Author Topic: Electroboom: How Right IS Veritasium?! Don't Electrons Push Each Other??  (Read 40958 times)

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Offline electrodacus

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Pick up your model and turn the wheels.  Make a quick video and show us the wheels won't turn.

Picking up the model means the chassis of the vehicle is no longer floating so I can apply a force between wheel and chassis.  That is not at all the same thing with applying force at the wheels (input and output) with chassis floating.
You pick up the vehicle with one hand on one wheel and the other on the other wheel (do not touch the vehicle body) just the low part of the wheels.
If one wheel is one palm and the other on the other hand just try and bring the hands together. If you do not allow any slip then you will feel that it is a locked system.

Offline gnuarm

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Pick up your model and turn the wheels.  Make a quick video and show us the wheels won't turn.

Picking up the model means the chassis of the vehicle is no longer floating so I can apply a force between wheel and chassis.  That is not at all the same thing with applying force at the wheels (input and output) with chassis floating.

Now you are just babbling.  What do you mean when you say the wheels are "gear locked"??? 


Quote
You pick up the vehicle with one hand on one wheel and the other on the other wheel (do not touch the vehicle body) just the low part of the wheels.
If one wheel is one palm and the other on the other hand just try and bring the hands together. If you do not allow any slip then you will feel that it is a locked system.

That is an absurd definition with no purpose.  However, with a gear ratio of other than 1:1, you can move the wheels by applying a force between them.  When it is 1:1, by definition, they must move together.  Not true for any other ratio.

This is BS.  You are never going to understand any of this.  Or, much more likely, you are just an Andrea Rossi troll. 

I'm done with you.  I have other things to do in life.  All the rest of us understand how the models work.  You don't.  It's that simple.  Bye.
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Offline electrodacus

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What is the voltage at the midpoint? Directly across from where it is fed? Unless it is 0V...

How will you measure the voltage without touching the wire ?
Also you measure the voltage between two points and you did not mentioned voltage relative to what.
If the multimeter is connected to the battery negative and the 1Ohm side is connected there you will measure 2V across that but if the 9Ohm is connected to battery negative you will measure 18V across it.
Your voltmeter is just another wire in parallel with one of them (one of the two halves). Yet since multimeter is maybe around 1MOhm way less power will be dissipated on the voltmeter.
It shows that even to make the measurement you need a wire with high resistance not to influence the conditions to much as current will be slightly larger when you connect the voltmeter.

Offline electrodacus

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I'm done with you.  I have other things to do in life.  All the rest of us understand how the models work.  You don't.  It's that simple.  Bye.

Have a good life.

Offline hamster_nz

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Electric field will also be symmetrical as is based on imbalance of charge and outside surface is the same for both halfs.
What is the voltage at the midpoint? Directly across from where it is fed? Unless it is 0V it will not be symmetrical.
How will you measure the voltage without touching the wire ?
You have a bigger problem - you stated that "Electric field will also be symmetrical" and then go on to measure the potential at 2V and 18V, showing that it isn't symmetrical at all, and in doing so explaining why different sides are dissipating different powers.
Gaze not into the abyss, lest you become recognized as an abyss domain expert, and they expect you keep gazing into the damn thing.
 

Offline electrodacus

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You have a bigger problem - you stated that "Electric field will also be symmetrical" and then go on to measure the potential at 2V and 18V, showing that it isn't symmetrical at all, and in doing so explaining why different sides are dissipating different powers.

You and Derek claim "energy doesn't flow in wires". There is zero evidence for that.
Both sides have the same metal surface area on the outside so what makes the higher resistance side receive 90% of the energy ?

 

Offline hamster_nz

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You have a bigger problem - you stated that "Electric field will also be symmetrical" and then go on to measure the potential at 2V and 18V, showing that it isn't symmetrical at all, and in doing so explaining why different sides are dissipating different powers.

You and Derek claim "energy doesn't flow in wires". There is zero evidence for that.
Both sides have the same metal surface area on the outside so what makes the higher resistance side receive 90% of the energy ?

 
Zero evidence? (glances as a wireless charger) yeah, if you say so.

And of course if you choose to use a stupid model so you can get 'counter intuitive' results... for example, where in the hardware store can I find this copper pipe you talk about which has 9 ohm resistance over 21 meters  :-//? Must have a cross-sectional area of a fraction of a mm.

One side has an 18V field over its 21m length, the other has a 2V field over the same length. Of course the charges in the stronger field will be doing more work (assuming both are made out of the same material, and the difference in resistance is due to cross-sectional area).

If you didn't want charges to do work in your wires, go get better wires, where the field strength in the wire will be lower for the same current, so less energy is doing work pushing changes around in those wire.

« Last Edit: July 22, 2022, 08:57:24 am by hamster_nz »
Gaze not into the abyss, lest you become recognized as an abyss domain expert, and they expect you keep gazing into the damn thing.
 
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Offline Naej

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Let's assume you do that by using say Coulomb gauge, which says that the divergence of the magnetic vector potential is always everywhere zero, ∇·A = 0.  This is not an arbitrary choice, as it leads to specific properties, which are not a result of Maxwell's laws, but the result of this specific gauge applied to the potential approach to Maxwell's model.
It's the gauge chosen by Maxwell if my memory is correct.
Finally, the term electric potential energy describes the potential energy associated with the electric field related to the charges present, those related to the Coulomb forces I mentioned; and those only.  It does not mean that all potential energy in a system that includes moving charges is in the electric field!

Can you see what I am trying to describe?  That I am not saying that anyone is wrong per se, I am saying that they are using such approximations that allow their viewpoint to be true, but that also change the system being considered so that the arguers are not actually talking about the same system anymore!
What's the approximation here? That it is not quantised?
 

Offline Nominal Animal

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What's the approximation here?
Those arguing different places for the majority of energy flow are making different approximations.

One uses the steady state model, ignoring everything that happens when the circuit is formed, and then argues with someone who only looks at the initial nonequilibrium state or an AC system saying they are "missing the entire point".

It is useless to try and unravel the argument, until they come to a mutual agreement of an exact experimental setup.

They won't, because each one wants to keep their own approximations that let them define the originally vague and fuzzy system in terms that make their chosen answer stick.  It's very, very silly.
 

Offline electrodacus

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You have a bigger problem - you stated that "Electric field will also be symmetrical" and then go on to measure the potential at 2V and 18V, showing that it isn't symmetrical at all, and in doing so explaining why different sides are dissipating different powers.

You and Derek claim "energy doesn't flow in wires". There is zero evidence for that.
Both sides have the same metal surface area on the outside so what makes the higher resistance side receive 90% of the energy ?

 
Zero evidence? (glances as a wireless charger) yeah, if you say so.

And of course if you choose to use a stupid model so you can get 'counter intuitive' results... for example, where in the hardware store can I find this copper pipe you talk about which has 9 ohm resistance over 21 meters  :-//? Must have a cross-sectional area of a fraction of a mm.

One side has an 18V field over its 21m length, the other has a 2V field over the same length. Of course the charges in the stronger field will be doing more work (assuming both are made out of the same material, and the difference in resistance is due to cross-sectional area).

If you didn't want charges to do work in your wires, go get better wires, where the field strength in the wire will be lower for the same current, so less energy is doing work pushing changes around in those wire.

Wireless chargers that you refer to are based on magnetic coupling and that is not DC.  In this circuit with 20V battery you have a constant magnetic field same as from a permanent magnet.
Just have you wireless charging enabled phone on a permanent magnet and see if you get any charge.
Both the magnetic and electric fields are constant over those 5 seconds in this circuit supplied by a 20Vdc battery.

The resistance values are for ease of calculation and yes it is possible to have 9Ohm for a 21m long pipe (I have no specified the diameter or thickness of the pipe wall) I also did not specified the metal that pipe is made out off.

The test is specifically setup to show that current travels through wires (the entire section) not coming from outside and traveling only at the surface.
There is nothing else exiting the battery other than charged particles (electrons) and those travel through wires.
The magnetic field you measure around the wire is due to the motion of those electrons and the electric field is due to charge imbalance.
So you get the electron first, then due to electron you get the electric and magnetic field not the other way around.

Offline bsfeechannel

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One uses the steady state model, ignoring everything that happens when the circuit is formed, and then argues with someone who only looks at the initial nonequilibrium state or an AC system saying they are "missing the entire point".

DC is an engineer that came late for the transient.
 
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Offline SiliconWizard

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What's the approximation here?
Those arguing different places for the majority of energy flow are making different approximations.

One uses the steady state model, ignoring everything that happens when the circuit is formed, and then argues with someone who only looks at the initial nonequilibrium state or an AC system saying they are "missing the entire point".

It is useless to try and unravel the argument, until they come to a mutual agreement of an exact experimental setup.

They won't, because each one wants to keep their own approximations that let them define the originally vague and fuzzy system in terms that make their chosen answer stick.  It's very, very silly.

Yep. This was pointed out ages ago already and it just keeps going on.
 
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Offline cbutlera

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What's the approximation here?
Those arguing different places for the majority of energy flow are making different approximations.

One uses the steady state model, ignoring everything that happens when the circuit is formed, and then argues with someone who only looks at the initial nonequilibrium state or an AC system saying they are "missing the entire point".

It is useless to try and unravel the argument, until they come to a mutual agreement of an exact experimental setup.

They won't, because each one wants to keep their own approximations that let them define the originally vague and fuzzy system in terms that make their chosen answer stick.  It's very, very silly.

Yep. This was pointed out ages ago already and it just keeps going on.

This brings to mind the interminable arguments that I witnessed many years ago on rec.bicycles.tech, about whether a bicycle stood on its lower spokes or hung from its upper spokes.  Back then I almost invariably sided with the views expressed by the great Jobst Brandt - I wouldn’t have dared to do otherwise.  But there are interesting parallels with this question.

When the wheels are turning, the path taken by the variation in the stress seems reasonably clear - the AC case.  The spokes can be treated as if they are able to resist a compressive load up to a point, because the wheel is a prestressed structure.  So up to a point, the spoked wheel behaves in much the same way as a disc wheel.  But when the wheels are not turning, things are not so clear - the DC case.

It becomes even less clear if the stationary wheel is a disc wheel that may or may not have an internal pre-stress, and we have no way of telling.  Once again it becomes essential to be very precise about the experiment, and exactly what question is being asked.

If the bicycle is inside a box being accelerated through space in a direction normal to the floor and is stationary with respect to that box, then to an observer in an inertial reference frame, kinetic energy appears to be flowing through the wheel.  But what is the route that that flow takes through the wheel?  Some would argue that it is passing up through the lower spokes, others that is is passing around the rim then down through the upper spokes.  Another group may argue that the apparent energy flow is through the entire structure, and it is meaningless to be any more precise.  A lone voice may then pipe up and say that since all of the relevant forces are in fact electrical, then every charge carrying particle in the box and bicycle is involved, along with many more outside.  Who is right – everyone – no one?

A brick suspended on a string is hanging in a room.  The string is attached to the ceiling with a suction cup that may or may not have adhesive on its suction face.  What is the route taken by the force holding the brick in the air?
« Last Edit: July 23, 2022, 07:49:13 am by cbutlera »
 
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Offline electrodacus

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It becomes even less clear if the stationary wheel is a disk wheel that may or may not have an internal pre-stress, and we have no way of telling.  Once again it becomes essential to be very precise about the experiment, and exactly what question is being asked.

The question I think is fairly clear.

How is the electrical energy delivered from the source (say a battery or charged capacitor) to the lamp?
To be even more clear a lamp is the same with a resistor or wire.

Derek take is that electrical energy travels from the battery trough that 1m air gap with his only proof being observing some small amount of current through the lamp in about the time it takes light to travel 1m.
He completely ignores the energy storage device formed by the transmission line.
Also is not about the distance from battery to lamp but about the distance from the switch.

Then there is absolutely no discussion about the steady state DC.  At DC you do not even need to understand or acknowledge energy storage and you can not explain energy not traveling through wires.

In both cases DC and AC electric current travels through wires which means electrical power is dissipated in the wires and electrical power integrated over time is electrical energy.
Both the electric field (as found between the capacitor plates) and magnetic field around a stream of moving charged particles (what electric current is) are conservative fields meaning they do not radiate into space and you can not collect energy from that.
So is the same as magnetic field around a permanent magnet is always there and there is no way for you to extract any energy from that.
You can move the magnet or a wire around a magnet but that is converting mechanical energy into electrical energy not getting energy from the magnet.

The lumped element model perfectly describes what happens as seen by my spice simulation that got the same results Derek got in his real world experiment.


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