General > General Technical Chat
Veritasium "How Electricity Actually Works"
electrodacus:
--- Quote from: aetherist on May 05, 2022, 10:57:41 pm ---
Yes i agree that the wire is a capacitor & has an initial field. But this duznt affect the early signal.
I insist that in theory there will be an early weak fast signal, but i grant that praps at 10 m (instead of the 1 m) i should have called it an early (very very) weak fast signal.
It would be easy to show that u are wrong. Lets place the switch at 5 m from the battery, then the distance switch to bulb on an angle would be say 5.1 m, & the early weak delay would be 5.1 m by 3.4 ns/m which is 17.0 ns. And your late strong delay would be 16 m along the wire at 3.4 ns/m which is 54.4 ns, or, if insulated, add 50% which makes it 81.6 ns.
I have been looking at the arguments re capacitors, & everyone has been wrong in every way.
--- End quote ---
I specifically mentioned that circuit is symmetrical. In this case both the charged and discharged capacitors are at the example same distance from the switch.
The so called small current when circuit is not symmetrical is the current to charge the capacitor made by the transmission line. In that case as it was in Derek's test your lamp is just an indicator to show the current used to charge that capacitor made from the transmission line wires.
As another example you can have the two capacitors circuit that I mentioned and add a lamp in series with the switch just next to the switch and you will see a current trough that lamp as soon as you close the switch but that current is provided by the energy stored in the transmission line so if you open the switch immediately after you closed it the charged in the transmission line will redistribute and the energy lamp used will be provided some nano seconds later by the main energy storage so that charged capacitor in this case.
Thus energy travels trough wires.
As long as your lamp is next to the switch it does not matter how far the battery is the lamp will see a small current immediately but that energy traveled trough wires and from wires (capacitor).
There is no evidence of energy traveling outside the wire just a wrong interpretation of the test result.
aetherist:
--- Quote from: electrodacus on May 05, 2022, 11:14:55 pm ---
--- Quote from: aetherist on May 05, 2022, 10:57:41 pm ---Yes i agree that the wire is a capacitor & has an initial field. But this duznt affect the early signal.
I insist that in theory there will be an early weak fast signal, but i grant that praps at 10 m (instead of the 1 m) i should have called it an early (very very) weak fast signal.
It would be easy to show that u are wrong. Lets place the switch at 5 m from the battery, then the distance switch to bulb on an angle would be say 5.1 m, & the early weak delay would be 5.1 m by 3.4 ns/m which is 17.0 ns. And your late strong delay would be 16 m along the wire at 3.4 ns/m which is 54.4 ns, or, if insulated, add 50% which makes it 81.6 ns.
I have been looking at the arguments re capacitors, & everyone has been wrong in every way.
--- End quote ---
I specifically mentioned that circuit is symmetrical. In this case both the charged and discharged capacitors are at the example same distance from the switch.
The so called small current when circuit is not symmetrical is the current to charge the capacitor made by the transmission line. In that case as it was in Derek's test your lamp is just an indicator to show the current used to charge that capacitor made from the transmission line wires.
As another example you can have the two capacitors circuit that I mentioned and add a lamp in series with the switch just next to the switch and you will see a current trough that lamp as soon as you close the switch but that current is provided by the energy stored in the transmission line so if you open the switch immediately after you closed it the charged in the transmission line will redistribute and the energy lamp used will be provided some nano seconds later by the main energy storage so that charged capacitor in this case.
Thus energy travels trough wires.
As long as your lamp is next to the switch it does not matter how far the battery is the lamp will see a small current immediately but that energy traveled trough wires and from wires (capacitor).
There is no evidence of energy traveling outside the wire just a wrong interpretation of the test result.
--- End quote ---
Capacitance/induction & induction/induction & radio/induction are all inductions.
Veritasium's 3.3 ns is due to induction (ie through the air), no matter whether u say that it is due to capacitance/induction or some other kind of induction.
However, Veritasium is of course wrong re his silly Poynting Vector having or transferring energy.
U & i agree that the main source of electric energy comes later via the wire (not via the silly Poynting field). But i don’t see how capacitance plays a part in that.
In any case the descriptions that i have seen re what happens with 2 capacitors in a circuit is all wrong.
electrodacus:
--- Quote from: aetherist on May 05, 2022, 11:49:46 pm ---Capacitance/induction & induction/induction & radio/induction are all inductions.
Veritasium's 3.3 ns is due to induction (ie through the air), no matter whether u say that it is due to capacitance/induction or some other kind of induction.
However, Veritasium is of course wrong re his silly Poynting Vector having or transferring energy.
U & i agree that the main source of electric energy comes later via the wire (not via the silly Poynting field). But i don’t see how capacitance plays a part in that.
In any case the descriptions that i have seen re what happens with 2 capacitors in a circuit is all wrong.
--- End quote ---
All energy transfer from source to load/lamp is done done trough wires and none of it radiated unless you want to take in consideration that some IR radiated photon from the wires heated the lamp filament.
aetherist:
--- Quote from: electrodacus on May 06, 2022, 12:02:52 am ---
--- Quote from: aetherist on May 05, 2022, 11:49:46 pm ---Capacitance/induction & induction/induction & radio/induction are all inductions.
Veritasium's 3.3 ns is due to induction (ie through the air), no matter whether u say that it is due to capacitance/induction or some other kind of induction.
However, Veritasium is of course wrong re his silly Poynting Vector having or transferring energy.
U & i agree that the main source of electric energy comes later via the wire (not via the silly Poynting field). But i don’t see how capacitance plays a part in that.
In any case the descriptions that i have seen re what happens with 2 capacitors in a circuit is all wrong.
--- End quote ---
All energy transfer from source to load/lamp is done done trough wires and none of it radiated unless you want to take in consideration that some IR radiated photon from the wires heated the lamp filament.
--- End quote ---
The energy is in the wire(s). Then some of the energy transfers from wire to wire throo the air via the em field, the delay being 3.3 ns, as proven by every X that i have ever seen.
We agree that the full electricity arrives via the wires (ie we agree not via the silly Poynting field).
Naej:
--- Quote from: Sredni on May 04, 2022, 09:39:22 pm ---
--- Quote from: Naej on May 02, 2022, 10:43:10 pm ---The relativistic answer (convention) is that a momentum density p of matter corresponds to an energy flux of pc².
If you want general relativity then the answer is, well, complicated. https://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html
Classical physics is simple and says you have a potential energy of -GMm/r, you're not moving 0 energy, you're just converting: potential->kinetic->heat.
--- End quote ---
In the quasistatic idealization I had in mind, the mass takes an infinite time to travel from point A to point B. So, velocity is basically zero along the path and only changes during 'generation' (the machine raises the weight) and 'dissipation' (the weight does work against the gravitational field and it all becomes heat).
In my view the energy is in the gravitational field of the system planet + weight. It is being added to the system during 'generation' and it is extracted from the system during 'dissipation'. The system occupies all space, so does it even make sense to ask if the energy is traveling?
--- End quote ---
It does not occupy all space, because potential energy is the sum of -Gm1m2/r.[
quote author=bsfeechannel link=topic=322795.msg4156429#msg4156429 date=1651721925]
I find it funny and at the same time annoying that these people who are reluctant to properly study math and physics, and end up struggling especially with electromagnetism, like to talk in the name of engineers, as if their misconceptions were the general mindset of our class so as to justify their position.
The hydraulic analogy, the origin of the energy in the wire idea, was dismissed right from the start by Maxwell himself in the 19th century. Derek only made this incontestable knowledge available to the masses.
[/quote]
Isn't it interesting how every argument for 'energy in vacuum' is an argument from authority? :o
Also, Maxwell wrote:
https://en.wikisource.org/wiki/Page:A_Treatise_on_Electricity_and_Magnetism_-_Volume_1.djvu/102
" We are thus led to a very remarkable consequence of the theory which we are examining, namely, that the motions of electricity are like those of an incompressible fluid, so that the total quantity within an imaginary fixed closed surface remains always the same."
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