Phase velocity and frequency of a matter wave

[Philip Koeck]

TL;DR Summary

If the total energy is proportional to the frequency the results are strange in the limit of small velocities

The relationships for matter waves are (see e.g. https://en.wikipedia.org/wiki/Matter_wave):

λ = h / p and E = h f, where E = m c²

From this the phase velocity can be derived and we get v_ph = c² / v.
v is the group velocity, which is also the velocity of the particle.

If I consider these expressions in the limit of v→0 (and p→0), I find that both λ and v_ph become infinite, whereas f remains finite.
So, a stationary particle with mass still has a finite frequency, but infinite wavelength and phase velocity!

If I, however, decide that E_kin = h f, where E_kin = m v²/2, I find that λ becomes infinite, f becomes zero and v_ph also becomes zero at the limit of v = 0.
Actually v_ph = v/2 for all non-relativistic velocities, in that case.
Somehow the latter would feel more natural.

What is the evidence or the reason that the energy in E = h f has to be m c²?
(And what is a stationary particle doing with a frequency?)

 

[hutchphd]

By "matter waves" do you mean the original concept from DeBoglie? The issue here is the rest mass energy?

 

[Philip Koeck]

By "matter waves" do you mean the original concept from DeBoglie? The issue here is the rest mass energy?

Yes, I guess so. For example electrons.
Yes the question is whether f should depend on the total energy m c² or only the kinetic energy.

 

[hutchphd]

I am not sure but I think DeBoglie included the rest energy in the frequency. Of course that will not affect energies of transition. Nor quantized angular momentum.
And it is of course absent in Schrodinger Equation.

 

[PeterDonis]

The relationships for matter waves

Depend on whether you want to use non-relativistic QM or relativistic QM (i.e., quantum field theory). These two different choices of model will give you different answers; that's really all you have discovered, and it's not a mystery at all, just a question of which model you want to use.

If you are using non-relativistic QM, relations like $E = mc^2$ are out of bounds; the rest mass of the particle is not considered part of its total energy. The "matter wave" equation would then only involve the particle's kinetic energy and you would get relationships something like the second set that you obtained.

If you are using relativistic QM, then the appropriate equation for relating energy and rest mass is not $E = mc^2$ but $E^2 = p^2 c^2 + m^2 c^4$, i.e., the correct relativistic dispersion relation for a particle of rest mass $m$. In the limit $p \to 0$, this becomes $E = mc^2$ and you get relationships something like the first set you obtained. The fact that you still have a finite frequency in this limit is simply telling you that the particle still has a finite energy: its rest mass.

 

[Philip Koeck]

Depend on whether you want to use non-relativistic QM or relativistic QM (i.e., quantum field theory). These two different choices of model will give you different answers; that's really all you have discovered, and it's not a mystery at all, just a question of which model you want to use.

So both choices of E are correct depending on the context?

 

[hutchphd]

"Correct" is a slippery term. For instance, the description that the planets are held in place by the platonic solid framework is correct for truly Keplerian analysis.
But I think in this context it is more a matter of defining the degrees of freedom that are of interest. There are no completely isolated systems, but one can usually ignore as inconsequential almost all of the rest of creation outside your system. The rest mass question falls here and as @PeterDonis points out the correct limiting forms are used. For instance a constant additive isolated term in the Hamiltonian for Schrodinger will produce no effect other than using up ink on your paper.

 

[Philip Koeck]

If you are using relativistic QM, then the appropriate equation for relating energy and rest mass is not $E = mc^2$ but $E^2 = p^2 c^2 + m^2 c^4$, i.e., the correct relativistic dispersion relation for a particle of rest mass $m$.

Just a quick comment: When I write m I mean the relativistic mass, not the rest mass.
For the rest mass I would write m₀.

 

[hutchphd]

I believe the term "relativistic" mass has been officially declared blasphemous. Probably with good reason because it leads to these confusions, and is not really useful.

 

[Philip Koeck]

I believe the term "relativistic" mass has been officially declared blasphemous. Probably with good reason because it leads to these confusions, and is not really useful.

What I mean is m = γ m₀.
With that mass E = m c² is the same as the expression given by Peter Donis.

 

[Philip Koeck]

"Correct" is a slippery term. For instance, the description that the planets are held in place by the platonic solid framework is correct for truly Keplerian analysis.
But I think in this context it is more a matter of defining the degrees of freedom that are of interest. There are no completely isolated systems, but one can usually ignore as inconsequential almost all of the rest of creation outside your system. The rest mass question falls here and as @PeterDonis points out the correct limiting forms are used. For instance a constant additive isolated term in the Hamiltonian for Schrodinger will produce no effect other than using up ink on your paper.

Is this maybe some sort of a conclusion: Whether or not I include the rest energy into E = h f only affects f and the phase velocity, but not the de Broglie wavelength.
So, no observable quantity is changed and the physics remains the same.
Is that correct?

 

[hutchphd]

What I mean is m = γ m0.

Understood. But folks here and generally in this century typically use m for m₀ and seldom use the term "relativistic mass" because it is not really useful. Just FYI.

Is this maybe some sort of a conclusion: Whether I include the rest energy into E = h f only affects f and the phase velocity, but not the DeBroglie wavelength.
So, no observable quantity is changed and the physics remains the same.
Is that correct?

I think so.

 

[Philip Koeck]

Understood. But folks here and generally in this century typically use m for m₀ and seldom use the term "relativistic mass" because it is not really useful. Just FYI.

I could blame it on the Austrian physics tradition, or maybe I'm just getting old.

 

[hutchphd]

Understood. I was taught "relativistic mass" and Einstein used it. But it is not really very useful and the Special Relativity experts (not me!) all seem to despise it.
At seventy I have already arrived at old! And indeed you are, regardless of your present age...

 

[PeterDonis]

When I write m I mean the relativistic mass, not the rest mass.

In the non-relativistic limit there is no such thing. In the relativistic case, as you point out, you end up with the same dispersion relation.

So both choices of E are correct depending on the context?

If you insist on "correct", then the relativistic equations are more "correct" because the non-relativistic equations are just approximations to the relativistic ones in the low velocity limit.

 

[Philip Koeck]

If you insist on "correct", then the relativistic equations are more "correct" because the non-relativistic equations are just approximations to the relativistic ones in the low velocity limit.

But E = m v²/2 is not an approximation of the relativistic expression, is it?

 

[PeterDonis]

But E = m v²/2 is not an approximation of the relativistic expression, is it?

It is if you consider kinetic energy only, which in the non-relativistic approximation, you are, since rest mass can't be converted to energy in the non-relativistic approximation so it drops out of the total energy that you are considering.

 

[Philip Koeck]

It is if you consider kinetic energy only, which in the non-relativistic approximation, you are, since rest mass can't be converted to energy in the non-relativistic approximation so it drops out of the total energy that you are considering.

At least to me that leads to a problem. For relativistic velocities you use E²=m₀²C⁴+p²c² in the expression for the frequency E=hf.
For non-relativistic velocities you use E=m₀v²/2.
The phase velocities you get are very different, c²/v in the first case and v/2 in the second.
How do you know when to switch from one model to the other?

There's also a follow-up thought: It seems that for matter waves f and therefore v_ph are not observables, whereas for light they clearly are.
Is that right? Or is there a way to measure f of an electron, for example?

 

[PeroK]

Or is there a way to measure f of an electron, for example?

Electron diffraction?

 

[Philip Koeck]

Electron diffraction?

Gives you the wave length, but not the frequency, or am I missing something?

 

[PeroK]

Gives you the wave length, but not the frequency, or am I missing something?

Aren't wavelength and frequency related?

 

[Philip Koeck]

Aren't wavelength and frequency related?

By the phase velocity, but that's not measurable either, or is it?

The strange situation, in my opinion is that when you switch from non-relativistic E to relativistic both f and v_ph change suddenly.
The de Broglie wavelength, however, changes smoothly.

If f and v_ph are not observable in principle I guess that's okay, though.

 

[PeroK]

By the phase velocity, but that's not measurable either, or is it?

Can that be derived from the particle's speed?

Note that diffraction is not directly measuring a wavelength but a distance from which the wavelength is inferred. It's the same for light. The wavelength (or frequency) is inferred from some measurement of energy or diffraction.

If you can infer a wavelength, why you can't infer a frequency?

 

[ergospherical]

In the relativistic region the phase velocity is $v_{\phi} = \omega/k = (E-m)/p$, i.e.\begin{align*}
v_{\phi} = \frac{\gamma - 1}{\gamma v}
\end{align*}For small $v$ expand $\gamma = (1-v^2)^{-1/2} \sim 1 + v^2/2$ (and similarly $1/\gamma \sim 1-v^2/2$) to get\begin{align*}
v_{\phi} &\sim \frac{(v^2/2)}{v}(1-v^2/2) \sim \frac{v}{2}
\end{align*}omitting terms $O(v^3)$.

 

[Philip Koeck]

In the relativistic region the phase velocity is $v_{\phi} = \omega/k = (E-m)/p$, i.e.\begin{align*}
v_{\phi} = \frac{\gamma - 1}{\gamma v}
\end{align*}For small $v$ expand $\gamma = (1-v^2)^{-1/2} \sim 1 + v^2/2$ (and similarly $1/\gamma \sim 1-v^2/2$) to get\begin{align*}
v_{\phi} &\sim \frac{(v^2/2)}{v}(1-v^2/2) \sim \frac{v}{2}
\end{align*}omitting terms $O(v^3)$.

That would sort things out, I think.
You use E - m₀c² = ħ ω = h f.

If I read it correctly v_Φ will not be larger than c at any velocity v then, it should always be between v/2 and c. Is that right.
(I would find it easier if you don't set c=1.)

So the wikipedia-text I referred to in the first post is wrong?

 

[Philip Koeck]

Can that be derived from the particle's speed?

That's exactly the problem. How is the speed related to the frequency?
In the expression E = h f, what is the E?

 

[PeroK]

That's exactly the problem. How is the speed related to the frequency?

I thought we were talking about measuring frequency? You measure the speed of the particle and measure the amount of diffraction, then infer the frequency, wavelength and phase velocity.

You seem to have developed a mental block about something here.

 

[Philip Koeck]

I thought we were talking about measuring frequency? You measure the speed of the particle and measure the amount of diffraction, then infer the frequency, wavelength and phase velocity.

You seem to have developed a mental block about something here.

To get the wavelength you simply use a grating with known periodicity and you place a detector at a certain distance from the grating and measure the distance of the first order diffraction spot.
I have no problems with that.

I don't see how to get the frequency or phase velocity though (maybe it's just me).
I can measure the speed of an electron or infer it from the accelerating potential I used, sure.
But how is the particle speed related to the frequency and phase velocity?
What is the E in E = h f or, alternatively, what is the relation between v and v_Φ?

 

[PeroK]

But how is the particle speed related to the frequency and phase velocity?
What is the E in E = h f or, alternatively, what is the relation between v and v_Φ?

All the equations are on the Wikipedia page you linked to, for example. If you measure the mass and speed of a particle, then everything else must be derivable from that. Surely?

 

[Philip Koeck]

All the equations are on the Wikipedia page you linked to, for example. If you measure the mass and speed of a particle, then everything else must be derivable from that. Surely?

Are the equations on Wikipedia right?
Post 25 in this thread suggests a different expression connecting E and f.

 

[PeroK]

Are the equations on Wikipedia right?
Post 25 in this thread suggests a different expression connecting E and f.

I think you are confused going between non-relativistic "energy", which is all kinetic, and relativistic "energy", which includes mass-energy. These are not the same quantities. One is not a low-speed approximation of the other.

 

[Philip Koeck]

I think you are confused going between non-relativistic "energy", which is all kinetic, and relativistic "energy", which includes mass-energy. These are not the same quantities. One is not a low-speed approximation of the other.

Yes, I know.
According to the Wikipedia text E = h f, where E is the total relativistic energy, including the rest mass term.
According to post 24 the relationship is E - m₀c²= h f, if I'm not mistaken.
I'm simply wondering which of the two is correct for relativistic velocities.

 

[PeroK]

Yes, I know.
According to the Wikipedia text E = h f, where E is the total relativistic energy, including the rest mass term.

This must be correct.

According to post 24 the relationship is E - m₀c²= h f, if I'm not mistaken.

I don't see that in post #24.

I'm simply wondering which of the two is correct for relativistic velocities.

It's the total relativistic energy. It must be.

 

[Philip Koeck]

I don't see that in post #24.

In post 24 it says v_Φ= ω/k, and then ω is replaced by E - mc², as I read it. (The h cancels or is set to 1, c is set to 1).

 

[ergospherical]

Particles are not waves and you should not confuse their equations - phase velocity isn't so meaningful for particles.