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Suppose a simple circuit with a DC voltage source and a resistor. The voltage of the source will be situated over the resistor. So the electric field (which is the gradient of the potential) will be constant in the resistor (if you assume a linear potential function in the resistor), and will be equal to zero in the conducting connections.

Since electrons are drifted because of an electric field (with Newton's second law and Lorentz' law for the force), what keeps them drifting in the ideal conductors? Or do they just keep their velocity they got in the resistor, and don't decelerate because there's no resistance there? What would mean that the electrons obtain their velocity within the resistor, which sounds a bit paradoxal...

Where's the loop in my argumentation, or am I just right?

BNJMNDDNN
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In idealistic schematics of circuit theory conductors with no voltage drop guide electrons to and from the elements. Electrons don't lose or gain speed here, that is they obey Newton's first law. In resistors there is an acceleration due to electric field and deceleration due to scattering from lattice sites, i.e. drift.

Therefore within this paradigm electrons don't drift in ideal conductors, they act inertially. Thus they will drift through the resistor and go losslessly through contacts into the battery where the electromotive force will bring them to the other side for the next lap. They won't obtain any velocity in the resistor, since they already had it exiting the battery; the electric field in the steady state sets it such that $v=\mu E$.

In reality however even in ballistic conductors, where scattering is largely eliminated and the voltage drop within the ballistic region is vanishingly small, the quantum contact resistance will appear as dominant (and unavoidable - http://www.ecs.umass.edu/ece697mm/Supplement_Lucent_fourt%20resist_Nature_2001.pdf). Therefore there are no such things as ideal conductors, and circuit theory is not particularly suited for analyzing the inner workings of resistivity, conductance, transmission etc.

mgphys
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  • Thanks! The minor points were idd little mistakes sorry. – BNJMNDDNN Jun 23 '13 at 16:24
  • You're welcome, I hope this settles the matter for you. No need to apologize, I trust they were just typos. – mgphys Jun 23 '13 at 16:28
  • But how will it gain its velocity in the battery? What is the mobility there? I know that in the resistor, it depends on the average scattering time, its mass and charge. And keeping conservation of charge in mind: how is it possible that the electrons lose velocity in the resistor? That would mean that there's a net electron flow out of the battery? – BNJMNDDNN Jun 23 '13 at 16:54
  • The workings of the battery is a whole other question, it might be helpful if you take a look at http://physics.stackexchange.com/questions/68817/battery-and-current-confusion. Keep in mind that electrons average speed remains constant. – mgphys Jun 23 '13 at 17:12
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Yes, in a resistor, the electric field accelerates the electrons. In a resistor, scattering collisions make the average velocity of the electrons in the resistor approximately constant (for a constant DC voltage source).

When the electrons some out of the resistor and enter a superconductor, they begin ballistic conduction. The electrons suddenly being moving in ways described by Newton's laws of motion -- in particular, they initially just keep going at the same velocity they were at when they left the resistor. Just as you suspected.

(A superconductor is more or less by definition a material packed full of movable electrons that electrons can travel through without scattering).

However, (still in accordance with Newton's laws), there can be a transient voltage across a physical superconductor. When that happens, the electrons inside accelerate faster and faster, since there is no scattering to slow them down, until they hit the boundary of (and in some cases exit) the physical superconductor.

I suspect the paradox you are alluding to occurs even without the superconductor. Say we have a battery in New York, and a light bulb in Los Angeles, connected by 2 very long copper wires. If there is a break in either wire near Los Angeles, the light bulb almost instantly goes dark. But after we fix that break, how does the battery miles away almost instantly "know" to start pumping again, pulling electrons out of one wire and pushing them into the other, and somehow those electrons -- creeping along at a few inches per minute -- almost instantly turn on the light bulb? One of my favorite explanations begins "The charges start out inside the light bulb filament." -- William Beaty 1995, and leads us to: What is electricity, really?

David Cary
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  • Note that you certainly don't need a superconductor for ballistic conduction. Also a superconductor is certainly not defined in a way you state it to be; you seem to be mixing two concepts - perfect conductors (idealistic. non-existent) and superconductors (real, with far more exotic properties than just movable electrons without scattering) – mgphys Jun 23 '13 at 17:25
  • @mgphys: Yes, real superconductors have many exotic and surprising properties. And yes, ballistic conduction does occur in non-superconductors. Are you implying that real superconductors are not filled with movable electrons that move without scattering? I would find that surprising, and would like to learn more. – David Cary Jun 23 '13 at 18:02
  • No, I'm just implying that there's a lot more to superconductors than just scatterless motion. – mgphys Jun 23 '13 at 18:14
  • Again, yes, the known superconductors have many other surprising and exotic properties that I would not expect from their definition. My understanding is that a "superconductor" is defined as "a conductor" (i.e., a material filled with movable electrical charges) "that has zero resistance" (i.e., electrical charges do not scatter). Do you have a better definition? – David Cary Jun 23 '13 at 19:40
  • Minimum definition for me would be: zero resistance (which would have to be mentioned in the context of Cooper pairs, not movable electrons), the Meissner effect (the expulsion of the magnetic field), phase transition below certain $T_c$ with the accompanying physical changes and quantization of magnetic flux lines. One could also mention the distinction between Type I and II superconductors, and the existence of a critical magnetic field for destroying this phase. – mgphys Jun 23 '13 at 20:09
  • Here is a nice free chapter about the subject http://www.cengage.com/resource_uploads/static_resources/0534493394/4891/SerwayCh12-Superconductivity.pdf, best regards. – mgphys Jun 23 '13 at 20:19