# Thread: If I turned on the bulb from a very distant place, could it be turned on in a second?

Furthermore, I think you may have noticed that I am not a person who speaks English as a first language.

1. I set up the bulb 2 light year away from the earth.

2. set up the switch and the battery here on earth.

3. connect two of them with the long wire. (Maybe I have to assume that the energy of the battery can reach there)

-> If I turn on the switch, how long will it take to turn on the bulb?

First of all, I know that electron's velocity is already known, but
the thing which I really don't know is that, aside from the velocity, turning on the switch means that I make the voltage difference, right?(I hope you understand that I have little knowledge about this.)
Then, if the distance is too too far away, then the voltage difference can be produced immediately? or does it take some time?

I hope you understand what I mean.

2.

3. Approximately 3 years (copper wire), you would observe it after 5.

4. Originally Posted by G O R T
Approximately 3 years (copper wire), you would observe it after 5.
Wouldn't that be wrong? The circuit would need to be active first? 4 light years of wire to get around before electrons get back to the source (how long would that take? Are you saying 3 years each leg?) Is that a necessity to have a circuit first?
And 2 light years for the light to reach the source.

5. Three years for the electrons to reach the light bulb.
Two years for the light to reach the Earth.

6. The electrons don't have to go from the earth to the light bulb. Each electron pushes the ones ahead of it. The effect is felt as something less than the speed of light.

7. More like a wave traveling up a river. The wave is the energy and it can travel much faster than the water does.
The energy can be carried by electrons simply shifting their orbit a bit instead of actually drifting, much like how the water in an ocean wave moves in a circle as a wave passes but does not actually travel with the wave.

For electrons in copper wire the actual electron flow is very slow, but the electrical energy wave in copper can travel at about 75% of the speed of light. (very roughly without calculations)

8. See this too, http://www.ultracad.com/articles/propagationtime.pdf because the author is explaining why it is not so much the copper wire itself that matters to the speed of propagation as it is the material around the wire.
I am not sure if I agree completely with his statements.

9. Originally Posted by Harold14370
The electrons don't have to go from the earth to the light bulb. Each electron pushes the ones ahead of it. The effect is felt as something less than the speed of light.
Indeed. I should have said - three years for the movement of electrons to reach the light bulb.

10. Originally Posted by John Galt
Originally Posted by Harold14370
The electrons don't have to go from the earth to the light bulb. Each electron pushes the ones ahead of it. The effect is felt as something less than the speed of light.
Indeed. I should have said - three years for the movement of electrons to reach the light bulb.
But what I was trying to understand is would that movement of electrons start if there was no complete circuit? So how do the electrons "know"if there is a circuit?
Like if after 5 years, say, you break the circuit that is before the return current has reached the Earth in the return wire, has the light already been working for 2 years?
Why even bother to connect the circuit at all then? Surely there is no way the electrons can know whether the circuit is complete at the beginning of the experiment!
I find this weird.

11. It would have to be 3yrs for the electricity to reach the bulb, 3yrs for the electricity to come back to earth to complete the circuit then another 2yrs ontop of that for you to see the light.

However, after the initial 6yr (approx) for the circuit to complete, you could turn the power off and just wait 2yrs in darkness for the flash to reach you.

12. Originally Posted by GoldenRatio
It would have to be 3yrs for the electricity to reach the bulb, 3 yrs for the electricity to come back to earth to complete the circuit then another 2yrs on top of that for you to see the light.

However, after the initial 6yr (approx) for the circuit to complete, you could turn the power off and just wait 2yrs in darkness for the flash to reach you.
That is what I thought at first but why do the electrons have to return to the Earth before the light will work? I don't think you even need a circuit.
But then that goes against the little bit that I thought I knew about electricity - it needs a circuit.

13. Why would it be different over a span of 2 light years when you need a circuit for a span of 20 feet? Aside from the distance, what has changed.

14. Originally Posted by Robittybob1
That is what I thought at first but why does the electrons have to return to the Earth before the light will work? I don't think you even need a circuit.
But then that goes against the little bit that I thought I knew about electricity - it needs a circuit.
The electrons do not have to return to Earth. Nothing in what has been said by others suggests this is the case and several members have clearly posted to the contrary.

Time for the current flow to reach the light - a high percentage of the speed of light - estimated as three years.
Time for light to reach the Earth - two years.

15. Originally Posted by John Galt
Originally Posted by Robittybob1
That is what I thought at first but why does the electrons have to return to the Earth before the light will work? I don't think you even need a circuit.
But then that goes against the little bit that I thought I knew about electricity - it needs a circuit.
The electrons do not have to return to Earth. Nothing in what has been said by others suggests this is the case and several members have clearly posted to the contrary.

Time for the current flow to reach the light - a high percentage of the speed of light - estimated as three years.
Time for light to reach the Earth - two years.
OK I accept that too, but do you need a circuit? You would have to have some place for the electrons to go after they reach the light, e.g. a capacitor, or another long wire. Are they both possibilities?

16. I imagine a capacitor could work, though once it was charged the light would go out. (I don't do analog circuits. I'm a digital logic man.)

17. Not going to go into capacitors & such, if you had a 2ly circuit. Why would you not have to wait for the circuit to complete before the light was to turn on?

18. You only need electrons flowing through the bulb for it to "make" light.

Once electrons are flowing (and this could be through capacitance) they don't need to get all the way back to Earth, they just need to flow through the bulb.

19. Originally Posted by pzkpfw
You only need electrons flowing through the bulb for it to "make" light.

Once electrons are flowing (and this could be through capacitance) they don't need to get all the way back to Earth, they just need to flow through the bulb.
Correct, the electrons need to pass through the filament in order to heat ir such that it starts glowing, this part is simple.
The part that has to do with how long it takes for the electrons to get to the bulb is much more complicated, this is a case of extremely long line and the theory of such lines is very complicated, see here. Since the line is obscenely long, it will exhibit an obscene amount of inductance and capacitance (as well as resistance). We cannot say that "the electrons travel at 75% speed of light, therefore they will get to the bulb in 2y/0.75=2.66y". It may very well be that the electrons never make it to the bulb, we need to solve the long transmission line equations, there is no easy way around it. Most likely, the electrons will never make it to the bulb, there is no realistic way to power a 2ly long transmission line. What we can say is that IF the long line equations have a solution showing that the time to get to the bulb is T[y], then we will see the light in T+2[y]. (here T may very well be ).

20. Originally Posted by Howard Roark
Originally Posted by pzkpfw
You only need electrons flowing through the bulb for it to "make" light.

Once electrons are flowing (and this could be through capacitance) they don't need to get all the way back to Earth, they just need to flow through the bulb.
Correct, the electrons need to pass through the filament in order to heat ir such that it starts glowing, this part is simple.
The part that has to do with how long it takes for the electrons to get to the bulb is much more complicated, this is a case of extremely long line and the theory of such lines is very complicated, see here. Since the line is obscenely long, it will exhibit an obscene amount of inductance and capacitance (as well as resistance). We cannot say that "the electrons travel at 75% speed of light, therefore they will get to the bulb in 2ly/0.75=2.66ly". It may very well be that the electrons never make it to the bulb, we need to solve the long transmission line equations, there is no easy way around it. Most likely, the electrons will never make it to the bulb, there is no realistic way to power a 2ly transmission line. What we can say is that IF the long line equations have a solution showing that the time to get to the bulb is T[ly], then we will see the light in T+2[ly]. (here T may very well be ).
I'm not able to solve those equations but I can still get the gist of your argument. What was going to force the electrons all that way? With inductance and capacitance on both sides, yes hard to see it work in real life.
Thank you (for a change).

21. Originally Posted by Robittybob1
Originally Posted by Howard Roark
Originally Posted by pzkpfw
You only need electrons flowing through the bulb for it to "make" light.

Once electrons are flowing (and this could be through capacitance) they don't need to get all the way back to Earth, they just need to flow through the bulb.
Correct, the electrons need to pass through the filament in order to heat ir such that it starts glowing, this part is simple.
The part that has to do with how long it takes for the electrons to get to the bulb is much more complicated, this is a case of extremely long line and the theory of such lines is very complicated, see here. Since the line is obscenely long, it will exhibit an obscene amount of inductance and capacitance (as well as resistance). We cannot say that "the electrons travel at 75% speed of light, therefore they will get to the bulb in 2ly/0.75=2.66ly". It may very well be that the electrons never make it to the bulb, we need to solve the long transmission line equations, there is no easy way around it. Most likely, the electrons will never make it to the bulb, there is no realistic way to power a 2ly transmission line. What we can say is that IF the long line equations have a solution showing that the time to get to the bulb is T[ly], then we will see the light in T+2[ly]. (here T may very well be ).
I'm not able to solve those equations but I can still get the gist of your argument. What was going to force the electrons all that way? With inductance and capacitance on both sides, yes hard to see it work in real life.
Thank you (for a change).
Yes, normally you troll my posts, I am glad that you understood one, for a change.

22. Originally Posted by GoldenRatio
Why would it be different over a span of 2 light years when you need a circuit for a span of 20 feet? Aside from the distance, what has changed.
Maybe there is a momentary pulse of electricity moving along a wire even when there isn't a circuit*. To get a long term effect from the light there needs to be a circuit, to keep providing a place for the electrons to move toward (the ground or the positive terminal).
(* I have not heard of this effect before. Does anyone know what this effect is called?)

23. Originally Posted by Robittybob1
But what I was trying to understand is would that movement of electrons start if there was no complete circuit? So how do the electrons "know"if there is a circuit?
How do electrons ever know if there is a circuit? Why does this question only arise in this scenario, and not in every circuit?

The answer is that the application of a voltage always generates a field locally first. That generates a magnetic field locally at first. As described by Maxwell's equations, a changing E field generates a changing B field, which generates a changing E field...and on it goes; that's a wave. A wave propagates along the line at a speed determined by geometry and materials constants. For typical wires, the velocity will be a good fraction of c (say, 0.6-0.8c).

24. Originally Posted by tk421
Originally Posted by Robittybob1
But what I was trying to understand is would that movement of electrons start if there was no complete circuit? So how do the electrons "know"if there is a circuit?
How do electrons ever know if there is a circuit? Why does this question only arise in this scenario, and not in every circuit?

The answer is that the application of a voltage always generates a field locally first. That generates a magnetic field locally at first. As described by Maxwell's equations, a changing E field generates a changing B field, which generates a changing E field...and on it goes; that's a wave. A wave propagates along the line at a speed determined by geometry and materials constants. For typical wires, the velocity will be a good fraction of c (say, 0.6-0.8c).
We can measure the voltage of a power source by touching both terminals e.g. 240 volts AC or 1.5 volts from a drycell battery, but what is the voltage at just one end of the battery and a long wire as the case would be if there was no circuit?

25. Originally Posted by tk421
Originally Posted by Robittybob1
But what I was trying to understand is would that movement of electrons start if there was no complete circuit? So how do the electrons "know"if there is a circuit?
How do electrons ever know if there is a circuit? Why does this question only arise in this scenario, and not in every circuit?

The answer is that the application of a voltage always generates a field locally first. That generates a magnetic field locally at first. As described by Maxwell's equations, a changing E field generates a changing B field, which generates a changing E field...and on it goes; that's a wave. A wave propagates along the line at a speed determined by geometry and materials constants. For typical wires, the velocity will be a good fraction of c (say, 0.6-0.8c).
If you were electrical engineer you would recognize a conventional “delay line”.

Roark is right again…. Time is wholly dependent on the inductance and capacitance. Because : T(y)>.75c.

26. Originally Posted by Robittybob1
We can measure the voltage of a power source by touching both terminals e.g. 240 volts AC or 1.5 volts from a drycell battery, but what is the voltage at just one end of the battery and a long wire as the case would be if there was no circuit?
You are missing the entire point. When you touch a voltmeter to the terminals of some source, there is, for a brief time, no circuit from the point of view of the source. Thus there is no qualitative difference at all between the everyday example with which you seem to be comfortable, and the less familiar, but essentially identical scenario of a multi-light-year long circuit. Until a wave has propagated to the destination and back, there is no way for the source to know that there's a circuit, is there?

27. Originally Posted by GTCethos
Originally Posted by tk421
Originally Posted by Robittybob1
But what I was trying to understand is would that movement of electrons start if there was no complete circuit? So how do the electrons "know"if there is a circuit?
How do electrons ever know if there is a circuit? Why does this question only arise in this scenario, and not in every circuit?

The answer is that the application of a voltage always generates a field locally first. That generates a magnetic field locally at first. As described by Maxwell's equations, a changing E field generates a changing B field, which generates a changing E field...and on it goes; that's a wave. A wave propagates along the line at a speed determined by geometry and materials constants. For typical wires, the velocity will be a good fraction of c (say, 0.6-0.8c).
If you were electrical engineer you would recognize a conventional “delay line”.

Roark is right again…. Time is wholly dependent on the inductance and capacitance. Because : T(y)>.75c.
And if you were a physicist or an electrical engineer who had studied Maxwell's equations, you would have known that I was in fact describing exactly how a delay line works. And inductance and capacitance per unit length of the line depends, as I said, on both geometry and materials constants (permittivity and permeability).

28. Originally Posted by tk421
Originally Posted by Robittybob1
We can measure the voltage of a power source by touching both terminals e.g. 240 volts AC or 1.5 volts from a drycell battery, but what is the voltage at just one end of the battery and a long wire as the case would be if there was no circuit?
You are missing the entire point. When you touch a voltmeter to the terminals of some source, there is, for a brief time, no circuit from the point of view of the source. Thus there is no qualitative difference at all between the everyday example with which you seem to be comfortable, and the less familiar, but essentially identical scenario of a multi-light-year long circuit. Until a wave has propagated to the destination and back, there is no way for the source to know that there's a circuit, is there?

I have been trying to think about it.
Until a wave has propagated to the destination and back, there is no way for the source to know that there's a circuit, is there?
I think that is wrong, but I don't have the words to argue the point. I think electrons will flow into the wire regardless of whether there is a circuit, the flow indicates there is some voltage there but (maybe) it isn't going to be the same voltage as across both terminals.
You get a twelve volt battery and a long insulated wire. You connect a voltage meter across from one end of the wire to the negative terminal and measure the voltage.
Will or won't there be a measurable voltage drop in the first instance?

29. If the wire is the right length it is an antenna and the wave just keeps going after it runs out of wire.
If the wave crosses another wire it induces a current in it too.

30. Originally Posted by dan hunter
If the wire is the right length it is an antenna and the wave just keeps going after it runs out of wire.
If the wave crosses another wire it induces a current in it too.
That is interesting.

31. Originally Posted by tk421
And if you were a physicist or an electrical engineer who had studied Maxwell's equations, you would have known that I was in fact describing exactly how a delay line works. And inductance and capacitance per unit length of the line depends, as I said, on both geometry and materials constants (permittivity and permeability).
Yes, there is no other way to look at the problem.
An electric field potential will advance out the two light year cable as a transmission line.
Typical cable propagates at around 2/3 c.
Time would be about 3 years.
Then electrons move through the light bulb.
Although the initial Dv/Dt is near infinite, current rise at the bulb would be very slow.
Electrons move a few mm in the bulb and you have light!

32. Originally Posted by Robittybob1
I have been trying to think about it.
Allow me to guide you toward a deeper understanding, then.

Originally Posted by tk421
Until a wave has propagated to the destination and back, there is no way for the source to know that there's a circuit, is there?
Originally Posted by rbob
I think that is wrong, but I don't have the words to argue the point. I think electrons will flow into the wire regardless of whether there is a circuit, the flow indicates there is some voltage there but (maybe) it isn't going to be the same voltage as across both terminals.
You get a twelve volt battery and a long insulated wire. You connect a voltage meter across from one end of the wire to the negative terminal and measure the voltage.
Will or won't there be a measurable voltage drop in the first instance?
My statement was not wrong, and I do have the words to argue the point. Part of what's likely tripping you up is the common misuse of water-in-pipe analogies to describe current flow. Let's dispose of that straightaway. The average velocity of electrons in a typical current-carrying wire is less -- often much, much less -- than a leisurely walking pace. How to reconcile that fact with the idea that propagation occurs at something like c? The answer is that the slow movement of electrons does not in any way imply slow propagation of a wave.

Next on the list of things that are probably getting in your way is your ignorance of Maxwell's equations. Those equations tell us that Ohm's law is not a law at all. Ohm's law is an approximation that holds only in the quasistatic regime. What the heck is that, you ask? That regime corresponds to circuit dimensions that are much smaller than the shortest wavelength of interest. Effectively, we ignore that the speed of light is finite. So if whatever intuition you have about electricity has derived from circuits for which Ohm's law is a good approximation, your intuition is very incomplete. In fact, it is so incomplete as to preclude your understanding the matter at hand.

Consider a superconducting cable long enough for the quasistatic approximation to fail. Now apply a voltage at, say, the left end of that cable. What happens? The voltage along the cable cannot instantly jump up. Instead, a wave propagates from one end to the other. It starts at the left end and proceeds to the right end at whatever the speed of light is for that cable. Ohm's law obviously fails spectacularly, as the voltage along the superconducting cable is not everywhere the same.

33. Originally Posted by dan hunter
If the wire is the right length it is an antenna and the wave just keeps going after it runs out of wire.
If the wave crosses another wire it induces a current in it too.
The term "right length" misleadingly implies that only a discrete, finite set of lengths will lead to radiation. In fact, radiation happens for virtually any length. It's just that wires that are very short compared to a wavelength have extremely small radiation resistances, so they don't couple efficiently into a radiation field.

And induction of current in a nearby conductor in no way requires wave propagation. Ordinary transformers do not depend on wave behaviour at all, for example. Just put two wires close (but not orthogonal) to each other, and pump a time-varying current through one. You'll get a current out of the other.

34. Originally Posted by tk421
Just put two wires close (but not orthogonal) to each other, and pump a time-varying current through one. You'll get a current out of the other.
Please clarify "time-varying current" for me.

35. Originally Posted by tk421
Consider a superconducting cable long enough for the quasistatic approximation to fail. Now apply a voltage at, say, the left end of that cable. What happens? The voltage along the cable cannot instantly jump up. Instead, a wave propagates from one end to the other. It starts at the left end and proceeds to the right end at whatever the speed of light is for that cable. Ohm's law obviously fails spectacularly, as the voltage along the superconducting cable is not everywhere the same.
Why are you making the cable superconducting? Is it the practicality of the resistance of a two light year length of wire?

36. Originally Posted by dan hunter
Originally Posted by tk421
Just put two wires close (but not orthogonal) to each other, and pump a time-varying current through one. You'll get a current out of the other.
Please clarify "time-varying current" for me.
Saw tooth current, alternating current, fluctuating current ...

37. Originally Posted by John Galt
Originally Posted by tk421
Consider a superconducting cable long enough for the quasistatic approximation to fail. Now apply a voltage at, say, the left end of that cable. What happens? The voltage along the cable cannot instantly jump up. Instead, a wave propagates from one end to the other. It starts at the left end and proceeds to the right end at whatever the speed of light is for that cable. Ohm's law obviously fails spectacularly, as the voltage along the superconducting cable is not everywhere the same.
Why are you making the cable superconducting? Is it the practicality of the resistance of a two light year length of wire?
The whole thought experiment is bizarre. I don't think any of our experts have told me whether it was essential to have a circuit maintained at all times.
I think resistance was going to be overcome by increasing the voltage.

38. I found the question of the circuit being open or closed interesting for a different reason.
Lets say the wire existed, and that it was a perfect ideal conductor.
Lets even say the electrons could zip through it at the speed of light.
Lets also accept that nothing, not even information can travel faster than light.

Now what would happen if you only connected the battery long enough to start the current flowing. Would you see the light flash after one year?
Would the current stop as soon as you disconnected the battery, or would that packet of electrons continue on their way?
How would they "know" the circuit had been broken before they got to the end of it?

Some of those questions are already answered in the posts, but I still think it is a good illustration of a basic difference in the classical and the relativistic view.

39. Originally Posted by tk421

Next on the list of things that are probably getting in your way is your ignorance of Maxwell's equations. Those equations tell us that Ohm's law is not a law at all. Ohm's law is an approximation that holds only in the quasistatic regime. What the heck is that, you ask? That regime corresponds to circuit dimensions that are much smaller than the shortest wavelength of interest. Effectively, we ignore that the speed of light is finite. So if whatever intuition you have about electricity has derived from circuits for which Ohm's law is a good approximation, your intuition is very incomplete. In fact, it is so incomplete as to preclude your understanding the matter at hand.
Yep.

40. Originally Posted by dan hunter
I found the question of the circuit being open or closed interesting for a different reason.
Lets say the wire existed, and that it was a perfect ideal conductor.
Lets even say the electrons could zip through it at the speed of light.
Lets also accept that nothing, not even information can travel faster than light.

Now what would happen if you only connected the battery long enough to start the current flowing. Would you see the light flash after one year?
Depends on the wire length. Not necessary one year.

Would the current stop as soon as you disconnected the battery,
Nope. The information needs time to travel, remember?

How would they "know" the circuit had been broken before they got to the end of it?
They won't.

41. Originally Posted by John Galt
Originally Posted by tk421
Consider a superconducting cable long enough for the quasistatic approximation to fail. Now apply a voltage at, say, the left end of that cable. What happens? The voltage along the cable cannot instantly jump up. Instead, a wave propagates from one end to the other. It starts at the left end and proceeds to the right end at whatever the speed of light is for that cable. Ohm's law obviously fails spectacularly, as the voltage along the superconducting cable is not everywhere the same.
Why are you making the cable superconducting? Is it the practicality of the resistance of a two light year length of wire?
The intent was to illustrate that even a superconducting cable would have different voltages at different points along it, owing to the finite speed of propagation. I wanted to remove any possible confounding effects of Ohm's law by removing any resistance along the cable. That way, any voltage drop along the cable would have to come from something other than an "Ohmic" effect.

42. Originally Posted by dan hunter
Originally Posted by tk421
Just put two wires close (but not orthogonal) to each other, and pump a time-varying current through one. You'll get a current out of the other.
Please clarify "time-varying current" for me.
Quite literally, di/dt <> 0.

The induced current will thus be a function of frequency, as well as of how well coupled the inducing magnetic field is to the second wire, among other things.

43. Originally Posted by Robittybob1
Originally Posted by John Galt
Originally Posted by tk421
Consider a superconducting cable long enough for the quasistatic approximation to fail. Now apply a voltage at, say, the left end of that cable. What happens? The voltage along the cable cannot instantly jump up. Instead, a wave propagates from one end to the other. It starts at the left end and proceeds to the right end at whatever the speed of light is for that cable. Ohm's law obviously fails spectacularly, as the voltage along the superconducting cable is not everywhere the same.
Why are you making the cable superconducting? Is it the practicality of the resistance of a two light year length of wire?
The whole thought experiment is bizarre. I don't think any of our experts have told me whether it was essential to have a circuit maintained at all times.
I think resistance was going to be overcome by increasing the voltage.
Stay focused on the thought experiment I've offered. The thing applying the voltage cannot have any information about what's on the other end of the cable until a round-trip delay of 2L/c has elapsed. So whether the cable is short-circuited or open-circuited on the distant end can't affect what the initial current is. So, not only is a closed circuit not required, the voltage source can't even know initially whether the circuit is closed.

What's propagating is a wave, not electrons.

As for "resistance overcome by voltage," that has nothing to do with the discussion. It's a linear system, so doubling the voltage only doubles the current. It doesn't change the speed, or any other qualitative aspect of the solution.

44. Originally Posted by tk421
Originally Posted by Robittybob1
Originally Posted by John Galt
Originally Posted by tk421
Consider a superconducting cable long enough for the quasistatic approximation to fail. Now apply a voltage at, say, the left end of that cable. What happens? The voltage along the cable cannot instantly jump up. Instead, a wave propagates from one end to the other. It starts at the left end and proceeds to the right end at whatever the speed of light is for that cable. Ohm's law obviously fails spectacularly, as the voltage along the superconducting cable is not everywhere the same.
Why are you making the cable superconducting? Is it the practicality of the resistance of a two light year length of wire?
The whole thought experiment is bizarre. I don't think any of our experts have told me whether it was essential to have a circuit maintained at all times.
I think resistance was going to be overcome by increasing the voltage.
Stay focused on the thought experiment I've offered. The thing applying the voltage cannot have any information about what's on the other end of the cable until a round-trip delay of 2L/c has elapsed. So whether the cable is short-circuited or open-circuited on the distant end can't affect what the initial current is. So, not only is a closed circuit not required, the voltage source can't even know initially whether the circuit is closed.

What's propagating is a wave, not electrons.

As for "resistance overcome by voltage," that has nothing to do with the discussion. It's a linear system, so doubling the voltage only doubles the current. It doesn't change the speed, or any other qualitative aspect of the solution.
It is very difficult to pry people away from Ohm :-)

45. Originally Posted by Howard Roark
It is very difficult to pry people away from Ohm :-)
He's solid enough in his domain, but things can get "funny" away from Ohm sweet Ohm.

46. Originally Posted by tk421
..... The thing applying the voltage cannot have any information about what's on the other end of the cable until a round-trip delay of 2L/c has elapsed. So whether the cable is short-circuited or open-circuited on the distant end can't affect what the initial current is. So, not only is a closed circuit not required, the voltage source can't even know initially whether the circuit is closed.

What's propagating is a wave, not electrons....
Agreed and understood (mostly).
However you have to establish a voltage difference on the wire to start with don't you?
Since the OP is using a battery wouldn't that mean you have to connect the wires at some point?

47. Originally Posted by dan hunter
Agreed and understood (mostly).
However you have to establish a voltage difference on the wire to start with don't you?
Since the OP is using a battery wouldn't that mean you have to connect the wires at some point?
Absolutely right. But the wires can be the cable; to separate them is to make an artificial distinction.

So, suppose I take an open-circuited cable -- and only a cable -- and then connect a 1V battery to, say, the left end. The voltage will be 1V instantaneously at the left end, and will eventually become 1V at the right end, as one would expect from an open circuit. "Eventually" here is L/v, where L is the cable length, and v is the propagation velocity (close to, but generally not equal to, c).

Now suppose I repeat the experiment, but with a short-circuit cable. Since the battery cannot know about the short-circuit condition right away, the initial behaviour is the same. A 1V wave propagates toward the right-hand end. Once it reaches that end, the short circuit imposes a boundary condition of zero voltage there. But energy must be conserved. The resolution to the seeming paradox is that a counter-propagating reflected wave must be produced at the shorted end (this is just ordinary reflection, as in a mirror, but perhaps in a less familiar context). A canceling wave propagates back toward the battery, which battery notices the short circuit finally after 2L/v seconds have elapsed. And then much sturm und drang as the battery pours coulomb after coulomb into the wire in a futile attempt at making the voltage nonzero.

The reason we rarely encounter the need to think about this is that light propagates really fast, so the action is over in nanoseconds for ~meter-long objects. But when the objects are of an extent that propagation delay is comparable to the wavelength of the highest frequency component of interest, one has to use the full set of Maxwell's equations, rather than the quasistatic Ohm's law subset.

48. Originally Posted by dan hunter
Originally Posted by tk421
..... The thing applying the voltage cannot have any information about what's on the other end of the cable until a round-trip delay of 2L/c has elapsed. So whether the cable is short-circuited or open-circuited on the distant end can't affect what the initial current is. So, not only is a closed circuit not required, the voltage source can't even know initially whether the circuit is closed.

What's propagating is a wave, not electrons....
Agreed and understood (mostly).
However you have to establish a voltage difference on the wire to start with don't you?
Since the OP is using a battery wouldn't that mean you have to connect the wires at some point?
That goes back to my question - What is the voltage drop into the wire if it is only connected to one side of the voltage source? Is it half the voltage of the power source?

49. Originally Posted by Robittybob1
That goes back to my question - What is the voltage drop into the wire if it is only connected to one side of the voltage source? Is it half the voltage of the power source?
A voltage is always defined between two points. You've only identified one. What is the other? Once you tell me that, I have a hope of answering.

50. Originally Posted by tk421
Originally Posted by Robittybob1
That goes back to my question - What is the voltage drop into the wire if it is only connected to one side of the voltage source? Is it half the voltage of the power source?
A voltage is always defined between two points. You've only identified one. What is the other? Once you tell me that, I have a hope of answering.
I thought someone (you) has already said it was the full voltage of the battery in If I turned on the bulb from a very distant place, could it be turned on in a second?. I want to know if there is a very long wire and a voltage source, when it is connected to just one terminal, is there a voltage difference between the wire and the battery terminal before it is connected.
There would have to be or else the power would not even flow, and it has to flow to discover if there is a circuit. If there is a circuit it keeps flowing, if there is an incomplete circuit it stops sometime afterwards.

51. Originally Posted by Robittybob1
I want to know if there is a very long wire and a voltage source, when it is connected to just one terminal, is there a voltage difference between the wire and the battery terminal before it is connected.
You realize that the way you phrased your question doesn't make sense, do you?

52. Originally Posted by Howard Roark
Originally Posted by Robittybob1
I want to know if there is a very long wire and a voltage source, when it is connected to just one terminal, is there a voltage difference between the wire and the battery terminal before it is connected.
You realize that the way you phrased your question doesn't make sense, do you?
I have edited that post now.
I want to know if there is a very long wire and a voltage source, when it is connected to just one terminal, is there a voltage difference between the wire and the battery terminal before it is connected.
There would have to be or else the power would not even flow, and it has to flow to discover if there is a circuit. If there is a circuit it keeps flowing, if there is an incomplete circuit it stops sometime afterwards.

53. Originally Posted by Robittybob1
Originally Posted by Howard Roark
Originally Posted by Robittybob1
I want to know if there is a very long wire and a voltage source, when it is connected to just one terminal, is there a voltage difference between the wire and the battery terminal before it is connected.
You realize that the way you phrased your question doesn't make sense, do you?
I have edited that post now.
I want to know if there is a very long wire and a voltage source, when it is connected to just one terminal, is there a voltage difference between the wire and the battery terminal before it is connected.
There would have to be or else the power would not even flow, and it has to flow to discover if there is a circuit. If there is a circuit it keeps flowing, if there is an incomplete circuit it stops sometime afterwards.
This is just the same as the one I pointed out to being meaningless. Try again.

54. Originally Posted by Howard Roark
Originally Posted by Robittybob1
Originally Posted by Howard Roark
Originally Posted by Robittybob1
I want to know if there is a very long wire and a voltage source, when it is connected to just one terminal, is there a voltage difference between the wire and the battery terminal before it is connected.
You realize that the way you phrased your question doesn't make sense, do you?
I have edited that post now.
If there is a very long wire and a voltage source, when it is connected to just one terminal, is there a voltage difference between the wire and the battery terminal before it is connected.
There would have to be or else the power would not even flow, and it has to flow to discover if there is a circuit. If there is a circuit it keeps flowing, if there is an incomplete circuit it stops sometime afterwards.
This is just the same as the one I pointed out to being meaningless. Try again.
I'll let someone else try first.
Slight edit to question:
If there is a very long wire and a voltage source, and you are unsure if you have a circuit or not, both ends of the wire are to be connected but there maybe a break. Is there a voltage difference between the wire and the battery terminal before it was fully connected?

My Logic:
There would have to be a voltage difference or else the power would not even flow, and it has to flow into the wire for us to discover if there is a complete circuit. If there is a circuit it keeps flowing, if there is an incomplete circuit it stops sometime afterwards.
Does anyone agree?

55. Originally Posted by Robittybob1
I thought someone (you) has already said it was the full voltage of the battery in If I turned on the bulb from a very distant place, could it be turned on in a second?.

Yes, but now you are asking about a different situation, so please clarify what you mean, rather than tellng me what I mean.

56. Originally Posted by Robittybob1
Slight edit to question:
.....If there is a very long wire and a voltage source, and you are unsure if you have a circuit or not, both ends of the wire are to be connected but there maybe a break. Is there a voltage difference between the wire and the battery terminal before it was fully connected?
.....
Are you asking if the potential in the wire (voltage) is 0 before it is connected to the battery terminal at 1.5 or whatever voltage it is?
I think they are assuming the wire to be at zero volts relative to the battery. In other words the same as if it was at ground potential.
Does that help?

57. Originally Posted by dan hunter
Originally Posted by Robittybob1
Slight edit to question:
.....If there is a very long wire and a voltage source, and you are unsure if you have a circuit or not, both ends of the wire are to be connected but there maybe a break. Is there a voltage difference between the wire and the battery terminal before it was fully connected?
.....
Are you asking if the potential in the wire (voltage) is 0 before it is connected to the battery terminal at 1.5 or whatever voltage it is?
I think they are assuming the wire to be at zero volts relative to the battery. In other words the same as if it was at ground potential.
Does that help?
If the wire does not touch the ground its potential is indeterminate and variable (floating), it can be millions of volts. ONLY if it is in contact with the ground its potential is zero (or close to it).

58. Originally Posted by dan hunter
Originally Posted by Robittybob1
Slight edit to question:
.....If there is a very long wire and a voltage source, and you are unsure if you have a circuit or not, both ends of the wire are to be connected but there maybe a break. Is there a voltage difference between the wire and the battery terminal before it was fully connected?
.....
Are you asking if the potential in the wire (voltage) is 0 before it is connected to the battery terminal at 1.5 or whatever voltage it is?
I think they are assuming the wire to be at zero volts relative to the battery. In other words the same as if it was at ground potential.
Does that help?
I was curious as to the answer so I went over to the electricians here and one of them sort of understood the problem but he reckoned there would be no voltage difference between one terminal of a battery and the ground. We never tested it though.
In vehicles one terminal is is connected to the chassis and that is called earth, usually negative to earth.
So if the other positive terminal is connected to the body of the car that is a dead short and sparks will occur as the connection is made.
So it is not as if all the power leaks out of the battery and seeps into the Earth through the tires (rubber on car tires conduct electricity did you know?), for within the battery it is electrons returning to the positive terminal that keeps the chemical reaction going.

So we have been talking about a wave going through this wire but behind that wave must be a current for the wave on its on won't fire up the light will it! So where is this supply of electrons coming from and is the source large enough to supply the number of electrons required?

59. Originally Posted by Robittybob1
...
I was curious as to the answer so I went over to the electricians here and one of them sort of understood the problem but he reckoned there would be no voltage difference between one terminal of a battery and the ground. We never tested it though.
In vehicles one terminal is is connected to the chassis and that is called earth, usually negative to earth.
So if the other positive terminal is connected to the body of the car that is a dead short and sparks will occur as the connection is made.
So it is not as if all the power leaks out of the battery and seeps into the Earth through the tires (rubber on car tires conduct electricity did you know?), for within the battery it is electrons returning to the positive terminal that keeps the chemical reaction going.

So we have been talking about a wave going through this wire but behind that wave must be a current for the wave on its on won't fire up the light will it! So where is this supply of electrons coming from and is the source large enough to supply the number of electrons required?
Do you want a real answer or not? If you do, then please define the question. Asking electricians for their opinion isn't going to help. Their background is not deep or broad enough to answer a question about electrodynamics.

So rather than drifting further afield as is your wont, please describe carefully the physical situation you are interested in. Identify the terminals between which you are going to measure the voltage. Do those things and someone can provide an answer. As it is, we can only speculate what your question might be, leading to answers that seem only to stimulate more random questions from you. To make progress, focus on one thing at a time, rbob.

60. Everyone has talked about the electricity traveling as a wave down the wire at around 0.75 c but what shape is this wave?
It is not just a single surge of electrons is it? Is there is a continuous wave that stays peaked behind the initial wave. It isn't like sinusoidal waves or is it?
I suppose if the power comes from a AC generator it will be sinusoidal but if it was a DC generator it could be pulses as well or a constant current like the output of a battery. The battery will only continue to operate if electrons are returned to the battery.

61. Originally Posted by Robittybob1
Everyone has talked about the electricity traveling as a wave down the wire at around 0.75 c but what shape is this wave?
It is not just a single surge of electrons is it? Is there is a continuous wave that stays peaked behind the initial wave. It isn't like sinusoidal waves or is it?
I suppose if the power comes from a AC generator it will be sinusoidal but if it was a DC generator it could be pulses as well or a constant current like the output of a battery. The battery will only continue to operate if electrons are returned to the battery.
It would help immensely if you were to read the thread with care. It's already been pointed out that electrons aren't the wave. The wave is in the field. The electrons in a DC circuit move very, very slowly, as I and others have already pointed out here.

What is the shape of the wave? It depends on what you're talking about. If we have a battery and a switch that connects it to a cable with infinite speed, then the voltage is a step that speeds along the cable at whatever the propagation velocity happens to be for that cable (whether 0.75c or some other value, depending on how the cable is configured).

62. Originally Posted by tk421
Originally Posted by Robittybob1
Everyone has talked about the electricity traveling as a wave down the wire at around 0.75 c but what shape is this wave?
It is not just a single surge of electrons is it? Is there is a continuous wave that stays peaked behind the initial wave. It isn't like sinusoidal waves or is it?
I suppose if the power comes from a AC generator it will be sinusoidal but if it was a DC generator it could be pulses as well or a constant current like the output of a battery. The battery will only continue to operate if electrons are returned to the battery.
It would help immensely if you were to read the thread with care. It's already been pointed out that electrons aren't the wave. The wave is in the field. The electrons in a DC circuit move very, very slowly, as I and others have already pointed out here.

What is the shape of the wave? It depends on what you're talking about. If we have a battery and a switch that connects it to a cable with infinite speed, then the voltage is a step that speeds along the cable at whatever the propagation velocity happens to be for that cable (whether 0.75c or some other value, depending on how the cable is configured).
OK so you say it is a field but the field is generated by the moving electrons. One electron moves and that in turn moves another so the actual electrons themselves don't move far but the ripple moves along the conductor but the field is only happening where the electrons are moving. So don't think I haven't followed or I don't understand that bit for I do.

63. Originally Posted by Robittybob1
OK so you say it is a field but the field is generated by the moving electrons. One electron moves and that in turn moves another so the actual electrons themselves don't move far but the ripple moves along the conductor but the field is only happening where the electrons are moving. So don't think I haven't followed or I don't understand that bit for I do.
Yes, but it is the ripple moving the electrons instead of the electrons moving the ripple.

Edit:
Well, there are a couple of other problems there but that is the most basic way of saying it.

64. @ robbity:
For this moment the analogy to waves in water might help.
Particle Motion in Deep Water

65. Originally Posted by dan hunter
Originally Posted by Robittybob1
OK so you say it is a field but the field is generated by the moving electrons. One electron moves and that in turn moves another so the actual electrons themselves don't move far but the ripple moves along the conductor but the field is only happening where the electrons are moving. So don't think I haven't followed or I don't understand that bit for I do.
Yes, but it is the ripple moving the electrons instead of the electrons moving the ripple.

Edit:
Well, there are a couple of other problems there but that is the most basic way of saying it.
How did you work that out? It seems a bit like the chicken or the egg problem.

66. Originally Posted by Robittybob1
OK so you say it is a field but the field is generated by the moving electrons. One electron moves and that in turn moves another so the actual electrons themselves don't move far but the ripple moves along the conductor but the field is only happening where the electrons are moving. So don't think I haven't followed or I don't understand that bit for I do.
The key point that I think you are still missing is that the wave moves much, much faster than do the electrons. I noted earlier that the average drift velocity of electrons is slower than the walking pace of an extraordinarily slow human. We know that waves propagate at near c, so obviously the ripple you speak of greatly outpaces the electrons. The reason that the distinction is important is to disabuse you of the idea of electrons-in-wires as being the same as water-in-pipes. The motions of these two entities have no fixed relationship to one another, and certainly no useful correlations for the purposes of the question in this thread.

Finally the field isn't "only happening" when the electrons are moving. You get fields from stationary electrons (which is why like charges repel, even if they aren't moving). The field is changing only when the electrons are moving. That's an important distinction.

67. Originally Posted by tk421
....
Finally the field isn't "only happening" when the electrons are moving. You get fields from stationary electrons (which is why like charges repel, even if they aren't moving). The field is changing only when the electrons are moving. That's an important distinction.
I am wondering if you really understand electricity.

With no current in a conductor all these stationary electron fields would cancel each other over the large numbers and there would be no net electromagnetic field around the conductor. Is that true?

In small regions you could get unbalanced attractive or repulsive forces. (Like Van der Waals forces)

68. Originally Posted by Robittybob1
Originally Posted by tk421
....
Finally the field isn't "only happening" when the electrons are moving. You get fields from stationary electrons (which is why like charges repel, even if they aren't moving). The field is changing only when the electrons are moving. That's an important distinction.
I am wondering if you really understand electricity.
I guess that's fair, as I am no longer wondering if you can understand the explanations offered to you.

With no current in a conductor all these stationary electron fields would cancel each other over the large numbers and there would be no net electromagnetic field around the conductor. Is that true?
No, not even close.

In small regions you could get unbalanced attractive or repulsive forces. (Like Van der Waals forces)
Eh?

I will try one more time to explain things to you, but you have to exert more effort than you've shown so far. The first step is to stop pretending to yourself that you understand much about electricity. The sooner you admit that to yourself, the more open you will be to the knowledge I am trying to impart to you. That sounds arrogant on my part, but given your arrogance above, that's fair. At least I'm able to demonstrate that my arguments are correct. You seem only to have uninformed, baseless opinion, but are unaware of how totally unreliable your knowledge is.

Now that you've moved goalposts yet again, let's address the latest problem you're having.

Take an open-circuited cable, apply a voltage and wait an extremely long time. The voltage across the cable will be uniform, as expected from Ohm's law. That voltage has associated with it an electric field. That electric field is supported by a net charge on each conductor (positive on the one connected to the positive terminal of the battery and a negative one on the other); the cable is a capacitor. If I put a small test charge in between the cable conductors, it will be repelled by one conductor and attracted by the other. That tells us that there is a field.

Note carefully that we have a static situation. No current is flowing, but we have an electric field. Your intuition that current is necessary for an electric field is just plain wrong. So forget your intuition; it is faulty.

For reference, see Maxwell's equations. One of them says that the divergence of the electric field is proportional to charge density. It doesn't say "time derivative of charge density," it says "charge density" period. So, you get an E-field even with no charge motion. If that weren't the case, capacitors wouldn't store energy.

69. Originally Posted by tk421
Originally Posted by Robittybob1
Originally Posted by tk421
....
Finally the field isn't "only happening" when the electrons are moving. You get fields from stationary electrons (which is why like charges repel, even if they aren't moving). The field is changing only when the electrons are moving. That's an important distinction.
I am wondering if you really understand electricity.
I guess that's fair, as I am no longer wondering if you can understand the explanations offered to you.

With no current in a conductor all these stationary electron fields would cancel each other over the large numbers and there would be no net electromagnetic field around the conductor. Is that true?
No, not even close.

In small regions you could get unbalanced attractive or repulsive forces. (Like Van der Waals forces)
Eh?

I will try one more time to explain things to you, but you have to exert more effort than you've shown so far. The first step is to stop pretending to yourself that you understand much about electricity. The sooner you admit that to yourself, the more open you will be to the knowledge I am trying to impart to you. That sounds arrogant on my part, but given your arrogance above, that's fair. At least I'm able to demonstrate that my arguments are correct. You seem only to have uninformed, baseless opinion, but are unaware of how totally unreliable your knowledge is.

Now that you've moved goalposts yet again, let's address the latest problem you're having.

Take an open-circuited cable, apply a voltage and wait an extremely long time. The voltage across the cable will be uniform, as expected from Ohm's law. That voltage has associated with it an electric field. That electric field is supported by a net charge on each conductor (positive on the one connected to the positive terminal of the battery and a negative one on the other); the cable is a capacitor. If I put a small test charge in between the cable conductors, it will be repelled by one conductor and attracted by the other. That tells us that there is a field.

Note carefully that we have a static situation. No current is flowing, but we have an electric field. Your intuition that current is necessary for an electric field is just plain wrong. So forget your intuition; it is faulty.

For reference, see Maxwell's equations. One of them says that the divergence of the electric field is proportional to charge density. It doesn't say "time derivative of charge density," it says "charge density" period. So, you get an E-field even with no charge motion. If that weren't the case, capacitors wouldn't store energy.
You got me there I should have said the conductor not only had no current but also had no net charge. Certainly a capacitor will have an electric field around it.
But that charge around or within a capacitor is not moving nor has much resemblance to the wave that is present with the passage of a current in a conductor.

70. Originally Posted by Robittybob1
You got me there I should have said the conductor not only had no current but also had no net charge.
Again, please read what I wrote. The conductors in a cable certainly each have a net charge when connected to a battery. There's no difference between the arrangement of the conductors of a cable and a capacitor.

Certainly a capacitor will have an electric field around it.
Not necessarily. But there will be an electric field between the conductors when connected to a battery.

But that charge around or within a capacitor is not moving nor has much resemblance to the wave that is present with the passage of a current in a conductor.
And no one said they did, rbob. It is really frustrating that you seem intent on not understanding things. Now you're just making up crap. Please stop.

You made a statement that a static charge arrangement has no field. That's the statement I was addressing. I proved you wrong. I have shown you that a field does not require current to exist. I never said, nor implied in any way, that a static field is, or resembles, a wave.

71. Originally Posted by tk421
....
You made a statement that a static charge arrangement has no field. That's the statement I was addressing. I proved you wrong. I have shown you that a field does not require current to exist. I never said, nor implied in any way, that a static field is, or resembles, a wave.
I have never said "that a static charge arrangement has no field". If you think I did show me where you think I did please?

I accepted "that a field does not require current to exist". So why are you going on about it? (Static electrical fields - is that what you are talking about?)
I have never said you that you have "said, nor implied in any way, that a static field is, or resembles, a wave". So why bring that up?

You must admit that over the months you and I don't really understand each other properly and this might be why we clash. I am trying my hardest to be clear, but it still doesn't seem to be good enough for. Let's try and be very basic and have some basic ideas in common in future please.

72. Originally Posted by Robittybob1
I have never said "that a static charge arrangement has no field". If you think I did show me where you think I did please?
You have an alarmingly poor memory. Worse, you're too flippin' lazy to look upward a few posts to see what you wrote. It is irritating that you spend so little effort and expect others to do all the work. You said this:

Originally Posted by rbob
With no current in a conductor all these stationary electron fields would cancel each other over the large numbers and there would be no net electromagnetic field around the conductor.
With that out of the way, let's proceed to your next bits:

I accepted "that a field does not require current to exist". So why are you going on about it? (Static electrical fields - is that what you are talking about?)
I have never said you that you have "said, nor implied in any way, that a static field is, or resembles, a wave". So why bring that up?
Sigh. I wrote that in response to this:

Originally Posted by rbob
But that charge around or within a capacitor is not moving nor has much resemblance to the wave that is present with the passage of a current in a conductor.
And then you continue:

You must admit that over the months you and I don't really understand each other properly and this might be why we clash. I am trying my hardest to be clear, but it still doesn't seem to be good enough for. Let's try and be very basic and have some basic ideas in common in future please.
And I'll say with great confidence that my ability to express myself clearly is far superior to yours. The problem is that you don't process logic well, and you're lazy, too, so you respond without thinking much first. Objective evidence is abundant. Look at the sheer number of posts you crank out per unit time. Look at the number of posts where you arrogantly offer advice, when you haven't spent even a second to see what the context of the ongoing conversation is.

So, while your proposal for detente seems reasonable on its face, it fails to comprehend the disproportionate burden you are placing on others for your own lack of diligence.

73. Originally Posted by tk421

And I'll say with great confidence that my ability to express myself clearly is far superior to yours. The problem is that you don't process logic well, and you're lazy, too, so you respond without thinking much first. Objective evidence is abundant. Look at the sheer number of posts you crank out per unit time. Look at the number of posts where you arrogantly offer advice, when you haven't spent even a second to see what the context of the ongoing conversation is.

So, while your proposal for detente seems reasonable on its face, it fails to comprehend the disproportionate burden you are placing on others for your own lack of diligence.
Was that a NO!

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