Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

The Lorentz model, describing the electron of the atom as an harmonic oscillator forced by an oscillating electric field $\vec{E}$, shows that the dipole moment $\vec{D}$ obeys the following equation in the stationary regime: $$\vec{D}=\frac{q^2}{2m\omega_0}\frac{\vec{E}}{\omega_0-\omega-i\gamma_d}$$ where $\omega_0$ and $m$ are the natural frequency and mass of the electron, $\gamma_d$ is the dissipation introduced by hand to take account of the energy emitted by the electron when accelerated (and also its collisions).

My trouble is that one finds that the energy transmitted by the electron to the field by unit of time, given by: $$P=-\vec{E}.\frac{d\vec{D}}{dt}$$ is always negative, implying that the electron cannot "feed" the field, however how much it is radiating. Can it be attributed to the fact that the electron energy is in first place coming from the field, and thus, our electron cannot give more than it receives? In what cases does the electron sources the background field?

share|improve this question
add comment

1 Answer 1

up vote 1 down vote accepted

In the driven oscillator approximation you gave above the external field is so strong that the electron motion "follows" $\vec{E}(t)$ (i.e. it moves with the same phase $+\pi$ = the opposite phase). It is possible when the initial (before switching $\vec{E}(t)$ on) oscillations are "absent" or their contribution is small in the solution $\vec{r}_e(t)$. The incident field "feeding" (amplifying) is possible when in the solution $\vec{r}_e(t)$ there is a velocity term along $\vec{E}$ that is not caused by the force $q\vec{E}$.

The radiated field $\vec{E}_{rad}$ is added to the external field $\vec{E}$ - they come in a superposition and get into equations of motion of the other charges.

If you speak of a "self-action", it boils down physically to $\gamma_d$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.