Understanding the Nature of Electricity: Flow of Electrons or Electric Charge?

[sophiecentaur]

Of course. The same way that an antenna can grab a range of signals out of the air from a whole range of EM waves, then a selective receiver can filter out the narrow band of frequencies it wants to look at from a single antenna downlead. Just think of the many instances where a single antenna feeds a large number of receivers (blocks of flats with a distribution amp on the roof) and where a number of transmitters can feed into a single transmitting antenna (UHF TV transmitting stations).
Superposition's the thing.

 

[toneboy1]

Of course. The same way that an antenna can grab a range of signals out of the air from a whole range of EM waves, then a selective receiver can filter out the narrow band of frequencies it wants to look at from a single antenna downlead. Just think of the many instances where a single antenna feeds a large number of receivers (blocks of flats with a distribution amp on the roof) and where a number of transmitters can feed into a single transmitting antenna (UHF TV transmitting stations).
Superposition's the thing.

Great!
How can an antenna have standing wave voltages on it, yet the charges don't 'resonate'?
So they're moving with the erratic signal, could you clarify what you mean? Maybe its the terminology 'resonate' that is throwing me.

Thanks

 

[sophiecentaur]

Great!
How can an antenna have standing wave voltages on it, yet the charges don't 'resonate'?
So they're moving with the erratic signal, could you clarify what you mean? Maybe its the terminology 'resonate' that is throwing me.

Thanks

Current can flow without there needing to be any resonance. An electric field can cause charge to flow in an antenna just as it can anywhere else. There is also the Magnetic field that will be causing currents. Alternating in direction isn't the same as resonance.

Are you trying to link this into QM, or something?

 

[toneboy1]

Are you trying to link this into QM, or something?

I assure you I am not, lol.

I'm just trying to get my mind around what is taking place in the antenna to create these standing wave voltages and currents.

 

[sophiecentaur]

You don't need a standing wave to pick up some energy from a passing EM wave. It's only a design detail to make it more effective.

Try this view of a dipole. If you have a two wire transmission line then a progressive EM wave can pass along it, bound to the two wires (this is how AC 'electricity' travels along wires). If you terminate the line with the appropriate resistor (the load) then the power is all dissipated in the load. If you terminate the line incorrectly (like an open circuit) then the wave is all reflected back (a nuisance, sometimes) and you will have a standing wave in it.
Now split the last metre or so of the line and separate the wires to form a straight line at right angles to the line (a dipole). You still have a transmission line and it's still terminated at an open circuit and you will still get a standing wave with NO CURRENT flowing at the tips. But there is still SPACE for some of the energy to be radiated into. This radiated energy 'looks' to the line like a resistance because Power is going somewhere. You can actually measure a resistance (an RF resistance, that is) if you connect an analyser / bridge / gizmo to the line and it's called (not surprisingly) the Radiation Resistance. But there is still a standing wave of current on the dipole and you will also measure a reactance across the line. By choosing the dipole to be a half wave long, the reactance goes to zero (that's your resonance). It isn't essential to resonate but you get a good 'match' if you do. All that remains to do is to choose your transmission line dimensions to have the same characteristic impedance as the dipole presents and all the power gets radiated. The dipole looks like about 70 Ohms so if the line is also 70 Ohms, Bob's your Uncle.
Most feeder is coaxial and not two wire (better isolation for the feeder) and you can commonly get 50 Ohm and 75 Ohm - not 70 Ohm. This is because of the way that real dipoles and real antennae are connected etc. etc. and you don't need to worry too much first time round about the discrepancy.
But the basic thing is that a dipole behaves very much like a continuation of the transmission line that feeds it.

 

[Averagesupernova]

An excellent post sophie!

 

[sophiecentaur]

Tvm.

 

[toneboy1]

You don't need a standing wave to pick up some energy from a passing EM wave. It's only a design detail to make it more effective.

Try this view of a dipole. If you have a two wire transmission line then a progressive EM wave can pass along it, bound to the two wires (this is how AC 'electricity' travels along wires). If you terminate the line with the appropriate resistor (the load) then the power is all dissipated in the load. If you terminate the line incorrectly (like an open circuit) then the wave is all reflected back (a nuisance, sometimes) and you will have a standing wave in it.
Now split the last metre or so of the line and separate the wires to form a straight line at right angles to the line (a dipole). You still have a transmission line and it's still terminated at an open circuit and you will still get a standing wave with NO CURRENT flowing at the tips. But there is still SPACE for some of the energy to be radiated into. This radiated energy 'looks' to the line like a resistance because Power is going somewhere. You can actually measure a resistance (an RF resistance, that is) if you connect an analyser / bridge / gizmo to the line and it's called (not surprisingly) the Radiation Resistance. But there is still a standing wave of current on the dipole and you will also measure a reactance across the line. By choosing the dipole to be a half wave long, the reactance goes to zero (that's your resonance). It isn't essential to resonate but you get a good 'match' if you do. All that remains to do is to choose your transmission line dimensions to have the same characteristic impedance as the dipole presents and all the power gets radiated. The dipole looks like about 70 Ohms so if the line is also 70 Ohms, Bob's your Uncle.
Most feeder is coaxial and not two wire (better isolation for the feeder) and you can commonly get 50 Ohm and 75 Ohm - not 70 Ohm. This is because of the way that real dipoles and real antennae are connected etc. etc. and you don't need to worry too much first time round about the discrepancy.
But the basic thing is that a dipole behaves very much like a continuation of the transmission line that feeds it.

So this transmission line for say, a block of flats. A coaxial cable coming off the antenna on the roof, terminated open-circuit with separate 'wires' coming off it at right angles to each unit's individual receiver? (it won't be 'matched'?)

ALSO, I assume the matter of the antenna DOESN'T operate from 'dipole moments', that's a separate concept isn't it?

THANKS

 

[sophiecentaur]

So this transmission line for say, a block of flats. A coaxial cable coming off the antenna on the roof, terminated open-circuit with separate 'wires' coming off it at right angles to each unit's individual receiver? (it won't be 'matched'?)

ALSO, I assume the matter of the antenna DOESN'T operate from 'dipole moments', that's a separate concept isn't it?

THANKS

What do you mean by that? The antenna will, ideally, be matched to the 50 Ohm feeder.
If you know about transformers at 50Hz AC, you will know that you can transform a 230V supply to suit a low resistance bulb (designed to operate at 12V, say). If the bulb is a 100W rating, it will draw about 8A. It will have a resistance of 1.5Ω. But the mains will only be supplying 100W, so it must be supplying less than 0.5A so it will be 'seeing' a resistance of more than 500Ω. You can ignore those actual details but you can see that the transformer is also transforming the load resistance. RF matching networks do the same sort of thing and it is easy to split a 50Ω line to feed two 50Ω loads with a suitable loss-less splitting network. (They would look like 25Ω if you just put them in parallel). The system can remain matched throughout and they get 50% each of the power. You are right to say that 'just hanging' random feeders across a feeder will spoil the match and will lose a lot of signal power.
In practice, there will, normally, be a distribution amplifier (on the roof) so that everyone gets a high level signal but the down-leads are isolated from each other.


BTW Dipole Moment is not part of this stuff.

 

[toneboy1]

What do you mean by that? The antenna will, ideally, be matched to the 50 Ohm feeder.
If you know about transformers at 50Hz AC, you will know that you can transform a 230V supply to suit a low resistance bulb (designed to operate at 12V, say). If the bulb is a 100W rating, it will draw about 8A. It will have a resistance of 1.5Ω. But the mains will only be supplying 100W, so it must be supplying less than 0.5A so it will be 'seeing' a resistance of more than 500Ω. You can ignore those actual details but you can see that the transformer is also transforming the load resistance. RF matching networks do the same sort of thing and it is easy to split a 50Ω line to feed two 50Ω loads with a suitable loss-less splitting network. (They would look like 25Ω if you just put them in parallel). The system can remain matched throughout and they get 50% each of the power. You are right to say that 'just hanging' random feeders across a feeder will spoil the match and will lose a lot of signal power.
In practice, there will, normally, be a distribution amplifier (on the roof) so that everyone gets a high level signal but the down-leads are isolated from each other.

You are a legend.
I feel like I've just taken a short course over the past couple days!

I didn't think so, out of interest, do you know what part of the EM wave causes the dipole moment (electron and proton to re-orientate) when a wave passes a molecule?
(maybe now I am "trying to link this into QM" :P )

 

[sophiecentaur]

You are a legend.
I feel like I've just taken a short course over the past couple days!

I didn't think so, out of interest, do you know what part of the EM wave causes the dipole moment (electron and proton to re-orientate) when a wave passes a molecule?
(maybe now I am "trying to link this into QM" :P )

Intense short course, definitely!
"Which part"? There is an E field and an H field. They are both changing, why should only one "part" be responsible. In any case, if you are talking about interaction with a single molecule then you can identify one photon that will be responsible - so that's QM :tongue2: back to you!

 

[DrZoidberg]

Like jam said, they do move, but in that process of moving they bump into each other, because they are tightly packed. That bumping is also form of movement. And that movement propagates at the speed, roughly half the speed of light. (Why half? Material reasons)

The thermal velocity of electrons in a wire is around 100,000 m/s. That's very fast but still 3000 times slower than light. If you close a switch the resulting change in electric potential propagates close to the speed of light. If an electric signal (i.e. a propagating change in potential) moved because electrons are "bumping" into each other, it would only move at 100,000 m/s. Just like a sound wave in air. The thermal velocity of air molecules is 340 m/s, that means that pressure waves (sound waves) move at 340 m/s. So electric signals are no pressure waves. They have nothing to do with electrons bumping into each other. An electric signal is mediated by electric fields. Let's say you have a battery and short it with a copper wire. The chemical reactions inside the battery create a charge imbalance which in turn creates an electric field which then pushes electrons through the wire. btw. as long as a current is flowing an electric field can exist inside the wire.

I repeat, current is NOT movement of electric charges, this is common misunderstanding. Current is a impulse of energy. And even that last one cannot be taken for granted.

"impulse of energy"? Did you take that expression from a Star Trek episode? It's certainly not a proper physics term.

 

[toneboy1]

Intense short course, definitely!
"Which part"? There is an E field and an H field. They are both changing, why should only one "part" be responsible. In any case, if you are talking about interaction with a single molecule then you can identify one photon that will be responsible - so that's QM :tongue2: back to you!

H'mm, ok well if there is a single particle of some charge, and a photon comes buy, would the particle move on one direction for half the wavelength then in the opposite for the other half? (and not get very far?)

Thanks

 

[toneboy1]

The thermal velocity of electrons in a wire is around 100,000 m/s. That's very fast but still 3000 times slower than light. If you close a switch the resulting change in electric potential propagates close to the speed of light. If an electric signal (i.e. a propagating change in potential) moved because electrons are "bumping" into each other, it would only move at 100,000 m/s. Just like a sound wave in air. The thermal velocity of air molecules is 340 m/s, that means that pressure waves (sound waves) move at 340 m/s. So electric signals are no pressure waves. They have nothing to do with electrons bumping into each other. An electric signal is mediated by electric fields. Let's say you have a battery and short it with a copper wire. The chemical reactions inside the battery create a charge imbalance which in turn creates an electric field which then pushes electrons through the wire. btw. as long as a current is flowing an electric field can exist inside the wire.

"impulse of energy"? Did you take that expression from a Star Trek episode? It's certainly not a proper physics term.

I wonder what Sophie will say, I recall sophiecentaur saying movement of electrons was in the mm/s, if there's a distinction I don't see it...

 

[truesearch]

I would agree with dr zoidberg's last comment.
I have read in the rules of these forums that content of posts should be traceable to accepted textbooks.
I have not met a textbook that does not recognise that current is a flow of charge.
If there is a book that suggests otherwise I would like the reference, I want a copy of that book.

 

[toneboy1]

studiot gave a reference to...some sort of textbook.
Anyway, I don't think the conclusion thus far was disputing that current was flow of charge/change in time
Just discussing the mechanics of it.

 

[Kholdstare]

I praise sophie's patience. @toneboy1 - read textbooks. Try to understand everything written there. That will save you a lot of time of what you're asking here.

 

[toneboy1]

I praise sophie's patience. @toneboy1 - read textbooks. Try to understand everything written there. That will save you a lot of time of what you're asking here.

As do I sophie's patience, as I have said. In fact at one point I had given up on a concept but sophie exclaimed not to.
Two things, first being that I can barely afford rent or three meals a day let alone text-books, I read everything relevant I can find on the internet not just forums, I spend a lot of as much time as I can learning about the natural world via the means available (khan academy, anything).
What I think is unfair Kholdstare is that you imply I could simply find things out elsewhere, as a matter of fact some of the time I want to cross-check that my model is actually the case, but it just so happens that I'm usually wrong :tongue2:

Second thing, If my questions (and the answers that follow) are of no interest or help to anyone else who finds the thread than I'll stop asking, I don't wish to selfishly consume people's time.

 

[Kholdstare]

@toneboy1 - I did not know about your economic problems. I am sorry to know that. But nothing can beat a well written textbook. It is typical among people to avoid reading textbooks and find quick answers and explanations. Sometimes this produces lengthy threads when the OP could read a little more and clear his doubts.

However, I deeply sympathize for your problem and I am sorry to suspect you as a lazy person. You are not selfishly consuming any people's time and everybody would love to help you.

 

[toneboy1]

@toneboy1 - I did not know about your economic problems. I am sorry to know that. But nothing can beat a well written textbook. It is typical among people to avoid reading textbooks and find quick answers and explanations. Sometimes this produces lengthy threads when the OP could read a little more and clear his doubts.

However, I deeply sympathize for your problem and I am sorry to suspect you as a lazy person. You are not selfishly consuming any people's time and everybody would love to help you.

I completely agree, I much preferred being able to bookmark things and it felt better on the eyes when I was at an institution that had a decent library.
Thank you for saying that. One problem I do find is that when you learn something small, that's wrong, early on, this little red herring ruins future models of how you see the world based on it, though I find when you are talking to another person they can easily say 'hey what you said is wrong or doesn't make sense' but when you're researching on your lonesome sometimes you just sit around getting more confused and you can't reason why.
(I hope people will still tell me If I am wasting someones time though in future)

 

[truesearch]

You are not wasting anybodies time!
No one is forced to post here! It is very satisfying to be able to help and there is so much expertise here that think we are all still learning.
We are all here for the same reason.

 

[sophiecentaur]

H'mm, ok well if there is a single particle of some charge, and a photon comes buy, would the particle move on one direction for half the wavelength then in the opposite for the other half? (and not get very far?)

Thanks

I think you mean half the period? But yes, basically.
This is the grey zone between QM and Classical.
There is a situation in the Ionosphere where there are free electrons that can be 'seen' to 'vibrate' as a radio wave passes. Again, this is because the energy gaps involved for a free electron in a very low density plasma are very small and a classical approach works fine by treating the plasma as a conductor with the electrons moving one way and the much heavier ions moving (a smaller distance) the other way, as the fields vary around them. Because the electrons are not in a metal, they actually do move a significant distance in the time period of a 1MHz radio wave.

If you don't have access to textbooks then trawl around the net for .org and .edu sites for more reliable opinions. Beware, there are some dreadful, cranky and harmful sites that may read as gospel. Look for a majority opinion if you get confused.

 

[toneboy1]

I think you mean half the period? But yes, basically.
This is the grey zone between QM and Classical.
There is a situation in the Ionosphere where there are free electrons that can be 'seen' to 'vibrate' as a radio wave passes. Again, this is because the energy gaps involved for a free electron in a very low density plasma are very small and a classical approach works fine by treating the plasma as a conductor with the electrons moving one way and the much heavier ions moving (a smaller distance) the other way, as the fields vary around them. Because the electrons are not in a metal, they actually do move a significant distance in the time period of a 1MHz radio wave.

If you don't have access to textbooks then trawl around the net for .org and .edu sites for more reliable opinions. Beware, there are some dreadful, cranky and harmful sites that may read as gospel. Look for a majority opinion if you get confused.

Thanks for the advice! There is an art to 'trawling', yes I've seen a lot of contradictory information on subjects where you would expect only people who know what they are talking about would contribute to.

I realize in hind-sight that I phrased my question poorly, intuitively I was thinking that only the E part of the photon would stimulate a dipole moment or electrons. I think you answered my queries never-the-less.

TYVM!