### Author Topic: Electroboom: How Right IS Veritasium?! Don't Electrons Push Each Other??  (Read 65666 times)

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#### electrodacus

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##### Re: Electroboom: How Right IS Veritasium?! Don't Electrons Push Each Other??
« Reply #700 on: July 20, 2022, 11:17:58 pm »
Newton 3'rd law applies to F1 and F2 the same way it applies to F3 and F4
Sure, but what does it say?

What do you mean but what does it say?
If the red box was perfectly slippery then F1 will be zero (that is after mass of the vehicle was accelerated from right to left).
Even will far less than perfect slip on the left wheel (just needs to slip before the right wheel has the chance to do that) the vehicle will just slide from right to left as I demonstrated here  https://odysee.com/@dacustemp:8/stick-slip-removed-from-front-wheels:0
« Last Edit: July 20, 2022, 11:20:49 pm by electrodacus »

#### gnuarm

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##### Re: Electroboom: How Right IS Veritasium?! Don't Electrons Push Each Other??
« Reply #701 on: July 20, 2022, 11:40:30 pm »
I've seen it many times.  I can draw it with my eyes closed.

I'm curious what you mean by that.
Can you draw that in your mind or on a piece of paper ?

Either

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If you can visualize that in your mind are you able to move one of the wheels and see how it affects the rest of the system ?

Yes, I don't typically draw things on paper because that is very, very limiting.  I see them in my mind and visualize how the operate.  In this case, I needed to attach some numbers to the pulley and wheel sizes to see which ratio was larger.  As it turns out, the wheel at F2 will turn more slowly, but with more force than the wheel at F1.   F1 and F2 are related by the leverage of the system, not Newton's laws...

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Maybe if you can imagine then imagine the wheels having spikes in the treadmill for the wheel on the right and in to the red box for the one on the left straight down.
This way wheels can not slip. Do you think anything can move now ?

Of course it can.  The wheels will turn according to the ratio of the various components and the car will move correspondingly.  In fact, that is the only way to properly understand what is going on.  If you move the treadmill to the left, the wheel at F1 will turn clockwise making the wheel at F2 turn clockwise, moving the car to the right.

The why I picture this more easily, is to imagine the treadmill is free wheeling and moving the car to the right.  Both wheels turn clockwise and the treadmill moves to the left because the right hand wheel turns faster than the left hand wheel.

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The treadmill will just stall unless is powerful enough to break those spikes.
Newton 3'rd law applies to F1 and F2 the same way it applies to F3 and F4

I'm sorry you don't understand simple mechanics.  Newtons law does not relate to forces at different points.  This is why you are so confused.  If F1 and F2 were always the same magnitude, but opposite directions, it would never move at all.  When you push on something, it moves because the opposing force is from inertia, which is another way of saying "acceleration".  That is an example of Newton's 3rd law.  Pushing on an fixed object is another example, where you push on the object and it pushes right back.  In all cases, the forces are on the same point, in opposite directions.

You think F1 and F2 are related, being the same magnitude and opposite direction because of some mysterious natural phenomenon transmitting the forces to one another.  This is not Newton's law.  It's also not true.  Newton's law says the treadmill pushes on the wheel and the wheel pushes back on the treadmill.  F1 and F2 are coupled through the mechanical leverage of the pulleys, etc.
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#### Naej

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##### Re: Electroboom: How Right IS Veritasium?! Don't Electrons Push Each Other??
« Reply #702 on: July 21, 2022, 12:03:42 am »
Newton 3'rd law applies to F1 and F2 the same way it applies to F3 and F4
Sure, but what does it say?

What do you mean but what does it say?
If the red box was perfectly slippery then F1 will be zero (that is after mass of the vehicle was accelerated from right to left).
Even will far less than perfect slip on the left wheel (just needs to slip before the right wheel has the chance to do that) the vehicle will just slide from right to left as I demonstrated here  https://odysee.com/@dacustemp:8/stick-slip-removed-from-front-wheels:0
Oh so what you mean by "Newton's 3rd law" is Coulomb friction. I see.

#### electrodacus

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##### Re: Electroboom: How Right IS Veritasium?! Don't Electrons Push Each Other??
« Reply #703 on: July 21, 2022, 01:03:11 am »

Yes, I don't typically draw things on paper because that is very, very limiting.  I see them in my mind and visualize how the operate.  In this case, I needed to attach some numbers to the pulley and wheel sizes to see which ratio was larger.  As it turns out, the wheel at F2 will turn more slowly, but with more force than the wheel at F1.   F1 and F2 are related by the leverage of the system, not Newton's laws...

Of course it can.  The wheels will turn according to the ratio of the various components and the car will move correspondingly.  In fact, that is the only way to properly understand what is going on.  If you move the treadmill to the left, the wheel at F1 will turn clockwise making the wheel at F2 turn clockwise, moving the car to the right.

The why I picture this more easily, is to imagine the treadmill is free wheeling and moving the car to the right.  Both wheels turn clockwise and the treadmill moves to the left because the right hand wheel turns faster than the left hand wheel.

I'm sorry you don't understand simple mechanics.  Newtons law does not relate to forces at different points.  This is why you are so confused.  If F1 and F2 were always the same magnitude, but opposite directions, it would never move at all.  When you push on something, it moves because the opposing force is from inertia, which is another way of saying "acceleration".  That is an example of Newton's 3rd law.  Pushing on an fixed object is another example, where you push on the object and it pushes right back.  In all cases, the forces are on the same point, in opposite directions.

You think F1 and F2 are related, being the same magnitude and opposite direction because of some mysterious natural phenomenon transmitting the forces to one another.  This is not Newton's law.  It's also not true.  Newton's law says the treadmill pushes on the wheel and the wheel pushes back on the treadmill.  F1 and F2 are coupled through the mechanical leverage of the pulleys, etc.

You are used with vehicle that have an obvious power source usually a motor or engine connected between the vehicle body and wheels.
This is about a special case and I will get why you are confused.
The body of this vehicle is isolated not in contact with the ground or the thing that powers the vehicle.
This vehicle has only two points of contact and no onboard energy source.
When treadmill applies a force F1 on the vehicle the only the only way the vehicle can push back is through the other point of contact with F2 that will be opposite and equal.

This is not a powered vehicle is a pushed vehicle from outside by the treadmill and normally it will move in the direction is being pushed but due to the way the wheels are connected by the belt it is locked so if wheel on treadmill was to slip with no energy storage the vehicle will just stay in place while treadmill will waste energy due to friction with the wheel but since there is also the internal energy storage (belt elasticity in this case) the vehicle can store the energy and move forward a little bit using the small amount if stored energy then things will repeat.

What do you think happens if you weld the wheel to the vehicle body (so they can no longer rotate)? Is in that case Newton's 3'rd law applicable and F1=F2 ?
Because the way that belt is connected between the wheels is about the same thing in this particular case.

If there is zero friction the wheel on the treadmill will just spin at the same speed as treadmill while the back wheel is stationary same as the vehicle body.
If you add a brake at the wheel on the right or an electric generator then vehicle will move from right to left so you will be able to produce only a limited amount of energy ideal case equivalent with vehicle potential energy based on treadmill speed and vehicle mass
Once the vehicle speed equal treadmill speed you can no longer produce any energy.
If you take that energy and apply to the wheel on the red box that will just be enough ideal case to bring the vehicle back to a stop.
This is a complicated way of saying that Pout can not be higher than Pin.

#### gnuarm

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##### Re: Electroboom: How Right IS Veritasium?! Don't Electrons Push Each Other??
« Reply #704 on: July 21, 2022, 06:20:10 am »
Newton 3'rd law applies to F1 and F2 the same way it applies to F3 and F4
Sure, but what does it say?

What do you mean but what does it say?
If the red box was perfectly slippery then F1 will be zero (that is after mass of the vehicle was accelerated from right to left).
Even will far less than perfect slip on the left wheel (just needs to slip before the right wheel has the chance to do that) the vehicle will just slide from right to left as I demonstrated here  https://odysee.com/@dacustemp:8/stick-slip-removed-from-front-wheels:0

This video shows the car working exactly as I described it.  In the beginning, you are doing something that makes the car move with the paper without the wheels turning.  But once you reposition the car, it moves opposite to the motion of the paper (the treadmill) which is exactly what I said it would do.

How can you be in denial of your own evidence???  The fact that your wheels are not very sticky and slip a lot, doesn't mean it's not working.  The slipping is what screws it up.  When the wheels don't slip, the car moves opposite the paper.  The wheels are turning and not slipping, proving your idea of the wheels being "gearbox locked" is pure bunkum.
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#### electrodacus

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##### Re: Electroboom: How Right IS Veritasium?! Don't Electrons Push Each Other??
« Reply #705 on: July 21, 2022, 06:51:03 am »
This video shows the car working exactly as I described it.  In the beginning, you are doing something that makes the car move with the paper without the wheels turning.  But once you reposition the car, it moves opposite to the motion of the paper (the treadmill) which is exactly what I said it would do.

How can you be in denial of your own evidence???  The fact that your wheels are not very sticky and slip a lot, doesn't mean it's not working.  The slipping is what screws it up.  When the wheels don't slip, the car moves opposite the paper.  The wheels are turning and not slipping, proving your idea of the wheels being "gearbox locked" is pure bunkum.

In the beginning the back wheels are on top of some solder wire 0.8mm diameter so that they have less contact surface (still fairly good as it is rubber wheels on metal) but worse than front wheels that are rubber on paper.
The video is specifically designed to show that is you remove the ability of front wheels to slip (the ones on the moving paper) the vehicle is just locked and it is dragred in the direction the paper moves as it is a locked gearbox by design.
The second half of the video shows what happens when front wheels (those on moving paper) are allowed to slip and that was also shown in slow motion video where you can better see how it works with belt being stretched (energy stored) then when wheels slip stored energy is discharged allowing a bit of motion against the direction of the treadmill then cycle repeats.
Any vehicle designed this way even with a chain will work the same way. You will just need a slow motion video to see the details.

You just do not get that any force F1 applied will result in F2 being equal and opposite direction. That is because you are not used with a vehicle being pushed from outside in that way. The vehicle will want to move in the direction is being pushed except the way the belt is connected it can't as it is locked so it will either slip in the direction is pushed (first part of the video) or energy is stored and then generator wheel slips allowing the motor wheel using the stored energy to move a bit forward then wheel locks again and cycle repeats (normally this is faster than you can see in a normal speed video).

#### gnuarm

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##### Re: Electroboom: How Right IS Veritasium?! Don't Electrons Push Each Other??
« Reply #706 on: July 21, 2022, 06:57:29 am »
This video shows the car working exactly as I described it.  In the beginning, you are doing something that makes the car move with the paper without the wheels turning.  But once you reposition the car, it moves opposite to the motion of the paper (the treadmill) which is exactly what I said it would do.

How can you be in denial of your own evidence???  The fact that your wheels are not very sticky and slip a lot, doesn't mean it's not working.  The slipping is what screws it up.  When the wheels don't slip, the car moves opposite the paper.  The wheels are turning and not slipping, proving your idea of the wheels being "gearbox locked" is pure bunkum.

In the beginning the back wheels are on top of some solder wire 0.8mm diameter so that they have less contact surface (still fairly good as it is rubber wheels on metal) but worse than front wheels that are rubber on paper.
The video is specifically designed to show that is you remove the ability of front wheels to slip (the ones on the moving paper) the vehicle is just locked and it is dragred in the direction the paper moves as it is a locked gearbox by design.
The second half of the video shows what happens when front wheels (those on moving paper) are allowed to slip and that was also shown in slow motion video where you can better see how it works with belt being stretched (energy stored) then when wheels slip stored energy is discharged allowing a bit of motion against the direction of the treadmill then cycle repeats.
Any vehicle designed this way even with a chain will work the same way. You will just need a slow motion video to see the details.

You just do not get that any force F1 applied will result in F2 being equal and opposite direction.

It's not a matter of not "getting it".  I am saying you are dead wrong about that.  There is zero reason to think that is correct.

At the 20 second mark, the car gets some traction with the paper and the other surface, and does exactly what I predicted, moving opposite the direction of the paper (treadmill).

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That is because you are not used with a vehicle being pushed from outside in that way. The vehicle will want to move in the direction is being pushed except the way the belt is connected it can't as it is locked so it will either slip in the direction is pushed (first part of the video) or energy is stored and then generator wheel slips allowing the motor wheel using the stored energy to move a bit forward then wheel locks again and cycle repeats (normally this is faster than you can see in a normal speed video).

You have no clue.  You make up all manner of silly stuff.

Make a video where you pick up the car and turn the wheels by hand.  That will show they are not "locked".  Put marks on the wheels, so we can count the turns.  You can turn one wheel some number of turns, and we can count the number of turns on the other wheel.  THAT is a constant.  Unless you use crappy surfaces where the wheels slip, that ratio determines where the car goes.

You have no clue about any of this.  Literally no clue.  I'm pretty sure you do not have a technical job.  It's not possible.
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#### Nominal Animal

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##### Re: Electroboom: How Right IS Veritasium?! Don't Electrons Push Each Other??
« Reply #707 on: July 21, 2022, 08:10:48 am »
Quote
when the circuit is first closed, regardless of the current being AC or DC.

For a very brief moment, it's AC even when it's DC, isn't it?
Yes, exactly.  The faster the rise, the higher frequency components you have in it.

Just like when doing logic circuits, you get all sorts of AC effects from the pulses, which need to be dealt with in real life circuits.

"Will move".  No, something makes them move.  That something is an electric field, which propagates through the circuit somewhat analogously to a shock wave when the circuit is first connected.  Also, some of the original "potential energy" is in the form of an electric field around the charged plate; it is not exactly correct to just lump it all into "potential energy" and call it good enough.
Or maybe it is correct.
Not exactly correct means it is an approximation.

When you are arguing about exactly where the majority of the energy flow in a system is, making such approximations is exactly how you "accidentally" manipulate it to fit your pre-selected model.

Consider this question:  Are the participants arguing where the energy flows, or what the setup being investigated exactly is?

To me, the disagreement stems from the latter, and is the reason why I'm not interested in their arguments and opinions, be they professors or internet celebrities or whatever else.  I am too familiar with both "garbage data + good model = garbage results" and "good data + garbage model = gargabe results" already; I want to drop the garbage parts and have people spend their time on the "good data + good model = good results" case instead.
In this case, each participant has their own model.  That's no good: OF COURSE their inner workings vary then, even if the results are exactly the same.

#### m k

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##### Re: Electroboom: How Right IS Veritasium?! Don't Electrons Push Each Other??
« Reply #708 on: July 21, 2022, 12:50:12 pm »
You are confusing some things.
No, you're leaving important bits out, oversimplifying the situation to fit your axiomatic model.

As soon as you close the switch the excess electrons from one plate will move in to wire which is neutrally charged and at the same time on the other plate electrons from the wire will migrate in to the plate with deficit of electrons.
"Will move".  No, something makes them move.  That something is an electric field, which propagates through the circuit somewhat analogously to a shock wave when the circuit is first connected.  Also, some of the original "potential energy" is in the form of an electric field around the charged plate; it is not exactly correct to just lump it all into "potential energy" and call it good enough.

When the circuit has stabilized, the electric field (potential difference along the circuit) has subsided to something small and stable, aside from thermal noise and such.  This does not mean it was small and stable and insignificant at the violent beginning.

While electron wave travels through wire
What electron wave? You need to specify that too, and not just give it a name and leave it at that.  Giving a thing a name is not the same as describing the thing.

Each individual electron is both a particle and a wave.  When an electron is bound to an atom, it is delocalized in the shape described by spherical harmonics.  When an electron is shared by a lattice (as they are in metal conductors), they are delocalized in various ways, and typically spread over or "shared" across multiple lattice atoms.

If we describe electron locations by the center or centroid of their delocalized volume, they really do move very slowly, something like a meter a second or so, often even slower, while the current and changes in the current propagate at over half the speed of light, or over hundred million times faster.

The electrons do not just push each other to move (as described in electrostatic approximation as the Coulomb force); they also interact via emitting and absorbing photons, and coupling to existing electromagnetic fields like the one caused by being matter not cooled to absolute zero.  Note that this EM field is NOT just "radiating outwards"; there is always both emission and absorption.

So, what you call "electron wave" is in reality a set of various possible interactions.  The majority (i.e., which kind of interaction is the most common or involves the most energy flow) depends on the exact configuration of the system; its geometry.

It is somewhat funny that the most complex phenomena occur when the circuit is first closed, regardless of the current being AC or DC.  This case has always wavelike properties, and being non-equilibrium situation, you have all kinds of energy flows all over.  Even if we assume a perfect switch, something that changes from 0 to 1 without any intermediate states in between, it still is a step-like pulse with a lot of higher frequency components, and thus definitely a wave.  Ramping the current smoothly has a nicer spectrum, but a time-discontinuous signal always contain lots of frequencies.

While the transmission line model does describe the observable voltages and currents at the ends of the line when the properties of the transmission line are known, it does not mean it is a complete picture of the interactions involved.  The fact that the model is based on electromagnetic waves, should make it obvious that it is not just about electron kinematics, but electromagnetic field interactions must play a significant role, too.
To explore the complete picture in a way consistent with our best understanding of physics, one needs to delve into quantum electrodynamics, which definitely belongs to the less intuitive section of physics.  I definitely have no idea how to even start describing it in laymans terms.

At the beginning QED is pretty simple, just follow the experiment and digest.

If you then put another light indicator beside the first one they both blink but not simultaneously.
If you then put many light indicators side by side they all blink but only one at the time and eventually they've all blinked equal amounts.
We've given that quantized random light a name photon.

Things become even more interesting if you put few slits between that light source and its indicators.
You can make the apparatus from a quality laser pointer, a comb and plain paper.
Block all but two adjacent comb spike intervals and point the laser to the wall through the comb.
Distances are not relevant, only clearness of the result will change.

You can also make the apparatus from a plain paper.
Make a diminishing monocular, construction must be light tight so that model type can be easier.
Punch a pin hole to the input, two close by holes to output and one hole to the middle.
Single holes must be so small that finally only coherent light goes through.
Double holes can be bigger but bigger ones need more light for better contrast.
Then just point a light to the wall through the decoration.

If the thing works you should see a string of bars or spots coming out from two slits or holes.
For explaining that all we need probability waves.

It's now a hundred years old discovery so many generations of scientists are educated and for some the probability part is dropped out.
So for them the wave is pretty real even that they argue that real can be measured.
The phenomena itself is very real of course, our computerized world is based on it.
Good thing to remember if you start arguing over the net.
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#### electrodacus

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##### Re: Electroboom: How Right IS Veritasium?! Don't Electrons Push Each Other??
« Reply #709 on: July 21, 2022, 03:35:01 pm »
It's not a matter of not "getting it".  I am saying you are dead wrong about that.  There is zero reason to think that is correct.

Let's look at the extreme case.

a) Wheels are not connected and free to rotate zero friction.   The vehicle will just stay in place while the wheel on treadmill will rotate.
b) Wheels are welded to vehicle frame.  The vehicle will either stay in place if the wheel on treadmill slips or it will move from right to left if wheel on red box slips.
c) Wheels are connected with a twisted belt (think at infinity symbol shape ∞)  and pulley have the same diameter 1:1 gear ratio. The vehicle will move easily from right to left so in the direction the treadmill moves.
d) Wheels are connected as in my diagram but pulley are equal gear ratio 1:1.  The vehicle will wheels will be locked no different than point b) so one of vehicle wheels will need to slip.

Hopefully you can agree with all 4 points.
If you do then the only thing you need to understand is that when gear ratio is say 2:1 the force F2 can not be larger than F1 not until wheel on treadmill slips.
Newton's 3'rd law applies in this particular case due to the place and direction that  F1 is applied to vehicle.
Maybe imagine treadmill moving in the other direction all things being equal as in my diagram. The gearbox will still be locked and one of the wheels will still need to slip.

#### cbutlera

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##### Re: Electroboom: How Right IS Veritasium?! Don't Electrons Push Each Other??
« Reply #710 on: July 21, 2022, 04:46:05 pm »
No, you're leaving important bits out, oversimplifying the situation to fit your axiomatic model.

What a concise and insightful sentence.  Like The Treachery of Images, the more one thinks about it the more there is in it.

Simplifying the situation to fit an axiomatic model is a technique that I use all the time.  I expect that we all use it.  As long as I realise that I am doing it and can justify doing so, then it is a very powerful tool.  But if I mistake that model for the truth and say that I know, I stop thinking.  As long as I keep thinking, I come to understand.  That way I might approach some truth.

(Text in italics from the 1985 film Insignificance)

#### Naej

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##### Re: Electroboom: How Right IS Veritasium?! Don't Electrons Push Each Other??
« Reply #711 on: July 21, 2022, 09:23:05 pm »
"Will move".  No, something makes them move.  That something is an electric field, which propagates through the circuit somewhat analogously to a shock wave when the circuit is first connected.  Also, some of the original "potential energy" is in the form of an electric field around the charged plate; it is not exactly correct to just lump it all into "potential energy" and call it good enough.
Or maybe it is correct.
Not exactly correct means it is an approximation.

When you are arguing about exactly where the majority of the energy flow in a system is, making such approximations is exactly how you "accidentally" manipulate it to fit your pre-selected model.

Consider this question:  Are the participants arguing where the energy flows, or what the setup being investigated exactly is?

To me, the disagreement stems from the latter, and is the reason why I'm not interested in their arguments and opinions, be they professors or internet celebrities or whatever else.  I am too familiar with both "garbage data + good model = garbage results" and "good data + garbage model = gargabe results" already; I want to drop the garbage parts and have people spend their time on the "good data + good model = good results" case instead.
In this case, each participant has their own model.  That's no good: OF COURSE their inner workings vary then, even if the results are exactly the same.
And what I mean by correct, is that it agrees with Maxwell's model of electromagnetism.

#### Nominal Animal

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##### Re: Electroboom: How Right IS Veritasium?! Don't Electrons Push Each Other??
« Reply #712 on: July 22, 2022, 01:17:56 am »
"Will move".  No, something makes them move.  That something is an electric field, which propagates through the circuit somewhat analogously to a shock wave when the circuit is first connected.  Also, some of the original "potential energy" is in the form of an electric field around the charged plate; it is not exactly correct to just lump it all into "potential energy" and call it good enough.
Or maybe it is correct.
Not exactly correct means it is an approximation.

When you are arguing about exactly where the majority of the energy flow in a system is, making such approximations is exactly how you "accidentally" manipulate it to fit your pre-selected model.

Consider this question:  Are the participants arguing where the energy flows, or what the setup being investigated exactly is?

To me, the disagreement stems from the latter, and is the reason why I'm not interested in their arguments and opinions, be they professors or internet celebrities or whatever else.  I am too familiar with both "garbage data + good model = garbage results" and "good data + garbage model = gargabe results" already; I want to drop the garbage parts and have people spend their time on the "good data + good model = good results" case instead.
In this case, each participant has their own model.  That's no good: OF COURSE their inner workings vary then, even if the results are exactly the same.
And what I mean by correct, is that it agrees with Maxwell's model of electromagnetism.
Maxwell's model of electromagnetism doesn't talk about "potential energy".  It describes the behaviour of electric fields (E), magnetic fields (B), charge density (ρ), and current density (J).

When moving from the vector field approach to the potential approach, the electric field (E) and magnetic field (B) are described in terms of an electric potential (φ, also called scalar potential) and magnetic vector potential (A) via E=-∇φ-∂A/∂t and B=∇×A, where ∇φ refers to the gradient of the electric potential, and ∇×A to the curl of the magnetic vector potential.

Thing is, the electric potential (φ) and magnetic vector potential (A) have gauge freedom.  Unless you fix them (as in stick to) with some gauge, their value doesn't really describe anything; only their behaviour does.

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.  There are other gauges that can give other values to the electric potential and magnetic vector potential, but exactly the same electric field (E) and magnetic field (B).  In other words, Maxwell's model does not help you in choosing the proper gauge.

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!

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#### electrodacus

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##### Re: Electroboom: How Right IS Veritasium?! Don't Electrons Push Each Other??
« Reply #713 on: July 22, 2022, 01:43:18 am »
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!

I think you forgot what the discussion is about.
How is the stored energy in say a charged capacitor transferred to the load. Is it through wires or outside the wires ?
Wire or resistor is one and the same thing so the transmission line itself is the load.
The constant electric field between the capacitor plates has no role it is due to difference in charge between the two plates that the electric field exists not the other way around.

The stored potential energy is just converted into kinetic energy. No different from air pressure storage where you have air molecules and sound speed limit in that medium vs electrons and speed of light.

#### Nominal Animal

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##### Re: Electroboom: How Right IS Veritasium?! Don't Electrons Push Each Other??
« Reply #714 on: July 22, 2022, 02:49:43 am »
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!
I think you forgot what the discussion is about.
No, absolutely not.

How is the stored energy in say a charged capacitor transferred to the load. Is it through wires or outside the wires ?
It depends on the geometry of the wires.

Wire or resistor is one and the same thing so the transmission line itself is the load.
The geometry of the transmission line determines how the energy flows!

If you use a coaxial cable, the initial pulse is a waveform with lots of high-frequency components, and they will propagate as an electromagnetic field in the dielectric insulator between the core and the shield.  As described by the transmission line model, you'll get reflections (of those waves) until the system reaches steady state.  Then, in the steady state, the electromagnetic field in the dielectric stays constant (except for thermal noise and external perturbations), and the energy flows within the core conductor, as current.

Something very similar happens when the circuit is disconnected, too.  Because of those constant fields, the circuit doesn't just immediately cease, as the remnant EM fields do their reflection stuff again.

If you ignore those initial non-equilibrium states, you do not have a full physical picture or model of what is happening.  By using approximations that ignore those, and by not defining the properties of the transmission lines in sufficient detail, you can make and prove any claim correct.  It is a useless argument.

#### electrodacus

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##### Re: Electroboom: How Right IS Veritasium?! Don't Electrons Push Each Other??
« Reply #715 on: July 22, 2022, 03:03:54 am »

It depends on the geometry of the wires.

The geometry of the transmission line determines how the energy flows!

If you use a coaxial cable, the initial pulse is a waveform with lots of high-frequency components, and they will propagate as an electromagnetic field in the dielectric insulator between the core and the shield.  As described by the transmission line model, you'll get reflections (of those waves) until the system reaches steady state.  Then, in the steady state, the electromagnetic field in the dielectric stays constant (except for thermal noise and external perturbations), and the energy flows within the core conductor, as current.

Something very similar happens when the circuit is disconnected, too.  Because of those constant fields, the circuit doesn't just immediately cease, as the remnant EM fields do their reflection stuff again.

If you ignore those initial non-equilibrium states, you do not have a full physical picture or model of what is happening.  By using approximations that ignore those, and by not defining the properties of the transmission lines in sufficient detail, you can make and prove any claim correct.  It is a useless argument.

We are discussing the exact setup that Derek made no coaxial cables involved.
there was about 21m of wire/pipe on each side of the circuit symmetrically.
Say we leave the 20V battery he had and will consider that a constant voltage supply.
The left side say it had 9Ohm and the right side 1Ohm exactly the same geometry just much thinner pipe but same external diameter and no 1kOhm reistor that is removed just this transmission line.
Say the switch was closed for 5 seconds (just some arbitrary amount of time) the circuit is opened.
Over those 5 seconds 20V/10Ohm = 2A * 20V * 5 seconds = 200Ws (200 Joules if you prefer) where delivered to this transmission line / load

How come 90% of the energy is on the 9Ohm side and just 10% on the 1Ohm side while you have the exact same geometry on both sides?

#### hamster_nz

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##### Re: Electroboom: How Right IS Veritasium?! Don't Electrons Push Each Other??
« Reply #716 on: July 22, 2022, 03:18:33 am »

It depends on the geometry of the wires.

The geometry of the transmission line determines how the energy flows!

If you use a coaxial cable, the initial pulse is a waveform with lots of high-frequency components, and they will propagate as an electromagnetic field in the dielectric insulator between the core and the shield.  As described by the transmission line model, you'll get reflections (of those waves) until the system reaches steady state.  Then, in the steady state, the electromagnetic field in the dielectric stays constant (except for thermal noise and external perturbations), and the energy flows within the core conductor, as current.

Something very similar happens when the circuit is disconnected, too.  Because of those constant fields, the circuit doesn't just immediately cease, as the remnant EM fields do their reflection stuff again.

If you ignore those initial non-equilibrium states, you do not have a full physical picture or model of what is happening.  By using approximations that ignore those, and by not defining the properties of the transmission lines in sufficient detail, you can make and prove any claim correct.  It is a useless argument.

We are discussing the exact setup that Derek made no coaxial cables involved.
there was about 21m of wire/pipe on each side of the circuit symmetrically.
Say we leave the 20V battery he had and will consider that a constant voltage supply.
The left side say it had 9Ohm and the right side 1Ohm exactly the same geometry just much thinner pipe but same external diameter and no 1kOhm reistor that is removed just this transmission line.
Say the switch was closed for 5 seconds (just some arbitrary amount of time) the circuit is opened.
Over those 5 seconds 20V/10Ohm = 2A * 20V * 5 seconds = 200Ws (200 Joules if you prefer) where delivered to this transmission line / load

How come 90% of the energy is on the 9Ohm side and just 10% on the 1Ohm side while you have the exact same geometry on both sides?

"exactly the same geometry just much thinner pipe "

Take away either pipe, and see how much energy is delivered by the remaining one...
« Last Edit: July 22, 2022, 03:26:19 am by hamster_nz »
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#### electrodacus

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##### Re: Electroboom: How Right IS Veritasium?! Don't Electrons Push Each Other??
« Reply #717 on: July 22, 2022, 03:24:14 am »
"exactly the same geometry just much thinner pipe "

Take away either pipe, and see how much energy is delivered by the reaming one...

Outside diameter of the pipe is the same. The thickness of the wall is lower so that resistance is 9x higher.
So line capacitace will be the same in both cases.

#### Nominal Animal

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##### Re: Electroboom: How Right IS Veritasium?! Don't Electrons Push Each Other??
« Reply #718 on: July 22, 2022, 03:39:33 am »
It depends on the geometry of the wires.

The geometry of the transmission line determines how the energy flows!
We are discussing the exact setup that Derek made no coaxial cables involved.
But Derek did not specify the setup exactly, he makes approximations and simplifications that makes "the exact setup" ambiguous and up to interpretation.

It really is like starting with "Let's model humans as toroids, with nutrition flow through the hole occurring in one direction only", and then explaining mental issues as stemming from having mouth and ass connected to the same tube, and from being jealous of turtles that can do cloacal respiration too.

To be specific, the situation when the circuit is closed or opened, between the two steady states (no energy flowing, and energy flowing steadily), is where the interesting stuff happens.  It is not an orderly progression marching through the system; it is a wavelike phenomena best described by quantum electrodynamics (in the physical sense), and well enough described by the transmission model if the properties of the system are specified in enough detail.

Even in a single conductor you get electromagnetic waves flowing through the conductor.  If there is anything nearby the conductor, in these situations the system can actually behave like (be a) waveguide for these waves, so that while the current flow in such parts of the system is minimal, a lot of energy can still flow; and only when the steady state is reached, does the current flow match the energy flow through the system.

I really don't know how to better explain this.  The transmission model is good, especially in the sense that you cannot just say that the conductor is a transmission line and get results that match the real world, because you do need to model the actual properties – geometry – of the system to apply the transmission line.  On the other hand, if you define your transmission line ignoring things that do affect it in the real world, you can get whatever results you want; it just doesn't match real world anymore, and you're just babbling on about an imaginary system.
« Last Edit: July 22, 2022, 03:49:15 am by Nominal Animal »

#### electrodacus

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##### Re: Electroboom: How Right IS Veritasium?! Don't Electrons Push Each Other??
« Reply #719 on: July 22, 2022, 03:54:45 am »
We are discussing the exact setup that Derek made no coaxial cables involved.
But Derek did not specify the setup exactly, he makes approximations and simplifications that makes "the exact setup" ambiguous and up to interpretation.

It really is like starting with "Let's model humans as toroids, with nutrition flow through the hole occurring in one direction only", and then explaining mental issues as stemming from having mouth and ass connected to the same tube, and from being jealous of turtles that can do cloacal respiration too.
[/quote]

I made it more clear saying it is symmetrical and you have 10m on each side plus 1m so 21m of wire/pipe on each side and the battery is a 20V constant and pipe resistance is total 10Ohm with 9Ohm for one half and the other half with 1Ohm.
I also specified a period for witch the switch is closed 5 seconds so we can discuss about all energy delivered by battery to circuit and where will that energy end up and in what form.

After those 5 seconds most of the energy that left the battery those 200Ws are still there as they increased the temperature of the pipe and so most of that was still not radiated to space.
But the half that has 9Ohm while having the same surface area (same external diameter for the pipe) will end up much warmer than the other part and will radiate over time much more electromagnetic radiation to space in the form of infrared photons as 90% of the energy from battery was delivered there.

Do you think that measuring anything external to wire like electric or magnetic field will help you know how much energy was dissipated to each side ? Since if energy was delivered from outside to the wire surface then you should be able to make measurements related to that.

#### hamster_nz

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##### Re: Electroboom: How Right IS Veritasium?! Don't Electrons Push Each Other??
« Reply #720 on: July 22, 2022, 04:42:11 am »
Do you think that measuring anything external to wire like electric or magnetic field will help you know how much energy was dissipated to each side ? Since if energy was delivered from outside to the wire surface then you should be able to make measurements related to that.

If somebody does show you how to do this, without being in electrical contact with the wire, you will just say it proves nothing, and change the goalposts again, or say "but by doing so you change the geometry of the system!". Fruitless.

Unless you say "If anybody can show me a method to determine how much energy was dissipated in each side without making electrical contact to the wires, then I accept that the energy flow is not in the wire" then why bother?

And even if you do say that (which you won't), then you will argue that "electrical contact" is loosely defined, such that the method given is invalid. You will start saying things like "electrical contact includes a capacitor!" or "it's a transmission line", or "but stored energy!"
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#### gnuarm

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##### Re: Electroboom: How Right IS Veritasium?! Don't Electrons Push Each Other??
« Reply #721 on: July 22, 2022, 04:54:58 am »
It's not a matter of not "getting it".  I am saying you are dead wrong about that.  There is zero reason to think that is correct.

Let's look at the extreme case.

a) Wheels are not connected and free to rotate zero friction.   The vehicle will just stay in place while the wheel on treadmill will rotate.

Not exactly, but close.  If we eliminate friction the inertia of the wheels will cause a force on the bearing of the axle, moving the car.  How much?  That depends on the masses and relative inertia of each.  Eliminate inertia and all extraneous forces, then maybe, yes, the wheel on the treadmill will only turn.

Quote
b) Wheels are welded to vehicle frame.  The vehicle will either stay in place if the wheel on treadmill slips or it will move from right to left if wheel on red box slips.
c) Wheels are connected with a twisted belt (think at infinity symbol shape ∞)  and pulley have the same diameter 1:1 gear ratio. The vehicle will move easily from right to left so in the direction the treadmill moves.

The twisted belt simply means the wheels turn in opposite directions.  Having a twisted belt connect the two axles, means they turn in opposite directions.  If the treadmill turns, that wheel will turn in that direction, and the other wheel will turn  in the opposite direction.  With a 1:1 ratio, this can not be accommodated without a wheel slipping.

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d) Wheels are connected as in my diagram but pulley are equal gear ratio 1:1.  The vehicle will wheels will be locked no different than point b) so one of vehicle wheels will need to slip.

You keep screwing up this idea of locked wheels.  YOU HAVE A MODEL.   PICK IT UP AND TURN THE WHEELS.  THEY ARE NOT LOCKED.  It would only be a problem if you put it on a single surface and tried to roll it.  Since the ratio between the wheels is not 1:1, they can't roll on a single surface.  With a 1:1 ratio, they will roll.  This is how 4 WD vehicles work.  They don't use a differential between front and rear wheels, yet, they roll along the ground.

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Hopefully you can agree with all 4 points.

LOL, hardly.  YOU HAVE A MODEL.  Until you try this with your model, you are ignorant of the facts.

Quote
If you do then the only thing you need to understand is that when gear ratio is say 2:1 the force F2 can not be larger than F1 not until wheel on treadmill slips.

The gear ratio does not determine the forces between the wheels and the surfaces, just as Newton's third law does not determine them either.

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Newton's 3'rd law applies in this particular case due to the place and direction that  F1 is applied to vehicle.

Rubbish

Quote
Maybe imagine treadmill moving in the other direction all things being equal as in my diagram. The gearbox will still be locked and one of the wheels will still need to slip.

Until you learn there is nothing locked about the belted wheels, you are doomed to failure.  Your rocket ship exploded on the launch pad.

What a buffoon!

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

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##### Re: Electroboom: How Right IS Veritasium?! Don't Electrons Push Each Other??
« Reply #722 on: July 22, 2022, 05:09:18 am »
If somebody does show you how to do this, without being in electrical contact with the wire, you will just say it proves nothing, and change the goalposts again, or say "but by doing so you change the geometry of the system!". Fruitless.

Unless you say "If anybody can show me a method to determine how much energy was dissipated in each side without making electrical contact to the wires, then I accept that the energy flow is not in the wire" then why bother?

And even if you do say that (which you won't), then you will argue that "electrical contact" is loosely defined, such that the method given is invalid. You will start saying things like "electrical contact includes a capacitor!" or "it's a transmission line", or "but stored energy!"

You can do a thing as simple as measure the temperature of the wires like with a thermal camera and you will not need to touch the wires but what that camera detects are infrared photons emitted from the wires so that is radiated emission.
You put in 200Ws from the battery in the wires you eventually get 200Ws worth of infrared photons and there is nothing else. 200Ws in and 200Ws out.
Both the magnetic and electric field around the wire's is conservative and it is due to moving charged particles (electrons).

Magnetic field around the wire will be the same around the 9Ohm as it will be around the 1Ohm pipe/wire.
Electric field will also be symmetrical as is based on imbalance of charge and outside surface is the same for both halfs.

#### Nominal Animal

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##### Re: Electroboom: How Right IS Veritasium?! Don't Electrons Push Each Other??
« Reply #723 on: July 22, 2022, 05:24:30 am »
No, you defined another system.  I don't care about the answer to your question about your system, I'm saying that you are each talking about different systems, which means there is no common argument at all.

#### hamster_nz

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##### Re: Electroboom: How Right IS Veritasium?! Don't Electrons Push Each Other??
« Reply #724 on: July 22, 2022, 05:29:30 am »
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.
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