# Circuit with superconducting wire 1 light year in length

1. Apr 11, 2007

### drpizza

Question regarding the propogation of the e-field.. would it propogate at nearly the speed of light?

If a lightbulb were put on a circuit, with half a light year of cable coming from each terminal on the bulb and connected to a sufficient battery/power supply, 1/2 a year until the bulb turned on?

What would be a sufficient power supply?

What if the bulb were very close to the positive side of the battery, with one light year of superconducting cable between the bulb and the negative terminal on the battery? Vice versa?

(This question seems to be annoying a few people on a physics listserv for high school teachers.)

2. Apr 11, 2007

### pixel01

I guess:
In the first case, the bulb will be on after 1/2 year and during that time, the power supply has to work all the time to push (or pull) the charges
In the second case, the bulb is on right away, but with reduced current. It will not operate normally until after 1/2 year

3. Apr 11, 2007

### Integral

Staff Emeritus
I believe that the signal will propagate through the circuit at about .3c, so it will be more like 1.5yr before the light comes on.

We must assume negligible wire resistance, as we do in normal household circuits.

I am interested it what our real physicist have to say.

Edit: I don't think it will make any difference where the bulb is.. but not real sure.

Humm... does a "hole" current move out to meet the electron current in middle? Where is the switch? Don't we run into some causality problems if the location of the switch is not taken into account.

Last edited: Apr 11, 2007
4. Apr 11, 2007

### Staff: Mentor

The light bulb lights as soon as the wavefront passes it. I don't see why it should be more complicated than that. I'm sure I'll be flamed for bringing it up again, but this is exactly the same as how water flow through a pipe works. Remember, water can't be flowing everywhere in a pipe the instant a valve is opened. But it is flowing (and doing whatever useful work you need it to do) as soon as the wavefront passes whatever section you are looking at, even if it isn't flowing yet somewhere else.

5. Apr 16, 2007

### drpizza

Okay, good enough.

But the other part of my question:
if the bulb is 1 inch from the positive terminal on the power supply vs. the bulb 1 inch from the negative terminal on the power supply...

Or, which way does the wavefront propogate? (Or does it propogate in both directions simultaneously?)

6. Apr 16, 2007

### Integral

Staff Emeritus
Isn't the switch the key here? Seems to me that that two wavefronts would propogate from the switch. While the switch is open the potential difference would be present across the open. Closing the switch would initiate 2 wave fronts, each traveling through the wire toward the terminals of the battery.

Am I making any sense?

7. Apr 17, 2007

### lpfr

Assumption: the two half year long wires are parallel and form a constant characteristic impedance transmission line.

I Agree.
The bulb will light normally (not reduced current) for 1 year. Then it will light double for half a year.
What happens after this depends on the impedance of the bulb + generator. If the sum matches the transmission line impedance, the bulb will come back to normal lighting. I left the others cases to the reader.

No, in a transmission line without dielectric and without magnetic material, the speed of signal is the same as in vacuum.

In the two cases the bulb alights as soon as you close the switch.

The front propagates away in the two cases. After (just after) the front you have the same current with opposite sign in each wire.

Do not confuse the direction of current with the direction of propagation of signal. You can think of what happens when you push or you pull a (very long) rod or a train.

No. When you close the switch, the current begin to pass in the two wires (opposite sense) and the front (just one) propagates away. If you ask how the other wire knows that the switch has closed, the answer is that the voltage in the other wire changes and that there is the charge of the capacity between the wires (formally you should call it "displacement current").

8. Apr 17, 2007

### pixel01

IPfr wrote: "The bulb will light normally (not reduced current) for 1 year. Then it will light double for half a year.
What happens after this depends on the impedance of the bulb + generator. If the sum matches the transmission line impedance, the bulb will come back to normal lighting. I left the others cases to the reader."

===================================

I disagree. Assume that the contact is in the middle (1/2 light year away) and the bulb is just 1 inch away from one of the electrodes. Before 1/2 year, the signal form + and - would not meet so the charges would be pushed by only one end. When they meet, charges are pushed by 2 ends. This can be understood easily by imagining a pumping system with one pump and one sucking pump. The pipe in this case is somewhat elastic. The flow of water will be changed after the two signal meet.
I think this example can make sense and I should not explain more.

9. Apr 17, 2007

### NeoDevin

Do we have to take into account the fact that the electron is a massive particle, and as such, cannot move at the speed of light?

10. Apr 17, 2007

### pixel01

The electron does not move at the speed of light, but (maybe) the current does. The two phenonmena are different.

11. Apr 17, 2007

### NeoDevin

I know they aren't the same, but I'm not familiar enough with the microscopic properties of currents to know if that has an effect.

12. Apr 17, 2007

### Integral

Staff Emeritus
Why would you assume this? This is a simple DC circiut, not clear to me that a tranmission line is a meaningful assumption. I assumed a single strand of copper, with no resistance of course.
Just how does that work? Where does the change in intensity come from?

No resistance, remember?
But this is not a transmission line, this is a simple circiut. I was just as wrong as you, according to http://en.wikipedia.org/wiki/Speed_of_electricity" [Broken] signals move at .6c in copper.

Absolutly impossible, there is a real causality problem with this statement. The signal propogates from the switch at a speed significantly less then the speed of light, the bulb cannot turn on before the information of the closed switch arrives thru the wire.

seems to me that you are one suffering confusion here.
The way I read what you have written is that you agree with me, the signal propagates away from the switch in both directions, one the current, the other the displacement current.

Last edited by a moderator: May 2, 2017
13. Apr 17, 2007

### pervect

Staff Emeritus
In order to make the question easy to answer, it would be good to revise the problem slightly.

Instead of connecting the lightbulb to a pair of wires, connect it to a ideal matched transmission line 1 light year long.

In that case, one can see that the speed of transmission of signal is just the speed of the transmission through the transmission line. If you have a vacuum dielectric, the speed will be nearly 'c'. If you have some other material for a dielectric, the speed may be lower. I think that the conductor/superconductor issue would have only a minor effect on the transmission speed, but it would impact losses.

I'm not sure I have a good handle on the loss situation for a superconducting transmission line that long. While DC signals have no loss in a superconductor, the same isn't true for an AC signal. It would not take much loss per meter before the signal was not detectable over such a distance.

Note that if there is an impedance mismatch between the bulb and the characteristic impedance of the transmission line, you will get a series of reflections along the line, of very long period.

14. Apr 18, 2007

### lpfr

I apologize to pixel01. He was right. "the bulb is on right away, but with reduced current." But it will not operate normally after ½ year. It will take 1 full year for the reflection to come back. An then, for a more year it will operate normally only in the case that the resistance of the bulb plus generator equals the characteristic impedance of the line.

What I wrote, about the current doubling would be true only if the characteristic impedance of the line is small compared to that of the bulb.

I'm not sure to have well understood what you mean. If the switch is at the far end of the line closes (that is the open line becomes a short-circuit terminated one), the front created, travels back to the bulb and it alights the bulb ½ year latter. If this is what you mean we agree.

This was about my assumption: the two half year long wires are parallel and form a constant characteristic impedance transmission line.

Obviously, you are not familiar with transmission lines. You can read:
http://en.wikipedia.org/wiki/Transmission_line

When two wires form a constant impedance transmission line, I can compute currents, voltages, speeds, etc. If the two wires are not parallel I do not know how to do this and, of course, neither you.
Obviously you are not familiar with transmission lines, reflected signals, characteristic impedances, etc. One you have read the wiki article, you can open a new thread to discuss your doubts.

You can found a lot of interesting and correct things in wikipedia. But, alas! you find also some bull**** (sorry). You can never be sure that what appears in WP is correct. You found one of those b.... If your search WP well, you can find others b... values for the speed of signals on copper, and even, the correct value.

My assertion holds for the bulb and the switch together (I neglected the some 80 ps that the signal takes to travel 5 cm).

Yes, this is a sensible suggestion. I just want to state that the way drpizza posted the problem was almost perfect. The only missing thing was the transmission line. The choice of 1 light-year of superconducting wire plus a bulb is perfect for a though experiment.

If someone is troubled with details about losses is superconducting wire, it is easy to transform the problem in a pulse generator, 10 meters of over the counter transmission line, a resistor instead the bulb, and a good oscilloscope.

I think that you mean reflections at the extremities of the line. To avoid this reflections it is the bulb plus generator which should match the impedance of the line.

15. Apr 18, 2007

### Integral

Staff Emeritus
Cool, you all have defined the problem in the configuration you want. Am I not allowed to do likewise?

So here is my configuration, which I feel is more how a high school student would visualize it.

I have a circle of wire 1 ly in circumference, equally spaced around the circle are a DC voltage source, a light bulb and a switch. Here we will assume a copper wire and that the voltage source is capable of driving 1 Amp thru the circuit. Now What happens when I close the switch?

16. Apr 19, 2007

### lpfr

This problem is much more difficult to SOLVE, than the problem posted by drpizza or the formulation by pervect.

If you want to find someone able to answer (not guess) your problem, you should create a new post.

You do not seem aware of the difficulties to solve this problem. I don't think that there is an analytical solution, and even a numerical one is very complex. It asks to solve Maxwell equations with very unsymmetrical limit conditions.

This is not as simple as you think it is, just as a pipe filling with water.

Of course, you can always guess what will happen. The music is similar to that of the pervect problem, but the lyrics can be very different. There will be fronts that propagate, but you do not know their variable speed and their variable current. yes, they should probably vary as the front position changes.