KJW Posted November 17, 2023 Posted November 17, 2023 1 hour ago, swansont said: What are the measurable effects of phase velocity? But isn't it true that λf = vp? Are you suggesting that neither λ nor f are measurable? I do accept that the phase of a De Broglie wave is unobservable, but it is not clear to me why vp is not at least indirectly measurable.
martillo Posted November 17, 2023 Author Posted November 17, 2023 (edited) 14 minutes ago, KJW said: But isn't it true that λf = vp? Are you suggesting that neither λ nor f are measurable? I do accept that the phase of a De Broglie wave is unobservable, but it is not clear to me why vp is not at least indirectly measurable. No, λf = vg the group velocity not vp the phase velocity. Edited November 17, 2023 by martillo
KJW Posted November 17, 2023 Posted November 17, 2023 1 hour ago, martillo said: 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: vp = c2/vg = c2/v The non relativistic approach: vp = vg/2 = v/2 Considering small velocities v << c for both, do you consider them compatible results? It seems to me that the non-relativistic approach is simply incorrect. For De Broglie waves of a non-zero mass particle, one considers wavelength, but I do not recall anyone considering frequency in a non-relativistic context. The frequency of a De Broglie wave is directly proportional the energy of the particle, and this energy quite simply includes the energy-equivalent of the mass. Thus, the relativistic approach is the only correct approach.
martillo Posted November 17, 2023 Author Posted November 17, 2023 (edited) 10 minutes ago, KJW said: It seems to me that the non-relativistic approach is simply incorrect. For De Broglie waves of a non-zero mass particle, one considers wavelength, but I do not recall anyone considering frequency in a non-relativistic context. The frequency of a De Broglie wave is directly proportional the energy of the particle, and this energy quite simply includes the energy-equivalent of the mass. Thus, the relativistic approach is the only correct approach. But the relativistic approach matches with the non-relativistic approach at very small velocities v << c. The problem in the discussion here in this thread is that the phase velocity of the De Broglie "matter wave" does not verify this. At very small velocities the relativistic and the non relativistic values for vp should be the same and is not the case. Edited November 17, 2023 by martillo
KJW Posted November 17, 2023 Posted November 17, 2023 14 minutes ago, martillo said: No, λf = vg the group velocity not vp the phase velocity. No, λf does not equal the group velocity. vg = dω/dk whereas λf = ω/k = vp.
joigus Posted November 17, 2023 Posted November 17, 2023 13 minutes ago, KJW said: but I do not recall anyone considering frequency in a non-relativistic context. Monochromatic-wave solutions of the non-relativistic Schrödinger equation.
martillo Posted November 17, 2023 Author Posted November 17, 2023 (edited) 7 minutes ago, KJW said: No, λf does not equal the group velocity. vg = dω/dk whereas λf = ω/k = vp. Seems right... So it would be observable afterall... Edited November 17, 2023 by martillo
KJW Posted November 17, 2023 Posted November 17, 2023 3 minutes ago, martillo said: You are wrong. Is the inverse. vg = ω/k whereas vp = dω/dk. You need to recheck this.
swansont Posted November 17, 2023 Posted November 17, 2023 34 minutes ago, KJW said: But isn't it true that λf = vp? Is it? Under what conditions do you have phase velocity ≠ group velocity? 34 minutes ago, KJW said: Are you suggesting that neither λ nor f are measurable? I do accept that the phase of a De Broglie wave is unobservable, but it is not clear to me why vp is not at least indirectly measurable. There are observable quantities whose product is not physically meaningful.
martillo Posted November 17, 2023 Author Posted November 17, 2023 3 minutes ago, KJW said: You need to recheck this. I did. I apologize. I was editing the post when you posted. So the phase velocity would be an observable afterall...
joigus Posted November 17, 2023 Posted November 17, 2023 1 minute ago, martillo said: So the phase velocity would be an observable afterall... Really? Please, tell me of an experiment to measure the phase velocity of the wave function of an electron.
martillo Posted November 17, 2023 Author Posted November 17, 2023 20 minutes ago, joigus said: Really? Please, tell me of an experiment to measure the phase velocity of the wave function of an electron. Ok, it is not directly observable but it is indirectly measurable. That is what I was thinking...
swansont Posted November 17, 2023 Posted November 17, 2023 2 minutes ago, martillo said: Ok, it is not directly observable but it is indirectly measurable. That is what I was thinking... No, it’s not observable and not measurable.
joigus Posted November 17, 2023 Posted November 17, 2023 3 minutes ago, martillo said: Ok, it is not directly observable but it is indirectly measurable. That is what I was thinking... How?
martillo Posted November 17, 2023 Author Posted November 17, 2023 (edited) 6 minutes ago, swansont said: No, it’s not observable and not measurable. 6 minutes ago, joigus said: How? Through the formulas we were discussing about! vp would be equal to c2/v or v/2 just depending on our approach. We just need to have the velocity v of the particle. Edited November 17, 2023 by martillo
KJW Posted November 17, 2023 Posted November 17, 2023 5 minutes ago, martillo said: Through the formulas we were discussing about! vp would be equal to c2/v or v/2 just depending on our approach. We just need to have the velocity v of the particle. Or alternatively, by measuring the frequency and wavelength by some form of diffraction experiment.
martillo Posted November 17, 2023 Author Posted November 17, 2023 (edited) 6 minutes ago, KJW said: Or alternatively, by measuring the frequency and wavelength by some form of diffraction experiment. Well, to measure and verify the wavelength of electrons you have the Davisson-Germer experiment... Edited November 17, 2023 by martillo
KJW Posted November 17, 2023 Posted November 17, 2023 2 minutes ago, martillo said: Well, to measure the wavelength of electrons you have the Davisson-Germer experiment... I don't know if the frequency is currently measurable, but in principle at least, one could use time crystals.
joigus Posted November 17, 2023 Posted November 17, 2023 11 minutes ago, martillo said: Through the formulas we were discussing about! vp would be equal to c2/v or v/2 just depending on our approach. We just need to have the velocity v of the particle. 5 minutes ago, KJW said: Or alternatively, by measuring the frequency and wavelength by some form of diffraction experiment. So if I have a "theory" that relates the size of the universe with the size of my nose --never mind how crazy that theory is--, when I'm measuring the size of my nose, can I claim to be measuring the size of the universe? 3 minutes ago, martillo said: Well, to measure the wavelength of electrons you have the Davisson-Germer experiment... Electron diffraction also measures electron wavelengths. How do you measure both wavelenght and frequency? --Thereby phase velocity. Huh? I'll be patiently waiting for any of your answers. This should be fun.
martillo Posted November 17, 2023 Author Posted November 17, 2023 (edited) 8 minutes ago, joigus said: So if I have a "theory" that relates the size of the universe with the size of my nose --never mind how crazy that theory is--, when I'm measuring the size of my nose, can I claim to be measuring the size of the universe? Well, if you have that crazy theory accepted as a valid one by the mainstream you can make the claim, yes. Not sure about the frequency. Wouldn't it be deduced from the wavelength and the velocity? Edited November 17, 2023 by martillo
joigus Posted November 17, 2023 Posted November 17, 2023 5 minutes ago, martillo said: Not sure about the frequency. Wouldn't it be deduced from the wavelength and the velocity? The particle's velocity is the group velocity. What laboratory measurement gives you the phase velocity? Didn't I say this before? On 11/14/2023 at 11:46 PM, joigus said: and the group velocity is the one that seems to correspond to the measurable degrees of freedom. And didn't you? On 11/14/2023 at 6:00 PM, martillo said: I wrote just v which is the velocity of the particle corresponding to the group velocity of the waves. They are not the same.
martillo Posted November 17, 2023 Author Posted November 17, 2023 9 minutes ago, joigus said: The particle's velocity is the group velocity. What laboratory measurement gives you the phase velocity? Didn't I say this before? And didn't you? Right. I'm getting a little confused, give me some time...
KJW Posted November 17, 2023 Posted November 17, 2023 12 minutes ago, joigus said: So if I have a "theory" that relates the size of the universe with the size of my nose --never mind how crazy that theory is--, when I'm measuring the size of my nose, can I claim to be measuring the size of the universe? A lot of science is based on indirect measurements based on some theory. For example, in chemistry, do we really have direct knowledge of chemical structure? Perhaps the most direct measurement of a molecule's structure is through x-ray crystallography, but even that involves the theory of x-ray diffraction along with dealing with Fourier transforms.
joigus Posted November 17, 2023 Posted November 17, 2023 2 minutes ago, KJW said: A lot of science is based on indirect measurements based on some theory. For example, in chemistry, do we really have direct knowledge of chemical structure? Perhaps the most direct measurement of a molecule's structure is through x-ray crystallography, but even that involves the theory of x-ray diffraction along with dealing with Fourier transforms. This sounds all very reasonable, and you do have a point. Nevertheless... We don't have a handle on the wave function itself. X-crystallography and the like is based on phase differences. Interference positions --and thereby wavelengths--... pretty much the same. No way to see the phase itself though. In fact, I'm working with some advantage here. I happen to know there is a very robust principle of physics --the gauge invariance principle-- that tells us it is impossible to know what that phase might be, as I can always gauge away any phase prescription that you take by locally re-defining the phase. So I have a pretty solid understanding of why what you claim cannot be true. Gimme any phase you like and I will "gauge it away" without breaking any known rules of quantum field theory, or classical electromagnetism, etc.
studiot Posted November 17, 2023 Posted November 17, 2023 3 hours ago, martillo said: 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 vp = vg = 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... Really ? Have your read his paper ? Then how do you explain this ? Taken directly from his paper. 1 hour ago, KJW said: For example, in chemistry, do we really have direct knowledge of chemical structure? Indeed we do. Here is a directly measured electron density map of hexamethylbenzene (after Spice 1964) I think it shows the structure rather well.
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