Matter waves (split from Photon is massless why?)

[martillo]

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.

Edited November 11, 2023 by martillo

[studiot]

40 minutes ago, martillo said:

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

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.

[swansont]

2 hours ago, martillo said:

 

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.

And why is their momentum different at the same velocity?

 

2 hours ago, martillo said:

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.

Yes. So close…

2 hours ago, martillo 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.

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.

[martillo]

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...

Edited November 12, 2023 by martillo

[joigus]

On 11/11/2023 at 9:39 PM, martillo said:

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.

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).

[martillo]

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.

[joigus]

8 minutes ago, martillo said:

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.

Ok. It's not that transversal waves must have a dispersion relation consistent with E=pc. Rather, massless waves, those that have a dispersion relation E=pc, must satisfy a transversality in the context of a gauge theory. You can prove this from Maxwell's equations in terms of the vector potential, if I remember correctly.

[martillo]

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?

[joigus]

3 hours ago, martillo said:

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?

"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". Namely: referring to a spacetime where time is observer-independent.

I don't understand your formulas, so I'l tell you what I know.

For a wave that is monochromatic (has just one frequency), the wavelength divided by the period is obviously the speed. What other speed is there? So,

\[ v_{p}=\frac{\omega}{k} \]

Is both the phase velocity and the group velocity.

More general waves have many frequencies. They're not monochromatic. When one such wave is concentrated around a particular region of space, ie, it has certain spatial "lumpiness", it is possible to prove it must be made out of a range of frequencies. It's not obvious, but not too difficult to see either, that the centre of such "lumps" move at a speed that's given by the derivative,

\[ v_{g}=\frac{d\omega}{dk} \]

In quantum mechanics, you must remember that the \( E \), \( p \) mechanical properties are related to the \( \omega \), \( k \) wave properties (frequency and wave number) by,

\[ E=\hbar\omega \]

\[ p=\hbar k \]

If you use the non-relativistic expression for kinetic energy, 

\[ E=\frac{p^{2}}{2m} \]

you get the dispersion relation,

\[ \omega\left(k\right)=\frac{\hbar k^{2}}{2m} \]

which produces a phase and group velocities,

\[ v_{p}=\frac{\omega}{k}=\frac{\hbar k}{2m} \]

\[ v_{g}=\frac{d\omega}{dk}=\frac{\hbar k}{m} \]

Now, for light waves (which are always relativistic, there's no non-relativistic approximation for photons), you have, \( c\omega=k \), which gives,

\[ v_{p}=\frac{\omega}{k}=c^{-1} \]

\[ v_{g}=\frac{d\omega}{dk}=c^{-1} \]

For relativistic matter waves, on the other hand, if you repeat these calculations, you get,

\[ v_{p}=\frac{\omega}{k}=c\sqrt{1+\frac{m^{2}c^{2}}{k^{2}\hbar^{2}}}\geq c \]

\[ v_{g}=\frac{d\omega}{dk}=\frac{1}{2\sqrt{k^{2}+\frac{m^{2}c^{2}}{\hbar^{2}}}}\left(2k\right)=\frac{c}{\sqrt{1+\frac{m^{2}c^{2}}{k^{2}\hbar^{2}}}}\leq c \]

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. This is very far from a rigorous analysis in terms of quantum fields, but it gives you an idea that the relativistic formalism implies some superluminal modes, which later we learn are non-observable, and the group velocity is the one that seems to correspond to the measurable degrees of freedom.

Quantum field have these virtual or non-observable degrees of freedom. They always do.

 

 

Edited November 14, 2023 by joigus
LateX rendering not working

[martillo]

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.

 

 

Edited November 15, 2023 by martillo

[mistermack]

Following on from my previous post, about a photon becoming trapped between two perfect mirrors, and producing an increase in mass of the system, it should be obvious that such a system could never move at the speed of light. If the system was moving at c, the photon could never travel from mirror to mirror because it would have to travel faster than c to do it. The only way anything can travel at c is in a perfect straight line. Any deviation, and you either are going slower than c overall, or have to exceed c in places to travel at c overall. 

[joigus]

8 hours ago, martillo said:

Right.

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

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.

 

 

Ok. First of all, sorry I wrote c^-1 on a couple of lines where I should have written c.

If you go over the equations I wrote you will see that, for v<<c,

\[ \sqrt{1+\frac{m^{2}c^{2}}{k^{2}\hbar^{2}}}\simeq\frac{mc}{\hbar k} \]

With this approximation,

\[ v_{p}=c\sqrt{1+\frac{m^{2}c^{2}}{k^{2}\hbar^{2}}}\simeq c\frac{mc}{\hbar k}=\frac{mc^{2}}{\hbar k} \]

\[ v_{g}=\frac{c}{\sqrt{1+\frac{m^{2}c^{2}}{k^{2}\hbar^{2}}}}\simeq\frac{c\hbar k}{mc}=\frac{\hbar k}{m} \]

So there seems to be a mismatch with the non-relativistic expressions directly obtained from \( E= \frac{1}{2}mv^{2} \). The latter being,

\[ v_{p}=\frac{\omega}{k}=\frac{\hbar k}{2m} \]

\[ v_{g}=\frac{d\omega}{dk}=\frac{\hbar k}{m} \]

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,

\[ mc^{2}\sqrt{1-v^{2}/c^{2}}\simeq mc^{2}\left(1+\frac{v^{2}}{2c^{2}}-\cdots\right)=mc^{2}+\frac{1}{2}mv^{2}-\ldots \]

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?

 

[martillo]

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.

Edited November 15, 2023 by martillo

[joigus]

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

Don't worry too much about the phase velocity. It's not very substantial (physical). It's very telling that in the relativistic case you get a phase velocity > c. Think about them as vacuum effects that don't do anything but shifting the ground state.

The group velocity is a more serious matter. It's related to the particle's actual velocity.

 

[martillo]

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...

[joigus]

2 minutes ago, martillo said:

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...

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

[martillo]

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...

[joigus]

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

[martillo]

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...

Edited November 16, 2023 by martillo

[joigus]

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,

\[ e^{⁻imc^{2}t/\hbar} \]

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

IOW, you should not get the dispersion relation,

\[ v_{p}=\frac{\omega}{k}=\frac{\hbar k}{2m} \]

But the one corrected for the rest mass,

\[ \hbar\omega=mc^{2}+\frac{1}{2}mv^{2}-\ldots=mc^{2}+\frac{\left(\hbar k\right)^{2}}{2m}-\ldots \]

\[ v_{p}=\frac{\omega}{k}=\frac{mc^{2}}{\hbar k}+\frac{\hbar k}{2m}-\ldots \]

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

\[ v_{p}=c\sqrt{1+\frac{m^{2}c^{2}}{k^{2}\hbar^{2}}}\simeq c\frac{mc}{\hbar k}=\frac{mc^{2}}{\hbar k} \]

keeping up to terms linear in \( \hbar k \)

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?

[martillo]

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.

Edited November 16, 2023 by martillo

[joigus]

3 minutes ago, martillo said:

I perfectly understood the mathematics giving those values for the phase velocity v_p in 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.

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?

[studiot]

58 minutes ago, martillo said:

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.

Isn't it ?

 

What happens to gamma as v tends to zero ?

[martillo]

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.

 

Edited November 16, 2023 by martillo

[joigus]

1 hour ago, martillo said:

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

I can't be more clear too.

 

It is not for the v_p phase velocity.

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.

 

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

v_p=mc²/v

How is that a velocity?