# Relativistic Currents

1. Jul 29, 2011

### Shark 774

I didn't know where to post this, but I figured since it has the word "relativity" in my question I would stick it here.

V = IR. In a super conductor R = 0 and I = V/R = V/0 which is infinity. This implies an infinite number of charges per second flowing, and hence that the charges are travelling with infinite speed. Obviously they are limited by the speed of light, so is there a formula for relativistic currents?

2. Jul 29, 2011

### ghwellsjr

Relativity has nothing to do with what happens in super conductors so this question really belongs somewhere else. But I will give you some answers anyway.

First off, in order for a current to flow in a super conductor, it has to be a loop and the charges flow in that loop forming an electromagnet, correct? It will take energy to create this magnet and so current doesn't just flow because the resistance is zero. You have to physically do something to get current flow and while you are doing that, the super conductor is no longer super. Once you get a current flowing, you can make the it super conducting but you can no longer change the amount of current flowing. So your first assumption that the current is infinite is incorrect.

Secondly, charges don't flow at infinite speed, where'd you get that idea? Even if the current were infinite (which it isn't) you can get that with an infinite number of electrons flowing at any speed.

3. Jul 29, 2011

### bcrowell

Staff Emeritus
I liked the first part of your answer, but this part doesn't make sense to me. We don't have an infinite number of electrons available in the circuit.

4. Jul 29, 2011

### ghwellsjr

I know that. I was addressing the logic of his statement:

5. Jul 29, 2011

### Staff: Mentor

V is also 0. 0/0 is not infiintiy, it is undefined. V=IR -> 0=I0 for any I. Ohm's law cannot be used to determine the current in a super conductor because it is satisfied for any current.

6. Jul 29, 2011

### bcrowell

Staff Emeritus
He was right and you were wrong. He said an infinite number of charges per second, which could only be achieved by having a finite number of charges (because that's what we have) going around the circuit at infinite speed. The same electron would have to pass through the ammeter infinitely many times in one second.

7. Jul 29, 2011

### ghwellsjr

Ben, some people think that electricity (whatever they think that means) travels down wires at the speed of light, which it doesn't; rather, electrical signals travel down wires at a significant fraction of the speed of light. Some people think that electrons travel down wires at a significant fraction of the speed of light, which they don't; rather they travel at normal very slow speeds. As currents increase in a wire, the individual electrons don't just go faster and faster; rather, more and more of them participate in the flow.

So for you to point out that we have a finite number of electrons and so they must be traveling at infinite speed in order to achieve infinite current when electrons never travel in wires at anywhere near the speed of light, let alone infinite speed seems to me to be adding to Shark's confusion rather than clearing it up. Note that he stated that he thinks the speed of electrons in wires is "limited by the speed of light" so he needs to learn that they go very slowly, even in a superconducter carrying a very large current.

8. Jul 29, 2011

### Shark 774

You have interpreted my question totally wrong. I know that superconductors have nothing to do with special (or general) relativity. However, I thought that as the resistance of materials approach zero, from Ohm's Law, the currents will approach infinity (I = V/R = V/0, as stated before). Obviously this can't happen because the electrons are restricted by the speed of light. So for electrons travelling at relativistic speeds, is there a modification that must be made to Ohm's Law, just as there are modifications that must be made to Newton's formula's when you include relativistic speeds?

9. Jul 29, 2011

### bcrowell

Staff Emeritus
Ghwellsjr answered your question in #2. You simply don't get infinite current because the circuit has inductance as well as resistance.

10. Jul 29, 2011

### Staff: Mentor

Your reasoning is incorrect, as pointed out in my previous post. Division by zero does not always imply infinity.

11. Jul 29, 2011

### Shark 774

Ok, well regardless of this small technicality, as the resistance APPROACHES zero the current, from Ohm's Law, will APPROACH infinity. And therefore my reasoning is not incorrect in that sense.

12. Jul 29, 2011

### bcrowell

Staff Emeritus
Dale's point is that $I=\Delta V/R$, and as R appeoaches zero, I does not necessarily approach infinity, since $\Delta V$ may also approach zero. Ghwellsjr has also pointed out that Ohm's law doesn't necessarily hold here. Neither of these points is a technicality.

[EDIT] Wrote "I does not necessarily approach zero" originally, should have said "infinity."

Last edited: Jul 30, 2011
13. Jul 29, 2011

### Staff: Mentor

No, your reasoning is again incorrect. It is only correct if V also approaches some non-zero value. Which is not the case. Regardless of if you are taking a limit, solving an equation, or just doing division, 0/0 is not always infinite. The math simply does not support your argument.

The mere fact that the denominator approaches 0 does not imply that the ratio approaches infinity as you have asserted. Consider sin(x)/x, as x approaches zero the ratio approaches 1.

Last edited: Jul 30, 2011
14. Jul 29, 2011

### ghwellsjr

Are thinking in terms of a length of superconducting wire that is attached to a voltage source like a battery or power supply? If so, you have to realize that every real source of voltage has some internal resistance and so when you short it out either with a very short piece of thick wire or with a length of superconducting wire, the measureable external voltage drops to zero and now the correct application of ohm's law is I=V/R where V is the internal voltage and R is the internal resistance. Typically, these real values will limit the current to a few amps or maybe hundreds of amps for a car battery. Most batteries intended for high current applications will tell you the maximum amperage that you can get out of them and most power supplies have a fuse that will protect them if the current is exceeded.

So even if you could hook up a piece of superconducting wire to a voltage source, you still would not be able to get infinite current through it. But this has nothing to do with relativity or the speed of the electrons, which, by the way, or pokey slow.

15. Jul 30, 2011

### pervect

Staff Emeritus
Not to mention that the superconductor stops superconducting at some critical current.

So to recap we have the following:

We connect a battery to a superconducting wire. The wire has some inductance L. So, I increases with time, such that I = (V/L)*time. I remains finite, but increases as time passes. With idealized batteries and superconductors, there's no obvious limit. But with real batteries and superconductors, at some point , as I increases with time, one of the following will happen:

1) the battery won't be able to put out more current (due to it's internal resistance)

2) The superconductor reaches its critical current and stops superconducting. Known as a quench, this will potentially release quite a lot of energy (look what happened at CERN, for instance).

16. Jul 31, 2011

### Per Oni

I have read that somewhere else in PF as well, however this is not true except perhaps in certain semi conductors. In my opinion, all the free conducting electrons take part in a normal Ohmic current and their speed will increase as the current gets bigger. However this electronic drift speed, in normal cases (excluding super conductors) will never be much more than 1 mm/sec because otherwise the wire will melt.

17. Aug 1, 2011

### ghwellsjr

OK, I was wrong, thanks for bringing that to my attention.

Maybe that's what Ben was trying to tell me, too.