What causes an electron to flow in a current?

[bhagwad]

Say I have a copper wire connecting the Earth and the moon (or some other large distance). The Earth side is touching a "negative" source and the moon is about to touch a positive one". As soon as the connection is made, how do the electrons at the Earth know about it? Obviously the signal can't travel faster than the speed of light, but if I'm an individual electron sitting at earth, what forces propel me to move?

 

[SteamKing]

IDK about wires connecting the Earth and the moon, but current flows in a conductor when a potential difference is established between two points.

 

[jtbell]

The electric field in the wire at the location of an electron, exerts an electric force on the electron.

Added: SteamKing just barely beat me to it. His potential difference ΔV and my electric field E are related by ΔV = -EΔx where Δx is the length of the wire, assuming the field is uniform along the length of the wire.

 

[bhagwad]

These are very "macro" answers. While I know about potential differences and electric fields, but I'm not certain as to the exact force felt. Does the electric field travel as fast as the speed of light? How does the electron on Earth even know about what's happening on the moon? What if the wire was connecting two galaxies? How and in what way will the electrons at one end "know" when to move and what propels them?

 

[jtbell]

Does the electric field travel as fast as the speed of light?

Changes in the electric field in a wire do not travel at the speed of light in a vacuum, but they do travel at a significant fraction of that speed. A Google search for "speed of electric signal in a wire" turns up a number of hits, including at least one old thread here:

https://www.physicsforums.com/showthread.php?t=360612

 

[sophiecentaur]

These are very "macro" answers. While I know about potential differences and electric fields, but I'm not certain as to the exact force felt. Does the electric field travel as fast as the speed of light? How does the electron on Earth even know about what's happening on the moon? What if the wire was connecting two galaxies? How and in what way will the electrons at one end "know" when to move and what propels them?

That's because it is a very "macro" business. There are ways of describing most macro phenomena in micro terms but they are nothing like as 'to the point' as Maxwell's Equations, the Gas Laws, Hooke's Law or Kirchoffs laws. Those laws involve a bit of Maths, from the start and people don't like that. They like the 'mechanical' feeling they can get from thinking particles. But a really good explanation in particle terms can't be had without loads of complications and, in the end, serious Maths. They are not any more valid and the macro approach is not a 'cop out'.

I would say that, only when you have fully investigated both macro and micro approaches to a given problem, can you claim to have a well informed opinion about which one is more 'real' or valid. People who can cope with both approaches do not tend to make such naive distinctions. They will use the one that works best for them for each particular situation.

Electrons in a semiconductor junction are best approached microscopically, the 'same' electrons in a length of wire or a resistor are best approached macroscopically.

 

[bhagwad]

I'm familiar with the macro way of looking at it already. Just like I studied thermodynamics in college and know about the general behavior of such systems, I also like to know what it is that an individual particle is "feeling". Basically I want to know what happens if I'm an electron and the circuit is completed one light year away. Do I feel an attraction to the opposite end? Repulsion from my end? I know the "drift" velocity is a statistical property and I can understand that. What I'm not clear about is how do I even know that a connection has been made one light year away after a certain point of time?

Whether people like maths or not is irrelevant to me as far as this is concerned.

 

[sophiecentaur]

The way to approach what the fields are, at a distance, is EM Wave theory (macroscopic). A particle (microscopic) in the field that you will then have calculated will "feel" the field as it is at a particular time. That's in the definition of a Field.
You would know about an event, 1 LY away, One Year 'later'. But the light / EM pulse you originally produced took One Year to get there, so you could say that you would know that you had caused something to happen 1LY away, 2Years later.

Circuit theory was really not designed around that sort of a problem. The nearest way to get an answer to that sort of question would be to approach it in terms of a transmission line, which is a macroscopic approach. That approach looks at the wave as it propagates and gets reflected at discontinuities. After sufficient time, the impedance, seen by the source at the transmitting end, will settle down to a steady value after all the significant internal reflections in the line have finally reached back to the source. In a 'sensible ' length of line, this could involve several transits of the line. Loss in a very long cable could mean that you would have no idea of just how far your signal got; it could have been dissipated on the way.
How practical or ideal did you want the model to be?

 

[bhagwad]

This explanation seems fine if you say that the electron feels the field after one year. I guess my question would be "How does the circuit know that it's been completed?"

 

[sophiecentaur]

The way to approach what the fields are, at a distance, is EM Wave theory (macroscopic). A particle (microscopic) in the field that you will then have calculated will "feel" the field as it is at a particular time. That's in the definition of a Field.
You would know about an event, 1 LY away, One Year 'later'. But the light / EM pulse you originally produced took One Year to get there, so you could say that you would know that you had caused something to happen 1LY away, 2Years later.

Circuit theory was really not designed around that sort of a problem. The nearest way to get an answer to that sort of question would be to approach it in terms of a transmission line, which is a macroscopic approach. That approach looks at the wave as it propagates and gets reflected at discontinuities. After sufficient time, the impedance, seen by the source at the transmitting end, will settle down to a steady value after all the significant internal reflections in the line have finally reached back to the source. In a 'sensible ' length of line, this could involve several transits of the line. Loss in a very long cable could mean that you would have no idea of just how far your signal got; it could have been dissipated on the way.
How practical or ideal did you want the model to be?

This explanation seems fine if you say that the electron feels the field after one year. I guess my question would be "How does the circuit know that it's been completed?"

I thought my answer implied that already. All that stuff about transmission lines says it in detail. It 'knows' when it gets enough echoes with the right timing so that it can unequivocally tell they must be coming from a completed circuit an open circuit or a perfect termination. For more detail, you would need to provide a precise problem.

The down side for you is that transmission lines have a characteristic impedance, if you terminate a transmission line with this value of resistance (often this is 50Ω). Unless you actually knew the length of the line, you wouldn't know whether 'no echo' meant the 50Ω was added or the line was long enough for an echo not to have arrived back.

I think it may best to come to terms with the macroscopic approach for most of the answer and then mentally put your particle at the far end for it to receive a little 'kick' when the pulse gets there.

 

[Dale]

This explanation seems fine if you say that the electron feels the field after one year. I guess my question would be "How does the circuit know that it's been completed?"

If you want to look at the micro view then there are no circuits. There are only fields and charges. The fields and charges are governed by Maxwell's equations if you are not so micro that you care about quantum effects, or by QFT if you do care about quantum effects.

Circuits are a macro approximation to Maxwell's equations which are only valid under certain assumptions, so they don't exist at the micro level. One of the assumptions is the so called "small circuit" approximation. A 1 ly circuit violates that approximation.

 

[jtbell]

This explanation seems fine if you say that the electron feels the field after one year. I guess my question would be "How does the circuit know that it's been completed?"

Changes in the electromagnetic field propagate on a local, point-by-point basis. A change in the field at point A produces a change in the field at a neighboring point B, which in turn produces a change in a next-neighboring point C, etc. This is described by a set of coupled partial differential equations for $\vec E$ and $\vec B$, namely Maxwell's Equations.

At the level of classical electrodynamics, that's as far as you can go in the way of "explanation," because Maxwell's Equations are to classical electromagnetism what Newton's laws of motion are to classical mechanics.

 

[bhagwad]

Right ho. Thanks!

 

[analogdesign]

Keep in mind that electrons are quantum mechanical objects so intuition about electrons is very hard to come by. One way to think about it in a rough analogy is to imagine a tube filled with ball bearings. The tube can be quite long. The tube is completely full of ball bearings so there is no space between the individual bearings. When you hit the bearing at one end of the tube then you see a bearing flying out of the side. You wouldn't have trouble imagining that the signal is communicated through the tube, correct?

It's not so different (in a very rough sense) with electrons in a metal. They are thermally agitated and moving very fast, being constantly repulsed through interactions with other electrons and vibrating lattice sites. Now if there is a potential difference the electron at the "start" of the tube tend to hit electrons more often further down the tube and transfer some kinetic energy. This process repeats. The potential then drops as the kinetic energy is continually transferred down the tube. This is Ohm's law.

Was that any help?

 

[cabraham]

What cuases electrons to flow in a current is - proximity to other charges, i.e. Coulomb force.

Claude

 

[bhagwad]

Keep in mind that electrons are quantum mechanical objects so intuition about electrons is very hard to come by. One way to think about it in a rough analogy is to imagine a tube filled with ball bearings. The tube can be quite long. The tube is completely full of ball bearings so there is no space between the individual bearings. When you hit the bearing at one end of the tube then you see a bearing flying out of the side. You wouldn't have trouble imagining that the signal is communicated through the tube, correct?

It's not so different (in a very rough sense) with electrons in a metal. They are thermally agitated and moving very fast, being constantly repulsed through interactions with other electrons and vibrating lattice sites. Now if there is a potential difference the electron at the "start" of the tube tend to hit electrons more often further down the tube and transfer some kinetic energy. This process repeats. The potential then drops as the kinetic energy is continually transferred down the tube. This is Ohm's law.

Was that any help?

YES, that is excellent! Thank you!

 

[sophiecentaur]

YES, that is excellent! Thank you!

Not so excellent when you actually try to calculate the non-thermal KE of all the electrons. The drift velocity is a few mm per second and the free electrons constitute perhaps 1/100000 of the mass of the metal*. What does that mean in terms of energy available to 'transfer'?
I know you want a good, 'particle' explanation but you should make sure you pick one that holds water.

*Copper has an atomic mass of 63 and an electron (one available per atom ) is 1/1800