Energy conservation in Electrodynamics

[Bob_for_short]

Let us suppose that we have a known electromagnetic wave-train of finite size propagating in a certain direction. On its way there is a probe charge. This EMW is an external field for the charge. The EMW has a certain energy-momentum (integral over the space). After action on the probe charge the wave continues its way away. In the end we have the energy of the initial wave (displaced somewhere), the kinetic energy of the charge (hopefully it starts moving), and the energy of the radiated EMF. Thus the total energy is not conserved, is it?

[swansont]

The energy of the EM wave has to change. Isn't this just Compton scattering?

[Bob_for_short]

No, it is not Compton. Just a regular electrodynamics problem.

 

How energy can change? Via destructive interference? How to show it?

[Bob_for_short]

EDIT: I can emit a half-period long wave from a radio-transmitter: [math]E(t)=E_0 sin(\Omega t), 0 < t < \pi/ \Omega [/math]. Then the final charge velocity will be clearly different from zero:

 

[math]ma=F(t), v(t>\pi/ \Omega)=\int_{0}^{t}F(t')dt'=\frac{2qE_0}{m\Omega}[/math].

 

In addition, the charge itself radiates some new wave during acceleration period. The radiated energy is only a small fraction of [math]\frac{mv^2}{2}[/math]. What can guarantee that the total energy remains the same?

Edited May 12, 2011 by Bob_for_short

[J.C.MacSwell]

EDIT: I can emit a half-period long wave from a radio-transmitter: [math]E(t)=E_0 sin(\Omega t), 0 < t < \pi/ \Omega [/math]. Then the final charge velocity will be clearly different from zero:

 

[math]ma=F(t), v(t>\pi/ \Omega)=\int_{0}^{t}F(t')dt'=\frac{2qE_0}{m\Omega}[/math].

 

In addition, the charge itself radiates some new wave during acceleration period. The radiated energy is only a small fraction of [math]\frac{mv^2}{2}[/math]. What can guarantee that the total energy remains the same?

 

Isn't that just the way it works? If the final charge velocity is not zero, then the initial wave train has less energy. It has done work and transferred momentum.

[Bob_for_short]

Isn't that just the way it works? If the final charge velocity is not zero, then the initial wave train has less energy. It has done work and transferred momentum.

 

Yes, the incident wave-train can get weaker if it is accompanied with the radiated wave and the resulting wave amplitude (=> energy-momentum) becomes smaller. I just do not see it immediately.