# What happens when the frequency of AC is very high?

Saptarshi Sarkar
If I consider a wire carrying AC current, I know that at an AC frequency of 0Hz, the current will always in the same direction. If I change the frequency to 1Hz, the current will flow left to right for 1 second and then right to left for 1 second.

I guessed that at these higher frequencies, as the voltage is the same, the velocity of the electron will not increase but the time period will decrease, so an single electron will move in an SHM whose amplitude will decrease as frequency increases. If this is correct, what will happen as the frequency becomes extremely high? Does the electron become stationary?

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It does not become stationary, but still moves, even though its movement may be less than its diameter.

Dr_Nate and Saptarshi Sarkar
Saptarshi Sarkar
It does not become stationary, but still moves, even though its movement may be less than its diameter.

Will I be able to measure any current then?

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Will I be able to measure any current then?
When a current starts to flow in a wire, all the electrons move along slowly, like water in a pipe. So you will still get the same flow rate (or current). Of course, it keeps changing direction. When an alternating voltage is applied to a wire, or when you first switch on, a ripple travles down the wire at the speed of light, and all the electrons then march along slowly together.

Saptarshi Sarkar
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@Saptarshi Sarkar I think you have a mistaken impression regarding the relationship between the speed of electrons and an electrical current. Google "drift velocity"

vanhees71, Klystron and Saptarshi Sarkar
It does not become stationary, but still moves, even though its movement may be less than its diameter.
The electron doesn't have a diameter. Are you referring to the diameter of an atom in a metal?

Saptarshi Sarkar
Summary:: How is the motion of electrons in very high frequency AC?

If I consider a wire carrying AC current, I know that at an AC frequency of 0Hz, the current will always in the same direction. If I change the frequency to 1Hz, the current will flow left to right for 1 second and then right to left for 1 second.

I guessed that at these higher frequencies, as the voltage is the same, the velocity of the electron will not increase but the time period will decrease, so an single electron will move in an SHM whose amplitude will decrease as frequency increases. If this is correct, what will happen as the frequency becomes extremely high? Does the electron become stationary?

The optical conductivity (and hence the AC current response) of both free and core electrons can be modeled quite well with simple harmonic oscillators. In the case of free electrons the restoring-force term is from the scattering of electrons (Drude model), whereas the core electrons' restoring force can be considered to be from the nucleus (Lorentz model). And, yes, this does go to very high frequencies. Right into the visible spectrum and beyond.

To understand what is going on with the velocity of the electrons you will need to understand electronic band structures.

vanhees71, tech99 and Saptarshi Sarkar
Saptarshi Sarkar
@Saptarshi Sarkar I think you have a mistaken impression regarding the relationship between the speed of electrons and an electrical current. Google "drift velocity"

I know about drift velocity and that current through a conductor is represented as $$I = neAV_d$$ where
I = current
e = electronic charge
A surface area of conductor perpendicular to which current flows
V_d = drift velocity

But, if the frequency is very high and the electron is not able to move which makes the drift velocity negligible, will I still be able to get a current?

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But, if the frequency is very high and the electron is not able to move which makes the drift velocity negligible, will I still be able to get a current?
Define "negligible" and reread post #2

Saptarshi Sarkar
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Even a straight conducting wire has some nonzero inductance and doesn't let an AC current of arbitrarily high frequency pass through it.

The drift velocity of electrons in a metal wire with a potential difference across its ends is usually much slower than the thermal motion of electrons, so if you could follow the trajectories of the electrons (assuming a classical mechanical model), you probably couldn't see the drift velocity when the random motion is superimposed on it.

Saptarshi Sarkar
But, if the frequency is very high and the electron is not able to move which makes the drift velocity negligible, will I still be able to get a current?
Let's assume you have a cubic material so that all directions have the same conductivity response. Then the current density ##\vec J## is determined by $$\vec J(\omega)=\hat \sigma(\omega)\vec E(\omega)$$, where ##\vec E## is your alternating electric field and ##\hat \sigma## is the optical (AC) conductivity and ##\omega## is the frequency. The hat on the ##\hat \sigma## is telling us that the optical conductivity is a complex-numbered quantity.

In the Drude model of free electrons, the shape of the real part of the optical conductivity is a Lorentzian function centred at zero frequency. So, you can see that as frequency increases the real part of the current density will decrease in lock step with the optical conductivity.

When you are in the frequency range around infrared light and higher, this all occurs on the surface of the conductor where the light is shining on it.

tech99 and Saptarshi Sarkar
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Even a straight conducting wire has some nonzero inductance and doesn't let an AC current of arbitrarily high frequency pass through it.
That does demonstrate how careful we need to be when trying to apply intuitive, mechanical descriptions to Electricity. It can be educative to estimate just how far a charge could move during a half cycle of a frequency of 1GHz with a drift velocity of 1mm/s (just assume a sawtooth waveform, for simplicity). That gets even more ridiculous at optical frequencies and I haven't even included Inductance.

Saptarshi Sarkar
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But, if the frequency is very high and the electron is not able to move which makes the drift velocity negligible, will I still be able to get a current?

Of course, how do you think a circuit at high RF frequencies works ? eg a radio transmitter or receiver
Pick a freq. 10MHz, 100MHz, 1GHz, 10GHz, 100GHz etc. None of those RF circuits would work if there was
no AC current.

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phinds and Saptarshi Sarkar
I think a few points of clarification are needed. Hopefully, I don't misrepresent what anybody has said.
Even a straight conducting wire has some nonzero inductance and doesn't let an AC current of arbitrarily high frequency pass through it.
To me, this makes it sound like you need a particular electron to completely cross a wire to get a current. That is incorrect. Current is determined by charges crossing a surface area (see the necessary variables in the drift equation above).

That does demonstrate how careful we need to be when trying to apply intuitive, mechanical descriptions to Electricity. It can be educative to estimate just how far a charge could move during a half cycle of a frequency of 1GHz with a drift velocity of 1mm/s (just assume a sawtooth waveform, for simplicity). That gets even more ridiculous at optical frequencies and I haven't even included Inductance.
Using your model, it wouldn't be very difficult and would take only a few lines if you know the density of states, voltage, cross-sectional area, and the resistance of the wire.

There is an alternative method to yours. In solid-state physics, we regularly model the free electrons via the Drude model. If you know scattering rate and the Fermi velocity you can get a mean free path of electrons. But that assumes a single type of carrier and quickly breaks down outside of the alkali metals and some semiconductors, due to the complexities of electronic band structures. It might be good enough for an order of magnitude estimate though.

There is still an ongoing electron drift over and above the frequency of oscillation of the charges
( electrons)
The net drift of the electrons at any moment is caused by the electric field (constant or alternating) [more precise: it also depends on the history of the electric field]. For your statement to be true you would need a non-zero constant electric field with the alternating field on top of it.

Saptarshi Sarkar
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The net drift of the electrons at any moment is caused by the electric field (constant or alternating) [more precise: it also depends on the history of the electric field]. For your statement to be true you would need a non-zero constant electric field with the alternating field on top of it.

Yeah, I will let you away with that

After all in years gone by, I have always told people that the charges ( electrons) in the wires of the power station dot get to the load
or alternatively, the electrons in the filament of the lightbulb load never leave the filament

I will edit my post

Saptarshi Sarkar
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To me, this makes it sound like you need a particular electron to completely cross a wire to get a current. That is incorrect. Current is determined by charges crossing a surface area (see the necessary variables in the drift equation above).

Yes, I meant that if you try to apply a too rapidly changing voltage between the endpoints of the wire/inductor, an ammeter inserted somewhere in between won't measure any significant current at any time. AC current is also said to "pass through a capacitor when the frequency is high enough", even though there are no electrons actually going through the dielectric layer.

Saptarshi Sarkar
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There's no current going through a capacitor. It's a bit unfortunate that one calls ##1/c \partial_t \vec{E}## still the "displacement current". A current is defined as moving charges going through a cross sectional area per unit time. Through a capacitor no charges go. Thus there's no current!

Saptarshi Sarkar and davenn
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There's no current going through a capacitor. It's a bit unfortunate that one calls ##1/c \partial_t \vec{E}## still the "displacement current". A current is defined as moving charges going through a cross sectional area per unit time. Through a capacitor no charges go. Thus there's no current!
If the capacitor contains a dielectric, though not with a vacuum, I would have expected electrons to be moving back and forth in response to the applied electric field, constituting an electric current.

vanhees71, davenn and Saptarshi Sarkar
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If the capacitor contains a dielectric, though not with a vacuum, I would have expected electrons to be moving back and forth in response to the applied electric field, constituting an electric current.
If so, the number is trivial compared to the actual current in the circuit. If there are very many, you would be having dielectric breakdown.

davenn and Saptarshi Sarkar
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If so, the number is trivial compared to the actual current in the circuit. If there are very many, you would be having dielectric breakdown.
Just to be difficult, I could point out that any net movement of charge through a metal conductor with a high frequency AC would be more or less the same as the movement of charges when the molecules in a capacitor's dielectric are polarised cyclically.
This thread is a good example of how "what is really happening" can be a fruitless conversation.

tech99 and phinds
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If the capacitor contains a dielectric, though not with a vacuum, I would have expected electrons to be moving back and forth in response to the applied electric field, constituting an electric current.

As phinds said, ONLY if there was a dielectric (solid, gas, vacuum) breakdown or unwanted leakage

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The ideal mass-spring system consists of a particle of mass ##m## attached to a spring with spring constant ##k##. The frequency is ##\omega=\sqrt{\frac{k}{m}}##.

What happens to the speed of the particle as ##\omega## increases beyond all bounds?

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If the capacitor contains a dielectric, though not with a vacuum, I would have expected electrons to be moving back and forth in response to the applied electric field, constituting an electric current.
Usually in macroscopic electrodynamics we consider as current (densities) only the "free" current (densities), and describe the dielectric via the polarization. Of course you can lump the polarization to the free charges and currents, also only partially. At the end you get the same result, because there's only one (microscopic) electromagnetic field and only one (microscopic) charge-current distribution.

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As phinds said, ONLY if there was a dielectric (solid, gas, vacuum) breakdown or unwanted leakage
It would apply to the distortion of the charges on the metal surface of the plates, that is if you really really want to pursue this. In which case the polarization could still be regarded as present in the vacuum plus surfaces.
In any case, the ‘matter of principle’ that there is no current becomes of less and less consequence. Show me a total vacuum and I would say that 'awkwardness' could actually apply. What is the actual problem with using the name Displacement Current?

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If the capacitor contains a dielectric, though not with a vacuum, I would have expected electrons to be moving back and forth in response to the applied electric field, constituting an electric current.
My evidence for suggesting that electrons move in a dielectric is that dielectrics can radiate, so that charges must be accelerating. On the other hand, a vacuum capacitor does not radiate.

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Of course, the electrons in a dielectric can move and radiate, but usually you don't describe them by free charge-current distributions but by polarization, because they are bound. As I said, in macroscopic electromagnetics you can always shuffle parts from the "sources" to the "fields" and vice versa without changing the physics. Of course, the one or the other description might be easier to handle.

Dale, Paul Colby and davenn
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My evidence for suggesting that electrons move in a dielectric is that dielectrics can radiate

they may well move a bit due to the electric field ?
but they are still not part of the overall current in the circuit, if you think they are, show some links

phinds
artis
apart from leakage current and dielectric polarization , if there was net circuit current through a capacitor it would stop being a capacitor and become a wire or an electrolyte , it could not hold charge for any useful amount of time , isn't this a good enough proof that there is no "real" current through a capacitor?

I guess I would describe a capacitor as any element in a circuit that doesn't pass DC current, so I guess a simple hand operated switch becomes a capacitor in the off position.

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I guess I would describe a capacitor as any element in a circuit that doesn't pass DC current, so I guess a simple hand operated switch becomes a capacitor in the off position.
No, that would be a grossly incomplete description of a capacitor.

Klystron, davenn and tech99
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apart from leakage current and dielectric polarization , if there was net circuit current through a capacitor it would stop being a capacitor and become a wire or an electrolyte , it could not hold charge for any useful amount of time , isn't this a good enough proof that there is no "real" current through a capacitor?

I guess I would describe a capacitor as any element in a circuit that doesn't pass DC current, so I guess a simple hand operated switch becomes a capacitor in the off position.
The question related to extremely high frequencies and whether an electron executed SHM. I respect people's explanations here (I am outgunned and outnumbered), but find it easier to consider that it has this motion, in both conductors and dielectrics, superimposed on a large thermal motion. Further, for conductors, the SHM is disturbed by collisions, hence we see shot noise accompanying the signal.
Some further evidence for the SHM motion of electrons is the existence of plasmon resonances and kinetic inductance ().

If the capacitor contains a dielectric, though not with a vacuum, I would have expected electrons to be moving back and forth in response to the applied electric field, constituting an electric current.
Yes, this is true.
If so, the number is trivial compared to the actual current in the circuit. If there are very many, you would be having dielectric breakdown.
The numbers are comparable. The electrons don't need to move far in an oscillation. Dielectric breakdown occurs because the electrons are ripped from their ions; this usually occurs due to a large voltage.
Just to be difficult, I could point out that any net movement of charge through a metal conductor with a high frequency AC would be more or less the same as the movement of charges when the molecules in a capacitor's dielectric are polarised cyclically.
In a sense, you are right. The free and bound charges contribute to the AC conductivity ##\sigma(\omega)## in the generalized form of Ohm's rule: ##\vec J(\omega)=\hat \sigma(\omega)\vec E(\omega)##. If I measured the conductivity at a single frequency, I wouldn't know how much was from free charge and how much was from bound charge. But, if I knew the conductivity over a large range of frequencies, I can see characteristic shapes in the graph due to bound and free charges.

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I guess I would describe a capacitor as any element in a circuit that doesn't pass DC current, so I guess a simple hand operated switch becomes a capacitor in the off position.

Almost anything that fits this description probably has a capacitance, but it doesn't tell what makes something a good capacitor and something else a bad one.

davenn
The question related to extremely high frequencies and whether an electron executed SHM. I respect people's explanations here (I am outgunned and outnumbered), but find it easier to consider that it has this motion, in both conductors and dielectrics, superimposed on a large thermal motion. Further, for conductors, the SHM is disturbed by collisions, hence we see shot noise accompanying the signal.

I should correct something I said. Earlier I described electrons as following simple harmonic motion when discussing the Drude and Lorentz models. I should have called it damped-driven harmonic motion, which is what the Drude and Lorentz models model. I know this very well, but I just used the wrong words.

The difference is that a simple harmonic oscillator will freely vibrate at one frequency, whereas we are using an alternating field to drive the oscillations and there is also a mechanism to dissipate this energy. The response of the oscillator then has a shape of a Lorentzian function.

You are pretty much spot on, if you replace SHM with damped-driven harmonic motion.

Graduate students of solid-state physics are taught exactly this. You can see it in the notes of the late Millie Dresselhaus: http://web.mit.edu/course/6/6.732/www/6.732-pt2.pdf.

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vanhees71
The ideal mass-spring system consists of a particle of mass ##m## attached to a spring with spring constant ##k##. The frequency is ##\omega=\sqrt{\frac{k}{m}}##.

What happens to the speed of the particle as ##\omega## increases beyond all bounds?