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I am self-studying electricity and magnetism, and I am confused about a point.

I have learnt that the drift speed of an electron is extremely small. However, according to Drude's model, the electron jumps around very fast and bounces against other particles in a conductor. I also read that this random movement happens even in the absence of electric field.

But how this is possible? Electrons should lose kinetic energy when bouncing off particles, and should become slower and slower. Maybe this process takes really long time and thus we can consider the slow down negligible?

But then what is making a lightbulb glow? I know it is heat generated by bouncing off particles. In my naive understanding, I thought this was dependent on the flow of electrons, but I am now doubting this since they are drifting really slowly. Is it the random scatter movement? That's also weird: if the electrons are scattered around even without a field, then the lightbulb should glow even without a voltage source, which doesn't happen.

So my questions:

  1. Does the "scattering speed" of the electrons depends on the presence of an electric field?

  2. What makes the lightbulb glow? Is it the electron slowly drifting? Is it the scatter speed increasing due to the electric field? Both?

What is really going on then?

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2 Answers 2

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When electrons bounce of off other electrons and/or molecules, they lose kinetic energy. At the same time the other party in the collision will absorb energy. Energy must always be conserved, so it is never "zero action" in an element which is not at $0K$. The drift of carriers in a material is proportional to the temperature.

An applied electric field (voltage) will put free electrons in the unidirectional motion required for a net current to form. These travelling electrons will then, as you say, impact with others with them giving and taking energy from one another. If the electron collides with something other than another free electron, the material structure itself absorbs this energy. In this case it is no longer translational energy but rather vibrational and/or rotational. This heats up the resistor and the resulting radiation (light) is best modelled with a blackbody-model where the spectrum and intensity is determined by the temperature of the filament.

When there is no applied voltage, there will be random movement of charges as you say. However, this is on a smaller, local scale. On the larger scale of the filament, there is no flow of electrons other than the random movement of electrons. A current is typically defined by the unidirection movement of electrons in one or several directions.

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  • $\begingroup$ @DakkVatter I am not sure this answer my question 1. You say that there is never "zero action" unless we are at $0K$, justifying this with conservation of energy. However, the bouncing of the electrons resulting in heat means that kinetic energy is transformed into thermal energy. If this goes on long enough, the bouncing should eventually stop, as all kinetic energy is converted into thermal energy. Is this correct? And if not, why not? $\endgroup$
    – Vaaal88
    Commented Mar 29, 2022 at 12:30
  • $\begingroup$ "An applied electric field (voltage) will put free electrons in motion and thus creating a current." - is this implying that the electrons are NOT moving without an electric field? I don't think this is correct. Surely they aren't drifting, but if they are moving, why isn't this movement enough to heat up the wire? $\endgroup$
    – Vaaal88
    Commented Mar 29, 2022 at 12:34
  • $\begingroup$ @Vaaal88 On a smaller local scale, electrons are always in motion and there may be a net transfer of charge, however macroscopically speaking, a current created by an applied voltage takes the form of a coherent motion of electrons, which simply isn't present when electrons are freely and randomly moving in all directions. Thus the conventional "current" is not present in that case $\endgroup$
    – DakkVader
    Commented Mar 29, 2022 at 12:53
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    $\begingroup$ @Vaaal88 There are what's called radiative and non-radiative transitions of energy. Radiative transitions of energy is essentially when a photon is spontaneously emitted when an electron gives up its charge. Higher temperature -> higher electron energy -> shorter wavelengths. But essentially it glows for the same reason that hot steel from a furnace glows. All matter with a temperature above 0K emits radiation with different intensities and spectrums. Electrons can either give their charge to another carrier, or it can simply 'give it up' as a photon $\endgroup$
    – DakkVader
    Commented Mar 29, 2022 at 13:37
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    $\begingroup$ @Vaaal88 Introductory courses more often chose to go with the wave-approximation for light. For small-scale applications and other fields, particle-approximation is mandated. The sub-field is called photonics which is my field of research (nonlinear photonics). $\endgroup$
    – DakkVader
    Commented Mar 30, 2022 at 10:07
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Assuming you are discussing an incandescent light bulb, the glow literally is a result of blackbody radiaton as the (usually) tungsten filament will heat to temperatures of ~3000K. The light bulb is acting as essentially a resistor, where the filament is slowing down the flow of electrons, emitting heat and light. But tungsten itself is not a normal resistor in that it does not follow Ohm's law, and you'll find that as the tungsten heats up, its resistance increases. Thus light bulbs are 'non-Ohmic resistors'.

In terms of your question regarding the relationship of electron drift velocity, this article may help: "the speed of electricity"

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  • $\begingroup$ You say that as a resistor it slows down the "flow of electron". This means it reduces the drift speed, right? $\endgroup$
    – Vaaal88
    Commented Mar 29, 2022 at 9:08
  • $\begingroup$ @Vaaal88 right, without resistance we'd have ballistic conduction. $\endgroup$
    – Ruslan
    Commented Mar 29, 2022 at 9:33
  • $\begingroup$ In the linked page I read "it's constantly bumping into the atoms of the filament [that] heats up the filament until it's hot enough to glow." But this bumping is there even without electric field, and yet only when we apply it the lightbulb glows. So either my assumption is wrong (no bumping without e. field) or it's not the random bumping. $\endgroup$
    – Vaaal88
    Commented Mar 29, 2022 at 9:54

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