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How non-ionizing radiation causes corona discharge?


BorisBoris

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Photoelectric effect doesn't work here. So do photons of radio waves (and lower frequencies) really exist?

In many high voltage applications, corona is an unwanted side effect. Corona discharge from high voltage electric power transmission lines constitutes an economically significant waste of energy for utilities. In high voltage equipment like Cathode Ray Tube televisions, radio transmitters, X-ray machines, and particle accelerators the current leakage caused by coronas can constitute an unwanted load on the circuit.
https://en.wikipedia.org/wiki/Corona_discharge

 

 

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It is not the radiation, it is the voltage. The words "high voltage" appear 3 times in that paragraph. And the first paragraph of that article explains what causes it. And the Mechanism section provides more detail:

Quote

Corona discharge results when the electric field is strong enough to create a chain reaction: electrons in the air collide with atoms hard enough to ionize them, creating more free electrons which ionize more atoms. The diagrams below illustrate at a microscopic scale the process which creates a corona in the air next to a pointed electrode carrying a high negative voltage with respect to ground. The process is: ...

https://en.wikipedia.org/wiki/Corona_discharge#Mechanism

 

(And, yes, obviously photons of radio waves exist.)

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But we can measure electric component od this radiations... if it is strong enough, causes corona?

And light does not have measurable electric component? Then why we are talking about low energy photons? They are simple electric + magnetic oscillating field (classical wave, not particles)...

Edited by BorisBoris
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9 minutes ago, BorisBoris said:

But we can measure electric component od this radiations... if it is strong enough, causes corona?

What radiation? And how do you measure the electric component?

I don't think radiation can cause a corona discharge. You need a high voltage.

11 minutes ago, BorisBoris said:

And light does not have measurable electric component? Then why we are talking about low energy photons? They are simple electric + magnetic oscillating field (classical wave, not particles)...

I have no idea why you are talking about low energy photons. Can you explain how they are relevant to high voltage discharge?

Light (which includes radio waves, in physics) can be modelled as either classical waves or particles. Depending on the application.

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49 minutes ago, Strange said:

What radiation? And how do you measure the electric component?

Any EMF meter gives resulsts in V/m, it is well known https://www.trifield.com/wp-content/uploads/2018/03/tf2_owners_manual.pdf

example

Quote

Electric Mode covers 40 Hz - 100 kHz with range of 1 - 1000 volts per meter (V/m)

Now imagine that frequency of a wave is 2 Hz... or 0.2 Hz.... it is radiation, and can do the same as light in Einstein's experiment. But formula E=hf does not have any sense.

https://www.researchgate.net/post/What_is_the_difference_between_am_electromagnetic_wave_and_a_photon

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@BorisBoris

Multimeters, oscilloscopes and electronic/electric devices used by electricians are typically measuring frequency of alternating current (AC) or pulsing DC.

  

34 minutes ago, BorisBoris said:

But formula E=hf does not have any sense.

This formula tells how much single photon has energy.

 

Edited by Sensei
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29 minutes ago, BorisBoris said:

Any EMF meter gives resulsts in V/m, it is well known https://www.trifield.com/wp-content/uploads/2018/03/tf2_owners_manual.pdf

example

Quote

Electric Mode covers 40 Hz - 100 kHz with range of 1 - 1000 volts per meter (V/m)

 

As it says, that is measuring the electric field (measured in volts/metre) not electromagnetic radiation.

The electromagnetic field measurement capability is on the following line:

Quote

RF Mode covers 20 MHz - 6 GHz with range of 0.001 - 19.999 milliwatts per square meter (mW/m2)

Which measure the power of the EM signal.

33 minutes ago, BorisBoris said:

Now imagine that frequency of a wave is 2 Hz... or 0.2 Hz.... it is radiation, and can do the same as light in Einstein's experiment. But formula E=hf does not have any sense.

Please provide some evidence to support his claim.

33 minutes ago, BorisBoris said:

And the opinions of a deluded idiot on the Internet do not count as evidence.

!

Moderator Note

Moved to Speculations. Please note the requirement to provide evidence.

 
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  • 2 weeks later...

Then why we are talking about photons, and not just oscillating electric field? Why light does not have voltage? Interesting debate: https://physics.stackexchange.com/questions/495219/can-air-be-ionized-using-microwaves

Yes, you can achieve air discharge in the atmosphere using high power microwave radiation. The mechanism is avalanche ionization (https://apps.dtic.mil/dtic/tr/fulltext/u2/a172227.pdf)
Just to be clear, the breakdown (ionization) of air, or any insulating material for that matter, is a voltage phenomenon. That voltage can be at any frequency, including microwave frequency. But if it not the electromagnetic energy (photon) of the microwaves that is causing the breakdown, but the magnitude of voltage. – Bob D Aug 4 '19 at 19:58
@BobD : I have not heard about electromagnetic radiation without voltage. If it is not electromagnetic energy of the microwaves that causes the breakdown, what is the source of the energy required for breakdown? (I hope it is obvious that air has higher energy after breakdown.) If your reasoning were correct, it would have been applicable to laser radiation as well, as visible light cannot cause one-photon ionization of air. 
 
Microwaves can cause hydrogen dissociation from my first hand experience so the argument that a single microwave photon does not have enough energy is not convincing. – my2cts Aug 4 '19 at 20:42
 
 
Edited by BorisBoris
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On 1/10/2020 at 2:03 PM, Strange said:

What radiation? And how do you measure the electric component?

I don't think radiation can cause a corona discharge. You need a high voltage.

I have no idea why you are talking about low energy photons. Can you explain how they are relevant to high voltage discharge?

Light (which includes radio waves, in physics) can be modelled as either classical waves or particles. Depending on the application.

I don't know the mechanism but if you put a fork in a microwave oven or other metallic or carbonized object with points (not safe!) there is a discharge between two tines/points.

 

Edited by StringJunky
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10 minutes ago, swansont said:

EM radiation has an electric field. If that’s a high enough amplitude, you can cause field ionization. 

Are changes in the pH level of water (or other liquids) associated with it?

Edited by BorisBoris
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1 hour ago, StringJunky said:

I don't know the mechanism but if you put a fork in a microwave oven or other metallic or carbonized object with points (not safe!) there is a discharge between two tines/points.

I think that is because of the (alternating) voltage induced in the metal by the microwaves.

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  • 3 months later...

https://www.researchgate.net/publication/258595634_Microwave_Triggered_Laser_Ionization_of_Air When irradiated with a high power microwave pulse these electrons would gain enough kinetic energy and further escalate avalanche ionization of air due to elastic electron-neutral collisions thereby causing an increased volumetric discharge region.

Edited by BorisBoris
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On 1/25/2020 at 9:33 AM, BorisBoris said:

'Microwave photon' is produced in nature only  by changing in spin of electron (21 cm, hydrogen). Does it affects spin of electron when it is absorbed?

220px-Hydrogen-SpinFlip.svg.png

https://link.springer.com/article/10.1007%2FBF01039308

 

That’s not the only way to produce microwave photons. But yes, the 21 cm transition (1.4 GHz) in hydrogen is a spin flip, just as the picture says. In Cs it’s at 9.2 GHz (a little more than 3 cm), and in Rb-87 it’s 6.85 GHz.

On 1/25/2020 at 6:49 AM, BorisBoris said:

Then why we are talking about photons, and not just oscillating electric field? Why light does not have voltage? Interesting debate: 

Who is “we”?

The units of electric field can be expressed as V/m. A strong electric field is associated with a large potential difference.

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