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| "Veritasium" (YT) - "The Big Misconception About Electricity" ? |
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| SandyCox:
I looked at the "Electronics World" paper by Ivor Catt and applied old-school transmission-line theory. The setup An 18 m long 75 Ohm transmission line. The reed switch is located at z=0m and the open end of the transmission line is at z=-18m. The transmission line is pre-charged to 8V. The reed switch is open-ended at t=0s. I chose a propagation speed of c=2.55e8 m/s. What transmission line theory tells us * The voltage wave is the sum of a wave travelling in the positive z direction (indicated by v+) and a wave travelling in the negative z direction (indicated by v-). * The reflection coefficient at the open end is equal to 1. The reflection coefficient at the reed switch is 0. * Prior to t=0s, the voltage across the entire length (z=-18m to z=0m) of the transmission line is equal to 8V. There is no current in the transmission line. We can use this information to calculate v+ and v- for t<0s. It turns out that both are constant and equal to 4V. * Closing the reed switch at t=0s initiates the transient. At z=0, v+ enters the 75 Ohm resistor, where it is terminated. v- is reflected from the open end of the transmission line and becomes part of v+ I attach figures for comparison with the results in Catt's paper. You do the comparison. A video showing the waves is available at The error in Catt's paper is that he views the transmission line as a capacitor, which it most certainly is not. A more detailed explain and the mathematical derivations will follow. There is nothing strange about this setup. I remember doing experiments like this when I was an undergraduate student and using transmission line theory to explain the measurements. Note that Catt's paper was published in the April 2013 issue. |
| SandyCox:
--- Quote from: aetherist on February 17, 2022, 09:22:09 pm --- --- Quote from: SandyCox on February 17, 2022, 12:23:22 pm ---Will you please explain how "new electricity" leads to the conclusion that a capacitor discharging into a transmission line leads to a step-like voltage? --- End quote --- The main thing here is that old (electron) electricity can't explain the discharge, ie the half voltage for double the time. The second thing here is that it might be possible to come up with lots of theories about what electricity is or isn’t, & for most of these to satisfactorily explain the discharge (ie to tick that box). I could come up with other theories that tick that box, but would they tick all of the boxes. My new (electon) electricity i think ticks all of the boxes (so far). But old (electron) electricity duznt tick the capacitor discharge box, as far as i can tell. But u or someone else might indeed be able to explain a way that drifting electrons tick the box. If so then i would not be able to use that box to falsify old (electron) electricity. But that would then leave me with all of the other boxes that falsify old (electron) electricity. And i only need one box. One strike & old (electron) electricity is out. Now to answer your question. New (electon) electricity says that in a charged capacitor the negative plate is full of (covered by)(saturated with) electons propagating in every direction, as is a short wire connected to the negative plate. On the short wire half of the electons are going one way & half are going the other way. When electons get to the end of the short wire they do a u-turn (in reality they go straight ahead as usual)(it is the surface of the wire that does a u-turn). When the short wire is connected to a new non-charged long wire the electons on the short wire that are already heading towards the long wire will instead of doing a u-turn will enter onto the long wire. Some of the electons going the "wrong way" on the short wire will have to go all the way to the end of that circuit, ie to the furthest end of the negative plate, & do a u-turn, & head back & enter the long wire. So, the time taken for the last electon to leave the capacitor & enter the long wire is double the average time. The average time is the time taken for an electon to travel from the farthest point on the negative plate to the nearest point on the long wire. That is the simple version of the electon discharge from the negative plate of a capacitor. The positive plate is different. I think that it has no electons. It has an induced positive charge, ie due to the repulsion of electrons due to the negative charge on the negative plate (ie due to the negatively charged electons on the negative plate). I need to have a think about how the positive plate might discharge, & how that would affect current in the long wire. The discharge involves the flow of electrons on the surface, & this will be very slow, ie much slower than the speed of light. This will produce i think a long slow weak discharge, in addition to the almost instantaneous electon discharge mentioned above. I need to have a think. I might be overplaying the importance of the short wire. If the capacitor holds say 100 times the number of electons on the short wire then the length of the short wire might not make much difference to the time of the (main) discharge, ie the geometry of the capacitor itself would be paramount. Here is a link to what Harry Ricker says. Harry has written lots of good articles. https://beyondmainstream.org/the-wakefield-experiments-background-and-motivation/ --- End quote --- Can new electricity answer the following question? What is the equation for the voltage across the capacitor, as a function of time, as it discharges through the resistor? |
| SiliconWizard:
In the end, the real underlying misconception is that electricity, as we usually define it, is not energy. :popcorn: |
| adx:
--- Quote from: SiliconWizard on February 18, 2022, 05:36:30 pm ---In the end, the real underlying misconception is that electricity, as we usually define it, is not energy. :popcorn: --- End quote --- Yes, electrons (with the r) become a mental crutch for something weirder. |
| adx:
--- Quote from: TimFox on February 17, 2022, 08:32:36 pm ---Mapping (imaging) electron density has been done for a long time using x-ray diffraction (crystallography) and different methods of electron microscopy. Strictly, this is a map of the probability spatial distribution of electrons in, say, a large molecule. --- End quote --- "Yes but" x-ray crystallography is more of a reconstruction, from a periodic structure and guesses of phase. AFM is more direct but also harder to argue it's "a picture" because it is done by feel even if it looks much the same. An ordinary camera photo is a map of the probability distribution of photons over a known integration time, complete with shot noise. The lens resolves and combines the phase onto the spatial array. It's hardly more intuitive, but is the sense of sight, so comes with instinctive understanding (clouds, motion blur etc). As far as I can tell, electron cloud densitometry produces an image directly and really is worthy of the usual meaning of "photo". The language is what struck me as useful - being able to point to a picture and say "this is the electron cloud". |
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