Particles as excitation of a field

[Butch]

Getting my head around this, a particle is an excitation of its appropriate field, but what is the nature of that excitation? 

It does not wash as an oscillation, as such it would either generate a continual wave or the particle would cease to exist. It seems to me it must be a perturbation of the field extending to infinity via the inverse square... am I on the right track?

[Silvestru]

29 minutes ago, Butch said:

Getting my head around this, a particle is an excitation of its appropriate field, but what is the nature of that excitation? 

It does not wash as an oscillation, as such it would either generate a continual wave or the particle would cease to exist. It seems to me it must be a perturbation of the field extending to infinity via the inverse square... am I on the right track?

Hmm, this is a complicated subject that I also had and still have trouble with. I recommend as a start the below paper.

https://arxiv.org/ftp/arxiv/papers/1204/1204.4616.pdf

If you still want to delve deeper in the subject you might also want to read about Zero-point_energy

Just to add to you post, an excitation does not create it's own field (for each particle).

An excitation in layman terms is when a region of a quantum field has more than the minimum energy density and that is what a particle is. (sometimes  )

 

Edited April 30, 2018 by Silvestru

[Butch]

1 minute ago, Silvestru said:

Just to add to you post, an excitation does not create it's own field (for each particle).

An excitation in layman terms is when a region of a quantum field has more than the minimum energy density and that is what a particle is. 

 

Thx, I was referring to fermion, boson fields etc.

There must be a threshold, otherwise there would be an infinite number of types and/or angular momenta... The inverse square can be plotted as a hyperbolic curve, unity on such a plot must have significance, no?

[Silvestru]

2 minutes ago, Butch said:

Thx, I was referring to fermion, boson fields etc.

There must be a threshold, otherwise there would be an infinite number of types and/or angular momenta... The inverse square can be plotted as a hyperbolic curve, unity on such a plot must have significance, no?

It doesn't work like that, you mention inverse square law but those are two entirely different kind of fields.

The electron field  is not the same as the electric field the electron creates. The electron field is everywhere across space-time.

 

 

[Strange]

43 minutes ago, Butch said:

Getting my head around this, a particle is an excitation of its appropriate field, but what is the nature of that excitation? 

It does not wash as an oscillation, as such it would either generate a continual wave or the particle would cease to exist. It seems to me it must be a perturbation of the field extending to infinity via the inverse square... am I on the right track?

It is a quantised oscillation: https://profmattstrassler.com/articles-and-posts/particle-physics-basics/fields-and-their-particles-with-math/

[Butch]

11 minutes ago, Strange said:

It is a quantised oscillation: https://profmattstrassler.com/articles-and-posts/particle-physics-basics/fields-and-their-particles-with-math/

It seems that the examples of quantized oscillation have points of reflection, thus COE. What limits the range of excitation?

[Silvestru]

13 minutes ago, Butch said:

What limits the range of excitation?

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.

Spoiler

 

 

[Butch]

47 minutes ago, Silvestru said:

Hmm, this is a complicated subject that I also had and still have trouble with. I recommend as a start the below paper.

https://arxiv.org/ftp/arxiv/papers/1204/1204.4616.pdf

If you still want to delve deeper in the subject you might also want to read about Zero-point_energy

Just to add to you post, an excitation does not create it's own field (for each particle).

An excitation in layman terms is when a region of a quantum field has more than the minimum energy density and that is what a particle is. (sometimes  )

 

 

2 minutes ago, Silvestru said:

What do you mean by range of excitation?

Hmm,  difficult to explain...  Example: If a lake were infinite and you dropped a pebble in it waves would travel away from the pebble and would never return,  thus no evidence of the pebble.  If however there were no energy loss and the lake had a shoreline that was perfectly reflective,  there would always be evidence of the pebble.  If the evidence of a particle is lost,  the particle no longer exists,  if the field that contains the excitation that is the particle has "shorelines", what are they? 

[Strange]

1 minute ago, Butch said:

 

Hmm,  difficult to explain...  Example: If a lake were infinite and you dropped a pebble in it waves would travel away from the pebble and would never return,  thus no evidence of the pebble.  If however there were no energy loss and the lake had a shoreline that was perfectly reflective,  there would always be evidence of the pebble.  If the evidence of a particle is lost,  the particle no longer exists,  if the field that contains the excitation that is the particle has "shorelines", what are they? 

Imagine a single ripple (centred on where the pebble went in) being self-sustaining. So that disturbance in the field is a "thing" (electron or whatever).  It doesn't continually require energy to keep it going because it is not a ripple in water (water has mass and requires energy to make it go up and down).

That wavelike disturbance can now move around and it represents the position of the electron. Actually, the square of the amplitude of the ripple represents the probability of finding the electron at that location (if I have extended your metaphor appropriately) so the electron is most likely to be in the middle but could be elsewhere. There is a really, really tiny probability that it will be detected on the other side of the galaxy.

You can then do calculations by analysing the paths of these ripples. If you add together every single possible path the ripple could take, you will find that some of the waves will constructively interfere and some will destructively interfere and the most probable path ends up being the one that corresponds to a "classical" description of the path of the electron (or photon, or whatever). This is how quantum theory, with all its probability based things, can describe the behaviour of photons and reproduce the results of classical optics. (See the Feynman lectures on QED if you haven't already.)

[Butch]

1 minute ago, Strange said:

Imagine a single ripple (centred on where the pebble went in) being self-sustaining. So that disturbance in the field is a "thing" (electron or whatever).  It doesn't continually require energy to keep it going because it is not a ripple in water (water has mass and requires energy to make it go up and down).

That wavelike disturbance can now move around and it represents the position of the electron. Actually, the square of the amplitude of the ripple represents the probability of finding the electron at that location (if I have extended your metaphor appropriately) so the electron is most likely to be in the middle but could be elsewhere. There is a really, really tiny probability that it will be detected on the other side of the galaxy.

You can then do calculations by analysing the paths of these ripples. If you add together every single possible path the ripple could take, you will find that some of the waves will constructively interfere and some will destructively interfere and the most probable path ends up being the one that corresponds to a "classical" description of the path of the electron (or photon, or whatever). This is how quantum theory, with all its probability based things, can describe the behaviour of photons and reproduce the results of classical optics. (See the Feynman lectures on QED if you haven't already.)

Such a ripple would not then be a wave but a static perturbation of the field that defined a waveform via the slope of the perturbation? This agrees with Pauli and coe for fermions, of course bosons need not meet those criteria.

[Silvestru]

Hmm this analogy is dangerous. I suspect the question: "How big is the ripple?" will follow.

[Butch]

9 minutes ago, Silvestru said:

Hmm this analogy is dangerous. I suspect the question: "How big is the ripple?" will follow.

My thinking is that the ripple would be a "well" probably absolute, with the horizon (unity in a hyperbolic)  playing a role related to Planck. Taken to infinite ends that horizon would be a right angle and represent a point source perturbation. Just speculative thinking of course. 

Edited April 30, 2018 by Butch

[Silvestru]

31 minutes ago, Butch said:

Such a ripple would not then be a wave but a static perturbation of the field that defined a waveform via the slope of the perturbation? This agrees with Pauli and coe for fermions, of course bosons need not meet those criteria.

This does not make sense. Excitation, ripple, wave it's all the same thing when talking about this subject. And what has the exclusion principle have to do with the subject?

22 minutes ago, Butch said:

My thinking is that the ripple would be a "well" probably absolute, with the horizon (unity in a hyperbolic)  playing a role related to Planck. Taken to infinite ends that horizon would be a right angle and represent a point source perturbation. Just speculative thinking of course. 

Like I said, analogies are dangerous when talking about QFT as there is still controversy on the subject and of course many unknowns. For example we don't have a theory for quantum field gravity.

I did not understand the rest of the sentence. what do you mean "well"? Also what do you mean "absolute"?

One thing I do know is that if we go further we need to be clear that there are two very different type of fields here but we are mixing-matching information about both.

 

 

To give an analogy of what you were asking is like me asking you how deep or long is the Earth's curvature of space-time? in meters preferably. Just how big is this hole?

 

Edited April 30, 2018 by Silvestru

[Butch]

4 hours ago, Silvestru said:

Like I said, analogies are dangerous when talking about QFT as there is still controversy on the subject and of course many unknowns. For example we don't have a theory for quantum field gravity.

I did not understand the rest of the sentence. what do you mean "well"? Also what do you mean "absolute"?

One thing I do know is that if we go further we need to be clear that there are two very different type of fields here but we are mixing-matching information about both.

 

 

To give an analogy of what you were asking is like me asking you how deep or long is the Earth's curvature of space-time? in meters preferably. Just how big is this hole?

OK, let's stick to fermions.

Simple answer,  it is as deep as it is wide.  Notice in your graphic,  the curvature never recurved,  if it did it would be oscillatory and would violate coe unless there were a reflective limit. The earth has mass density much less than an elementary particle.  My thought is that the well of a particle is very deep, maybe not infinitely deep,  maybe...

Certainly it is a well as deep as it gets in this universe.

Your illustration is a deformation of a 2 dimensional plane, we are actually dealing with a perturbation of a 3 dimensional field (flux density). Regardless we can represent this with a hyperbolic curve (or nearly hyperbolic depending on the "depth" of the hole.).

So you see "how big is the ripple?" is not the question, rather how great is the flux density at the point source. 

Also the earth has dimension, a particle really has no dimension. It has location defined by a single point. 

As far as Pauli goes, I was of course referring to exclusion. If two fermions that agreed with my "well" were to occupy the same Qn's the wells would merge and the result would be a single fermion,  such a merger wether possible or not is however outside the scope of this discussion... for now. 

Edited April 30, 2018 by Butch

[Strange]

5 hours ago, Butch said:

My thinking is that the ripple would be a "well" probably absolute, with the horizon (unity in a hyperbolic)  playing a role related to Planck. Taken to infinite ends that horizon would be a right angle and represent a point source perturbation. Just speculative thinking of course. 

To understand the shape of the "ripple" you would need to solve the wave equation (for the particular context). For example, you could look at the shapes of different electron orbitals in atoms, which will give you an idea of the range of possibilities.

[Mordred]

9 hours ago, Strange said:

Imagine a single ripple (centred on where the pebble went in) being self-sustaining. So that disturbance in the field is a "thing" (electron or whatever).  It doesn't continually require energy to keep it going because it is not a ripple in water (water has mass and requires energy to make it go up and down).

That wavelike disturbance can now move around and it represents the position of the electron. Actually, the square of the amplitude of the ripple represents the probability of finding the electron at that location (if I have extended your metaphor appropriately) so the electron is most likely to be in the middle but could be elsewhere. There is a really, really tiny probability that it will be detected on the other side of the galaxy.

You can then do calculations by analysing the paths of these ripples. If you add together every single possible path the ripple could take, you will find that some of the waves will constructively interfere and some will destructively interfere and the most probable path ends up being the one that corresponds to a "classical" description of the path of the electron (or photon, or whatever). This is how quantum theory, with all its probability based things, can describe the behaviour of photons and reproduce the results of classical optics. (See the Feynman lectures on QED if you haven't already.)

 

well described +1

Edited April 30, 2018 by Mordred

[geordief]

21 hours ago, Strange said:

Imagine a single ripple (centred on where the pebble went in) being self-sustaining. So that disturbance in the field is a "thing" (electron or whatever).  It doesn't continually require energy to keep it going because it is not a ripple in water (water has mass and requires energy to make it go up and down).

That wavelike disturbance can now move around and it represents the position of the electron. Actually, the square of the amplitude of the ripple represents the probability of finding the electron at that location (if I have extended your metaphor appropriately) so the electron is most likely to be in the middle but could be elsewhere. There is a really, really tiny probability that it will be detected on the other side of the galaxy.

You can then do calculations by analysing the paths of these ripples. If you add together every single possible path the ripple could take, you will find that some of the waves will constructively interfere and some will destructively interfere and the most probable path ends up being the one that corresponds to a "classical" description of the path of the electron (or photon, or whatever). This is how quantum theory, with all its probability based things, can describe the behaviour of photons and reproduce the results of classical optics. (See the Feynman lectures on QED if you haven't already.)

Can I  understand the vacuum as being a medium whose  permittivity limit approaches infinity?

 

I know that  it does have a finite number  but my natural inclination is to see it this way.

 

Is there a reason why it is finite ? Because the vacuum is never empty perhaps?

[Strange]

13 minutes ago, geordief said:

Can I  understand the vacuum as being a medium whose  permittivity limit approaches infinity?

 

I know that  it does have a finite number  but my natural inclination is to see it this way.

 

Is there a reason why it is finite ? Because the vacuum is never empty perhaps?

I think that is an interesting question. I don't think we know why vacuum permittivity and permeability have the (finite) values they do. 

[swansont]

1 hour ago, Strange said:

I think that is an interesting question. I don't think we know why vacuum permittivity and permeability have the (finite) values they do. 

I think that they are finite is much more interesting than what values they have; part of the reason for the latter is because we chose a value for the second and the meter, etc. 

[Butch]

18 hours ago, Strange said:

To understand the shape of the "ripple" you would need to solve the wave equation (for the particular context). For example, you could look at the shapes of different electron orbitals in atoms, which will give you an idea of the range of possibilities.

Not all particles are bound, I can understand the bound electron, the reflective limit is what Schroedinger is all about (Thanks again for beating that into me!). 

There are particles that are quite stable outside of atomic structure, just now a thought occurs to me... in the double slit experiment with electrons, can the wave function of the electron be determined? Does it change with the spacing of the slits? The direction of my thinking here is that the wave function might be determined by environment and that while the field perturbation is a simple well, it produces a wave function via outside influence.

[geordief]

11 minutes ago, swansont said:

I think that they are finite is much more interesting than what values they have; part of the reason for the latter is because we chose a value for the second and the meter, etc. 

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

 

 

 

 

[Strange]

5 minutes ago, Butch said:

in the double slit experiment with electrons, can the wave function of the electron be determined? Does it change with the spacing of the slits?

Calculating the wave function (and its evolution) is how the result of the experiment was calculated. Changing the slits will change the wave function because it changes the boundary conditions.

Doing something (anything) that determines which slit the wave function goes through also changes the wave function, which is why the interference pattern disappears in that context.

[Butch]

2 hours ago, geordief said:

Can I  understand the vacuum as being a medium whose  permittivity limit approaches infinity?

 

I know that  it does have a finite number  but my natural inclination is to see it this way.

 

Is there a reason why it is finite ? Because the vacuum is never empty perhaps?

Very interesting! Pretty much the crux of my question! 

Also at issue here, although it is quite convenient for us, why is c limited and constant?

4 minutes ago, Strange said:

Calculating the wave function (and its evolution) is how the result of the experiment was calculated. Changing the slits will change the wave function because it changes the boundary conditions.

Doing something (anything) that determines which slit the wave function goes through also changes the wave function, which is why the interference pattern disappears in that context.

This validates my understanding of the field excitation as an inverse square perturbation (A well)!

In other words the electronic has no wave function, only a proximity slope until it is influenced.

Edited May 1, 2018 by Butch

[Strange]

3 minutes ago, Butch said:

This validates my understanding of the field excitation as an inverse square perturbation (A well)!

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

[geordief]

3 minutes ago, Butch said:

Very interesting! Pretty much the crux of my question! 

Also at issue here, although it is quite convenient for us, why is c limited and constant?

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