0

The force on an electron due to electric field is given by $F=qE$. For a circularly polarized light, $E=E_x+iE_y$. So $F=q(E_x+iE_y)$ which means particle travels in a circle. As there is magnetic field associated with light, the particle experiences Lorentz force in the axial direction. This makes particle to travel in a helical path.

Please let me know if there is any mistake in my analysis.

SchrodingersCat
  • 4,923
  • 3
  • 23
  • 49
Pavan
  • 173

1 Answers1

0

No mistake, but if you are considering the Lorentz force on a charged particle due to an EM wave then consider this: The B-field amplitude is $c$ times less than the E-field amplitude. Therefore the magnitude of the force on the charged particle due to the B-field component of the Lorentz force ($=qvB$) is $v/c$ times less than the force due to the E-field component ($=qE$), because $E=cB$ for an EM wave in vacuum.

So unless you are making your electron move relativistically then the Lorentz force due to the B-field can be ignored. If the electron will be moving relativistically, then you'll need to solve a problem something along these lines.

[EDIT: There seems to be some confusion here: The E-field of circularly polarised light (e.g. travelling in the z-direction) could be written as $$ \vec{E} = E_0 \sin (\omega t - kz) \hat{i} + E_0 \cos(\omega t -kz) \hat{j}$$ Thus the acceleration of an electron at $z=0$ is given by $$\ddot{\vec{r}} = \frac{e}{m_e} E_0 \sin(\omega t)\hat{i} + \frac{e}{m_e} E_0 \cos(\omega t) \hat{j}$$ where the contribution of the B-field to the Lorentz force has been ignored for the reasons discussed above.

Integrating twice with respect to time and ignoring the constants of integration (assume the electron starts from rest at some coordinate with $z=0$ - or to put it another way, let's assume that $\omega$ is the frequency in the rest frame of the electron and that $\vec{r}$ is the displacement in that rest frame): $$\vec{r} = -\frac{e}{m_e \omega^2} \sin(\omega t) \hat{i} - \frac{e}{m_e \omega^2} \cos(\omega t) \hat{j},$$ which is circular motion in the $x,y$ plane of the electron's initial rest frame.]

Community
  • 1
ProfRob
  • 130,455
  • Isn't the force due to circularly polarized light given by average of forces due to two electric fields separately? And how do you explain the force due to unpolarized light? – Pavan Oct 10 '15 at 17:00
  • @Pavan It is not an "average", it is the vector sum. Do you want me to repeat the analysis twice? The relationship between E and B-fields is the same for both components. The force for unpolarised light is incoherent. – ProfRob Oct 10 '15 at 17:42
  • @RobJeffries Are you sure that there is no mistake while saying that circularly polarised light means that the electron subject to such field will move in circles? Solving the equation of motion does not in general bring that solution, does it? – gented Oct 10 '15 at 20:57
  • @GennaroTedesco Yes, it does - see edit. – ProfRob Oct 10 '15 at 21:16
  • @RobJeffries Sorry to nag again, but $z=z(t)$ as the motion along the $z$-axis is uniform (no $F_z$ component). This reflects into the equations for $x(t), y(t)$, namely, $\ddot{x}(t) = (e/m),E_0, \sin(\omega t-kz(t))$, with $z(t) = z_0 + v_0t$ (and likewise for the $y$ component). – gented Oct 10 '15 at 21:38
  • @GennaroTedesco As I clearly say - "and ignoring the constants of integration (assume the electron starts from rest at some coordinate with $z=0$". And what do you mean by "likewise for the y component"? The y-motion is oscillatory. – ProfRob Oct 10 '15 at 21:45
  • @GennaroTedesco OK, I see what you are saying. But isn't it the case that I can put the electron in a rest frame to begin with? All that changes is the frequency of the wave in that rest frame. – ProfRob Oct 10 '15 at 22:14
  • @RobJeffries If the electron is such that $v_0=0$, then yes, the solution ignores the $z$ component and moves in a circle (but saying $v_0=0$ seems like cheating a little :p); otherwise the motions should be some sort of helicoid as composition of a uniform motion plus a circle even without considering the magnetic component (that's what I was noticing as a footnote, otherwise all good). – gented Oct 10 '15 at 22:29
  • @GennaroTedesco Not cheating. Even in a gas of electrons at finite temperature and random velocities you can always consider the interaction in the rest frame of the electron. All that changes is the amplitude of the E-field of the EM wave and its frequency. You can then go back into the lab frame I suppose and yes, you will then have a more complicated helical motion. – ProfRob Oct 10 '15 at 23:01