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

• Saptarshi Sarkar
In summary, the conversation discusses the behavior of electrons in a wire carrying AC current at different frequencies. It is noted that at higher frequencies, the velocity of the electrons does not increase but the time period decreases, resulting in a decrease in amplitude. It is also mentioned that at extremely high frequencies, the electron does not become stationary but still moves, although its movement may be less than its diameter. The relationship between the speed of electrons and electrical current is also discussed, along with the concept of drift velocity. The conversation also touches upon the optical conductivity of free and core electrons and the effect of frequency on the electron's motion. Finally, the conversation addresses the question of whether a current can still be measured at very high frequencies and the role of the
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?

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

Dr_Nate and Saptarshi Sarkar
tech99 said:
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?

Saptarshi Sarkar said:
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
@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
tech99 said:
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
Saptarshi Sarkar said:
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
phinds said:
@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?

Saptarshi Sarkar said:
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
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
Saptarshi Sarkar said:
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
hilbert2 said:
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
Saptarshi Sarkar said:
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.

Last edited:
phinds and Saptarshi Sarkar
I think a few points of clarification are needed. Hopefully, I don't misrepresent what anybody has said.
hilbert2 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).

sophiecentaur said:
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.

davenn said:
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
Dr_Nate said:
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
Dr_Nate said:
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
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
vanhees71 said:
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
tech99 said:
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
phinds said:
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
tech99 said:
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

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?

tech99 said:
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.

davenn said:
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?

tech99 said:
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.

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
tech99 said:
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
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.

artis said:
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
artis said:
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 ().

tech99 said:
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.
phinds said:
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.
sophiecentaur said:
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.

artis said:
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
tech99 said:
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.

Last edited:
vanhees71
Mister T said:
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?

The magnetic field arises as the electric current vector changes. The AC could not be transmitted without an alternating magnetic field. As electric currents and magnetic fields alternate at higher frequencies, more of the energy is transmitted down the wire in the form of changing magnitudes than in the magnitudes themselves. It doesn't really matter how far the charge carriers travel. Building and collapsing a tiny magnetic field really fast can transmit as much energy as building and collapsing a large one slowly.

At sufficiently high frequencies the impedance of the conductor is so high the waves reflect off. That's why you can't use copper as a fiber optic cable. You can shine light on copper, and the electrons move, but the wave bounces off.

It's not so much that the electron drift reduces to zero at sufficiently high frequency as it is that transmission down the wire ceases.

• Electromagnetism
Replies
27
Views
1K
• Electromagnetism
Replies
2
Views
1K
• Electromagnetism
Replies
25
Views
3K
• Electromagnetism
Replies
17
Views
767
• Electromagnetism
Replies
24
Views
1K
• Electromagnetism
Replies
2
Views
2K
• Electromagnetism
Replies
3
Views
2K
• Electromagnetism
Replies
15
Views
2K
• Electromagnetism
Replies
6
Views
2K
• Electromagnetism
Replies
13
Views
2K