martillo

1 hour ago, studiot said:

Where is vp referred to in relativity ?

Nowhere. RT does not treat "matter waves" but "matter waves" also work for relativistic particles which have considerable high velocities when the non-relativistic equations do not work. We are discussing the "matter waves" comparing both cases, the relativistic and the non-relativistic ones. The problem is that there is a point where both approaches appear to be incompatible.

1 hour ago, studiot said:

What is v in your equation ?

Velocity v is the velocity of the particles. The group velocity of the associated "matter wave" is v_g = v. The phase velocity depends in the approach:

Non-relativistic approach: v_p = v/2

Relativistic approach: v_p = c²/v

They are not compatible. As v tends to zero the first approach gives v_p tending to zero while the second approach gives v_p tending to infinite.

1 hour ago, studiot said:

You introduced matter waves.

What is the dependent variable in such a wave and what variables does it depend on ?

I asked you about pilot waves do you know what they are ?

I'm considering the De Broglie "matter waves" as the "pilot wave" associated to a particle. The variables are the mass and the velocity of the particle.

1 hour ago, studiot said:

You have entirely missed the point I was making which is

Quantum waves cover all space at any one time and vary in time.

As such they do not 'travel' anywhere.

"Matter waves" have phase velocity v_p and group velocity v_g  and v_g = v.

You are entering the discussion now. If you had read more carefully the earlier posts I wouldn't have to explain all that.

On 11/11/2023 at 6:28 PM, swansont said:

On 11/11/2023 at 1:55 PM, martillo said:

Everything seems to work fine if you define p = E/c for both massive or massless particles except that the concept of mass is then just omitted.

A particle at rest has energy, so I don’t see how that would work. A particle at rest has a momentum of mc?

I was wrong: p = E/c works for massless particles only like the photon.

On 11/11/2023 at 6:28 PM, swansont said:

DeBroglie wavelength of an electron and proton moving at the same speed is not the same. Mass definitely play a role.

You are right. Mass plays a role in the "matter waves".

51 minutes ago, joigus said:

I'm in the middle of answering you, but first please clarify this:

v_p=mc²/v

How is that a velocity?

My mistake. I should have written v_p = c²/v in the relativistic case.

43 minutes ago, studiot said:

I had though your worry was that Lorenz formulae in general and gamma in particular did not reduce to newtonian mechanics as the velocity of a moving body reduces.

No, that is not the discussion.

43 minutes ago, studiot said:

Photons are not bodies in the mechanical sense any more than quantum probability waves are continuous travelling waves.

We are not discussing about photons at this point. It's about massive particles and their associated "matter waves" in both the relativistic and the non-relativistic case.

43 minutes ago, studiot said:

Pilot waves, although an interesting variation, have not been substantiated experimentally.

As @joigus says the phase velocity is considered a non observable magnitude, is not measurable. I have already seen the "matter waves" said to be not measurable in other places. 

1 hour ago, joigus said:

And it is, if only you just admit that in non-relativistic mechanics you can redefine the spectrum of energy to include this new "energy at zero velocity". It is. It is a particular case.

What the non-relativistic approach cannot give you is the mc² term, for obvious reasons. It cannot give you a formula that's wrong. It only slightly generalises it to be right, ie, to include the rest energy. Haven't you noticed that the second term exactly coincides with Schrödinger's term?

Am I not being clear?

You are being very clear. You don't need to continue repeating your viewpoint. 

I can't be more clear too.

45 minutes ago, studiot said:

Isn't it ?

It is not for the v_p phase velocity.

45 minutes ago, studiot said:

What happens to gamma as v tends to zero ?

Gamma tends to 1, so what? 

What happens to the relativistic phase velocity v_p when v tends to zero? Does it tends to the non-relativistic v_p? It doesn't. That is the point.

SUMMARYZING:

As I have already posted several times, the phase velocity v_p of the waves associated to massive particles is defined as (simple way): 

relativistic case: v_p = mc²/v

non-relativistic case: v_p = v/2

(Where v is the velocity of the particle coincident with the v_g group velocity of the waves.)

They give incompatible results. They are incompatible definitions.

More clear impossible.

4 hours ago, joigus said:

I insist: There's a physical reason not to expect the Taylor expansion to recover the Schrödinger dispersion relation. And that's the humongous rest mass factor in the phase of the wave function. That is,

 

e⁻imc2t/ℏ

 

which the Schrödinger theory cannot fathom. This "shifts" the Schrödinger dispersion relation.

IOW, you should not get the dispersion relation,

 

vp=ωk=ℏk2m

 

But the one corrected for the rest mass,

 

ℏω=mc2+12mv2−…=mc2+(ℏk)22m−…

 

 

vp=ωk=mc2ℏk+ℏk2m−…

 

And indeed, that's what you get, if you Taylor expand what I showed you before,

 

vp=c1+m2c2k2ℏ2−−−−−−−−√≃cmcℏk=mc2ℏk

 

keeping up to terms linear in ℏk

That is what I can live with. Namely: an otherwise Schrödinger-compatible dispersion relation that only corrects for the rest-mass term.

Can't you?

I perfectly understood the mathematics giving those values for the phase velocity v_p in both the non-relativistic and the relativistic cases. The point is that I cannot live with their incompatibility: the non-relativistic case should be a particular case of the relativistic one just considering v << c.

15 hours ago, joigus said:

I can live with that. The phase velocity of a Schrödinger wave is not an observable.

I can't...

I can't live with things like that.

A totally new theory must surge. No other way. Impossible to "weld" current ones. 

But too much things to question, too much things to demonstrate, a huge task...

4 minutes ago, joigus said:

Why? Unless you're doing gravitation, the zero of energy is immaterial, and all that matters is differences of energy.

Is just in v_p where a problem arises. It should be solved someway...

3 minutes ago, joigus said:

LOL. And again I made a mistake when LateXing the answer. I guess you caught me. The results are OK I think.

The root square is a quotient, no pb.

All the problem I think is in the different definition for the energy E used in each case:

For the non-relativistic case E = KE.

For the relativistic case E = mc² + KE.

Just one definition should be used...

5 hours ago, joigus said:

In particular, it's the phase velocity that is to blame. I understand that's your point. Is it. not?

Well there's a physical reason for that. All these expressions come from the assumption that it's a free particle we're talking about. Einstein's celebrated formula for the kinetic energy,

 

mc21−v2/c2−−−−−−−−√≃mc2(1+v22c2−⋯)=mc2+12mv2−…

 

gives a humongous rest-energy term mc². This enormous energy shift is a constant, so it is of no effect when it comes to energy considerations, but produces a contribution to the frequency that kicks the Schrödinger dispersion relation out of whack.

Is something like that what has you worried?

Yes, that is exactly the point.

Returning to the basic relation p = E/v_p we have for the phase velocity v_p = E/p where:

For the non-relativistic case E = KE and when v goes to zero v_p = E/p goes to zero.

For the relativistic case: E = mc² + KE and when v goes to zero v_p = E/p goes to infinite.

Then, for v_p, the non-relativistic case is not just the particular case of the relativistic one when v << c.

1 hour ago, joigus said:

"Classical" is an adjective that people use to mean "as opposed to quantum". I don't think you want to say "classical" there, but "non-relativistic".

Right.

1 hour ago, joigus said:

I don't understand your formulas

They are derived (quite at the end) in the paper I have already mentioned: https://as.nyu.edu/content/dam/nyu-as/as/documents/silverdialogues/SilverDialogues_Peskin.pdf

1 hour ago, joigus said:

So the phase velocity is generally greater than c, while the group velocity is less than c. That's what I meant when I said that the relativistic approach is apparently paradoxical.

Yes that sounds strange for me too. Not actually my point:

I'm considering that the non-relativistic approach is widely said to be a particular case of the relativistic approach obtained for low velocities (v << c) but this is not the case for the"matter waves" we are talking about. The phase velocity in the non-relativistic case cannot be derived from the relativistic case just considering v << c.

Quantum Mechanics predicts (as derived in the paper I mentioned) for the relativistic case a phase velocity = w/k = c²/v while for the non-relativistic case a phase velocity = w/k = v/2 which cannot be derived form the first just by the approximation v << c. I mean, in the non-relativistic case the phase velocity goes to zero with the velocity v going to zero while in the relativistic case it goes to infinite.

I don't know if this is just another "paradox" that could actually be explained someway. As for now it looks for me like an incompatibility that surges while trying to reconcile Relativity with Quantum Mechanics.

But may be we could be going out of topic with this now so don't worry too much about.

7 hours ago, swansont said:

Energy is conserved, so the total energy of the system remains constant. Any decrease in PE would show up as an increase in KE. There is no change in the binding energy. The minimum energy of the system at annihilation is the ground state energy of positronium.

Right. The total energy (PE plus KE) is then conserved and would be negligible in the case of the positronium.

I have made some calculations giving a total energy of about few 7.2 ev much less than the relativistic energy of the pair (2xmc²) about 1 Mev.

4 hours ago, joigus said:

Relativistic wave equations, eg, have still another different dispersion relation that's apparently paradoxical (if one tries to interpret it in terms of a 1-particle theory).

What is difficult to grasp for me is that the classical approach is said to be a particular case of the relativistic approach obtained for low velocities (v << c). In the case of the relativistic wave associated to a particle the phase velocity is v_p = c²/v while in the classical approach is v_p = v/2 what cannot be obtained from the first one for v << c. The two approaches seem to be incompatible. Is that what you mean apparently paradoxical?

1 hour ago, joigus said:

Here I explained how that is not the case in terms of frequency (inverse period and proportional to energy) and wave number (inverse wavelength and proportional to momentum):

Different waves have different dispersion relations. Relativistic wave equations, eg, have still another different dispersion relation that's apparently paradoxical (if one tries to interpret it in terms of a 1-particle theory).

Right. The article in the link I provided says the same but I wrote it wrongly.

Quite all the waves (except transversal ones) can be equated in the form p = E/v_p where v_p is the phase velocity of the waves. I wrote just v which is the velocity of the particle corresponding to the group velocity of the waves. They are not the same.

2 hours ago, Engineeer said:

Move those somewhere in business or ethics

I'm not a moderator to be able to move posts to other threads.

16 minutes ago, Engineeer said:

You can't even define light. If you could find where I did in my comments you'd have the edge on me in physics. I only know the graph to define when and how in my untorturable mind. 

 

??? I don't get the point. Seems out of the discussion in this thread.

Could this be actually an answer for another thread?

6 hours ago, swansont said:

The positronium ground state energy is known, which allows for some knowledge of the potential energy.

From Wikipedia: "An electron and positron orbiting around their common centre of mass. An s state has zero angular momentum, so orbiting around each other would mean going straight at each other until the pair of particles is either scattered or annihilated, whichever occurs first. This is a bound quantum state known as positronium."

I think that for "ground state" you mean the possible configuration (considering even their spins) of minimum energy between them. They would stay orbiting for some time and they do not annihilate at this state. The state is unstable and the attractive force between the electron and the positron accelerates them towards each other until the annihilation takes place at some distance less than the orbiting radius.

Wikipedia article says the orbiting radius and binding energy can be roughly estimated through an analogy to the hydrogen atom but this is not the binding energy precisely at the annihilation which is what I'm referring to. The annihilation distance is less than the orbiting radius and the binding energy at the annihilation is higher than that while orbiting.

The binding energy increments as the distance diminishes and they both seems to remain still unknown at the final state of the pair real annihilation.

No, it is about the Electric Potential Energy between the electron and the positron which depends in their distance: kq₁q₂/r.

It does exist but the problem is that the distance r at which the annihilation takes place is unknown and the Heisenberg's uncertainty makes it uncertain so actually we cannot determine it. What I observe is that the energy tends to infinite as the distance r tends to zero.

Just for instance, if the annihilation distance would be about the Compton-wavelength of the electron (2.42631x10^-12 m) then such energy would be about 6x10^-17 Joules = 375 ev (electron-volts) what would be about a thousandth of the mass-energy of the electron (mc² approximately 500 Kev ) and would be negligible. 

Actually the annihilation distance and the associated Electric Potential Energy are unknown. Nowhere is mentioned about them.

4 hours ago, studiot said:

I would help you to think about what is the meaning of momentum.

Momentum implies the exertion of a classical force.

A simple classical (wave) explanation as to why light exerts a force when it impinges upon a massive body can be had direct from maxwell and the orientation of the E and B fields of the wave.
The wave exerts a Lorenz force on the body at it impinges.

But force is rate of change of momentum or momentum is the time integral of force.

Yes, I do understand how an electromagnetic wave is able to infringe a force and so how it can carry momentum while being a massless propagating "perturbation" of the E/M field.

Actually my problem was in the relativistic approach to photons but I think I was already clarified now why photons have zero mass also in RT. 

2 hours ago, swansont said:

And you lost it.

If it plays a role, it plays a role. It’s not omitted. It just isn’t explicitly written out, because it’s simple to write “p” than the equation for it.

I meant by "omit" exactly what you say but seems a misspelling could take place and so I must not call it that way.

I hope that, with the answers I have received to my questionings, my participation in the thread could have also helped the OP in better understanding the photons...

1 hour ago, swansont said:

A particle at rest has energy, so I don’t see how that would work. A particle at rest has a momentum of mc?

I have already admitted I was wrong in generalizing for massive particles.

1 hour ago, swansont said:

DeBroglie wavelength of an electron and proton moving at the same speed is not the same. Mass definitely play a role.

In the wave-like behavior actually the momentum p appears, not the mass. De Broglie wavelength is actually defined by their momentum p which is different for them (λ = h/p) at same velocity.

But for massive particles there is a relation between the momentum p and the mass m... May be I should rethink about what I have said. The point is that in general, waves are not associated to a mass but De Broglie law precisely combines both. Definitely I must rethink about...

Better to say that the wave-like behavior also omits the concept of mass although it plays a role in the determination of the wavelength.

12 minutes ago, KJW said:

As for why a photon is massless, (IIRC) this has something to do with the Higgs field and the symmetry-breaking of the four electroweak bosons.
 

Nothing simple...

25 minutes ago, joigus said:

No. Each wave has a different dispersion relation. E=pc gives you a photon D.R. Matter waves are a different matter. 😬

In the link I posted after, quite all the waves relate their energy and momentum in the form p = E/v excepting for transversal waves only.

2 minutes ago, joigus said:

Sorry. A c⁴ should be c² inside the square root of what I wrote there.

No problem for me.

51 minutes ago, martillo said:

Electromagnetic waves would be massless so p = E/c would hold for them.

With analogy, just a question: Would the definition p = E/v apply to any other wave like sound and water waves?

I have already found a good answer to my question at: https://as.nyu.edu/content/dam/nyu-as/as/documents/silverdialogues/SilverDialogues_Peskin.pdf

Seems the exception is on pure transversal waves which do not propagate. Would be the case of standing waves.

2 hours ago, joigus said:

No, it's not E=|p|c for massive particles. As @MigL said, it's,

for massive particles. So E=|p|c1+m²c4|p|2−−−−−−−√ . With this factorisation maybe it's clearer how and why it's not the same for massive particles?

Right.

The general equation to be applied ever is E² = (Mc²)² + (pc)² and for massive particles p = mv = γMv while for massless ones, like the photon, the factor γ = (1 - v²/c²)^-1/2 is undefined for v = c and so must be defined p = E/c.

Electromagnetic waves would be massless so p = E/c would hold for them.

With analogy, just a question: Would the definition p = E/v apply to any other wave like sound and water waves?

1 hour ago, swansont said:

Why on earth would the momentum change if you observed wave vs particle behavior?

p = E/c applies to both

 

Everything seems to work fine if you define p = E/c for both massive or massless particles except that the concept of mass is then just omitted.

In the "waves-like" behavior mass does not play any role. Waves would not have mass. In the concept of "particles-like" behavior frequency has no meaning. There's no frequency associated to any particle in the Standard Model of particles.

I think Relativity Theory is about the "particles-like" behavior. When we begin to associate the frequency f for the energy E = hf we are entering in "Quantum Mechanics" (QM) and so we would be talking about "Relativistic Quantum Mechanics" (RQM), is that right?

Everything would work fine while not entering in the concept of mass.

By the way, with Relativity Theory we would need to talk about "longitudinal" and "transversal" masses, isn't it?

10 minutes ago, joigus said:

 

|p|=E/c

 

with,

 

E=hν

 

with h Planck's constant, and |nu being the photon's frequency in Hertzs.

Direction of momentum given by,

 

p|p|=k

 

So, for massless objects you define p = E/c and define E = pc. Wouldn't that be a cyclic indetermination?