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Today I learned that energy is transferred to a lightbulb through electromagnetic waves produced by the movement of electrons and according to Poynting's law, the direction of this energy is perpendicular to the electric field.

On the other hand, from what I previously know, light is produced due to the interaction between the light bulb's atoms and the moving electrons which means energy is carried by the moving electrons and not by the electromagnetic waves produced by the electrons.

I'm new to the this subject, so I'm confused. Does the electromagnetic waves turn on the light bulb or the moving electrons in the wires?

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You are confusing two different electromagnetic waves which are involved here. So I will elaborate in a more detailed way.

  • The cables between the power plant in your province and the light bulb in your home are surrounded by an electromagnetic field of low frequency ($50$ or $60$ Hz, depending on your country). It is the Poynting vector of this low-frequency electromagnetic field which delivers energy into the filament of the light bulb (the blue arrows in the image below).
    enter image description here
    image based on the one from Poynting vectors of DC circuit, showing the electric field (read arrows), magnetic field (green arrows) and Poynting vector (blue arrows)
  • Due to this electromagnetic field the electrons inside the filament move back and forth (with frequency $50$ Hz). When moving they collide with the metal atoms, thus transferring kinetic energy to the atoms, so that their unordered movement will increase. I.e. the temperature rises very much ($\approx 2400 °$C).
  • According to this temperature the filament emits electromagnetic radiation with a broad range of very high frequencies (roughly from $10^{14}$ to $10^{15}$ Hz), i.e. infrared and visible light. See also Black-body radiation for the physics behind that.

so I'm confused , does the electromagnetic waves turn on the light bulb or the moving electrons in the wires ?

As seen above it is both: The low-frequency electromagnetic waves turns on the electrons in the light bulb. And the moving electrons in the filament turn on the light waves.

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  • $\begingroup$ Nice. Quick question though, the em field that makes.the electrons going back and forth inside the wire does create an em field inside the wire, right? $\endgroup$ Feb 5, 2022 at 14:11
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    $\begingroup$ @untreated_paramediensis_karnik Yes, there is also an electric field inside the wire (without this the electrons would not move). But most of the field energy is located outside the wire. $\endgroup$ Feb 5, 2022 at 14:24
  • $\begingroup$ Your answer is correct (and I did upvote) but it is rare for people to think of 60Hz current and voltage in terms of E&M wave since the system is basically in the conduction (rather than radiation) regime. $\endgroup$ Feb 5, 2022 at 15:26
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    $\begingroup$ electrons inside the filament move back and forth - "back and forth" is not why the filament heats up. A DC voltage would not generate any EM waves, but the bulb will still glow. Maybe you can add that to the answer. It might clarify OP's confusion. $\endgroup$ Feb 5, 2022 at 17:24
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    $\begingroup$ @ArchismanPanigrahi Yes, of course the filament will also heat up when driven by DC instead of AC. I have updated the answer to make it more clear. $\endgroup$ Feb 5, 2022 at 18:26
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The light bulb glows due to a multi-stage process of energy conversion:

  1. magnetic to electric: in the turbines/generators of the power plant, a magnetic field is converted to an electric field in a coil due to induction (well, the voltage in the power plant could also be photovoltaic, but let's consider the probability of that origin a lot lower as of today)
  2. electric to kinetic: as soon as you close the light switch, the electric field accelerates electrons relative to atom bodies in the circuit wires all the way from the power plant, through the light bulb, and back to the power plant; since this cannot happen faster than light, this is conveyed by electromagnetic waves, and since the electrons cannot sustain a high velocity (see stage 3), the involved electromagnetic energy may be considerable in comparison to the kinetic energy of the electrons
  3. kinetic to thermal: the accelerated electrons collide again with atom bodies after what is called the mean free path length; thereby, their directed kinetic energy (current) is converted into the undirected motion of the atoms (heat); since the resistance in the light bulb's wire is higher than in the public electricity network, this conversion preferably happens in the bulb
  4. thermal to electromagnetic: the undirected motion of the atoms of the bulb causes the electrons to wiggle so irregular and fast that they emit what is called (in idealization) black-body radiation, a mixture of all sorts of frequencies (i.e., colors), which is clustered around a certain frequency; the position of this distribution of light colors in the electromagnetic frequency spectrum is determined by the temperature of the light bulb's wire, which is why it is called the color temperature of the emitted light; the higher the temperature, the higher the most weighted frequency of the radiation; if the temperature is below I think ~400 °C you will not see the radiation because it is in the invisible infrared; if the temperature could get above 5800 °C, the emitted light would be more blue than the natural sunlight (but of course above 5800 °C even tungsten wires start to evaporate, so that it is technically not possible to get a lightbulb emit bluish light)

At any stage, there is an input energy form, an output energy form, and something that does the conversion (a technical device, part, etc.). In the last stage, the input is heat, the output is an electromagnetic field, and the wire is the converter.

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  • $\begingroup$ Thanks for the answer but still confused , what is the role of the EM waves produced by the moving electrons in the wires and and the Poynting vector in heating the bulb ? $\endgroup$
    – firas
    Feb 6, 2022 at 1:51
  • $\begingroup$ When current is flowing the energy is predominantly carried in electromagnetic waves surrounding the wires. The kinetic energy of the electrons is entirely negligible. $\endgroup$
    – Jagerber48
    Feb 6, 2022 at 6:07
  • $\begingroup$ Jagerber48 But according to this answer the kinetic energy of the free electrons is transferred to thermal energy and that's happen due to the atoms of the bulb colliding with the free electrons and there is no mention of EM waves interacting with atoms of the bulb $\endgroup$
    – firas
    Feb 6, 2022 at 7:13
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    $\begingroup$ AT THE BULB what happens is the EM wave which have travelled from the source deliver energy the electrons in the bulb filament causing them to accelerate within the filament body. Because of collisions, the (low frequency) EM energy is transfered through the electrons, into phonons in the lattice of the filament leading ot heating up of the lattice. This heat then excites valence and conductance electrons in the filament which then decay and emit light (high frequency EM energy). $\endgroup$
    – Jagerber48
    Feb 7, 2022 at 0:22
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    $\begingroup$ So I would say this answer is mostly correct except for step 2. There should be a step before 2 which explains that energy is transferred from the power plant to the bulb via low-frequency EM waves, rather than kinetic energy of electrons in the wires. The EM waves then delivery kinetic energy to the electrons at the bulb (this is essentially the step 2 as written with some modification). The rest of the answer is good. $\endgroup$
    – Jagerber48
    Feb 7, 2022 at 0:25
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The free electrons are given energy (over and above their random thermal energy) by an electric field propagating as a wave. That's why they get this extra energy almost simultaneously throughout the light bulb's filament when you close the switch. This extra energy is shared with atoms (or ions) of the filament through collisions between the electrons and the (vibrating) atoms. So the filament gets hot.

I think that the confusing word in your question is "carried". I would say that the energy is carried by e-m waves but given to free electrons and thence to the filament itself as thermal energy.

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  • $\begingroup$ Thanks for the answer , but what about the EM waves produced by the moving free electrons and travel with the direction of the Poynting vector which is perpendicular to the electric field ? aren't they what gives the atoms of the bulb the energy so that the bulb start to heat ? $\endgroup$
    – firas
    Feb 6, 2022 at 2:02
  • $\begingroup$ @NazareneChristianSoldier: No, actual collisions with the moving electrons directly are the primary means of energy transfer from them to atoms in the filament. Everything works basically identically for the DC case, where the average velocity of electrons in the filament as a whole stays constant with time. (Individual electrons still speed up and then collide, after travelling the mean free path, on average.) $\endgroup$ Feb 6, 2022 at 19:34
  • $\begingroup$ @ Peter Cordes : So what's the role of EM and the Poynting vector ? $\endgroup$
    – firas
    Feb 7, 2022 at 1:12

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