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I have been told a many times that in a region with oscillating electric and magnetic field, a free electron if left will also oscillate. But I don't think its true. I actually asked this to my instructor and he also couldn't give a satisfactory answer.


Lets say there is an electron and a light beam is approaching it (as per the figure). When the light "hits" the electron, it will accelerate in the upward direction.

enter image description here

The speed of the electron will keep on increasing till the first two quadrants pass through the point reaching a maxima (as the direction of force is along the direction of motion). In the third and fourth quadrant, the speed will decrease but it will just make the gained speed going back to zero.

But the point to be noted here is that "having a zero speed at the end doesn't imply that the net displacement is 0". The electron is displaced permanently from its initial position. And similarly as more and more wave pulses cross through the point, it will keep on being displaced. So it never oscillates.

Can someone explain what's wrong with this argument? Why should the charge oscillate?


P.S :- I didn't consider the effects of magnetic field as I think it will just make the path a bit curved. I don't think it will have any such notable effect (right ?)

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  • $\begingroup$ What do you mean by an electron at rest? That would indicate you know both it's position and momentum. $\endgroup$
    – mcodesmart
    Commented Jan 2 at 18:43

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If you write an expression for magnitude of the force on the electron, you get something that is oscillatory; and that is what is meant by the "electron oscillates". The electron doesn't oscillate in the sense that it is rigidly fixed at some point in space and oscillates about that exact point.

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  • $\begingroup$ so doesn't that mean the electron will never come back to the location from where it was displaced ? And regarding your last statement :- In what sense does it oscillate ? Because we won't be able to see the oscillating force acting on it , all we could see is its motion right ? $\endgroup$
    – Ankit
    Commented Jun 2 at 8:22
  • $\begingroup$ There is no guarantee that it won't come back to its original spot at some particular time. The concept of oscillation can be applied to motion without regard to linear restoring forces. For example, if I walk back and forth across the street, I am "oscillating", but there is no spring pulling me back and forth. $\endgroup$ Commented Jun 2 at 13:41
  • $\begingroup$ Sorry but It means there must be some flaw in the way I am thinking but still your answer doesn't point out the mistake for which I had asked the question. Could be give an example in support of your statement :- "There is no guarantee that it won't come back to its original spot at some particular time." $\endgroup$
    – Ankit
    Commented Jun 2 at 15:19
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Let us assume a monochromatic electromagnetic plane wave propagating in the positive $x$-direction, $$ \vec{E}(t,x,y,z) = \vec{e}_y E_0 \sin[\omega(t-x/c)], \quad \vec{B}(t, x,y,z)= \vec{e}_z E_0 \sin[\omega(t-x/c)],$$ where I am using the Gauss system. The equation of motion for the position vector $\vec{r}(t)$ of the electron is given by $$ m \ddot{\vec{r}}(t)=q \left[\vec{E}(t, \vec{r}(t))+ \frac{\dot{\vec{r}}(t)}{c} \times \vec{B}(t,\vec{r}(t) )\right],$$ and we take the initial conditions $\vec{r}(0)=0$ and $\dot{\vec{r}}(0)=0$. As long as $|\dot{\vec{r}}(t)| \lt \! \!\lt c$, the contribution from the magnetic field can safely be neglected and the equation of motion simplifies to $$ m \ddot{\vec{r}}(t) = q \vec{E}(t, \vec{r}(t)).$$ Inserting the electric field of the monochromatic plane wave given above, we arrive at the system of differential equations for $\vec{r}(t)=x(t) \vec{e}_x+y(t)\vec{e}_y+z(t)\vec{e}_z$ given by $$m \ddot{x}(t)=0, \quad m \ddot{y}(t) =q E_0 \sin[\omega(t-z(t)], \quad m \ddot{z}(t)=0 $$ subject to the initial conditions $x(0)= y(0)=z(0)=0 $ and $\dot{x}(0)= \dot{y}(0)=\dot{z}(0)=0$. The solution is easily found as $$x(t)=z(t)=0, \quad y(t)= -\frac{q E_0}{m\omega^2} \sin \omega t +\frac{qE_0}{m\omega} t, $$ i.e. the electron performs an oscillation with angular frequency $\omega$ superimposed by a linear motion with constant drift velocity $qE_0/m\omega$.

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  • $\begingroup$ Your solution is not an oscillation in the sense of the coordinates changing sign. The latter would be required if you want to explain for instance the radiation pressure on a free charge by means of the Lorentz force, which otherwise would permanently change sign as $B$ changes sign but not $v$. $\endgroup$
    – Thomas
    Commented Dec 27, 2023 at 16:22
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    $\begingroup$ @Thomas The term $\sim \sin \omega t$ is, of course, an oscillation. What else? $\endgroup$
    – Hyperon
    Commented Dec 27, 2023 at 16:25
  • $\begingroup$ You are just describing in mathematical equations what the OP described already in words. His concern is that the velocity never reverses (contrary to what is assumed for instance in textbooks when 'deriving' the radiation pressure on a free charge) $\endgroup$
    – Thomas
    Commented Dec 27, 2023 at 16:36
  • $\begingroup$ Note that the drift term depends strongly on the initial conditions, or the phase of the wave when it's "turned on". There is no drift if $\vec E\propto \cos(\omega t)$. $\endgroup$
    – Puk
    Commented Dec 27, 2023 at 16:58
  • $\begingroup$ @Puk Yes, of course. I was just referring to the the situation described in the original question. $\endgroup$
    – Hyperon
    Commented Dec 27, 2023 at 19:53
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Your observation is correct. The important point here is that for a free electron initially at rest, its velocity always has the same sign, varying only between 0 and some maximum positive value. The explanation of radiation pressure on a free charge by virtue of the Lorentz force $q/c\cdot\vec{v}\times\vec{B}$ (which one can find in many physics textbooks) is therefore incorrect, as the direction of radiation pressure would be be changing sign all the time. There is a publication discussing this issue, where various approaches to resolve this inconsistency are suggested. I personally think though that the resolution to this apparent paradox is simply the fact that there are in practice no free electrons. They are always bound either in atoms/molecules or by external electromagnetic fields, either of which would prohibit any translatory motion of the electron.

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I think what you have wrong is thinking about light as a small thing going on in one spot. If you accelerate an electron, you kick it out of that spot. Light isn't like that.

Light is a spread out wave. Kind of like waves at the beach. If an electron gets pushed up, the wave there will push it back.

Here is a video with some good animations and great explanations of some really curious optical phenomena. The first one sets up some questions. The sugar water barber pole effect | Optics puzzles 1.

The second begins to explain why they work this way. You will find the answer to your question in that. Explaining the barber pole effect from origins of light | Optics puzzles 2

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    $\begingroup$ "If an electron gets pushed up, the wave there will push it back." – Are you saying that if the electromagnetic field pushes an electron up (in the positive $y$ direction, say), then it will inevitably enter an area where the field pushes it down again (negative $y$)? That seems to contradict the fact that it's possible for an electromagnetic field to be a plane wave that's invariant with respect to the $x$ and $y$ coordinates (and thus moves in the $z$ direction). $\endgroup$ Commented Dec 28, 2023 at 7:01
  • $\begingroup$ @TannerSwett - No. I should have said the wave will advance until the electron is in a region where the wave will push it back. $\endgroup$
    – mmesser314
    Commented Dec 28, 2023 at 14:23
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Let's say we have an electron sitting in an electric field that is approximately constant in space and sinusoidal in time.

From this we can write the Force on the electron from the electric field as:$$F=ma \approx F_{E}\sin{(\omega t)}$$

This can be transposed to find its acceleration:$$a \approx \frac{F_{E}\sin{(\omega t)}}{m}$$

This can be integrated relative to time twice to find the displacement of the electron as:$$x=\int\int a \,dt^{2} \approx -\frac{F_{E}\sin{(\omega t})}{m \omega^{2}} + v_{0}t + x_{0}$$

Now consider that basically all real perturbations will have a valid Fourier transform, meaning they can be written as a linear combination of some amount of sinusoids (likely infinite). Finally, the second integral of a linear combination of sinusoids is going to also be a linear combination of sinusoids of the initial conditions are taken as 0, luckily special relativity rigorously shows us that we can do that.

If our motion is purely a linear combination of sinusoids, which are inherently oscillatory, then our motion must also be oscillatory.

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