Matter waves (split from Photon is massless why?)

[studiot]

2 hours ago, martillo said:

It is not for the v_p phase velocity.

Actually I don't know why you are discussing phase or group or any continuous wave velocity in relation to special relativity.

SR has nothing to say about such things. It is stated in the title that it is about moving bodies ( not moving waves. ).

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

 

2 hours ago, martillo said:

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.

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.

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

 

Edited November 16, 2023 by studiot

[martillo]

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. 

Edited November 16, 2023 by martillo

[studiot]

3 hours ago, martillo said:

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

 

No, that is not the discussion.

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.

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. 

Where is vp referred to in relativity ?

What is v in your equation ?

 

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 ?

 

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.

[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.

 

Edited November 16, 2023 by martillo

[joigus]

Ok, @martillo, you must agree that the description of a mechanical problem doesn't change if we just re-define the energy by a constant shift E₀.

Right?

E-->E+E₀

Equivalent problem. Same solutions. Everything the same.

Right?

Do we agree on that?

[martillo]

3 minutes ago, joigus said:

Ok, @martillo, you must agree that the description of a mechanical problem doesn't change if we just re-define the energy by a constant shift E₀.

Right?

E-->E+E₀

Equivalent problem. Same solutions. Everything the same.

Right?

Do we agree on that?

Yes. I have no problem that in RT is defined the total energy as E = mc² + KE  for m being the "rest mass".

[studiot]

Watch these animations.

There are lots of youtube animations, this one is good because you can play with it, setting the vp and vg parameters to all possibilities to see the effect on the 'particle' carried.

https://www.surendranath.org/GPA/Waves/GroupPhase/GroupPhase.html

[martillo]

22 minutes ago, studiot said:

Watch these animations.

There are lots of youtube animations, this one is good because you can play with it, setting the vp and vg parameters to all possibilities to see the effect on the 'particle' carried.

https://www.surendranath.org/GPA/Waves/GroupPhase/GroupPhase.html

Watching the animation but I don't get how it can help in the discussion.

Note that for matter waves we have:

with the relativistic formula v_p = c²/v_g and so v_p > v_g always.

with the non-relativistic formula v_p = v_g/2 so v_p < v_g always.

 

Edited November 16, 2023 by martillo

[joigus]

15 minutes ago, martillo said:

Yes. I have no problem that in RT is defined the total energy as E = mc² + KE  for m being the "rest mass".

Then \( E=mc^{2}+\frac{p^{2}}{2m} \) once one corrects for the zero of energies by adding the rest energy, right?

[martillo]

53 minutes ago, joigus said:

Then E=mc2+p22m once one corrects for the zero of energies by adding the rest energy, right?

Something seems to not work for me.

The non-relativistic KE verifies KE = p²/(2m) but seems to me is not valid for the relativistic KE.

You are adding the constant relativistic term mc² to the non-relativistic KE. It would be a strange energy, half relativistic and half non-relativistic.

I don't know where you can go this way.

 

Edited November 17, 2023 by martillo

[joigus]

57 minutes ago, martillo said:

Something seems to not work for me.

The non-relativistic KE verifies KE = p²/(2m) but seems to me is not valid for the relativistic KE.

You are adding the constant relativistic term mc² to the non-relativistic KE. It would be a strange energy, half relativistic and half non-relativistic.

I don't know where you can go this way.

 

It's a Taylor-series expansion. Remember?:

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

Didn't you understand when I said it, or don't you understand now?

It is neither strange nor non-strange. It's a Taylor-series expansion. A Taylor-series expansion is... well. A Taylor-series expansion.

What's strange about that?

[martillo]

1 hour ago, joigus said:

It's a Taylor-series expansion. Remember?:

 

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

 

Didn't you understand when I said it, or don't you understand now?

It is neither strange nor non-strange. It's a Taylor-series expansion. A Taylor-series expansion is... well. A Taylor-series expansion.

What's strange about that?

Well, you cut the Taylor series staying with the first two non zero terms. It becomes an approximation of the relativistic energy E valid for small values of the variable v²/c² only. It is then an approximation of the relativistic energy E for v << c. It is an approximation of the relativistic energy E in the non-relativistic range.

Fine, go on.

 

[joigus]

7 hours ago, martillo said:

Well, you cut the Taylor series staying with the first two non zero terms. It becomes an approximation of the relativistic energy E valid for small values of the variable v²/c² only. It is then an approximation of the relativistic energy E for v << c. It is an approximation of the relativistic energy E in the non-relativistic range.

Fine, go on.

 

Ok. So the quantity that's "under the obligation" of giving us back the non-relativistic kinetic energy \( \frac{1}{2}mv^{2} \) is not \( E \), but \( E - mc^{2} \). That is, when \( v \) is very small as compared to \( c \), 

\[ \frac{1}{2}mv^{2}\simeq\sqrt{p^{2}c^{2}+m^{2}c^{4}}-mc^{2} \]

Do you agree so far? I just want to make it very clear that I'm not trying to pull wool over your eyes in any way.

This technique of "incremental agreement" I learned from @studiot, by the way.  

Edited November 17, 2023 by joigus
Latex editing

[martillo]

40 minutes ago, joigus said:

Ok. So the quantity that's "under the obligation" of giving us back the non-relativistic kinetic energy 12mv2 is not E , but E−mc2 . That is, when v is very small as compared to c , 

 

12mv2≃p2c2+m2c4−−−−−−−−−−√−mc2

 

I agree in the validity of that formula. E can also be expressed as E = γmc² = mc² + KE where γ is the relativistic factor (1-v²/c²)^-1/2 and so: KE = (γ -1)mc² 

Expanding γ in the Taylor series as you did and for v << c then KE = (γ -1)mc² approximates to mv²/2.

But I guess you would have a reason to consider the relation E = (p²c² + m²c⁴)^1/2 so go ahead.

40 minutes ago, joigus said:

Do you agree so far? I just want to make it very clear that I'm not trying to pull wool over your eyes in any way.

This technique of "incremental agreement" I learned from @studiot, by the way.  

Yes, I have discussed with @studiot in this step by step way other times. It takes its time but could be good to find the key point of agreement or disagreement.

 

Edited November 17, 2023 by martillo

[joigus]

59 minutes ago, martillo said:

I agree in the validity of that formula. E can also be expressed as E = γmc² = mc² + KE where γ is the relativistic factor (1-v²/c²)^-1/2 and so: KE = (γ -1)mc² 

Expanding γ in the Taylor series as you did and for v << c then KE = (γ -1)mc² approximates to mv²/2.

But I guess you would have a reason to consider the relation E = (p²c² + m²c⁴)^1/2 so go ahead.

Yes, I have discussed with @studiot in this step by step way other times. It takes its time but could be good to find the key point of agreement or disagreement.

 

Ok, let's take it from there then.

This \( \frac{1}{2}mv^{2}=p^{2}/2m \) is what should be identified with the Einstein quantum relation ℏω (Planck's constant times the angular frequency), as the time-independent Schrödinger equation says, \( \left(P^{2}/2m\right)\psi=E\psi \). Then, it's a matter of doing some algebra after plugin in De Broglie's \( p=\hbar k \),

\[ \hbar\omega\simeq\sqrt{\left(\hbar k\right)^{2}c^{2}+m^{2}c^{4}}-mc^{2} \]

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

Taylor expanding the square root in powers of the small quantity \( \frac{\hbar^{2}k^{2}}{m^{2}c^{2}} \) (as \( mc\gg\hbar k \)),

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

So that,

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

while,

\[ v_{g}=\frac{d\omega}{dk}\simeq\frac{d}{dk}\left(\frac{\hbar k^{2}}{2m}\right)=\frac{\hbar k}{m} \]

So indeed,

\[ v_{p}=\frac{v_{g}}{2} \]

Please, mind the \( \simeq \) in the next-to-the-last line.

What about \( v_{g}=c²/v_{p} \) Mind you, this is an exact identity that comes from,

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

Of course you cannot get that from the Schrödinger dispersion relation, as is clear from the fact that it has no \( c \) in it. So you're mixing an exact identity there with an approximation, which could be part of the reason of your confusion.

Does any of that address any of your concerns? Does it make sense?

[martillo]

Well, your reasoning is a bit complicated to follow for me... 

3 hours ago, joigus said:

So you're mixing an exact identity there with an approximation, which could be part of the reason of your confusion.

You arrive to two expressions for the phase velocity and say one is the exact one while the other an approximation.

The "exact" one: v_p = c²/v_g = c²/v

The "approximated" one: v_p = v_g/2 = v/2

I can understand now how you justify the difference.

I will only say that for me there is a too big difference. The same big difference as to consider E = mc² + mv²/2 or to consider just E = mv²/2 what is the original cause of the difference. So big difference as for me to consider them incompatible results. 

But you can say is just my perception. Fine, I will not try to convince you.

As for me I just will continue thinking about a better theory that could be able to solve such kind of differences between current theories. Please, just don't try to convince me to not do it. I will not give up anyway.

 

Edited November 17, 2023 by martillo

[joigus]

23 minutes ago, martillo said:

As for me I just will continue thinking about a better theory that could be able to solve such kind of differences between current theories. Please, just don't try to convince me to not do it. I will not give up anyway.

 

You can do as you please, of course.

I will just finish by telling you that a theory that fixes a problem that doesn't exist is not a better theory. The phase velocity is not an observable quantity, so any theory that "fixes" the phase velocity to conform to a particular theoretical prejudice of what it should look like is deeply misguided.

And please understand that I mean well. I don't want you to waste your time in something that I think is hopeless, and that's why I'm telling you.

[studiot]

17 hours ago, martillo said:

Something seems to not work for me.

The non-relativistic KE verifies KE = p²/(2m) but seems to me is not valid for the relativistic KE.

You are adding the constant relativistic term mc² to the non-relativistic KE. It would be a strange energy, half relativistic and half non-relativistic.

I don't know where you can go this way.

 

It would help if you took note of the steps I am offering.

 

I think (and I may be wrong about this) that you have been following the wrong exposition of De Broglie's work (1924 and 1927 papers).

Far too many (some well respected) texts offer this explanation of what he supposedly said.

 

Consider a beam of EM radiation with energy E = hf, where f is the frequency.

The wavelength. λ,  is thus λ  = w/f where w is the velocity of the wave.

For EM radiation w = c so

λ = c/f  = hc/E

 

This is all fine and dandy but

But he did not say this.

By analogy take the pair of equations

k = 1/λ = E/hc

and

f = E/h

because this false derivation does not lead to appropriate results, in particular matter waves as pilot waves, with a correctly defined group and phase velocity.

They lead to a phase velocity, w = v/k = c

and

a group velocity

g = df/dk = (dE/h) / (dE/hc)  = c

 

This is correct for a quantum which always equals c but not correct for a particle which moves with a velocity, different from c.

 

[martillo]

28 minutes ago, joigus said:

You can do as you please, of course.

I will just finish by telling you that a theory that fixes a problem that doesn't exist is not a better theory. The phase velocity is not an observable quantity, so any theory that "fixes" the phase velocity to conform to a particular theoretical prejudice of what it should look like is deeply misguided.

And please understand that I mean well. I don't want you to waste your time in something that I think is hopeless, and that's why I'm telling you.

You know, current theories are full of "non observable" things like this we are talking about in the "matter waves" and "undetectable" things like the "virtual particles" or the "dark matter". So "undetectable" that let me wonder if at the end they would actually exist at all. I'm looking for a theory without any of such kind of things. Something that would look hopeless for you, I know, but not for me. I'm considering some possibilities on some things but I can't discuss them in the forum because of the lack of some theoretical and experimental demonstrations at this time. They just look as possible for me now and I will continue working on them as I could. May be I could come here with something really demonstrable in the future, not sure, just may be...

Meanwhile, is important for me to discuss some things in the forum and I appreciate your time dedicated in our discussions. Thanks a lot.

[swansont]

20 minutes ago, martillo said:

You know, current theories are full of "non observable" things like this we are talking about in the "matter waves" and "undetectable" things like the "virtual particles" or the "dark matter". So "undetectable" that let me wonder if at the end they would actually exist at all. I'm looking for a theory without any of such kind of things.

Not being an observable is distinct from something that isn’t directly observable. And observable means it can be measured, or the resulting effect can be measured. So virtual photons can be confirmed because the model makes testable predictions. Dark matter in inferred because of its gravitational effects, which can be measured. At the end of the day, agreement with experiment is what matters.

What are the measurable effects of phase velocity?

 

[martillo]

3 minutes ago, swansont said:

Not being an observable is distinct from something that isn’t directly observable. And observable means it can be measured, or the resulting effect can be measured. So virtual photons can be confirmed because the model makes testable predictions. Dark matter in inferred because of its gravitational effects, which can be measured. At the end of the day, agreement with experiment is what matters.

What are the measurable effects of phase velocity?

 

Yes the phase velocity is totally non observable, not directly nor indirectly by any related physical effect. There's no physical effect related to it. It can't be experimentally verified anyway. But being non observable is not a justification for the different incompatible values that different approaches give to it. If we find incompatible different values something is wrong somewhere. I would not know precisely what is wrong at this time. For @joigus the different values are not incompatible but for me they are so different as to be able to consider them theoretically incompatible:

The relativistic approach: v_p = c²/v_g = c²/v

The non relativistic approach: v_p = v_g/2 = v/2

Considering small velocities v << c for both, do you consider them compatible results?

[joigus]

14 minutes ago, martillo said:

For @joigus the different values are not incompatible but for me they are so different as to be able to consider them theoretically incompatible:

 

I think you are misrepresenting what I said. One is fully relativistic (v_p=c²/v); the other (v_p=v/2) is a non-relativistic approximation.

I see now I must have done a very poor job of explaining that. I thought it was clear for you when you said,

3 hours ago, martillo said:

You arrive to two expressions for the phase velocity and say one is the exact one while the other an approximation.

 

Is e^x=1+x for small x incompatible with e^xe^-x=1? No, if instead of e^x=1+x you write \( e^{x} \simeq 1+x\) instead of just saying they are equal.

Edited November 17, 2023 by joigus
minor correction

[martillo]

1 hour ago, studiot said:

It would help if you took note of the steps I am offering.

 

I think (and I may be wrong about this) that you have been following the wrong exposition of De Broglie's work (1924 and 1927 papers).

Far too many (some well respected) texts offer this explanation of what he supposedly said.

 

Consider a beam of EM radiation with energy E = hf, where f is the frequency.

The wavelength. λ,  is thus λ  = w/f where w is the velocity of the wave.

For EM radiation w = c so

λ = c/f  = hc/E

 

This is all fine and dandy but

But he did not say this.

By analogy take the pair of equations

k = 1/λ = E/hc

and

f = E/h

because this false derivation does not lead to appropriate results, in particular matter waves as pilot waves, with a correctly defined group and phase velocity.

They lead to a phase velocity, w = v/k = c

and

a group velocity

g = df/dk = (dE/h) / (dE/hc)  = c

 

This is correct for a quantum which always equals c but not correct for a particle which moves with a velocity, different from c.

 

De Broglie formula is λ = h/p. According to both the Electromagnetic Theory of light and Relativity Theory photons do not have mass and so De Broglie theory would in principle not apply but they have moment p and so  the equation λ = h/p would be valid for photons. With p = E/c you can derive λ = hc/E and with λf = c it can be derived E = hf. Also, for photons the phase and group velocities are v_p = v_g = c and there's no problem with them.

This way De Broglie relation appear straightforward for photons.

I agree that for massive particles the De Broglie relation becomes something much more complicated...

 

31 minutes ago, joigus said:

I think you are misrepresenting what I said. One is fully relativistic (v_p=c²/v); the other (v_p=v/2) is a non-relativistic approximation.

I see now I must have done a very poor job of explaining that. I thought it was clear for you when you said,

Is e^x=1+x for small x incompatible with e^xe^-x=1? No, if instead of e^x=1+x you write ex≃1+x instead of just saying they are equal.

In which way I would be misrepresenting what you said? At the end, the two different expressions for v_p are compatible for you or not?

Edited November 17, 2023 by martillo

[swansont]

58 minutes ago, martillo said:

Yes the phase velocity is totally non observable, not directly nor indirectly by any related physical effect.

Something like when you have two possible answers for a square root. If one isn’t physical, you ignore it.

[joigus]

20 minutes ago, martillo said:

In which way I would be misrepresenting what you said? At the end, the two different expressions for v_p are compatible for you or not?

They are incompatible, only if you interpret they are both exact.

They are compatible, if you interpret one is approximate and the other is exact.

If you just say,

53 minutes ago, martillo said:

For @joigus the different values are not incompatible

Then you're misrepresenting what I said.

If you juxtapose an approximate statement and an exact one and you take both at face value, of course you would be expected to find contradictions!

x-posted with @swansont