The effect of a radio wave on an electron

[brotherbobby]

TL;DR

Taking a oscillating electric field $E=E_0\sin\omega t$, we use kinematics to find the position of an electron under its influence : $x(t) = \dfrac{a_0}{\omega}t-\dfrac{a_0}{\omega^2}\sin\omega t$. The second term is obvious - the electron oscillates too. But how do we explain the first term, which shows that the electron also "drifts" with uniform velocity along the same direction as that of the electric field. $\text{How to understand this drift?}$

Statement of the problem : The problem is a solved example in Kleppner and Kollenkow's book, p. 21. The example is titled : "The Effect of a Radio Wave on an Ionospheric Electron". I copy and paste the example below and underline the last bit in red, which is my area of doubt. I hope the text is readable.

Question : The authors use the an oscillating field in one direction $E=E_0\sin\omega t$ to derive the position of the electron which is located at the origin at rest when time starts. The position of the elextron as a function of time turns out to be : $x(t) = \underline{\dfrac{a_0}{\omega}t}-\dfrac{a_0}{\omega^2}\sin\omega t$.

The underlined term in the equation is the puzzle, which the authors refer to towards the end. They say that it "corresponds to motion with uniform velocity, so in addition to the jiggling motion the electron starts to drift away. Can you see why?" (my emphasis)

I am afraid I don't see why. Far as I can tell, the electron should oscillate back and forth about the origin $x=0$.

Guess : (and of course I could be wrong), is that this is connected to the time-dependence of the electric field. As the electron oscillates, the field changes during that time. This keeps changing the mean position of oscillation for the electron, further ahead with time. But why is such a drift unform, even assuming I am correct? No clue as of yet.

Many thanks.

 

[Dale]

Look at $\dot x$ instead of $x$. What is the range of $\dot x$?

 

[A.T.]

I am afraid I don't see why. Far as I can tell, the electron should oscillate back and forth about the origin $x=0$.

This seems similar to this question you had recently:

Doesn't the deflection due to the coriolis force while going up cancel the defection due to the force while going down? If it does, wouldn't the body drop at the same point?

Consider for simplicity what happens to a body initially at rest that experiences acceleration $a$ for 1s, and then $-a$ for 1s. Does it end up in the same place it started?

 

[Sagittarius A-Star]

$x(t) = \dfrac{a_0}{\omega}t-\dfrac{a_0}{\omega^2}\sin\omega t$

$v(t) = \dfrac{a_0}{\omega} (1 -\cos\omega t)$

in addition to the jiggling motion the electron starts to drift away. Can you see why?"

Assume you repeat periodically:
accelerate your car from 0 mph to 65 mph, then press the brake pedal to decelerate from 65 mph to 0 mph.

Then your car drifts in average.

 

[Sagittarius A-Star]

Far as I can tell, the electron should oscillate back and forth about the origin $x=0$.

That would be in average the case for electrons in a resistor, but in the OP screenshot, the electron is - according to the problem - in the ionosphere.

In the ionosphere, the acceleration of the electron is proportional to E(t).
In a resistor, the drift-velocity of the electrons is proportional to E(t).

 

[brotherbobby]

Look at $\dot x$ instead of $x$. What is the range of $\dot x$?

Yes, let me calculate. We have the position of the electron relative to the origin O : $x(t) = \dfrac{a_0}{\omega}t-\dfrac{a_0}{\omega^2}\sin\omega t$. Upon differentiating, $\dot x(t) = \dfrac{a_0}{\omega} - \dfrac{a_0}{\omega}\cos\omega t$. To check, this matches the initial condition of "at rest initially" : $\dot x(0)=\dfrac{a_0}{\omega}-\dfrac{a_0}{\omega}=0$.

We find that the velocity of the electron $v_x (=\dot x)$ has a drift component, independent of time. It would be good to plot the velocity function with time. I have taken $a_0=1\;,\;\omega = 2$. The cosine curve is pushed up by the amount $a_0/\omega$.
At time $t=2n\pi/\omega$, the cosine term vanishes but the first term remains : $x(2n\pi/\omega)=\dfrac{a_0}{\omega}$.
Is this the term responsible for the drift speed of the electron?

 

[brotherbobby]

Consider for simplicity what happens to a body initially at rest that experiences acceleration $a$ for 1s, and then $-a$ for 1s. Does it end up in the same place it started?

Imagine the body to start from the origin. After 1 s, the body is at a position of $x(1)=\dfrac{1}{2}a$ with a velocity $v(1) = a$. After another second of deceleration by the same amount, the position of the particle is $x(2) = a-\dfrac{a}{2}= \dfrac{a}{2}$. As for i ts velocity : $v(2) = a-a=0$.
We find that while the particle returns to rest, it is not at the position where it began, but "ahead" of it. Clearly, this is the result of the velocity given to the body during the initial acceleration.

 

[Sagittarius A-Star]

After another second of deceleration by the same amount, the position of the particle is $x(2) = a-\dfrac{a}{2}= \dfrac{a}{2}$.

That's wrong. Besides the fact, that the physical units on the left and right side of the equation are different, $x(2s)$ cannot be equal to $x(1s)$, because the body continues to move while the deceleration.

 

[brotherbobby]

That's wrong. Besides the fact, that the physical units on the left and right side of the equation are different, $x(2s)$ cannot be equal to $x(1s)$, because the body continues to move while the deceleration.

Yes, forgive me, I realised my mistake just after you posted. I was thinking of editing my response.
Let me do it again.

The velocity of the electron after 2 s is fine, $v(2) = 0$.
The position of the electron : $\boxed{x(2)} = \dfrac{a}{2}+a-\dfrac{a}{2} = \boxed{a}$. The electron moves a distance $a$ ahead.

Please note the physical units are taken care of. They seem out of place because everywhere there is the time duration of $t=1\,\text{s}$.

 

[Sagittarius A-Star]

The electron moves a distance $a$ ahead.

That's almost correct. It moves a distance $a (1 \text{s})^2$ ahead.

Please note the physical units are taken care of. They seem out of place because everywhere there is the time duration of $t=1\,\text{s}$.

Please see above.

 

[A.T.]

We find that while the particle returns to rest, it is not at the position where it began

Yes, exactly. Acceleration is the derivative of velocity, so with equal but opposite acceleration phases, only the velocity changes cancel, but not necessarily the position changes. This also applies to the Coriolis accelerations on the way up and down.

 

[Dale]

Is this the term responsible for the drift speed of the electron?

Yes.

Note that each cycle accelerates and then decelerates the electron. The acceleration portion accelerates from zero to some positive velocity. The deceleration portion is symmetric. So it decelerates from the positive velocity back to zero. So the velocity never becomes negative. Without the velocity becoming negative the particle cannot return to its original location.

 

[brotherbobby]

Yes, exactly. Acceleration is the derivative of velocity, so with equal but opposite acceleration phases, only the velocity changes cancel, but not necessarily the position changes.

Thank you. The point may be elementary but deep. Far as I know, books don't say these things. One has to "discover" them on one's own, or with help.

 

[Dale]

Far as I know, books don't say these things. One has to "discover" them on one's own, or with help.

Well, K&K’s book didn’t give you the answer, but they did prepare you with the skills that you needed to discover it, and they did point out that there was something interesting to discover there. So I wouldn’t exactly say that this is something that books don’t say. More like they chose to present it in a way to encourage a student’s active participation in learning.

 

[sophiecentaur]

Yes.

Note that each cycle accelerates and then decelerates the electron. The acceleration portion accelerates from zero to some positive velocity. The deceleration portion is symmetric. So it decelerates from the positive velocity back to zero. So the velocity never becomes negative. Without the velocity becoming negative the particle cannot return to its original location.

I have a small problem / comment with this. If the phase of the first cycle of incoming wave is inverted then won't the electron move down instead of up? The resulting drift could vary along the path of the wave. Or what?

 

[Dale]

If the phase of the first cycle of incoming wave is inverted then won't the electron move down instead of up?

Yes. The result depends strongly on the initial velocity of the particle and the initial phase of the wave. The discussion above is about the specific conditions of the problem and does not generalize.

 

[sophiecentaur]

Yes. The result depends strongly on the initial velocity of the particle and the initial phase of the wave. The discussion above is about the specific conditions of the problem and does not generalize.

It implies that the momentum change of the electron will be in a direction appropriate to the E field of the wave. So the scattering of the wave would also be a direction appropriate to the E field. That's not a problem as I see it; it's just a fact that's good tp know but the difference in the actual amount of scattering might be very small. For a wave from an MF transmitting mast the 'sign' of the scattering would depend on the +sign on the transmitter and, from the shape of the radiation pattern, would be greater in the vertical direction.

The induced free electron motion is relevant in mf ionospheric propagation (mentioned previously) and is also affected by the Earth's magnetic field (producing rotation of polarisation). Maybe that's a bigger effect.
The displacement effect would also be finite for the positive ions up there too but many thousandths smaller, I think.

 

[A.T.]

Far as I know, books don't say these things.

Books? I learned that before any formal physics education from this:

Give it a try:
https://freeasteroids.org/

There is no friction in space, so you get an intuitive feel for what it means to apply just one force at a time.

 

[sophiecentaur]

Books? I learned that before any formal physics education from this:

View attachment 372209

Give it a try:
https://freeasteroids.org/

There is no friction in space, so you get an intuitive feel for what it means to apply just one force at a time.

Hmm. Maybe on a subjective level but do those processors really care about cumulative errors? The player would be compensating for the sort go thing (what did you actually learn, whilst having fun?)

 

[A.T.]

(what did you actually learn, whilst having fun?)

Learning while having fun is the most effective way to learn. The point here is to get the right intuitions about how things behave, so you don't get confused by a correct math result that contradicts your wrong intuitions.

But after trying the online version of asteroids linked above, I noticed that it does have some drag. Which makes sense game-play wise, since you have infinite fuel, and could achieve unplayable speeds otherwise. I must have confused it with my own versions of the game, where I could play around with the drag amount, including zero drag.

This one should have zero drag, but has gravity:
http://moonlander.seb.ly/

And this is another great app, where you can toggle/adjust drag and gravity:
https://www.algodoo.com/

 

[sophiecentaur]

Learning while having fun is the most effective way to learn. The

Very true as long as what you are being presented with is accurate Science (in this case). A good simulator is great value but I wonder how much was spent on avoiding 'quirks' in the system.
I do take your point though.

 

[tech99]

Maybe I miss the point of this discussion, but I cannot understand how an EM wave can induce a direct current in the ionosphere. EM waves cannot have a unidirectional component. If the wave is pulsed, then Fourier tells us it is a continuous carrier modulated by the pulse waveform.

 

[sophiecentaur]

Half of the electrons will be displaced in the other direction. The mean current should be zero.

 

[sophiecentaur]

If the wave is pulsed, then Fourier tells us it is a continuous carrier modulated by the pulse waveform.

Actually , for a 'pulsed' waveform there is no propagated DC component. The mean amplitude of E will be zero so there will be positive and negative E fields, again causing displacement in both directions.

 

[A.T.]

Half of the electrons will be displaced in the other direction.

The mean amplitude of E will be zero so there will be positive and negative E fields, again causing displacement in both directions.

Not in the idealized case analysed in the book, as already explained:

Acceleration is the derivative of velocity, so with equal but opposite acceleration phases, only the velocity changes cancel, but not necessarily the position changes.

 

[sophiecentaur]

Not in the idealized case analysed in the book, as already explained:

The printed passage starts by mentioning the Ionosphere so it's not really an idealised situation unless it's a sinusoidal wave. In the case of a standing wave between two plates, it would be possible to add a DC component to the wave for balance but, if the wave is launched, there would be no DC component so there would be E fields of both polarities. It's a 'AC coupled" situation, for want of a better term.

Acceleration is the derivative of velocity, so with equal but opposite acceleration phases, only the velocity changes cancel, but not necessarily the position changes.

My point has been that a question was raised about a pulsed waveform. The integral of the velocity variation gives an equal and opposite deflection for each sign of the RF wave. The effect of the short, high pulse would be the same as the reverse effect during the (inevitable) low, long negative section of the waveform. I have to admit that I haven't done the integration (it would be easy for rectangular pulses) but my argument at least suggest a way of account for the claimed paradox; it would be a dispersion effect as with a sine wave.

 

[Sagittarius A-Star]

If God had intended us to use analogies, he would not have released Mathematics for our use.

You can find an analogy in posting #4.

The math you can find in the OP, when considering the initial conditions in the OP's attachment
($x_0=v_0=0$).

Taking a oscillating electric field $E=E_0\sin\omega t$, we use kinematics to find the position of an electron under its influence : $x(t) = \dfrac{a_0}{\omega}t-\dfrac{a_0}{\omega^2}\sin\omega t$.

So there is a drift.

Incorrect is the statement in the problem description, that there would be no acceleration in $y$- or $z$-direction, because a radio wave includes also a magnetic field.

 

[tech99]

The printed passage starts by mentioning the Ionosphere so it's not really an idealised situation unless it's a sinusoidal wave. In the case of a standing wave between two plates, it would be possible to add a DC component to the wave for balance but, if the wave is launched, there would be no DC component so there would be E fields of both polarities. It's a 'AC coupled" situation, for want of a better term.

My point has been that a question was raised about a pulsed waveform. The integral of the velocity variation gives an equal and opposite deflection for each sign of the RF wave. The effect of the short, high pulse would be the same as the reverse effect during the (inevitable) low, long negative section of the waveform. I have to admit that I haven't done the integration (it would be easy for rectangular pulses) but my argum

The printed passage starts by mentioning the Ionosphere so it's not really an idealised situation unless it's a sinusoidal wave. In the case of a standing wave between two plates, it would be possible to add a DC component to the wave for balance but, if the wave is launched, there would be no DC component so there would be E fields of both polarities. It's a 'AC coupled" situation, for want of a better term.

My point has been that a question was raised about a pulsed waveform. The integral of the velocity variation gives an equal and opposite deflection for each sign of the RF wave. The effect of the short, high pulse would be the same as the reverse effect during the (inevitable) low, long negative section of the waveform. I have to admit that I haven't done the integration (it would be easy for rectangular pulses) but my argument at least suggest a way of account for the claimed paradox; it would be a dispersion effect as with a sine wave.

Assuming a pulsed "sinusoidal" waveform, as you rightly say, this is an AC situation and there can be no DC component that could cause a drift. It is the same as passing any wave shape through a blocking capacitor - no DC component.

 

[Sagittarius A-Star]

@tech99 As I wrote in posting #5,
In the ionosphere, the acceleration of the electron is proportional to E(t).
In a resistor, the drift-velocity of the electrons is proportional to E(t).

 

[Roberto Pavani]

As others have noted, the drift direction is phase-dependent, so it's really an artifact of the initial conditions rather than a general physical effect. If there's any real (tiny) net drift, I'd look at radiation pressure, the v × B force gives a push along the propagation direction. For a wave arriving at an angle, the horizontal component of that push might produce some drift along the ionospheric layer, while the vertical component would be opposed by the layer itself.