How can the power (in Watts) absorbed by the electron be calculated, knowing the incident electric field amplitude $ E_0 $, wavelength $ \lambda $, and the electron momentum relaxation time $ \tau $ in the medium ?

The units seem to check out for $ \frac {|q_e| E_0 \lambda }\tau $ (result is in Watts), but the correct answer is off by several orders of magnitude. What is missing?


1 Answer 1


I guess you're also doing the Plasmonics course of edX. (Great course by the way, although I find the exercises quite difficult.)

I solved the task using the formula for power that was shown in the lecture video:
$$ P = <\cfrac{\vec{F}+\vec{F}^*}{2}\cdot \cfrac{\vec{v}+\vec{v}^*}{2}> = \cfrac{1}{2}\Re (\vec{F}^* \cdot \vec{v}) $$

The velocity is time derivative of position $\vec{r}$. To get $\vec{r}$ let's solve the equation of motion that is accounting for collisions (this was done in the edX course):
$$ m_e\ddot{\vec{r}}=-\gamma m_e \dot{\vec{r}} - |e|\vec{E_0}e^{-i\omega t}$$ which yields the solution:
$$\vec{r}(t)=\cfrac{|e|}{m_e\, \omega \,(\omega + i\gamma)}\vec{E_0}e^{-i\omega t}$$ $\vec{F}$ is Lorentz force due to electric field:
$$\vec{F} = |e| \vec{E_0}e^{-i\omega t}$$

So that gives me: $$ P = \cfrac{1}{2}\Re (\vec{F}^* \cdot \vec{v}) = \cfrac{|e|^2}{2} \Re \left(E_0^2 \cfrac{-i\omega}{m_e \omega(\omega + i\gamma)} \right) = \cfrac{E_0^2\,e^2 \gamma}{2 m_e \,(\omega^2 + \gamma^2)} $$

Now we now that $\lambda = \frac{2 \pi c}{\omega}$, from this we can calculate $\omega$ knowing the wavelength. And following again the defitions from edX course $\gamma = \frac{1}{\tau}$.

As you can see, the unit is still OK although the expression changed a lot.

I would like to know if there is a simpler way but I cannot think of any right now.

  • $\begingroup$ Incredibly clear and straightforward explanation. Thank you so much. $\endgroup$ Feb 4, 2019 at 21:54

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