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bobdavis

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In summary: The de Broglie wavelength seems a bit artificial IMO. The wavelength for a massive particle is$$\lambda = \frac h p$$where ##p = \gamma mv## is the (relativistic) momentum.As a particle has zero momentum in its rest frame, it has infinite de Broglie wavelength in its rest frame. Whether that makes physical sense or not, it doesn't allow an analysis based on simple length contraction.

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bobdavis

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Dale

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Almost. In relativity the de Broglie relations are given by $$P= \hbar K$$ where in units with ##c=1## we have ##P=(E,\vec p)## and ##K=(\omega,\vec k)##bobdavis said:Does this correspond to a dilation in the de Broglie wavelength of the muon from the perspective of the observer?

##K## transforms like a four-vector with ##\vec k## being the spacelike part. So it is more closely related to length contraction than time dilation. And of course the wavelength is the inverse of the wave vector.

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bobdavis

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I'm not sure I understand yet. Maybe a simpler question, is it the case that all observers would agree on the number of oscillations of the de Broglie wave across the spacetime interval traversed by the muon or is this observer-dependent?

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Dale

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I don’t know how this is measured or determined, so I cannot say.bobdavis said:is it the case that all observers would agree on the number of oscillations of the de Broglie wave across the spacetime interval traversed by the muon

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PeterDonis

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In any correct relativistic model of a wave, the amplitude and phase of the wave will be invariant functions on spacetime--i.e., at a given spacetime event, the wave will have an amplitude and phase that is independent of any choice of reference frame.bobdavis said:is it the case that all observers would agree on the number of oscillations of the de Broglie wave across the spacetime interval traversed by the muon or is this observer-dependent?

It should be evident that this implies that the number of cycles of any wave along a given curve in spacetime (in this case, a given segment of the worldline of the muon) will be an invariant.

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PeterDonis

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Mathematically you would look at the 1-form corresponding to the wave vector, and its inner product with the tangent vector to the muon's worldline.Dale said:I don’t know how this is measured or determined

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Dale

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So ##n=K_\mu u^\mu##?PeterDonis said:Mathematically you would look at the 1-form corresponding to the wave vector, and its inner product with the tangent vector to the muon's worldline.

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PeterDonis

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I think the inner product gives the derivative of ##n##, so to speak; the "rate" at which cycles occur along the worldline, i.e., something like ##dn / d \tau##. (Just as the wave vector itself gives frequency and wave number, not wavelength and cycle time.)Dale said:So ##n=K_\mu u^\mu##?

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It does indeed, by definition of the wave 1-form as the exterior derivative of the phase function.PeterDonis said:I think the inner product gives the derivative of n, so to speak; the "rate" at which cycles occur along the worldline

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The de Broglie wavelength seems a bit artificial IMO. The wavelength for a massive particle is$$\lambda = \frac h p$$where ##p = \gamma mv## is the (relativistic) momentum.bobdavis said:

I'm not sure I understand yet. Maybe a simpler question, is it the case that all observers would agree on the number of oscillations of the de Broglie wave across the spacetime interval traversed by the muon or is this observer-dependent?

As a particle has zero momentum in its rest frame, it has infinite de Broglie wavelength in its rest frame. Whether that makes physical sense or not, it doesn't allow an analysis based on simple length contraction.

Note that the de Broglie hypothesis, although still widely taught it seems, was superseded by modern QM in the 1920s.

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Just to work this out for an example: Take the electromagnetic field. For simplicity let's consider the four-potential to describe it. For the discussion here gauge dependence does play no significant role anyway. The four-potential transforms as a four-vector field, i.e., under Lorentz transformationsPeterDonis said:In any correct relativistic model of a wave, the amplitude and phase of the wave will be invariant functions on spacetime--i.e., at a given spacetime event, the wave will have an amplitude and phase that is independent of any choice of reference frame.

It should be evident that this implies that the number of cycles of any wave along a given curve in spacetime (in this case, a given segment of the worldline of the muon) will be an invariant.

$$A^{\prime \mu}(x')={\Lambda^{\mu}}_{\nu} A^{\nu}(x)={\Lambda^{\mu}}_{\nu} A^{\nu}(\hat{\Lambda}^{-1}x').$$

For a plane wave

$$A^{\mu}(x)=a^{\mu} \exp(\mathrm{i} k \cdot x).$$

Here ##a^{\mu}=\text{const}##, and ##k=(\omega,\vec{k})## is the wave four-vector. The above transformation laws tells us that

$$A^{\prime \mu}(x')={\Lambda^{\mu}}_{\nu} a^{\nu} \exp(\mathrm{i} k \cdot \hat{\Lambda}^{-1} x').$$

Since now

$$k \cdot \hat{\Lambda}^{-1} x' = (\hat{\Lambda} k) \cdot (\hat{\Lambda} \hat{\Lambda^{-1}} x') = k' \cdot x',$$

this indeed implies that the phase ##k \cdot x## is indeed an invariant (it's already written as a scalar, and thus that's obvious even without the above derivation). The amplitude ##a^{\mu}## transforms as components of a four-vector and thus the four-vector itself is invariant.

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If ##u^{\mu}## is the four-velocity of an observer of the wave, ##u_{\mu} k^{\mu}## is the angular frequency ##\omega=2 \pi f## of the wave as measured by the observer.PeterDonis said:I think the inner product gives the derivative of ##n##, so to speak; the "rate" at which cycles occur along the worldline, i.e., something like ##dn / d \tau##. (Just as the wave vector itself gives frequency and wave number, not wavelength and cycle time.)

The transformation behavior of the wave four-vector describes the Doppler effect and aberration. For more on relativistic waves, see

https://itp.uni-frankfurt.de/~hees/pf-faq/rela-waves.pdf

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I’d argue it is even more fundamental. For any function on spacetime or a general manifold, ie, in particular any phase function ##\phi##, the one-form ##k = d\phi## describes changes in ##\phi## around each point. With ##\dot \gamma## being the tangent vector of some curve (in particular the world line of an observer in spacetime), the change of ##\phi## with the curve parameter ##s## is given by ##d\phi/ds = d\phi(\dot\gamma) = \dot\gamma^\mu \partial_\mu \phi = \dot\gamma^\mu k_\mu##. This is by definition of the exterior derivative.vanhees71 said:If ##u^{\mu}## is the four-velocity of an observer of the wave, ##u_{\mu} k^{\mu}## is the angular frequency ##\omega=2 \pi f## of the wave as measured by the observer.

The transformation behavior of the wave four-vector describes the Doppler effect and aberration. For more on relativistic waves, see

https://itp.uni-frankfurt.de/~hees/pf-faq/rela-waves.pdf

The phase change per proper time in the case of waves in spacetime is the observed frequency (up to factors of 2 and pi which are only relevant for publication purposes according to Feynman )

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bobdavis

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PeterDonis said:In any correct relativistic model of a wave, the amplitude and phase of the wave will be invariant functions on spacetime--i.e., at a given spacetime event, the wave will have an amplitude and phase that is independent of any choice of reference frame.

It should be evident that this implies that the number of cycles of any wave along a given curve in spacetime (in this case, a given segment of the worldline of the muon) will be an invariant.

I see, thank you for the response.

It is evident to me that if phase is independent of reference frame then the number of cycles along a given curve will be independent as well.

I have two follow up questions:

1) Have there been any experiments which directly verify this frame-independence of de Broglie wave phase?

2) Does this imply that we can express a particle's half-life in a frame-independent way in terms of the average number of cycles of the de Broglie wave that will occur before particle decay?

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Vanadium 50

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The de Broglie wavelength is more of a step leading to QM than QM itself. I would go further and say it's not much of a theory as it has little predictive power. Once we had the Schroeding Equation, anout a decade later, this idea became obsolete - of historic interest more than physics.

Next, even the Schroeding Equation is non-relativistic. Taking a non-relativistic concept (either Schroeding or de Broglie) and expecting it to make sense relativisticlly (e.g. time dilation) is going to be a lesson in futility.

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bobdavis

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Is it not ok for me to ask here about what the logical consequences would be of using the de Broglie model and the special relativity model together?Vanadium 50 said:

The de Broglie wavelength is more of a step leading to QM than QM itself. I would go further and say it's not much of a theory as it has little predictive power. Once we had the Schroeding Equation, anout a decade later, this idea became obsolete - of historic interest more than physics.

Next, even the Schroeding Equation is non-relativistic. Taking a non-relativistic concept (either Schroeding or de Broglie) and expecting it to make sense relativisticlly (e.g. time dilation) is going to be a lesson in futility.

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I think you are going astray here. The wavelength of a muon is a theoretical construction that explains diffraction. Likewise its frequency (cycles per second) is not something directly measurable. In the sense that the muon is not oscillating between two states with this frequency.bobdavis said:1) Have there been any experiments which directly verify this frame-independence of de Broglie wave phase?

2) Does this imply that we can express a particle's half-life in a frame-independent way in terms of the average number of cycles of the de Broglie wave that will occur before particle decay?

The observed half-life of the muon can be calculated directly from the proper time that elapses for the muon, which in relativity is the length of its spacetime path.

One of the things that modern QM achieved was to explain experimental wave-particle duality. The de Broglie model in a sense created the wave-particle duality - which QM then explained. It was a big step in the right direction, but you shouldn't take it too literally.

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Vanadium 50

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Yes, it is not. It would be like trying to understand chemistry using only the elements earth, fire, air and water.bobdavis said:Is it not ok for me to ask here about what the logical consequences would be of using the de Broglie model and the special relativity model together?

In both cases, we've moved beyond this.

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bobdavis

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I see, so if I'm understanding correctly, basically the only measurable prediction we get from the de Broglie model is the diffraction patterns at the detector, and it would be unjustified to extrapolate this to some kind of actual sinusoid wave pattern of the muon during the interval between creation and detection.PeroK said:I think you are going astray here. The wavelength of a muon is a theoretical construction that explains diffraction. Likewise its frequency (cycles per second) is not something directly measurable. In the sense that the muon is not oscillating between two states with this frequency.

The observed half-life of the muon can be calculated directly from the proper time that elapses for the muon, which in relativity is the length of its spacetime path.

One of the things that modern QM achieved was to explain experimental wave-particle duality. The de Broglie model in a sense created the wave-particle duality - which QM then explained. It was a big step in the right direction, but you shouldn't take it too literally.

So the answer to (1) would be partly yes, in the sense that all observers would predict/measure the same diffraction patterns at the detector, and partly N/A in the sense that the model gives no measurable predictions to verify/refute in the interval between creation & detection anyway; and the answer to (2) would be yes, but only mathematically, and only because using the number of cycles of a theoretical de Broglie wave gives a roundabout way of expressing the length of a spacetime interval, and in that case we could just as well use the length of the spacetime interval directly instead of overcomplicating it with these extra considerations that go beyond the actual applicability of the de Broglie model?

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I think you're flogging an extinct horse here. The de Broglie model was a transient theory that was a major step on the way to QM; but it's not part of modern QM.bobdavis said:I see, so if I'm understanding correctly, basically the only measurable prediction we get from the de Broglie model is the diffraction patterns at the detector, and it would be unjustified to extrapolate this to some kind of actual sinusoid wave pattern of the muon during the interval between creation and detection.

So the answer to (1) would be partly yes, in the sense that all observers would predict/measure the same diffraction patterns at the detector, and partly N/A in the sense that the model gives no measurable predictions to verify/refute in the interval between creation & detection anyway; and the answer to (2) would be yes, but only mathematically, and only because using the number of cycles of a theoretical de Broglie wave gives a roundabout way of expressing the length of a spacetime interval, and in that case we could just as well use the length of the spacetime interval directly instead of overcomplicating it with these extra considerations that go beyond the actual applicability of the de Broglie model?

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bobdavis

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1) does make verifiable predictions and has a domain of applicability in which it is valid to use

2) is still used in standard introductory physics education as an educational stepping stone

So to me it's not immediately obvious why I should have thought my questions were more akin to asking about 4 element chemistry than to asking about classical electromagnetism in Newtonian physics, even though the world isn't actually Newtonian and many aspects of electromagnetism are now understood to be purely observer-dependent relativistic effects.

It may very well be the case that there are issues with my questions but just telling me that my questions are so stupid that I shouldn't even have asked them here and saying I should use "modern QM" (which, clearly since I'm asking the questions I'm not quite at that point in my education on quantum physics) doesn't tell me anything that would improve my understanding instead of just shutting down my attempt to inquire, such as: whether I'm extrapolating the model beyond the domain of applicability; whether I'm reasoning incorrectly within the domain of applicability; whether I'm reasoning correctly but incorporating unnecessary theoretical baggage (apparently all 3 of these have been the case?); whether any analogous questions could be asked in the modern QM model of the scenario that might lead to more useful insights or whether the modern QM model is such a departure that there's not even any analogous questions to be asked.

If any of these things were obvious to me ahead of time, I would have asked different questions.

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You're not staying within its domain of applicability. You want to use it to analyse and understand high-energy muon lifetimes.bobdavis said:1) does make verifiable predictions and has a domain of applicability in which it is valid to use

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Vanadium 50

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Stated another way, he is using a low-velocity approximation to study a high velocity system. And it's not working.PeroK said:You're not staying within its domain of applicability. You want to use it to analyse and understand high-energy muon lifetimes.

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Dale

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@bobdavis your questions seem fine to me

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Vanadium 50

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The question was asked, and answered. We can't make people accept the answer.

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Dale

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My hesitance above was more because I don’t know if the phase of a de Broglie wave is measurable. If not, then it would not need to be invariant. But in any case you are clearly right that since it has a wave four vector it makes sense that the phase should be an invariant also. I didn’t think it through carefullyvanhees71 said:this indeed implies that the phase k⋅x is indeed an invariant (it's already written as a scalar, and thus that's obvious even without the above derivation)

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Mister T

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You are confused because a remnant of it remains in physics, and that is simply to refer to the quantity ##h/p## as the deBroglie wavelength.

The upshot of it all is that there is no deBroglie wave, but there is a deBroglie wavelength. It's an unfortunate linguistic remnant.

Time dilation is a phenomenon in which time passes at a different rate for objects moving at different speeds. It is a consequence of Einstein's theory of relativity, and it states that time moves slower for objects moving at high speeds compared to those at rest.

Muons are subatomic particles that are created in the upper atmosphere and have a very short lifespan. Due to their high speeds, they experience time dilation, which allows them to travel longer distances before decaying. This allows muons to reach the Earth's surface, which they would not be able to do without time dilation.

The equation for time dilation is t' = t / √(1 - v^2/c^2), where t' is the time experienced by the moving object, t is the time experienced by the stationary observer, v is the velocity of the moving object, and c is the speed of light.

The de Broglie wavelength is a measure of the wave-like behavior of particles, and it is inversely proportional to their momentum. As muons travel at high speeds, their momentum increases, causing their de Broglie wavelength to decrease. This is a direct result of time dilation, as the muons experience slower time and therefore appear to have a shorter wavelength.

Time dilation and muon de Broglie wavelength are important concepts in understanding the effects of relativity and the behavior of subatomic particles. They have been experimentally verified and have practical applications in fields such as particle physics and space exploration. They also provide evidence for the validity of Einstein's theory of relativity.

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