[General] On photon momentum

John Duffield johnduffield at btconnect.com
Mon Jan 30 11:43:16 PST 2017


Chip:

 

When an ocean wave moves over another ocean wave, the curvature of the “up and over” path depends on the amplitude and wavelength of the other wave. If however all ocean waves were 1m high, the curvature of the path waves would depend only on the wavelength. Given what I said about h, when an electromagnetic wave moves through itself, the curvature of its path depends on the wavelength. So for the Dirac spinor, there’s only one wavelength where that curved path is a closed path: 

 



 

Regards

John D

 

From: General [mailto:general-bounces+johnduffield=btconnect.com at lists.natureoflightandparticles.org] On Behalf Of Chip Akins
Sent: 30 January 2017 14:31
To: 'Nature of Light and Particles - General Discussion' <general at lists.natureoflightandparticles.org>
Subject: Re: [General] On photon momentum

 

Hi John D

 

The amplitude of the wave not being the size of the wave makes sense in this context.

 

But if the electron’s mass is somehow dependent on the amplitude always being the same, then how does that relate to…

If you’re going to “wrap up” a wave into a spin ½ spinor to make a stable standing-wave standing-field particle, only one wavelength will do. 

 

Wavelength is size.  

 

So how do we equate amplitude and wavelength to make this electron with the size and mass it has in nature?

How do we show that only one wavelength will work?

 

Chip

 

 

From: General [mailto:general-bounces+chipakins=gmail.com at lists.natureoflightandparticles.org] On Behalf Of John Duffield
Sent: Sunday, January 29, 2017 5:16 PM
To: 'Nature of Light and Particles - General Discussion' <general at lists.natureoflightandparticles.org <mailto:general at lists.natureoflightandparticles.org> >
Subject: Re: [General] On photon momentum

 

Chip:

 

My thoughts? The amplitude of a wave isn’t the size of the wave.

 

Think of a seismic wave with an amplitude of 1 metre. It moves from West to East. As it does, your house shakes 1 metre to the North, then 1 metre to the South. At the same time a house 10km North shakes 10cm to the North, then 10cm to the South. A house 100 km North shakes 1cm to the North then 1cm to the South. Et cetera.  

 

Regards

JohnD

 

From: General [mailto:general-bounces+johnduffield=btconnect.com at lists.natureoflightandparticles.org] On Behalf Of Chip Akins
Sent: 29 January 2017 22:58
To: 'Nature of Light and Particles - General Discussion' <general at lists.natureoflightandparticles.org <mailto:general at lists.natureoflightandparticles.org> >
Subject: Re: [General] On photon momentum

 

Hi John D

 

Yes Planck’s constant applies to all wavelengths.  However experimental evidence and experience tell us that the transverse physical size of a wave gets smaller as the longitudinal wavelength gets smaller with energy. 

 

An opening which will allow a high frequency wave to pass through, will also completely block a significantly lower wavelength from passing.

 

So it seems that all wavelengths do not have the same physical transverse extents.

(My thoughts are that the wave extents are the wavelength / 2 pi. This seems to match the evidence and works well in the RF spectrum for system design considerations. Openings in Faraday shielding, unshielded trace lengths etc. need to be kept within a prescribed limit (fraction of a wavelength) based on the expected interfering frequency and the attenuation required.)

 

Your thoughts?

 

Chip

 

From: General [mailto:general-bounces+chipakins=gmail.com at lists.natureoflightandparticles.org] On Behalf Of John Duffield
Sent: Sunday, January 29, 2017 2:37 PM
To: 'Nature of Light and Particles - General Discussion' <general at lists.natureoflightandparticles.org <mailto:general at lists.natureoflightandparticles.org> >
Subject: Re: [General] On photon momentum

 

Chip:

 

Planck’s constant h is common to all photons regardless of wavelength. Look at those pictures of the electromagnetic spectrum. Irrespective of wavelength, the depicted amplitude is always the same. 

If you’re going to “wrap up” a wave into a spin  ½ spinor to make a stable standing-wave standing-field particle, only one wavelength will do. 

 

As for which characteristic of space, I’m not sure. Perhaps it’s something like an elastic limit.   

 

Regards

JohnD

 

From: General [mailto:general-bounces+johnduffield=btconnect.com at lists.natureoflightandparticles.org] On Behalf Of Chip Akins
Sent: 29 January 2017 14:36
To: 'Nature of Light and Particles - General Discussion' <general at lists.natureoflightandparticles.org <mailto:general at lists.natureoflightandparticles.org> >
Subject: Re: [General] On photon momentum

 

Hi John D

 

I am not understanding your idea.  Can you explain how you feel that h contributes to the specific rest mass of the electron and not some other mass value? To which characteristic of space are you referring?

 

Chip

 

From: General [mailto:general-bounces+chipakins=gmail.com at lists.natureoflightandparticles.org] On Behalf Of John Duffield
Sent: Sunday, January 29, 2017 8:25 AM
To: 'Nature of Light and Particles - General Discussion' <general at lists.natureoflightandparticles.org <mailto:general at lists.natureoflightandparticles.org> >; 'Hodge John' <jchodge at frontier.com <mailto:jchodge at frontier.com> >
Subject: Re: [General] On photon momentum

 

Chip:

 

I think the electron has the mass that it has because h is what it is, because space has a particular characteristic: 

 



 

Some people liken it to a crystal. 

 

Regards

JohnD

 

From: General [mailto:general-bounces+johnduffield=btconnect.com at lists.natureoflightandparticles.org] On Behalf Of Chip Akins
Sent: 29 January 2017 13:45
To: 'Hodge John' <jchodge at frontier.com <mailto:jchodge at frontier.com> >; 'Nature of Light and Particles - General Discussion' <general at lists.natureoflightandparticles.org <mailto:general at lists.natureoflightandparticles.org> >
Subject: Re: [General] On photon momentum

 

Hi John Hodge

 

Thank you.  I think you have made a good point here.  For diffraction to work the way it does it seems the “photon” must have momentum.

 

Hi Chandra.  

 

It seems to me that the simplest explanation of all we observe is to suspect that momentum is inherent in the motion of energy in space, and the cause for inertia.  This approach allows us to derive E=mc^2 from the circulating energy in a particle.  This would keep the particle stationary until it is acted on by an outside force. It would then also explain the property of inertia. It helps us to understand why light wants to travel a straight line unless deflected (diffracted).

 

Like John D I feel space waves as energy propagates. However unlike a water wave, which is a simple displacement of particles of mass, a wave in space is a differential displacement of a transverse wave, with one part moving one way and the other part moving in the opposite direction.  This differential displacement is what can give us part of the Chandra CTF type behavior of space.  It yields things like electric charge naturally. It also causes things like the type of confinement in elementary fermions which Albrecht talks about.

 

But in all this discussion I think we, and physics in general, have missed something important.  Space cannot be a linear medium.  Our equations generally describe space in “linear” relationships, like E=hf. But this ignores the resonant conditions which cause the specific masses of stable particles. It seems that resonances must be included in our physics before we really understand why the electron at rest is the specific mass and energy level which it possesses. I also think that once we identify and quantify the non-linear resonances of space, and their causes, we will be able to see better how all the pieces fit.

 

Hi Andrew

 

I have been able to detect EM radiation which is slower than 1Hz, so I am having a bit of trouble accepting the integer approach to the solution of quantization of waves. But I understand your example and appreciate its simplicity, and the smallest value of n could be whatever nature has chosen.

 

Chip

 

From: General [mailto:general-bounces+chipakins=gmail.com at lists.natureoflightandparticles.org] On Behalf Of Hodge John
Sent: Saturday, January 28, 2017 10:10 PM
To: Nature of Light and Particles - General Discussion <general at lists.natureoflightandparticles.org <mailto:general at lists.natureoflightandparticles.org> >
Subject: Re: [General] On photon momentum

 

I do. And it explains diffraction.

Hodge

 

On Saturday, January 28, 2017 7:12 PM, Dr Grahame Blackwell < <mailto:grahame at starweave.com> grahame at starweave.com> wrote:

 

Dear All,

[Notably Chandra & Chip],

 

I'm having a bit of a problem over this question of: 'How does a photon carry momentum'? (or similar words.)

It seems to me that in order to even beginning to address this question, one needs a clear definition of 'momentum' that's applicable to the momentum carried by a photon.

I may be looking in the wrong places (if so please advise), but the only definitions of momentum that I can find either refer to 'mass' or refer to some other phenomenon which in turn refers to momentum - i.e. circular references.

If I'm going to figure, or be persuaded, how a photon carries momentum I first need to know what momentum IS in respect of a photon (yes, I know it's E/c, that's a measure it's not a definition).

Of course I'm aware of the paper "Light is heavy", but I don't feel it's appropriate just to extract from that some sort of mass-equivalence of a photon.  If we do, we get the result that 'm'=E/c^2, so 'm'c = E/c - gives the right result, but appears to be some sort of convoluted self-confirmation (i.e. a circular argument dressed up in fancy clothes).  It certainly doesn't DEFINE a photon's momentum, just evaluates it.

 

Does anyone have a convincing definition of momentum that's applicable to a photon?  One that can be used as a firm basis for theorising?

(I'd be glad if colleagues didn't use this as an excuse to yet again present their own personal theory/model - I'm looking for a definition that would be agreed by all, or at least most, physicists.)

 

Thanks in anticipation,

Grahame


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