# How do electrons start to travel though a wire?

• antistrophy
In summary, the conversation discusses the possibility of setting up a wire one light year long with light bulbs connected to a battery, and how the bulbs would light up in different scenarios. The conversation also addresses the issue of resistance, voltage, and the propagation of potential through the wire. It is determined that the lights would not light up due to the high resistance of the circuit and the inability to put current in the circuit.
antistrophy
In trying to understand electricity I thought of the following: If I set up a wire one light year long, stretching out into space ½ a light year out and ½ a light year back, and connected a series of light bulbs all the way around the wire, and then connected a battery (large enough to handle the resistance obviously), how will the light bulbs light up? I thought of three different scenarios;

a) All the bulbs light up instantaneously. (in which case I don’t understand anything)

b) The bulbs on the negative side of the battery start lighting up one by one all the way around taking just over one year to light the last one next to the positive side. (I don’t like this idea either).

c) Both bulbs on the positive side and on the negative side of the battery light up as electrons are being pushed from the negative and pulled by the positive, with the last bulb to light up being the furthest away from the battery at just over ½ year. (I’m going to go for this one)

d) I just thought of a forth scenario that nothing happens for 1 light year and then all of them come on slowly, or something.

Hrmm...I don't really know for sure. I thought about it for about 5-10 minutes and wasn't really able to come up with a good guess. I'm just going to have to go with C.

They would never light up.

The resistance of 18 gauge wire is .0075 ohms/foot, so the resistance of the circuit is 2.3 x 1014 ohms. If I need 100 mV to light a bulb, my "battery" needs a voltage of 23 trillion volts. Of course that's impossible, but even if it were possible, the terminals would arc over.

Basically, a light year is so long that you can always find a path of less resistance. So you won't be able to put any current in the circuit.

If you used wires that could handle it the answer would be c) except that the wave might move slower than the speed of light. It depends on the arrangement of the wires.

So basically what we are saying here is that the electrons closest to the battery are affected first?

Forget the problem about the resistance; just think of it as an extremely short time period on a normal circuit.

They would never light up.

The resistance of 18 gauge wire is .0075 ohms/foot, so the resistance of the circuit is 2.3 x 1014 ohms. If I need 100 mV to light a bulb, my "battery" needs a voltage of 23 trillion volts. Of course that's impossible, but even if it were possible, the terminals would arc over.

Basically, a light year is so long that you can always find a path of less resistance. So you won't be able to put any current in the circuit.

I think the OP is simply interested in a circuit that is long enough to observe a sequence of events happening on a human scale.

Say we shorten it dramatically and use high speed cameras to observe the events.

The OP simply wants to understand if the lights all light up at once or in sequence.

There is a voltage wave traveling c or slower from the battery/switch out to the end of the wire so sequential lightup is the answer.

The potential between each battery terminal and wire is less than the potential between the terminals of the battery. Depending on the capacitance of the wire, it would seem that the lights would dimly light in sequence in parallel with distance from the battery, then full current would occur when the two potentials meet at the far end of the wire, and the lights would then go to full intensity in sequence in parallel starting at the far end of the wire back towards the battery.

okay so this voltage wave will travel down the wire (from both +ve and -ve ends of the battery?) and if they meet (closed circuit) then electrons move all at once?

In other words at the far side of the circuit put a switch. How long before we see a voltage difference and did any electrons next to the battery move even though the switch is open.

Still other words if it is sequential how do the electrons next to the battery "know" that the circuit is closed and that they can start moving if the voltage hasn't completed the circuit yet?

rcgldr said:
t would seem that the lights would dimly light in sequence in parallel with distance from the battery, then full current would occur when the two potentials meet at the far end of the wire

I was afraid this might be the case but I still have the problem of the lights lighting before the voltage has gone around the circuit. (and determined that it is closed)

I forgot to mention inductance is also an issue, affecting the rate of potential propagation.

antistrophy said:
In other words at the far side of the circuit put a switch. How long before we see a voltage difference and did any electrons next to the battery move even though the switch is open.
The time it takes depends on how long it takes for the voltage to propagate thorugh the wire. Capacitance and inductance will be an issue here.

antistrophy said:
Still other words if it is sequential how do the electrons next to the battery "know" that the circuit is closed and that they can start moving if the voltage hasn't completed the circuit yet?
They don't. The the far end of the wire could be open or closed and the initial reaction near the terminals would be the same.

antistrophy said:
I was afraid this might be the case but I still have the problem of the lights lighting before the voltage has gone around the circuit. (and determined that it is closed)
The initial voltage will be some fraction (probably 1/2) of the final voltage once the circuit fully completes. This may or may not be enough to cause the lights to get lit up.

Still other words if it is sequential how do the electrons next to the battery "know" that the circuit is closed and that they can start moving if the voltage hasn't completed the circuit yet?

I THINK this is kind of how it works. Someone correct me if I'm wrong.

The voltage source, in this case a battery, applies a force on the electrons, causing their random movements through the conductor to have a net movement away from the voltage source. Since the ones at the end of the wire cannot move further away the force equalizes and nothing happens. When the positive terminal is connected the electrons now have a destination to get to. The ones down towards the negative terminal don't immediately know that they now have a source, but over the course of a short amount of time the charge that had built up opposing their movement disappears and they again start their way towards the positive end. Note that a great many of these electrons aren't traveling away from the negative source. Most still take random paths through the conductor leading to no net direction overall, however as you increase the voltage more and more end up taking a net path towards the positive terminal. In the end you have a steady current through the whole conductor.

DaveC426913 said:
I think the OP is simply interested in a circuit that is long enough to observe a sequence of events happening on a human scale.

Say we shorten it dramatically and use high speed cameras to observe the events.

The OP simply wants to understand if the lights all light up at once or in sequence.

yes exactly...
But the big Q Dave is... do you have an answer ? :) was hoping to see your take on the problem

I remember this same Q on a forum a very long time ago. Not sure if it ever did get answered satisfactorily.

now the Voltage isn't travelling/flowing the path around the cct ? it doesn't in any normal cct. Current flows around the cct, voltage is just a potential difference between 2 points in the cct. its that potential difference that is "pushing" the electrons (current) out of the power source.
pretty sure I have that right ;)

I don't have an answer for sure, really looking forward to some insightful responses :)

I suspect there's going to be a lag if the switch is closed immediately. We also have velocity factors of the wires to take into account. ie electron propagation in a wire or other conductor isn't at the speed of light.
eg... 50 Ohm 1/4 inch coax 55% of c, hi quality 50 Ohm coax ~ 80 - 85% of c
wire or aluminium dipole antenna ~ 95% of c.

The other scenario I thought of is ...

the power source is connected to the loop, the switch is still open, after any time lags taken into account, there is going to be + and - charges present on either side of the switch contacts. as per normal cct. will all lights then light up instantly when the switch is closed ?

Dave

davenn said:
Will all lights then light up instantly when the switch is closed?
You end up with the same propagation delays starting from the just closed switch.

The change in electric field cannot travel faster than the speed of light, and the voltage on a charge depends on the distance the charge(electron) is in the field (integral of the field with respect to distance). When the switch is closed, this is a step change, or as close to as possible, which means that the spectrum of the EM wave generated should be made up of an infinite sum of sinusoids. This will look like a wave front of a new voltage level going through the conductor. If the voltage source is DC, then after the EM wave has looped, the magnetic component of the change in electric field goes away and only the magnetic field of the current is left, and a new static electric field is present.

Also, think of high speed digital circuits, when a pulse is sent from one circuit to the next, there is a time delay before the information is transferred; there is not an instantaneous transfer of charge and data, and, circuits closer will see the signal before circuits further away which is a practical concern in high speed digital and RF circuit board layouts. If it were possible for all the lights to light up at the same time, high speed circuit designers would have already taken advantage of that effect to solve these issues where the data on the pins do not arrive at the same time due to conductor length differences.

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DaveC426913 said:
I think the OP is simply interested in a circuit that is long enough to observe a sequence of events happening on a human scale.

Say we shorten it dramatically and use high speed cameras to observe the events.

That's a different problem, and we need a different abstraction. In the original problem, you have a system that is big and slow. It doesn't work, for essentially the same reason you can't blow through a mile long straw. If you say, OK, let's scale it down so we have a system that is big and slow, now the abstraction that this is well modeled by DC theory breaks down. You can't think "wire" any more. You have to think "transmission line".

That's a different problem, and we need a different abstraction. In the original problem, you have a system that is big and slow. It doesn't work, for essentially the same reason you can't blow through a mile long straw. If you say, OK, let's scale it down so we have a system that is big and slow, now the abstraction that this is well modeled by DC theory breaks down. You can't think "wire" any more. You have to think "transmission line".

Not necessarily a different problem, everything is ultimately a transmission line, even if it is in the big and slow scale, you still can apply the lumped circuit models, its just that the effects are so negligible that it looks the same as the "DC theory". The original question is actually asking about instantaneous changes which is very far away from a DC theory.

when we connect a voltage source to the wire(assuming dc)..the electrons adjacent to the + polarity are attracted to the positive polarity and neutralizes the + charge..,so the atoms adjacent to this becomes positively charged and they attracts the electrons from their neighboring atoms and the process continues ...,similarly electrons of atoms which are adjacent to - polarity are repelled from the negative charge of the voltage source...and the process continues ...so i think lights adjacent to the battery terminals glow first...and in the conduction process electrons do not move with speed of light

hanii said:
when we connect a voltage source to the wire(assuming dc)..the electrons adjacent to the + polarity are attracted to the positive polarity and neutralizes the + charge..,so the atoms adjacent to this becomes positively charged and they attracts the electrons from their neighboring atoms and the process continues ...,similarly electrons of atoms which are adjacent to - polarity are repelled from the negative charge of the voltage source...and the process continues ...so i think lights adjacent to the battery terminals glow first...and in the conduction process electrons do not move with speed of light

The electrons may not move at the speed of light, but the EM wave they generate when they move from the higher to lower potential will travel at or near the speed of light which is the field acting on the electrons further away from the source.

It sounds like (in simpler language) we can think of the EM wave as a longitudinal pulse traveling down both sides of the wire (somewhere around c) away from the battery. This initially moves electrons but as the pulse reaches the far side if it can’t continue around the circuit the electrons do not move any further.

It is almost as though we can treat the wire as a small capacitor that is filling one side while the EM wave is passing. This makes sense in that as the wave passes along the wire there will be a voltage difference in front of it and behind it. This means the electrons are moving without a complete circuit but only as much as the wire can “hold”. Whether this lights up the bulbs in the original thought depends on how long the wire is I suppose.

Thank you everyone for your great responses so far.

antistrophy said:
It sounds like (in simpler language) we can think of the EM wave as a longitudinal pulse traveling down both sides of the wire (somewhere around c) away from the battery. This initially moves electrons but as the pulse reaches the far side if it can’t continue around the circuit the electrons do not move any further.

It is almost as though we can treat the wire as a small capacitor that is filling one side while the EM wave is passing. This makes sense in that as the wave passes along the wire there will be a voltage difference in front of it and behind it. This means the electrons are moving without a complete circuit but only as much as the wire can “hold”. Whether this lights up the bulbs in the original thought depends on how long the wire is I suppose.

Thank you everyone for your great responses so far.

This sounds like a new 'classic' question that Feynman would have loved. Like his rotary lawn-sprinkler sucking water, submerged in a swimming pool.

antistrophy said:
It sounds like (in simpler language) we can think of the EM wave as a longitudinal pulse traveling down both sides of the wire (somewhere around c) away from the battery. This initially moves electrons but as the pulse reaches the far side if it can’t continue around the circuit the electrons do not move any further.

It is almost as though we can treat the wire as a small capacitor that is filling one side while the EM wave is passing. This makes sense in that as the wave passes along the wire there will be a voltage difference in front of it and behind it. This means the electrons are moving without a complete circuit but only as much as the wire can “hold”. Whether this lights up the bulbs in the original thought depends on how long the wire is I suppose.

Thank you everyone for your great responses so far.

Not exactly accurate for a few reasons:

The electric field generated by the switch is not a pulse, it is a step. The field in the circuit changes from no field, up to the field generated by the battery. Since a battery is DC, there will be a new static field in the wire after the initial step wavefront has carried through.

The reason I think this is important to correct is because you say the electrons do not move any further after this initial wavefront, which is false because then the lights would only blink rather than stay on once the switch is turned on. The electrons keep moving after the step change in the field has made its way through the circuit, but now the electrons are moving in a DC static field.

Also keep in mind the light bulbs in this circuit, which change the field as it goes through them as voltage drops along the path.

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Drat. For a second there I thought I understood. Can I ask simpler questions?

1 Is the voltage difference moving the electrons?

2 Does the voltage difference move around the circuit (i.e. if I connect the wires to the battery the voltage difference (in the battery) has moved to the end of the wires)?

3 is there a voltage difference between the battery and the wires when first connected?

I am going to have to think about what DragonPetter said as I think my problem is in there somewhere.
I think I meant to say earlier "the voltage wave" not the EM wave

antistrophy said:
Drat. For a second there I thought I understood. Can I ask simpler questions?

1 Is the voltage difference moving the electrons?

2 Does the voltage difference move around the circuit (i.e. if I connect the wires to the battery the voltage difference (in the battery) has moved to the end of the wires)?

3 is there a voltage difference between the battery and the wires when first connected?

I am going to have to think about what DragonPetter said as I think my problem is in there somewhere.
I think I meant to say earlier "the voltage wave" not the EM wave

1 No.

2 Yes.

3 Yes, the battery terminals where the wires are connected wil exhibit the PD of the battery.

I hate to say this, but I don't think the electrons next the battery move first, I think they have to move all at once because of this:
If instead of wires, I touch a capacitor to the positive side and a capacitor to the negative side but do not touch them together, do either of them get the slightest charge? both? but they would have had to if any electrons moved.

Back to the original question: Do the electrons next to the battery move first before the voltage has traveled around the circuit?

antistrophy said:
Drat. For a second there I thought I understood. Can I ask simpler questions?
Well, your question has me second guessing myself too. Hopefully someone can confirm or refute my explanation.
antistrophy said:
1 Is the voltage difference moving the electrons?
You should try thinking of this in terms of electric fields rather than voltage differences, since voltage differences are dependent on the electric field. The electric field can only change through an electromagnetic wave, and these waves have a finite speed. Before the connection of the wire to the terminals, there is no electric field across the wire, but once a connection is made and electrons start to move, and EM waves are generated as this change occurs.

antistrophy said:
2 Does the voltage difference move around the circuit (i.e. if I connect the wires to the battery the voltage difference (in the battery) has moved to the end of the wires)?
Yes, the voltage difference is a result of position difference in an electric field. As the electric field changes through the wire, the positions in the wire will have a change in voltage.

antistrophy said:
3 is there a voltage difference between the battery and the wires when first connected?

There is not a voltage difference between the battery and the circuit components until they are connected between the terminals. There is only a voltage difference between the battery terminals internally inside the battery. When you connect the ends of the circuit to the terminals, the electric field will change and propagate through the wire and as the field changes at each point along the circuit, a voltage difference results.

you could think of it as pushing a metal square loop into a B field. The bar moving at a speed v into the B field, the electrons will experience a force due to the Lorentz force F=qvB.
And when one electron moves it pushes the electron in front of it because of its E field.

antistrophy said:
I hate to say this, but I don't think the electrons next the battery move first, I think they have to move all at once

If that were true, then they would be exceeding the speed of light.

Okay, I get it now. The initial movment of the electrons near the battery before the EM wave has gone around is very small and stops immediately. Something like "compressing" rather than moving. And as rcgldr says I suppose when the waves cross at the back side the full current travels back down the wire toward the battery. I imagine if you touch a capacitor to one end of the battery you may get a very^2 small charge from this initial movment (but that's another argument).

Thank you everyone, I really appreciate the help.

hey guys i got a doubt...! do electrons emit or produce EM wave while they are in conduction?? i thought that only magnetic field is associated with them...! (according to amperes law i think)

hanii said:
hey guys i got a doubt...! do electrons emit or produce EM wave while they are in conduction?? i thought that only magnetic field is associated with them...! (according to amperes law i think)

When a charge accelerates, it produces an EM wave. This is how they create waves with antennas.

but in a conductor ...electrons travel with a constant speed called drift speed...they don't accelerate...this is due to resistivity of the material used as conductor

hanii said:
but in a conductor ...electrons travel with a constant speed called drift speed...they don't accelerate...this is due to resistivity of the material used as conductor

The drift speed depends upon the current and that is not the same all along the wire at the same time..

If you apply a step function to the end of the wire then the nearby electrons will start to move, producing a charge imbalance which will be producing an electric field which will start to move the next electrons in line. All this takes time because 1. The electrons take time to get moving and 2. Time is needed for the electric field change to propagate across each step along the wire (speed of light). The process gives a moving wavefront along the wire. The distant parts of the wire are undisturbed until this wavefront reaches them.

As the wire is long, the electrons will move 'easily' because the potential on their far side starts off at zero. The pulse will travel along the wire until it reaches a discontinuity (say an unconnected end). At that point, the potential at the end will reach that of the pulse and electrons will stop flowing - things will have 'backed up' (in mechanical terms) and no more current can flow from the power supply. This is very much like the effect of charging a capacitor - except that the capacitance is distributed along the length of line and there is also an inductance involved, which has current flowing through it. At the open end this current will stop and that change in current in the inductance will generate a Voltage which will then propagate back to the source (an echo).
All this is a rather naff way of describing some of what goes on in a transmission line. The theory may be too mathematical for some tastes but it predicts accurately what happens in real life transmission lines. They can make open circuits look like short circuits and vice versa.

There is another point. If the wire is in a loop, the same thing is happening at the negative terminal of the battery - a negative step function sets of from the negative terminal at just under c and at the same time as the positive step starts off from the + terminal. The two waves will meet half way round.

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