Particles as excitation of a field

[Butch]

2 minutes ago, Strange said:

I said nothing about inverse square. As far as I know that is totally irrelevant.

I said something about it early on in this discussion, I think it is quite relevant.

1 minute ago, geordief said:

I think c and the permittivity/permeability  are the "same thing";the 3  hang together in one equation.  unrelated to any other variable.

 

 

I agree.

14 minutes ago, geordief said:

Is it bound up with the universal speed limit (of causation)?

 

 

 

 

I agree.

On 4/30/2018 at 9:57 AM, Silvestru said:

What do you mean by range of excitation?

Apologies, I want to clarify as I still think you are imagining a field similar to the magnetic one.

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No, not stuck on that... can you create a standing wave in a rope if an end is free? No, you must have reflection and if you don't keep shaking the end of the rope the wave will travel away no matter how long the rope even with a magic rope that has no energy losses.

[studiot]

7 minutes ago, geordief said:

I think c and the permittivity/permeability  are the "same thing";the 3  hang together in one equation.  unrelated to any other variable.

 

 

Clasically yes indeed.

 

Wiki is good on this.

 

https://en.wikipedia.org/wiki/Impedance_of_free_space

 

I think the finiteness comes form the fact that in the classical wave equation we are equating a function of space to a function of time (as a second order differential equation)  and we would not have a wave if the constants were not finite.

 

Note that the correct term for the excitation of the field representation is a wave packet not a wave.

 

Again Wiki has a good picture.

 

https://en.wikipedia.org/wiki/Wave_packet

[Butch]

2 minutes ago, studiot said:

 

Note that the correct term for the excitation of the field representation is a wave packet not a wave.

Not possible without reflection, this is fine for describing a particle in a very small system (with reflective limits), not for an unbound particle.

[studiot]

Why not?

[Butch]

3 minutes ago, studiot said:

Why not?

In the wiki both waves are travelling, think in three dimensions, variations in flux density of the field,  the wave would be spherical and traveling away from the point source as well as diminishing via the inverse square...  Your particle is doomed. 

[Strange]

Just now, Butch said:

In the wiki both waves are travelling, think in three dimensions, variations in flux density of the field,  the wave would be spherical and traveling away from the point source as well as diminishing via the inverse square...  Your particle is doomed. 

You are still thinking of it as a classical wave that travels away from the source. That is not a good analogy.

[Butch]

6 minutes ago, Strange said:

You are still thinking of it as a classical wave that travels away from the source. That is not a good analogy.

How is this wave different?  You could say that it is a wave function, not a wave...  and I will agree. However the perturbation(I think a better description than excitation.) is not a wave of any sort, rather interaction of the perturbation with outside influence will produce a wave nature. 

Edited May 1, 2018 by Butch

[studiot]

3 minutes ago, Butch said:

variations in flux density of the field,

 

 

2 minutes ago, Strange said:

You are still thinking of it as a classical wave that travels away from the source. That is not a good analogy.

 

Agreed but instead of the word travels doesn't this sound better?

You are still thinking of it as a classical wave that travels spreads away from the source.

 

The point of a wave packet is that it doesn't spread. That is how it models a travelling particle.

I thought the moving graphics on Wiki described this rather well.

 

A travelling EM (classical) wave takes its field along with it as it goes.

That is how it spreads.

 

The quantum field always occupies all the space available to it.

3 minutes ago, Butch said:

How is this wave different?  You could say that it is a wave function, not a wave...  and I will agree. However the perturbation(I think a better description than excitation.) is not a wave of any sort, rather interaction of the perturbation with outside influence will produce a wave nature. 

Perturbation is another word for excitation and one mathematical way of approaching this is perturbation theory.

 

But like everything else about quantum theory it is hydra like - multiheaded so no single one of our methods covers it all.

[geordief]

10 minutes ago, Strange said:

You are still thinking of it as a classical wave that travels away from the source. That is not a good analogy.

Is that because it is only found when detected and so does not "travel" between detections?

[Strange]

7 minutes ago, geordief said:

Is that because it is only found when detected and so does not "travel" between detections?

The fact that the wave function is not localised is what allows for effects like tunnelling, where the particle doesn't move through a barrier (it just "appears" on the other side). But when a particle moves from a source to a detector, that is represented by the (peak of the) wave packet moving from one location to the other.

[Butch]

4 minutes ago, studiot said:

The point of a wave packet is that it doesn't spread. That is how it models a travelling particle.

I thought the moving graphics on Wiki described this rather well.

Not without reflection to conserve energy. 

Look at it this way,  what is the energy of a free electron? 

What is the energy delta of a free electron that interacts with an outside influence. 

Of course the first answer is another question...  Relative to what? A wave has energy relative to itself. A static perturbation has no energy unless it reacts to an outside influence. 

[Strange]

Just now, Butch said:

Not without reflection to conserve energy. 

Because the wave packet doesn't spread, nothing is required to conserve energy. Free electrons don't evaporate, do they.

[Butch]

4 minutes ago, Strange said:

The fact that the wave function is not localised is what allows for effects like tunnelling, where the particle doesn't move through a barrier (it just "appears" on the other side). But when a particle moves from a source to a detector, that is represented by the (peak of the) wave packet moving from one location to the other.

I agree...  But!  I would say the point source,  not the wave peak. 

[Strange]

Just now, Butch said:

I agree...  But!  I would say the point source,  not the wave peak. 

What is the "point source"?

[Butch]

7 minutes ago, Strange said:

Because the wave packet doesn't spread, nothing is required to conserve energy. Free electrons don't evaporate, do they.

No they don't! A wave packet can give up all of its energy. A perturbation as I describe can give up or accept some energy temporarily,  but will return to the "well" form. In the case of a bound electron even the ground state is temporary influence of the well. 

5 minutes ago, Strange said:

What is the "point source"?

The point where the flux density of the field is greatest. 

Edited May 1, 2018 by Butch

[Strange]

2 minutes ago, Butch said:

No they don't! A wave packet can give up all of its energy.

If it did that, then the particle would have disappeared. The fact that the particle is represented by a standing wave that doesn't disperse is what makes it a particle.

3 minutes ago, Butch said:

The point where the flux density of the field is greatest. 

I don't really know what that means in this context. 

[Butch]

15 minutes ago, Strange said:

If it did that, then the particle would have disappeared. The fact that the particle is represented by a standing wave that doesn't disperse is what makes it a particle.

I don't really know what that means in this context. 

The perturbation of the field that I speak of is akin to the curvature of space-time around a mass.  In a fermion field the perturbation increases flux density to its maximum at a single point this is our particle, the flux density decreases via the inverse square.  Note that this is a static entity.

The wave function of such an entity would be dependant upon physicality of the outside Influence(s).

This structure could accept energy, but would seek to return to its original state. Since the point source is at maximum flux density, another particle could not have the same quantum description. 

My point about the particle as a standing wave is that if it is a wave of any sort standing or not,  it could easily give up all of its energy and cease to exist! 

Edited May 1, 2018 by Butch

[Strange]

14 minutes ago, Butch said:

In a fermion field the perturbation increases flux density to its maximum at a single point this is our particle, the flux density decreases via the inverse square.  Note that this is a static entity.

It sounds as if by "flux density" you mean the wave function of the particle? If so, I don't think this decreases with an inverse square law (in general).

Or maybe you are just mixing up different concepts.

After all, even neutral particles have a wave function so the term "flux density" doesn't really apply there.

[Butch]

13 minutes ago, Strange said:

It sounds as if by "flux density" you mean the wave function of the particle? If so, I don't think this decreases with an inverse square law (in general).

Or maybe you are just mixing up different concepts.

After all, even neutral particles have a wave function so the term "flux density" doesn't really apply there.

No,  I am speaking of the perturbation/exitation of the fermion field. Regardless if the field exists because the particle does or the particle exists in the field (chicken or the egg) am I correct to say all fermion particles share the same field? 

I am saying that the wave function of the particle is a result of the perturbation interacting with outside influence. 

For example two electrons in proximity would produce a common recurvature of the field between them,  this would produce a sinus curve that would describe a wave. The closer they were the higher the implied wave frequency.  You can extend this process to the two slit experiment or to Schroedinger equation in an atom. 

And yes this would occur without charge. You can do the same with two neutrinos or an electron and a neutrino etc. 

Edited May 1, 2018 by Butch

[Strange]

5 minutes ago, Butch said:

No,  I am speaking of the perturbation/exitation of the fermion field. 

So you are talking about the ave function of the particle which defines the (probability of the) location of the particle. This is not "flux density". And it doesn't have an inverse square law.

Quote

am I correct to say all fermion particles share the same field?

I don't believe so. Every particle type is an excitation of a particular type of field.

[Butch]

7 minutes ago, Strange said:

So you are talking about the ave function of the particle which defines the (probability of the) location of the particle. This is not "flux density". And it doesn't have an inverse square law.

I don't believe so. Every particle type is an excitation of a particular type of field.

First start thinking in terms of a single unbound particle, for now. 

Maybe the term "particle well" would be better to describe the perturbation of the field.

A fermion particle is an excitation of a fermion field or maybe the fermion field (chicken or the egg)  either way works here. 

The wave function is apparent only as the particle interacts with outside influences. 

Picture a well, you have seen the graph,  two hyperbolic curves representing the depth and width of the well. The wave function is dependant upon what the slope of the curve is where the perturbation is "bounded" to a system. 

Edited May 1, 2018 by Butch

[Strange]

Just now, Butch said:

Maybe the term "particle well" would be better to describe the perturbation of the field.

What is wrong with the standard terminology "wave function"?

[Butch]

3 minutes ago, Strange said:

What is wrong with the standard terminology "wave function"?

The wave function of the particle is undefined until it meets outside influence,  such as an electron being bound to an atom. I elaborated in my previous post,  please take a look. 

The particle does not have a wave function, it manifests a wave function. 

Edited May 1, 2018 by Butch

[Strange]

Just now, Butch said:

The wave function of the particle is undefined until it meets outside influence,  such as an electron being bound to an atom. 

You are suggesting that the particle ceases to exist when it is free?

Wave function of a free particle: https://en.wikipedia.org/wiki/Free_particle#Quantum_free_particle

[Butch]

4 minutes ago, Strange said:

You are suggesting that the particle ceases to exist when it is free?

Wave function of a free particle: https://en.wikipedia.org/wiki/Free_particle#Quantum_free_particle

This is a consequence of a particle being a wave phenomenon... 

This is what I have been saying, review previous content! 

If a particle were a wave of any sort, it would cease to exist without reflective boundaries...  Such as a free electron.