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

[toneboy1]

They are not different, only different levels of describing the same concept. Kholdstare is explaining the physical description of how it comes about.

In that case I'm intrigued and confused. So it is kind of like a 'compression wave' of charge/time on a grand level but, so on a small level where does Kholdstare's physical description incorporate this?
thanks

 

[DragonPetter]

In that case I'm intrigued and confused. So it is kind of like a 'compression wave' of charge/time on a grand level but, so on a small level where does Kholdstare's physical description incorporate this?
thanks

Simply work out the units of the equation kholdstare provided and see that it gives the units in the definition sophiecentaur gave. That is a good first step to seeing how they are related.

I don't know if I would use phrases like grand level to describe it, but current can be given by kholdstare's equation within the context that it is derived, which are materials with charge carriers.

 

[toneboy1]

Ok, I still don't see how the model of 'slow moving charge carriers' (electrons), moving in some way fits into what is apparently the same model.

As far as I can see from following your instructions:
I=∆Q/∆t=qnAvd = (units) q . N/m^3 . m^2 . m/s = qN/s

this seems to imply that IT IS the slow moving charge carriers responsible for the fast speed of current...?

 

[sophiecentaur]

Ok, I still don't see how the model of 'slow moving charge carriers' (electrons), moving in some way fits into what is apparently the same model.

As far as I can see from following your instructions:
I=∆Q/∆t=qnAvd = (units) q . N/m^3 . m^2 . m/s = qN/s

this seems to imply that IT IS the slow moving charge carriers responsible for the fast speed of current...?

This would seem to be your problem. You are assuming that the current 'moves fast'. Why? Consider the water in the middle of a lake, with two fast-moving streams - one in one out. How fast is it moving in the middle? How many litres per second are going through the lake, though? Is it speed or volume flow that counts?
What DOES move fast is the effect, in an electric circuit, of turning it on. A pulse of EM travels through / along it and will cause current to flow into the light bulb at the end in a matter of a few nanoseconds. That doesn't mean that the 'current flowed all the way' to the bulb from the battery in order to light it up.
When you stamp on a bicycle pedal, the bike moves forward 'instantly'; you don't wait for the chain to move round from the chain wheel to the sprocket before you move off. Also, by choosing an appropriate gear, you can deliver different amounts of power with the same pedalling speed. Chain speed / electron speed are not the relevant quantities here.
I think some people here are trying to 'bend' reality to fit their own personal models of 'Electricity'. It would be better to start with the very basics and move towards a fuller understanding. The accepted model really does work well and I think you need to believe it's the best way to look at things.

 

[toneboy1]

...I think some people here are trying to 'bend' reality to fit their own personal models of 'Electricity'. It would be better to start with the very basics and move towards a fuller understanding. The accepted model really does work well and I think you need to believe it's the best way to look at things.

Ok, making sense again. So some EM field propagates through the wire making them all drift, first in (from battery) last out (to circulate back around).

Thanks!

 

[sophiecentaur]

Furthermore - you don't actually need a connection. The same thing happens between a transmitting antenna and a distant receive antenna. No charges can flow between them at all. Charges flowing in one place can cause charges to flow in another place.

 

[toneboy1]

Furthermore -...No charges can flow between them at all. Charges flowing in one place can cause charges to flow in another place.

Great answer before btw.

So trying not to use my own terms but the model I gather from what you've told me, there is this EM wave propagating through the antenna (at say 10MHz) and the charge throughout the antenna is sort of just oscillating where it is, because it moves too slowly to really go anywhere. (?) We are really seeing VOLTAGES along the antenna as the discernible change. (?)

Given I=∆Q/∆t=qnAvd What sort of current can we expect to see in the inductor if there is a bandpass filter connected to the antenna?
Would it be like a regular 2μA AC?

THANKS!

 

[sophiecentaur]

What sort of current can we expect to see in the inductor if there is a bandpass filter connected to the antenna?
Would it be like a regular 2μA AC?

THANKS!

If the bandpass filter presents the right impedance (matched) to the feed point of the antenna then you would expect an AC signal just like was transmitted. Mostly, of course, the signal would have modulation (information) on it (the only reason for transmitting in the first place, usually!)

As I mentioned before, most antennae (except for some AM broadcast transmitting antennae) do not have resonant circuits in them because they are required to transmit and receive over a range of frequencies.

If you look at the volts and current at different parts of a simple dipole, they follow the pattern of a standing wave on the antenna - Zero current at the ends, for instance and maximum volts at the ends for dipoles around half a wavelength long. Google imnages of current distribution on a dipole.

 

[toneboy1]

If the bandpass filter presents the right impedance (matched) to the feed point of the antenna then you would expect an AC signal just like was transmitted. Mostly, of course, the signal would have modulation (information) on it (the only reason for transmitting in the first place, usually!)

As I mentioned before, most antennae (except for some AM broadcast transmitting antennae) do not have resonant circuits in them because they are required to transmit and receive over a range of frequencies.

If you look at the volts and current at different parts of a simple dipole, they follow the pattern of a standing wave on the antenna - Zero current at the ends, for instance and maximum volts at the ends for dipoles around half a wavelength long. Google imnages of current distribution on a dipole.

Right, so the actual modulation of the inductance / capacitance on the transmission circuit antenna is how they send information for say FM?

(Google imaging λ/2...was helpful)

Standing wave for reception but not for transmission? (all the energy being reflected back otherwise)

If unless you have any objections to that I think I may have 'got it' ;D

As always, Thanks!

 

[sophiecentaur]

Right, so the actual modulation of the inductance / capacitance on the transmission circuit antenna is how they send information for say FM?

Owch. That's a step too far, I'm afraid. You're inventing things in your head! The FM signal is produced and amplified and it's then fed to the antenna (which is probably up at the top of a mast, tens of metres away). The matching network and antenna have to be designed to be broad band enough to take several signals at once so you can't do the modulation at the antenna.

 

[toneboy1]

Owch. That's a step too far, I'm afraid. You're inventing things in your head! The FM signal is produced and amplified and it's then fed to the antenna (which is probably up at the top of a mast, tens of metres away). The matching network and antenna have to be designed to be broad band enough to take several signals at once so you can't do the modulation at the antenna.

A couple dosen steps too far apparently!
Well how about Standing wave for reception but not for transmission? (all the energy being reflected back otherwise) Was I right there?

 

[sophiecentaur]

So you buy this sexy expensive new receiver and then you have to climb up onto your chimney to connect it up to the aerial? This is killing me!
The antenna installation needs to be entirely separate from the the transmitter (except in the case of an MF receiver with a ferrite rod, where the coil / rod antenna is an integral part of the first stage of the receiver and is tuned as you turn the tuning knob - so your not as daft as you may look ). For HF, VHF or UHF, however you may want to connect several receivers to an elevated antenna and they may each want to receive entirely separate channels.
Is this helping? I'm beginning to enjoy it.

 

[toneboy1]

So you buy this sexy expensive new receiver and then you have to climb up onto your chimney to connect it up to the aerial? This is killing me!
The antenna installation needs to be entirely separate from the the transmitter (except in the case of an MF receiver with a ferrite rod, where the coil / rod antenna is an integral part of the first stage of the receiver and is tuned as you turn the tuning knob - so your not as daft as you may look ). For HF, VHF or UHF, however you may want to connect several receivers to an elevated antenna and they may each want to receive entirely separate channels.
Is this helping? I'm beginning to enjoy it.

I appreciate your help and I am amazed at your patience but I'm more confused now than when I started :P
I'll understand if you want to give up. God knows it's 4:15am here and I don't think I'm ever going to understand how you can get multiple signals off a piece of non resonating metal.

 

[sophiecentaur]

Resonance only means you get a bit more power out. Any passing em wave will induce a current into anything that conducts. Proper engineering just maximises the amount.
Two RF signals will produce two outputs from the device. (It's all linear.)
Don't give up.

 

[toneboy1]

Resonance only means you get a bit more power out. Any passing em wave will induce a current into anything that conducts. Proper engineering just maximises the amount.
Two RF signals will produce two outputs from the device. (It's all linear.)
Don't give up.

Ok, well I suppose could we PLEASE go over:

-If most don't resonate how is there a standing wave?
-Aren't standing waves bad? (because all the energy is being reflected back, or is that not true for receivers?)
-So if you've just got a conductor in the air, there is all manner of current in it (crazy mess) from all the different EM waves passing through it?

Edit- This picture is the type of thing I'm thinking of, what's going on inside the antenna (receiver) as far as voltages and currents?
How does the inductance and capacitance play a role?
Thanks

 

[sophiecentaur]

You don't need resonance for a standing wave. If a wave is reflected at a barrier, the reflected wave will interfere with the incident wave and the resultant will be a standing wave as the relative phases change over the distance from the reflection. You are right to pick this up because most times you are told about standing waves, you are dealing with strings at resonance - when there are reflections at both ends and, when the spacing is right, the energy builds up in the interference pattern and you get resonance (But for friction losses, the amplitude would go infinite). If you take a short string, fixed at the other end, and waggle it from side to side. you will get a standing wave, even for a very slow oscillation but it isn't exactly obvious - except for the fact that the displacement of the string goes from max at your hand and zero at the other end. The shape of the string will be triangular for very frequencies but it will become more and more sinusoidal as you go faster, until you reach the resonant frequency, then the string shape will be a sinusoid.
This ties in with the images you have seen. Think of the wave along a short dipole as an interference pattern - min current at the end because the 'I' can't go anywhere!. If the dipole happens to be half wavelength long then the current will be in just the right phase to slosh up and down in time with the input signal and you will get resonance. Because the current is high, then this will couple best into a radiated wave. The antenna is 'matched' into free space and appears to be just a resistor of about 70Ω at the feed point aamof. The amplitude of the resonance is, of course, limited by the loss of energy into space.

You will, by now, be suffering from indigestion, I'm sure. There's an AWFUL LOT of this to take in.

 

[sophiecentaur]

OH yes - "reflection is bad" when it occurs in a transmission line because some of the power you want to be transmitted will be reflected and cause high voltages on the line - compromising the electronics and also introducing echos and non flat frequency response. Reflection IN an antenna is part of the way it works; you could say that the currents flowing due to the resonance will mean more power is radiated. Not all antennas are resonant, however - you can launch a wave from a very long wire and, by the end of it, most of the energy has been radiated.

 

[toneboy1]

OH yes - "reflection is bad" when it occurs in a transmission line because some of the power you want to be transmitted will be reflected and cause high voltages on the line - compromising the electronics and also introducing echos and non flat frequency response. Reflection IN an antenna is part of the way it works; you could say that the currents flowing due to the resonance will mean more power is radiated. Not all antennas are resonant, however - you can launch a wave from a very long wire and, by the end of it, most of the energy has been radiated.

OH! The transmission line! So is that meant to have a constant voltage along it?
(maybe if we were looking at an instant in time)

If you've just got a conductor (an unused antenna) sitting in the air, with multiple EM waves going through it, are there multiple standing waves due to these EM passing?

 

[sophiecentaur]

OH! The transmission line! So is that meant to have a constant voltage along it?
(maybe if we were looking at an instant in time)

If you've just got a conductor (an unused antenna) sitting in the air, with multiple EM waves going through it, are there multiple standing waves due to these EM passing?

The Peak to Peak voltage variations should be the same at all points along the line.

In pretty well every situation we are dealing with a linear system. The currents / waves along the antenna are a superposition of the effects of all the different passing EM waves. Multiple standing waves, if you like.

 

[toneboy1]

In pretty well every situation we are dealing with a linear system. The currents / waves along the antenna are a superposition of the effects of all the different passing EM waves. Multiple standing waves, if you like.

THANK YOU I've been looking for that answer for ages!
So they can sort of filter out different frequencies of this one, superposition-ed wave, with different circuits to use the same antenna for multiple signals at once?

 

[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 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 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

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!