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|Hello, everyone, I have a question regarding the relation between photon emitting in terms of electricity. I asked the question in the Electrical Engineering forum but I was told that here is the appropriate place.

Here's what I know

A photon is emitted whenever an electron goes from higher energy level to lower energy level(goes from upper to lower valence shell), thus lowering its own internal energy and because of the difference between its current state of energy and its former one has to go somewhere(law of energy conservation) a photon is produced.Also, according Newton's law - an object with constant velocity and direction remains moving without losing energy unless some external force is exerted upon it.So, my questions is when a particular electron pass from an environment with lower resistance to an environment with higher resistance(flowing through conductive wire with given R1 and reaches a resistor R2 where R2 > R1 - a DC circuit) some force is being exerted upon its direction of movement but electromagnetic wave is not produced, why is that, since in AC current produce such waves if electrons change their direction totally, so is this the necessary condition(180 degrees of change) for an electron to lose energy or is there something else?

Thanks in advance.

Best regards, Nina

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  • $\begingroup$ Please link to the EE post. $\endgroup$ – Qmechanic Dec 30 '16 at 20:23
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Also, according Newton's law - an object with constant velocity and direction remains moving without losing energy unless some external force is exerted upon it.

When one is talking of electrons and photons, one is in the quantum mechanical regime, and also in the special relativity regime. Newton's laws apply to the macroscopic classical regime. Both the electron and the photon are elementary particles following quantum mechanical equations.

So, my questions is when a particular electron pass from an environment with lower resistance to an environment with higher resistance(flowing through conductive wire with given R1 and reaches a resistor R2 where R2 > R1 - a DC circuit)

Resistance is an emergent quality , it emerges from the underlying quantum mechanical behavior and describes the macroscopic behavior of circuits. Electrons in solids are described by the quantum mechanical model of the band theory of solids..

In a conductor the electrons belong in an energy band that ties them to the whole lattice, and the attraction of the electric field gives to each individual electron a drift velocity which will build up the current. In an insulator most electrons are tied in their locations in the lattice and very few are in the conduction band . More energy is needed to give a drift velocity to electrons.

some forces is being exerted upon its direction of movement but electromagnetic wave is not produced,

the concept of force at the quantum level is a change in momentum, a dp/dt . All such changes for an electron in an electric field will give electromagnetic radiation, i.e. a photon will carry off some momentum and energy. This energy will be in the infrared frequencies and will appear as heat when the current is high. For a conductor where the electrons are in the conduction band very little radiation is released because the dp/dt of the individual electrons is small.

Note that resistors heat up, that is photons in the infrared frequencies.

why is that, since in AC current produce such waves if electrons change their direction totally, so is this the necessary condition(180 degrees of change) for an electron to lose energy or is there something else?

No , it is not necessary to have a 180 degrees reversal. Any acceleration/deceleration will give off radiation. Applying a voltage to a circuit induces accelerations on the drifting electrons, including the statistical scatterings due to their motion. This radiation is in the infrared frequencies, appearing macroscopically as heat.

Now to address this:

A photon is emitted whenever an electron goes from higher energy level to lower energy level(goes from upper to lower valence shell), thus lowering its own internal energy and because of the difference between its current state of energy and its former one has to go somewhere(law of energy conservation) a photon is produced

For electrons in the conduction band the quantum mechanical energy differences in a conductor (where a large number of the electrons of the lattice are), are so small as to be considered a continuum. For electrons in insulators there will be scatterings of electrons moving in the conduction band with electrons bound strongly in the atoms of the lattice and those will take up energy and momentum in the form you are describing.

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  • $\begingroup$ I managed to grasp the concept thanks to you, let me ask one more question for even further clarification.I understand that in any given DC electrial circuit an electromgnetic wave is being produced with a certain frequency, so my guess is that in AC circuits the same thing happens , BUT this time the friquency of the emitted wave is determined by the frequency of the AC source, therefore a 60Hz AC current will produce 60Hz electromagnetic wave, right? $\endgroup$ – Nina Vladimirova Dec 30 '16 at 19:59
  • $\begingroup$ The radiation from DC has a frequency distribution, not a single frequency, and because it appears as heat it is in the infrared. If you put a very high voltage across a gap and a continuous spark breaks out, it is a similar mechanism, except it is in the visible. The AC current generates a coherence in the motion of the electrons on top of the statistical drift velocity by the changes in the voltage and so there will be both the infrared and the specific frequency of the AC radiated. of course in a conductor it will be the main frequency. $\endgroup$ – anna v Dec 30 '16 at 20:05
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It is important to understand the nature of what is going on in the wire while the electrons are moving through. The picture that the electron is moving freely through the wire is too simplified. On a quantum scale, the electron is continuously interacting with the wire material (copper in most cases) and other electrons in the wire. There is a huge number of scattering processes which may happen inside the wire and some of them will be capable to produce photons. But the dominant process will be a scattering of the electrons with the wire atoms which transfers energy to the wire atoms. The increase of kinetic energy of the wire atoms is nothing else than the increases temperature of the wire. I said there are some processes which might produce light but which wavelength are we expecting? For red light we would need approx 1,7eV (~750 nm). With the typical voltages and the very short length between two scattering events the electrons will not have enough energy to produce visible light. However you could reduce the density of scattering atoms and then ... now we are about to invent gas discharge tubes and soon we will redo the Franck–Hertz experiment.

On a very high level the resistance of a wire is the boiled down version of the microscopic description. This shows you that the resistance cannot be a quantum state and thus a change of the resistance is nothing were you magically change the energy of an elementary particle. It is just describing that the energy loss rate due to scattering processes changes when you go from one wire to the next one. If you want more physical background on this, it might be worth to look up the Drude moddel.

About the part with Newton mechanics, remember that there is a potential gradient present in the wire, so what you expect is that the electrons should continuously increase their speed. Obviously this is not the case in a wire, as the carrier scattering is constantly dissipating energy from the electrons to the wire atoms.

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If we look on a live stationary DC circuit via idealized macroscopic EM theory where there is no noise and the current density $\mathbf j$ is perfectly constant in time everywhere, the system does not produce EM radiation. This follows from Maxwell's equations: if $\partial\mathbf j/\partial t = 0$ everywhere, there is no EM radiation connected to the system.

Of course, in real world current density is not perfectly static and it actually fluctuates in time around some mean value. The electrons and nuclei move erratically and the higher the temperature, the more intense the erratic motion. According to Maxwell's equations, this means the system actually bathes in erratic EM field and also creates it own erratic EM field. This field has "radiation component", it can be detected from far away and it is called thermal radiation.

Often this thermal radiation is ignored, because it has negligible influence on the operation of the circuit.

Finally, to get back to your example of an electron passing from low resistivity to high resistivity space: the electron radiates erratically based on its erratic motion all the time; passage from one space to another takes place in very small portion of the system and quite quickly; at any time, the number of electrons in this transition is much lower than the number of electrons either in the low resistivity region or high resistivity region. It would probably be quite difficult to detect any features of the EM radiation from the transition region, as most of the radiation intensity comes from the low and high resistivity parts, which contain most of the electrons in the system.

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