Assume a battery and resistor are connected in a series. If we double the resistance, the current gets cut in half. This is a macroscopic effect.

How do those electrons know that resistance has increased? Does every electron travel through that resistor and find out that it has high resistance?

Do atoms in the resistor and electrons in the battery (which pumps those electrons) communicate via space?

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    $\begingroup$ |in this article motls.blogspot.com/2011/11/… Lubos Motl explains how classical fields emerge from the underlying quantum mechanical layer. Voltages and resistances are in the domain of classical fields, and the article should give an understanding of the complexity of describing the meta level of voltages using the mathematics of the underlying level. An analogy is temperature emerging from the underlying statistical mechanics. This hyperphysics.phy-astr.gsu.edu/hbase/electric/miccur.html#c1 may help $\endgroup$ – anna v Oct 16 '15 at 5:38
  • $\begingroup$ resistance would mean in the last figure that the electron drift velocity gets lower due to interactions with the non metalic lattice of the resistor, and thus the current gets lower. $\endgroup$ – anna v Oct 16 '15 at 5:40
  • $\begingroup$ @annav : so current should be high at entry and low at exit of electrons from resistor? $\endgroup$ – Ramki Oct 16 '15 at 5:52
  • $\begingroup$ No, a balance is immediately reached, the particle interactions (electron electron , hole hole, hole electron)happen with virtual photons and with the velocity of light so one cannot see a discontinuity. Like pressure, which is a statistical effect from zillions of molecules, and which equalizes in a circuit in times comparable with the velocity of sound ( the velocity of the transmission of the interactions) $\endgroup$ – anna v Oct 16 '15 at 6:19
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    $\begingroup$ No, sorry. If one wants to really understand this one has to take the corresponding courses, quantum field theory. Lubos' article was the first time this was explained clearly for me, though I had taken a qft course back in 1961. here is another article phys.ksu.edu/personal/wysin/notes/quantumEM.pdf . I think people just hand waved the process of emergence , because classical EM works well and does not need the underlying QM frame $\endgroup$ – anna v Oct 16 '15 at 6:36

Resistance and current are macroscopic concepts. While obviously useful in daily life (depending on your day job :)), if you want to understand what's happening at a microscopic level with electrons you want to talk about the corresponding microscopic quantities resistivity and current density.

Overview of Resistivity (electrons are sensitive to local properties of a material)

Current density is a local quantity, it is a vector telling you how much current flows at each point in space. contrast this with current, which is the integral of current density over a cross sectional area and is thus not a local quantity--current includes information about the thickness of the wire. Current density is related to the applied electric field by the resistivity $\rho$ \begin{equation} \vec{E}=\rho \vec{J} \end{equation}

Resistivity is a property of a material. It depends on how the electrons scatter off of the lattice of the material. A simple model that gives qualitatively the right picture (but makes several wrong quantitative predictions--if you want the full story you need quantum mechanics and stat mech) is the (classical) Drude model, which says \begin{equation} \rho=\frac{nq^2}{m} \tau \end{equation} where $n$ is the number density of the charge carriers, $q$ is the charge of the charge carriers, $m$ is the mass of the charge carriers, and $\tau$ is the average time between scattering events between the electron and the material in the wire. $\tau$ tells you about the material. individual electrons certainly "know about" $\tau$. So electrons "know about" the resistivity.

Caveat: I'm lying slightly to simplify things... a more precise statement is that averaging over a small region the electrons know about the resistivity.

Also note: Quantum mechanics isn't strictly needed to understand resistivity, unless you want a more quantitatively accurate description of the scattering processes involved which affects for example the frequency dependence of resistivity.

The resistance is related to the resistivity by \begin{equation} R=\frac{\rho L}{A} \end{equation} The factors of $L$ and $A$ are purely geometrical, related to your particular macro physical setup, and have nothing to do with microphysics. It just comes from integrating the effects of many individual electrons each obeying the microscopic version of Ohm's Law $\vec{E}=\rho\vec{J}$. Electrons don't know about the length of the wire. So electrons know about resistivity, but they don't know about resistance.

Hooking Resistors up in parallel

What about hooking resistors up in parallel? Well, obviously this is another macroscopic property of the system and electrons don't know how the resistor they are travelling through is hooked up to a circuit. They only know about applied field and local properties of the material like $\tau$.

What happens is that the battery works to keep the voltage across the resistors fixed, and in order to do that it supplies as much current as needed. It's admittedly hard to see when you first go through things how this doesn't require some electron somewhere to know about the global topology of the circuit (ie, know that the resistors are in parallel). But each individual component is just doing its job locally. The battery naturally acts to keep the voltage the same across its terminal. The electrons moving through the resistor only know about the applied field and scattering off the material. It's the combination of all these different components, each acting locally, that build up to the notion of currents moving through a given circuit. But this isn't related to quantum mechanics per se.

A way to understand it is that when we write down circuit laws for batteries + resistors we typically assume that the system is in equilibrium. If you take a circuit that has one resistor, then suddenly add a second in parallel, then at that instant there is no current through the second resistor and system is not in equilibrium. In particular, electrons will begin to flow through the second resistor (because they can), but this decreases the voltage at one end of the battery. So in order to remain in equilibrium the battery pumps out more current until equilibrium is reached and Ohm's laws are satisfied.

Hooking Resistors up in series (added do to comments below)

A way to understand what's going on is to imagine the circuit approaching equilibrium. We start with a battery + 1 resistor. Then we add a second resistor in series. Very shortly after adding the second resistor, the system is not in equilbrium because the current through the first resistor will not change. However, charges will not be able to flow at the same rate through the second resistor as they could through a wire. So there will be a build up of charge around the second resistor. This will generate a field that opposes current flow through the first resistor, decreasing the current. The buildup of charge will also act to increase the current through resistor two. Eventually (actually an extremely short time on human timescales), the system will reach equilibrium when current through resistor equals the current through resistor two.


Electrons "know about" an increased resistivity because this is determined by properties of the material the electrons are traveling in. However they may or may not "know about" an increase in the resistance because the resistance is not a purely local quantity, it depends for example on the length of the wire. Since electrons don't know about resistance, it might not be surprising to learn they don't know directly about equivalent resistance either (which depends on exactly how you hook a circuit up).

  • $\begingroup$ thanks for some new insight.You are bringing me close to understanding voltage and current in terms of pure underived quantities. Suppose we add resistor in series now(to already batter and resistor), what is happening to the Electric field already there? what battery and the new resistor did to the field? $\endgroup$ – Ramki Oct 17 '15 at 6:07
  • $\begingroup$ Added a short section about adding a resistor in series. $\endgroup$ – Andrew Oct 18 '15 at 8:00

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