Quetzalcoatl

Yes, I realize that, but I'm still going to give it a try. The mathematics of it looks similar to a symmetry, but one in phase space. I'm going to try to reformulate this using a simple relativistic Lagrangian. I'm also going to look at the resources you'd provided above, to see what others did in phase space in perhaps similar situations. No promises of course, but I'll keep posting as this evolves and as I find time to work on this... As an aside, thinking again about the Hamiltonian commutator, for the Fractional FT generator I might try using the formula: [latex]genrator=\frac{d}{d\alpha}\mathfrak{F}(\alpha)\mid _{\alpha=0}[/latex] Who knows, it might just end up being the regular FT, the same way as i is the generator of the U(1) group: [latex]i=\frac{d}{d\alpha}\mathfrak{e^{i\alpha}}\mid _{\alpha=0}[/latex]

The key point in the idea is to have q & d/dq on the same footing in the Lagrangian, and that the action is invariant under FT of the q & d/dq operators, meaning if we FT all wave functions and operators, the physics (and action) shouldn't change. Since there is also a Fractional FT, with a rotation parameter [latex]\alpha[/latex], also under that transformation of all wave functions and operators, the physics (and action) shouldn't change. I don't know how to do the calculations in Lagrangian mechanics, so I tried Hamiltonian mechanics instead. In Hamiltonian mechanics we have q & p coordinates in phase space. In QM we promote them to Q & P operators. There is a relation between Q & P: [latex]P=\mathfrak{F}^{-1}(\hbar Q)\mathfrak{F}[/latex] [latex]\hbar Q=\mathfrak{F}P\mathfrak{F}^{-1}[/latex] where [latex]\mathfrak{F}[/latex] is the FT. (taken from http://gregnaber.com/wp-content/uploads/Simple-Harmonic-Oscillator.pdfon page 154) This looks similar for example to the relation between [latex]\vec{e_x}[/latex] & [latex]\vec{e_y}[/latex]: [latex]\vec{e_y}=R_{\frac{\pi}{2}}[\vec{e_x}][/latex] The idea is that FT is a symmetry of the Hamiltonian operator. For that we demand that the commutator of FT with the Hamiltonian is zero. For example, for a free particle, we have: [latex]H=\frac{P\cdot P}{2m}[/latex] But we also demand it be invariant under a 90 degree rotation of the x-y coordinates, R: [latex]H=\frac{R_{\frac{\pi}{2}}P\cdot R_{\frac{\pi}{2}}P}{2m}[/latex] Similarly, we want to demand invariance under a FT (90 degree rotation of the q-p coordinates): [latex]H'=\mathfrak{F}H\mathfrak{F}^{-1}=\mathfrak{F}\left(\frac{P\cdot P}{2m}\right)\mathfrak{F}^{-1}=\mathfrak{F}\left(\frac{\mathfrak{F}^{-1}(\hbar Q)\mathfrak{F}\cdot\mathfrak{F}^{-1}(\hbar Q)\mathfrak{F}}{2m}\right)\mathfrak{F}^{-1}=\frac{(\hbar Q)^2}{2m}[/latex] [EDIT: The following is wrong. Fractional FT generator should commute with H...] The FT operator is a symmetry of the mechanics if it commutes with H: [latex]\left[\mathfrak{F},H\right]\psi=\mathfrak{F}H\psi-H\mathfrak{F}\psi=\mathfrak{F}\frac{P^2}{2m}\psi-\frac{P^2}{2m}\mathfrak{F}\psi=\frac{1}{2m}\left(\mathfrak{F}P^2-P^2\mathfrak{F}\right)\psi[/latex] So I need to show that: [latex]\mathfrak{F}P^2-P^2\mathfrak{F}=0[/latex] Calculating both terms: [latex]\mathfrak{F}P^2\hat{\psi}=\mathfrak{F}\left[p^2\hat{\psi}(p)\right]=+\frac{d^2\psi(-q)}{dq^2}[/latex] [latex]P^2\mathfrak{F}\hat{\psi}=P^2\left[\psi(-q)\right]=P\left[-i\frac{d\psi(-q)}{dq}\times -1\right]=-\frac{d^2\psi(-q)}{dq^2}[/latex] But then I actually get: [latex]\left[\mathfrak{F}P^2-P^2\mathfrak{F}\right]\hat{\psi}=2\frac{d^2\psi(-q)}{dq^2}\ne 0[/latex] EDIT: Looking at this some more, I realized I shouldn't have used the FT in the commutator, which is a discrete rotation of a quarter loop in the q-p domain. I should have used the generator of that transformation. This would be derived somehow from the Fractional FT with [latex]\alpha[/latex] very small. This is getting too complicated. What is the way to do this in Lagrangian mechanics???

Thank you! I just realized that a local gauge symmetry over phase-space will probably not quantize to a "particle" per-se, but to some weird analogue of one, as it won't have an energy in the usual sense. That is, I think it won't. I have to admit I don't know enough about phase-space geometry, but remember hearing about things such as, the action behaving like a metric connecting [latex]\dot{q}[/latex] and [latex]p[/latex] like a space and its co-space dual, and about the Poincare form on phase-space. You got me interested though, because it sounds like other symmetries exist, and the oddity of it being in phase-space may add some insight to the Fourier rotation idea. If I have any, I'll post it for others to review. Btw, the thing that forced me to think about Fourier transformations was a Computer Vision fellow student who asked how odd it is that the transform looks very similar to its inverse, as she had no background in the subject and was seeing it for the first time! If anyone else can elaborate on the subject, I'd love to hear what you have to say!

Hi, I recently had to think a bit about the property of the Fourier Transform being sort of a rotation in the frequency-time (or more generally, q-dq) domain. That got me thinking of symmetry groups. I also recalled that the Fourier operator is unitary. This fits very nicely into gauge theory. This reminded me of the Fractional Fourier transform. So it's even a continuous rotation, a Lie group. Moreover, this transformation is extremely connected to Hilbert space, the underpinning of QM, connecting the position operator with the momentum operator (also later for fields in QFT). So we have under our noses a continuous-unitary-Hilbert rotation operator, which transforms not spacetime, but phase-space. Due to these properties, I imagine there should be an invariant way to describe physics, whether it's in the "time domain" or "frequency domain" OR anywhere in between. This set of transformation of the wave-function, for instance, over phase-space, would define a global symmetry group for the action, and so we can look for a conserved quantity (phase-space's own "angular momentum", for rotations in the q-dq plane). If it sometimes isn't a global symmetry, but only a local one, "a new field can be defined, the quanta of which can be described as particles..." Was this subject ever studied as a proposed gauge theory? Note that an important aspect of this is that we have unitarity. You don't see that every day.

Mordred, Thanks for the link. I'm reading it now, and it looks promising! ajb, Thank you so much for the quick answers. I've looked at KK in the past, but not in any detail. I'll also review Cartan formalism as you suggest. How much time does it take physicists to learn all this stuff in academia???

Hi All, Sorry for being so lengthy, I was trying to be as precise as possible, and efficiently bunch up all of my questions on this topic in this one post. Most of this post is just a review. I am not a physicist, so if there are errors, please point them out so I can learn from them. My questions follow. [latex]\tilde{A}[/latex] is the EM potential one-form such that [latex]F_{\mu\nu}=(d\tilde{A})_{\mu\nu}[/latex] using the exterior derivative. [latex]\vec{\Psi}[/latex] is a wave-function (wave-vector). It is a function of space-time coordinates, with its "vectorness" values being in fiber coordinates. This distinction in coordinates means that [latex]\mu[/latex] and [latex]\nu[/latex] below are somewhat different. In QED, this is the equation for the EM connection: [latex](\nabla_\mu \vec{\Psi})^\nu=(\partial_\mu-ieA_\mu)\Psi^\nu[/latex] Via some algebraic manipulations, we get: [latex](\nabla_\mu \vec{\Psi})^\nu=\partial_\mu\Psi^\nu+\frac{eA_\mu}{i}\Psi^\nu=\partial_\mu\Psi^\nu+(\frac{e\tilde{A}}{i})_\mu\delta^\nu_\xi\Psi^\xi[/latex] Now trying to put this into the form of a connection, in a General Relativitic fashion: [latex](\nabla_\mu \vec{\Psi})^\nu=\partial_\mu\Psi^\nu+\Gamma^\nu_{\mu\xi}\Psi^\xi[/latex] Where [latex]\Gamma[/latex] is the EM connection in this context and not gravity (i.e. not Levi-Civita). Equating the two formulas, we get: [latex]\Gamma^\nu_{\mu\xi}=(\frac{e\tilde{A}}{i})_\mu\delta^\nu_\xi[/latex] This is different from the Levi-Civita connection, as it is not symmetric in space-time index [latex]\mu[/latex] and fiber index [latex]\xi[/latex]. My questions: Does the EM connection not being symmetric mean that this connection has torsion? If there is torsion, is that why we have [latex]\frac{1}{4}F_{\mu\nu}F^{\mu\nu}[/latex] in the QED Lagrangian? Does [latex]\Gamma^\nu_{\mu\xi}=(\frac{\partial{\vec{e}_\mu}}{\partial{x^\xi}})^\nu[/latex] still hold, for some vector/fiber field frame [latex]\vec{e}_\mu[/latex] analogously to how the Levi-Civita connection is defined in GR for gravity? Are these [latex]\vec{e}_\mu[/latex] some of the extra dimensions in M-Theory? Is there a "greater connection" that includes both the gravity connection and this EM connection? How would it handle the difference between the fiber dimensions and the space-time dimensions? Is this usually called [latex]\omega[/latex] by any chance? It is well known that [latex]\tilde{A}[/latex] has gauge freedom of [latex]\tilde{A}+df[/latex] where d is the exterior derivative and f is a scalar field. Defining [latex]f=\frac{1}{e}\beta[/latex], a change of gauge would look like: [latex]\vec{\Psi}'=e^{i\beta}\vec{\Psi}[/latex] [latex]\tilde{A}'=\tilde{A}+\frac{1}{e}d\beta[/latex] Plugging [latex]\tilde{A}[/latex] into the EM covariant derivative [latex]\tilde{\nabla}[/latex] we get: [latex]\tilde{\nabla}'=\tilde{\nabla}-id\beta[/latex] This finally gives us (after some algebra): [latex]\nabla'_\mu\vec{\Psi}'=...=e^{i\beta}\nabla_\mu\vec{\Psi}=(\nabla_\mu\vec{\Psi})'[/latex] As I understand it, [latex]\beta[/latex] would be the [latex]U(1)[/latex] coordinate parameter for the EM gauge group. Is the fiber of the Levi-Civita connection (i.e. gravity) just the tangent vector space which is the 4D space-time? This would be analogous to the fiber of the EM connection being the Lie algebra [latex]\mathfrak{u}(1)[/latex], with the [latex]\beta[/latex] coordinate for the corresponding Lie group, [latex]U(1)[/latex], right? Do theories exist which treat space-time dimensions as being finite cyclic dimensions, like [latex]U(1)[/latex], [latex]SU(2)[/latex], etc? This way all dimensions, space-time dimensions (x, y, z, t) and QFT gauge dimensions (such as EM's [latex]\beta[/latex]), work the same way in principle. Is there a [latex]``\Gamma'_{Levi-Civita}=\Gamma_{Levi-Civita}+dG"[/latex] gauge transformation for the Levi-Civita connection? It seems like this should somehow be related to gauge freedom of the Lorentz group on space-time and space-time tensors. If true, would this mean that the space-time dimensions (x, y, z, t) are merely the gravity potential's (i.e. the Levi-Civita connection's) gauge freedoms, in a way analogous to the EM potential's ([latex]\tilde{A}[/latex]) gauge freedom, the [latex]\beta[/latex] coordinate? If you read thus far, thanks!

ajb, thanks for replying! But now I'm not sure I understand. When you say standard QFT adheres to invariance under the Poincare group, do you just mean that standard QFT is a relativistic (tensorial) theory? and the tensors, mathematically "hide" the invariance stuff making the equations look neater? I understand there is a problem combining GR with QFT in a fully satisfactory manner, but I didn't realize it was the relativistic Poincare invariance that was the problem. I thought it had more to do with the use of continuous manifolds. Can you please elaborate on this a bit? I thought I understood, but I'm not sure I follow anymore. You are definitely more technical than I am, but I'll try to keep up! (I kind of enjoy the understanding that comes with the technical stuff)

What do you mean? Does the universe in QFT not look the same to different observers? Or is this a statement about the difference between SR and GR?

Thanks ajb (and imatfaal for trying! i meant you're in constant orbit, so technically accelerating, yes). That's what I was expecting. And you added the clear connection to the non-agreed upon vacuum, which I liked. I was just struck by how this is not very much talked about, but yet is very mysterious, while Hawking radiation specifically from black holes is quite a popular topic, although it seemed to me a black hole wasn't really necessary. Maybe the reason is that with just a regular sun/planet/etc there isn't the effect of having negative energy particles falling back down and evaporate the sun/planet/etc. Or is there? I wouldn't think so, but not sure.

Hi All, Looking again into the Casimir/Hawking/Unruh effects, made me think that: 1. Take any gravity source. 2. Use general relativity's gravity-acceleration equivalence principle. 3. The Unruh effect applies. 4. We get radiation in a similar way to Hawking radiation around a black hole? So, are black holes special in that they give enough energy for virtual particles to emit as radiation, or does every massive object (meaning one with a nonzero stress-energy tensor) have a "Hawking" radiation? To be more specific, do the Earth or Sun produce radiation in a similar mechanism as does a black hole (in addition to their more conventional radiation spectrum)?

I see your point. It is true, if we say rest mass is something that distinguishes between particles (which is what we say) then one would get many many particles using this method... I guess my thought was that some of these definitions date back way before QFT and the Higgs mechanism were established, and in some sense could be revised. Maybe it's not bad to have many quasi-particles, as long as you have a small set of massles/"bare" particles. But only if that would prove useful. Also, I haven't thought through about how binding energy comes into this picture. Probably as another interaction, generating more waves in the particle's field, adding up using superposition again, to give rise to a heavier new quasi?-particle. But that is just conjecture on my part, as I'd like it to be a similar mechanism. I'm not sure what picture I prefer yet, but I'll go ahead and look at this paper you found. I already like the preface Thanks for your patience with me. As I am not a formal physicist I find it is always wise to keep a healthy bit of humility at heart when speaking about the subject. But I did learn a lot on my own, from books and lecture videos, so I don't come empty handed. The thing is, with a book, or a video, one can't ask questions. That is how I started using this forum. I always had envy toward people who do this for a living (not for the pay of course)...

Hi Mordred, I took a look at the quasi-particle list, and I think what I am thinking about in terms of my attempted analogy, is closest to "electron quasi-particle" relating to my electron example. "Electron quasi-particle" being an electron interacting with a solid. The "electron quasi-particle"'s mass would be different than the mass of an electron in a vacuum. You say we certainly cannot think of a solid as a vacuum, but I was thinking maybe we can think of a vacuum as being analogous to a solid, because of the Higgs field the vacuum seems to encompass. It's not a perfect analogy, but why would it not fit here? The analogy (glass <->Higgs) is probably not perfect as I said, but how do the greater degrees of freedom make the situation *inherently* different? At the core, I'm trying to understand, if the same mechanism that is responsible for giving the polaritons mass (photons in a glass solid) and thus makes us consider them to be quasi-particles, is also the same mechanism that gives massles "electrons" in the Higgs field vacuum a mass. If so, they should be considered quasi-particles, on an equal footing with polaritons. My reasoning for all of this is that for simplicity's sake (aka occam's razor), I would expect there to be only one way things (quantum fields) acquire rest-mass. Also, I am not familiar with confinement. What does that mean? Thanks! btw, this is in the Britannica link you supplied: "There is reason to suspect, however, that all particles may actually be disturbances in some underlying medium and, hence, are themselves quasiparticles." -- http://www.britannica.com/EBchecked/topic/486549/quasiparticle

Is it right to say that a particle interacting with Higgs is in "free space"? I would think not, and so I would call it a quasi-particle, because it is not in free space. It is in space that has Higgs, and it is interacting with it. Maybe the definition of free space should be modified when we talk about a non zero Higgs field being everywhere and interacting with massive particles?

To stress what others here were saying, in the balloon analogy, you should think of the 2D surface of the balloon as being analogous to the 3D universe/space that you see around you. The gas inside of the balloon is not part of the analogy. Neither is the volume of the balloon at all part of the analogy. Only the 2D surface of the balloon and any dots (locations) on it.

After much searching, I found this video about the subject. Apparently there is a quasi-particle called a "polariton" which is the sum of the original photon together with the changes introduced by the atom lattice: From Wikipedia, it sounds like the reference is made to "exciton polaritons", which result from the coupling of photons (coming from outside the material for instance) with excitons. Excitons are excited (energetic) dipoles as I understood it. So basically the bound electrons in the material in this example. From this, by analogy, I get that all particles that travel slower than c, are in fact quasi-particles in this sense. So the "bare" "electron", does travel at c. What we decided to call an "electron" in physics - the one with the rest mass, the one travelling slower than c - that particle is actually in fact a quasi-particle (like the polariton). It is the quasi-particle which is the "bare" "electron" coupled with the Higgs field!