If the charge oscillates at relativistic velocities

[arcTomato]

Hello PF.
I'm just curious.
I found the following description in a textbook I am reading.

If the charge oscillates at nonrelativistic velocities, $u<<c$, then we may neglect magnetic forces, since $E = B$ for an electromagnetic wave.

What I'm interested in here is what would $E$ and $B$ look like if the charge was oscillating at close to the speed of light ( means relativistic velocities) ?

Thank you.

 

[vanhees71]

I guess you are considering a charge in an external plane em. wave. The latter indeed has $|\vec{E}|=|\vec{B}|$ (when using good old Gaussian or Heaviside-Lorentz units ;-)). Now the force on the charge is given by the Lorentz formula, i.e.,
$$\vec{F}=q \left (\vec{E} + \frac{\vec{v}}{c} \times \vec{B} \right).$$
thus the force due to the magnetic field is by $|\vec{v}/c| \ll 1$ smaller for non-relativistic velocities.

For relativistic velocities you must use the relativistic equation of motion and then the electric and magnetic forces get of the same order of magnitude, because then $|\vec{v}/c| \lesssim 1$ (it's always smaller than 1 but can get close to one if $|\vec{v}|$ comes close to $c$).

 

[arcTomato]

Sorry for the late reply.

I guess you are considering a charge in an external plane em. wave.

I am thinking about that very situation.

I see. Thank you for the detailed explanation.

The latter indeed has |E→|=|B→| (when using good old Gaussian or Heaviside-Lorentz units ;-)).

However, this part bothered me a little.
From which equation exactly do you get that the electric field is equal to the magnetic field?

 

[vanhees71]

$$\newcommand{\ii}{\mathrm{i}}$$

Just use the Maxwell equations for $\rho=0$ and $\vec{j}=0$. Then you get (in Heaviside-Lorentz units with $c=1$)
$$\vec{\nabla} \times \vec{E}=-\partial_t \vec{B}, \quad \vec{\nabla} \cdot \vec{B}=\vec{\nabla} \cdot \vec{E}=0, \quad \vec{\nabla} \times \vec{B}=\partial_t \vec{E}.$$
Taking the curl of the first equation and then the other Maxwell equations leads to
$$(\partial_t^2-\Delta) \vec{E}=0.$$
Now make the plane-wave ansatz
$$\vec{E}(t,\vec{x})=\vec{E}_0 \exp[-\ii \omega t+\mathrm{i} \vec{k} \cdot \vec{x}].$$
Plugging this in the wave equation yields
$$\omega=|\vec{k}|$$
From $\vec{\nabla} \cdot \vec{E}=0$ you get $\vec{k} \cdot \vec{E}_0=0$.
Further it's clear that also
$$\vec{B}(t,\vec{x})=\vec{B}_0 \exp[-\ii \omega t+\mathrm{i} \vec{k} \cdot \vec{x}].$$
With Faraday's Law you get
$$\partial_t \vec{B}=-\ii \omega \vec{B}=-\vec{\nabla} \times \vec{E}=-\ii \vec{k} \times \vec{E},$$
i.e.,
$$\vec{B}=\frac{\vec{k}}{\omega} \times \vec{E}.$$
Since $\vec{k} \perp \vec{E}$ you get
$$|\vec{B}|=\left |\frac{\vec{k}}{\omega} \right| |\vec{E}|=|\vec{E}|,$$
because of $|\vec{k}|=\omega$.

 

[arcTomato]

@vanhees71 Thank you for your very detailed answer.
It made a lot of sense to me. It's very helpful.