especially since you'd need a 3-d diagram as wiki also mentions, but the field line expression gives the correct qualitative answer.

 

[vanhees71]

Again, please make sure to also write down the arguments. The complete notation of the above statement is
$$E_{\parallel}'(x')=E_{\parallel}(x),$$
i.e. with the Lorentz transformation matrix $\hat{\Lambda}$, $x'=\hat{\Lambda} x$ you have
$$E_{\parallel}'(x')=E_{\parallel}(\hat{\Lambda}^{-1} x').$$