# Poynting vector from 1st term in Lienard-Wiechert field

$$\vec{E} = q (1-v^2) \frac{\vec{R_{t'}} - \vec{v}R_{t'}}{(R_{t'} - \vec{v}\vec{R_{t'}})^3}$$

$$\vec{H} = - q (1-v^2) \frac{\vec{R_{t'}} \times \vec{v}}{(R_{t'} - \vec{v}\vec{R_{t'}})^3}$$

for particle with charge $q$ and constant velocity $v$, where $R_{t'}$ is radius-vector from position of a particle to observer at retarded moment $t'$.

The particle moves along $z$-axis with $x=0$, $y=0$.

These vectors can be written using present time $t$:

$$\vec{E} = q (1-v^2) \frac{\vec{R}}{(R^*)^3}$$

$$\vec{H} = - q (1-v^2) \frac{\vec{R} \times \vec{v}}{(R^*)^3}$$

where

$$R^* = \sqrt{(1-v^2)(x^2 + y^2) + (z-vt)^2}$$ and everything is taken at $t$.

So, we have cylindrical symmetry and we can parametrize:

$$x = r \cos{\phi}$$ $$y = r \sin{\phi}$$

then

$$\vec{E} = q (1-v^2) \frac{1}{(R^*)^3} (r\cos{\phi} \vec{e_x} + r\sin{\phi} \vec{e_y} + (z-vt) \vec{e_z})$$

$$\vec{H} = - q (1-v^2) \frac{1}{(R^*)^3} (vr\cos{\phi} \vec{e_x} - vr\sin{\phi} \vec{e_y} + 0 \vec{e_z})$$

Poynting vector $\vec{S} = [\vec{E} \times \vec{H}]$ is (in cartesian coordinates):

$$\vec{S} = \frac{q^2(1-v^2)^2}{(R^*)^6} (-vr\cos{\phi} (z-vt) \vec{e_x} - vr\sin{\phi}(z-vt) \vec{e_y} + vr^2 \vec{e_z})$$

Finally, in cylindrical coordinates we have:

$$\vec{S} = \frac{q^2(1-v^2)^2}{(R^*)^6} (-v(z-vt) r \vec{e_r} + vr^2 \vec{e_z})$$

The fact that $r$-component of $\vec{S}$ has different signs in different space-points is strange for me, because for some space-regions $\vec{S}$ directs towards the particle. Of course, this term is non-radiative, but I have some doubt in this expression for $\vec{S}$.

Question: If my expression is right, how to explain it?

• ...and your question is? – ACuriousMind Apr 16 '15 at 16:02
• If my expression is right, how to explain it? – newt Apr 16 '15 at 16:08

Ah, yes, I think I got an answer. With moving particle energy also moves in the same direction. It was strange for me that we have $r$-component of $\vec{S}$ but it seems to be normal.