6 ms·

> Canonically, electric current results from the collective movement of electrons, each carrying one indivisible chunk of electric charge. But the dead steadine

by jocaal 3y ago

> Canonically, electric current results from the collective movement of electrons, each carrying one indivisible chunk of electric charge. But the dead steadiness of Chen’s current implied that it wasn’t made of units at all. It was like finding a liquid that somehow lacked individually recognizable molecules In maxwell's equations, current density J is defined in terms of the E-field. When talking about electricity, people make the typical quantum mechanical wave-particle mistake. Electricity refers to two things, photons and electrons and how they interact with eachother. Both act as wave-particles, but photons act more like waves and electrons more like particles. The thing that gets people is that photons are the things that move energy around. A photon is an electromagnetic wave. In a wire, you can have an electromagnetic wave traversing the wire at some proportion of the speed of light, while the electrons are moving at speeds closer to meters per second. We defined current to be proportional to the E-field (because that is what is moving the energy) and thus we shouldn't refer to the movement of electrons as current.

orra 3y ago
> while the electrons are moving at speeds closer to meters per second Plus, with AC, the electrons move back and forth, instead of just moving forward!
- XorNot 3y ago
  It is however still movement, which is important because static electric fields don't produce magnetism, whereas moving charges do - which is the principle benefit of AC current.
  - andyferris 3y ago
    Well, technically that depends on your reference frame. If the _observer_ starts moving, they will observe a magnetic field where before there was only electric. It’s not that static electric fields “produce” magnetism. Magnetism is a result of relativity. (But yes, it is still movement…).
swayvil 3y ago
Does the thing at one end of the wire get heavier and the thing at the other end lighter? Because of the mass of the electrons moved from one thing to the other.
- mike_hock 3y ago
  An electron has .05% the mass of a proton, and only a small imbalance of electrons and protons is necessary to generate extremely strong electric fields by earthly standards.
  - swayvil 3y ago
    That's fabulous man, but it doesn't actually answer my question. If the answer is yes then we have moving electrons. If the answer is no then maybe we don't.
    - senseiV 3y ago
      Bruh its simple physics, does one end or the other get lighter, by all measures we care about, not really, the mass of a proton or electron is beyond any consumer hardware measurement. I doubt it would matter beyond extreme scenarios or controlled experiment
    - ben_w 3y ago
      Imagine a loop of pipe filled with water, and a pump pushing the water: The water moves around the pipe. The movement doesn't cause any part of the pipe to get heavier. Imagine a different scenario, where the pipe ends in a big box: This time the box does fill up with water, and gets heavier. Mapping the analogy from water back to electrons: a loop of pipe is like a loop of wire and a battery; while the pipe ending with a box becomes a capacitor or antenna, and that will leak[0] before you can measure the mass change — but technically yes the the mass of any given wall of a capacitor or of an antenna will be very slightly changed by this sort of thing. For a sense of scale, to get a total charge of 1 coulomb using electrons, the mass of those electrons will be about 5.7 nanograms, and trying to squeeze that much charge into the last millimetre of some length of a wire 1mm in cross section diameter, involves about 60% of the energy in this explosion: https://www.youtube.com/watch?v=wqKn_3iJOP4 https://www.youtube.com/watch?v=wqKn_3iJOP4 As nothing gets close to being able to hold that kind of energy, even if you're trying to accumulate a lot of excess electrons, those electrons leak well before even coming close to nanograms of excess mass. [0] The bit on the top with the sparks coming out is a capacitive electrode: https://en.wikipedia.org/wiki/Tesla_coil https://en.wikipedia.org/wiki/Tesla_coil
blix 3y ago
One of Maxwell's other equations links the E-field to charge density. You can't talk about the time derivative of the E-field without also talking about the time derivative of charge density. It's true that interactions between charged particles invoke photons, but you can have charged particles (and their associated E-fields) travelling at constant velocity in a vaccum and still define an associated current density without considering photons. I'm not sure your interpretation produces a useful intuition for this situation. One of the underlying ideas of the paper is that, in conventional metals where conduction band electrons are much more wave than particle, you can still measure when they 'enter' and 'exit' the material through perturbations in the field, that is bursts of photons that occur when electron waves interact with an obstacle.
- deleted 3y ago
  [deleted]
oh_my_goodness 3y ago
It's true that waves carry the energy. But shot noise is still determined by the amount of charge on an electron.
jstanley 3y ago
> you can have an electromagnetic wave traversing the wire at some proportion of the speed of light, while the electrons are moving at speeds closer to meters per second. This isn't as unintuitive as it sounds though. In a water pipe you can have a pressure wave reach from one end to the other very quickly, even as the individual water molecules move very slowly. If water pressure is voltage, water molecules are electrons, and the pipe is the wire, then it's easy to see how a "voltage" can reach the other end of a "wire" without any of the "electrons" having to move very far at all.
- jocaal 3y ago
  The problem with this analogy is that pressure waves can only travel in a medium. The pressure wave exists because the particles in the medium bump against eachother to propagate energy so the analogy gives people the impression that electrons are bumping against eachother to propagate energy. Photons are distinct entities that can travel without a medium. See michelson-morley experiment. Electricity is nothing like what we experience on our macro scale. Analogies never convey the full story.
  - jstanley 3y ago
    > the analogy gives people the impression that electrons are bumping against eachother to propogate energy. If this is not approximately how electricity works, why does the current stop flowing when you disconnect the wires, or equivalently when you "connect" them via something that doesn't have free electrons (i.e. an insulator)?
    - jocaal 3y ago
      Unfortunately the explanation to your question is something that takes years of education to understand and even then, most people just come out understanding some parts. The best I can do on an internet forum is tell you that the analogy is wrong, because experiments told us its wrong. My advice is if you want to understand this stuff, do the math! ;)
      - 3y ago
rubberpoliceman 3y ago
This is not even approximately true. j is not “defined in terms of E”. It occurs as a source term in one of the inhomogeneous Maxwell equations, and consistency with the other inhomogeneous one (div D=rho) requires that div j = dot rho, which means that j is the flow of charge density. So, I guess if electron movement shouldn’t be referred to as current, we shouldn’t refer to electrons as charges, either…
- jocaal 3y ago
  When solving for the wave equation, the substitution made for J is sigma * E. So defined in terms of the E field and some conductivity constant. When solving the wave equation, there is no mention of charge density, everything is fields. This predicted the existence of electromagnetic waves. In reality, we know there is only electrons and photons. So thus the wave equation is the source of truth and what we call fields are just a consequence of the complex interactions between electrons and photons and this is studied in quantum electrodynamics.

by jocaal 3y ago

orra 3y ago

> while the electrons are moving at speeds closer to meters per second Plus, with AC, the electrons move back and forth, instead of just moving forward!

XorNot 3y ago
It is however still movement, which is important because static electric fields don't produce magnetism, whereas moving charges do - which is the principle benefit of AC current.
- andyferris 3y ago
  Well, technically that depends on your reference frame. If the _observer_ starts moving, they will observe a magnetic field where before there was only electric. It’s not that static electric fields “produce” magnetism. Magnetism is a result of relativity. (But yes, it is still movement…).

swayvil 3y ago

Does the thing at one end of the wire get heavier and the thing at the other end lighter? Because of the mass of the electrons moved from one thing to the other.

mike_hock 3y ago
An electron has .05% the mass of a proton, and only a small imbalance of electrons and protons is necessary to generate extremely strong electric fields by earthly standards.
- swayvil 3y ago
  That's fabulous man, but it doesn't actually answer my question. If the answer is yes then we have moving electrons. If the answer is no then maybe we don't.
  - senseiV 3y ago
    Bruh its simple physics, does one end or the other get lighter, by all measures we care about, not really, the mass of a proton or electron is beyond any consumer hardware measurement. I doubt it would matter beyond extreme scenarios or controlled experiment
  - ben_w 3y ago
    Imagine a loop of pipe filled with water, and a pump pushing the water: The water moves around the pipe. The movement doesn't cause any part of the pipe to get heavier. Imagine a different scenario, where the pipe ends in a big box: This time the box does fill up with water, and gets heavier. Mapping the analogy from water back to electrons: a loop of pipe is like a loop of wire and a battery; while the pipe ending with a box becomes a capacitor or antenna, and that will leak[0] before you can measure the mass change — but technically yes the the mass of any given wall of a capacitor or of an antenna will be very slightly changed by this sort of thing. For a sense of scale, to get a total charge of 1 coulomb using electrons, the mass of those electrons will be about 5.7 nanograms, and trying to squeeze that much charge into the last millimetre of some length of a wire 1mm in cross section diameter, involves about 60% of the energy in this explosion: https://www.youtube.com/watch?v=wqKn_3iJOP4 https://www.youtube.com/watch?v=wqKn_3iJOP4 As nothing gets close to being able to hold that kind of energy, even if you're trying to accumulate a lot of excess electrons, those electrons leak well before even coming close to nanograms of excess mass. [0] The bit on the top with the sparks coming out is a capacitive electrode: https://en.wikipedia.org/wiki/Tesla_coil https://en.wikipedia.org/wiki/Tesla_coil

blix 3y ago

One of Maxwell's other equations links the E-field to charge density. You can't talk about the time derivative of the E-field without also talking about the time derivative of charge density. It's true that interactions between charged particles invoke photons, but you can have charged particles (and their associated E-fields) travelling at constant velocity in a vaccum and still define an associated current density without considering photons. I'm not sure your interpretation produces a useful intuition for this situation. One of the underlying ideas of the paper is that, in conventional metals where conduction band electrons are much more wave than particle, you can still measure when they 'enter' and 'exit' the material through perturbations in the field, that is bursts of photons that occur when electron waves interact with an obstacle.

deleted 3y ago
[deleted]

oh_my_goodness 3y ago

It's true that waves carry the energy. But shot noise is still determined by the amount of charge on an electron.

jstanley 3y ago

> you can have an electromagnetic wave traversing the wire at some proportion of the speed of light, while the electrons are moving at speeds closer to meters per second. This isn't as unintuitive as it sounds though. In a water pipe you can have a pressure wave reach from one end to the other very quickly, even as the individual water molecules move very slowly. If water pressure is voltage, water molecules are electrons, and the pipe is the wire, then it's easy to see how a "voltage" can reach the other end of a "wire" without any of the "electrons" having to move very far at all.

jocaal 3y ago
The problem with this analogy is that pressure waves can only travel in a medium. The pressure wave exists because the particles in the medium bump against eachother to propagate energy so the analogy gives people the impression that electrons are bumping against eachother to propagate energy. Photons are distinct entities that can travel without a medium. See michelson-morley experiment. Electricity is nothing like what we experience on our macro scale. Analogies never convey the full story.
- jstanley 3y ago
  > the analogy gives people the impression that electrons are bumping against eachother to propogate energy. If this is not approximately how electricity works, why does the current stop flowing when you disconnect the wires, or equivalently when you "connect" them via something that doesn't have free electrons (i.e. an insulator)?
  - jocaal 3y ago
    Unfortunately the explanation to your question is something that takes years of education to understand and even then, most people just come out understanding some parts. The best I can do on an internet forum is tell you that the analogy is wrong, because experiments told us its wrong. My advice is if you want to understand this stuff, do the math! ;)
    - 3y ago

rubberpoliceman 3y ago

This is not even approximately true. j is not “defined in terms of E”. It occurs as a source term in one of the inhomogeneous Maxwell equations, and consistency with the other inhomogeneous one (div D=rho) requires that div j = dot rho, which means that j is the flow of charge density. So, I guess if electron movement shouldn’t be referred to as current, we shouldn’t refer to electrons as charges, either…

jocaal 3y ago
When solving for the wave equation, the substitution made for J is sigma * E. So defined in terms of the E field and some conductivity constant. When solving the wave equation, there is no mention of charge density, everything is fields. This predicted the existence of electromagnetic waves. In reality, we know there is only electrons and photons. So thus the wave equation is the source of truth and what we call fields are just a consequence of the complex interactions between electrons and photons and this is studied in quantum electrodynamics.