What Stabilizes The Electron Orbit In An Atom?

Motanache

Joined Mar 2, 2015
540
post #81
I do not know what we are interested in the chemists' opinion.

Chemists use here notions from physics that they do not really love.

When I see the Glenn Hollandl opinion, I regret that I went to school.
That pulls me down.
We know to think only in school patterns.

We store and reproduce information. And we thought this is all.
 

nsaspook

Joined Aug 27, 2009
13,315
post #81
I do not know what we are interested in the chemists' opinion.

Chemists use here notions from physics that they do not really love.

When I see the Glenn Hollandl opinion, I regret that I went to school.
That pulls me down.
We know to think only in school patterns.

We store and reproduce information. And we thought this is all.
Chemistry is applied physics just like electrical engineering. The simplified tables of chemistry (based on the current physics theories) are just as valid as using Ohms law in circuit theory instead of calculating the energy flow using EM field equations. You, Glenn or anyone else are free to use the method you chose to solve problems if the answer is correct within reasonable error requirements. Hell, assuming the world is flat still works for most local navigation problems.
The theories are developed to solve structured problems, without going to school or having some sort of structured education it's usually very hard to see what that problem is and how to solve it in a logical manner. 99% of education is learning what's crap and how to easily detect crap so you don't waste a lot of time with the wrong patterns.
 

Thread Starter

Glenn Holland

Joined Dec 26, 2014
703
Getting back on the main road of the original question, I'm wondering if matter waves are transverse or longitudinal?

In my layman's opinion, it seems that matter waves would be longitudinal since De Broglie's formula for the wave length of a particle is Planck's constant divided by the momentum of the particle. Planck's constant is 6.626 X 10 exp -34 Joule-Seconds which represents an amount of energy applied to the particle over a period of time.

My intuition is that Planck's constant represents a condition where something is pushing the particle and compressing it in the direction of its travel. This is similar to what happens when a volume of a compressible fluid is pushed, so the particle's wave length is analogous to the wave length of sound in a fluid.

This is probably wrong as hell, but I'm trying to keep the discussion going. :)
 

Motanache

Joined Mar 2, 2015
540
Getting back on the main road of the original question, I'm wondering if matter waves are transverse or longitudinal?
Nobody is thinking of that. They use a model to explain a physical phenomenon.
They have a simple formula that explains what they see in the experiment.

EDIT: I want to say they existed over time a lot of books and theories about it. They died slowly that they were not useful.
https://en.wikipedia.org/wiki/Toroidal_ring_model
http://www.oocities.org/xplorer2x/Ring.htm
www.cybsoc.org/electron.pdf

http://blackholeformulas.com/files/RingElectron.html

Getting back on the main road of the original question,
Do not you want to talk about pair generation?
When a gamma-frequency electromagnetic wave passes through the proximity of an atom gives birth to an electron-positron pair. What explanation the phenomenon has without energy conservation ? What trigger that ?
 
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Thread Starter

Glenn Holland

Joined Dec 26, 2014
703
If an atomic electron is actually a wave phenomenon, then what are other atomic electron waves interacting with when a chemical bond (such as a covalant bond) is formed?

Seems there must be some kind of "center of charge" in the so called "electron cloud" that's interacting with other charge centers. Or does this electron cloud have a homogeneous charge that's spread evenly around the nucleus?
I've done more reading (I've got 4 chemistry books in my collection) and believe I've an found an answer to this particular question.

The "electron" in the wave is actually the peak value of charge as the magnitude of the charge varies sinusoidally as it's plotted with distance around the nucleus.
 

Motanache

Joined Mar 2, 2015
540
You have a very interesting conception of electromagnetic waves.
We have to do at least something practical to show this.

<<The "electron" in the wave is actually the peak value of charge>>
If the electromagnetic waves correspond with an electric charge,
Why do they not interact with each other with theirs electrical charge?


We know that the interference of electromagnetic waves is quite similar to mechanical waves: interference and diffraction.

Where the electrical charge is used ?
I am ready to ask ham radio colleagues to do an experiment.
 

Motanache

Joined Mar 2, 2015
540
We neglect resonant phenomena in the antenna.

The curve line represents the amplitude of the electric field in different point of space.
1.An electric charge generates an electric field
2. This electric field interact with another electric charge......

The antenna has an electric polarization alternating positive and negative.

When the antenna possesses positive electric charge generates positive electric field around.

But in tis time at specific point of space there is also the negative electric field.

In other words, something can interact as if there was a negative electrical charge present, when the antenna already has a positive charge.

It's what Einstein said: if the sun suddenly disappears, the Earth would still be attracted half a hour.

So we're talking about electrical charge that does not exist, but the electric field exists. Electrical interaction exists.

That is why I consider the very good idea to consider the peak of that graph corresponding of a positive charge, and the minimum of that graph of a negative charge.
 
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Motanache

Joined Mar 2, 2015
540
Some people just don't have that level of understanding yet.
That's in religion.
If there are people who have the level I look gladly forward explaining to us the issue raised by TS.

Do you think this discussion has a purpose or we wasting time with it?
Something useful can be obtained from this discussion?

About chemical bonding. The nucleus of an atom attracts electrons of another atom. But to "stay" together, the electrons must be at the permissible energy levels (The covalent bond)About chemical ion beam and Wan der Waals bond:

Imagine water molecule. Hydrogen Hydroxide:
The oxygen atom has a multiproton nucleus, so it attracts the electrons more powerfully than the hydrogen nucleus.
The whole electronic cloud moves to oxygen.

Thus the positive charge occurs at hydrogen side and the negative charge occurs at the oxygen side.
The molecule is polar. Now, the molecule rotates. These electric charges being rotated generate a magnetic field.
Molecules can interlace each other magnetic and electric. These are the van der waals forces.
But these forces also appear to nonpolar molecules. Because the polarization of a molecule can polarize the other by induction.

Note: do not rush to put problem of smart. I did not write because I think I am smart.

I'm not really smart guy, but I am trying to solve this question.
I guess those who suggest others are cleverer want to stop our discussion.

Anyway we are few interested in this discussion.
The problem is very difficult and we are about to quit the discussion.
We still needed a little hit and we quit.

I still think that if we answer this problem we will be able to use the knowledge gained in the technique
 
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Thread Starter

Glenn Holland

Joined Dec 26, 2014
703
Some else posted a question about whether gravitational waves existed and I believe the Wave/Particle Duality may be applicable to that problem.

Photons do not have rest mass, however they do carry momentum as if they did have rest mass. Therefore, gravity can also exert a force on a photon much like a particle that has a rest mass. Using De Broglie's idea that photons and particles can behave like each other, I'm wondering if photons can also exert a gravitational attraction on matter?

On another question about the Uncertainty Principle (which says that the position and momentum of a particle cannot be simultaneously determined), I'm wondering about this scenario:

If a particle (such as an electron) with a known velocity strikes a stationary target, the momentum can be determined by conservation of momentum (m1 X v1 = m2 X v2). However isn't the point of impact on the target also the position of the electron? In that case, it seems the momentum and position of the electron can be reasonably determined to be at the same location.
 

Motanache

Joined Mar 2, 2015
540
Photons do not have rest mass,
We put 0 at the photon rest mass, not because it is. It is not the result of a measurement.
They wanted to show through that 0 only that it does not have sense photon at rest.
You can use E = mc^ 2 to calculate the photon mass. Then Newton's formula to calculate the force of gravity attraction.
 

nsaspook

Joined Aug 27, 2009
13,315
On another question about the Uncertainty Principle (which says that the position and momentum of a particle cannot be simultaneously determined), I'm wondering about this scenario:

If a particle (such as an electron) with a known velocity strikes a stationary target, the momentum can be determined by conservation of momentum (m1 X v1 = m2 X v2). However isn't the point of impact on the target also the position of the electron? In that case, it seems the momentum and position of the electron can be reasonably determined to be at the same location.
How to you determine the point of impact? Does the electron 'touch' another particle and leave a infinitely small mark we can observe later or do the repulsive forces of like charges at a distance deflect electrons somewhere before the quantum position probability cloud of both particles?
 

Papabravo

Joined Feb 24, 2006
21,228
How to you determine the point of impact? Does the electron 'touch' another particle and leave a infinitely small mark we can observe later or do the repulsive forces of like charges at a distance deflect electrons somewhere before the quantum position probability cloud of both particles?
Nice try, but if you know the momentum, you don't know where the particle is and have no idea if it will hit the target or not. If it does get close enough to be scattered, then the momentum is continuously changing and you don't know where the particle will be. Either way you are stuck.
 

nsaspook

Joined Aug 27, 2009
13,315
Nice try, but if you know the momentum, you don't know where the particle is and have no idea if it will hit the target or not. If it does get close enough to be scattered, then the momentum is continuously changing and you don't know where the particle will be. Either way you are stuck.
Exactly, it's a hard concept to accept that the universe is not deterministic (because of probability amplitude distributions) to observers in this universe.
 

Thread Starter

Glenn Holland

Joined Dec 26, 2014
703
How to you determine the point of impact? Does the electron 'touch' another particle and leave a infinitely small mark we can observe later or do the repulsive forces of like charges at a distance deflect electrons somewhere before the quantum position probability cloud of both particles?
As an example, in an X-Ray tube electrons are accelerated to a given velocity by a high voltage between a negative cathode and a positive anode which is the point of impact.

On a "macroscopic" viewpoint, it is generally known where the electron beam strikes the anode and the momentum imparted to the anode can also be calculated. However from a microscopic viewpoint, when we're dealing with a small particle (like an electron which is also moving as a wave), the resolution of exact the point of impact cannot be known with any degree of accuracy and my proposition was merely a philosophical exercise.
 

nsaspook

Joined Aug 27, 2009
13,315
As an example, in an X-Ray tube electrons are accelerated to a given velocity by a high voltage between a negative cathode and a positive anode which is the point of impact.

On a "macroscopic" viewpoint, it is generally known where the electron beam strikes the anode and the momentum imparted to the anode can also be calculated. However from a microscopic viewpoint, when we're dealing with a small particle (like an electron which is also moving as a wave), the resolution of exact the point of impact cannot be known with any degree of accuracy and my proposition was merely a philosophical exercise.
Sure, I know that you mean but the "macroscopic" view is a group approximation (spot size) of the reality of individual charges with measurement tolerances to beam angle and charge density measurements over the impact region.
 
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