joigus

Good advice.

It does, because elementary particles are characterised in QFT as irreducible representations of the Poincaré group (Lorentz groups and translations). If you study the Lorentz group for its own sake, you will be led purely mathematically to objects that represent faithfully the properties of spinors, because they rotate by SU(2), and locally SU(2) looks like SO(3). That as to rotations. The version that includes boosts (the whole Lorentz group) are objects that boost and rotate with SL(2,C), which looks locally like SO(1,3) (Lorentz group).

I don't think there's anything to understand that hasn't been understood very long ago. You seem to relate linear momentum (which is proportional to mass) with this "capacity for linear reappearance" (linear?), which should be a purely kinematic factor, independent of mass. How come?

@Kartazion. Here's the point. Apparently you're clueless. Your question ( \( Q \) ), rests on a notion. Let's call that notion \( W \), \( Q = \) Can science prove the existence of an intelligent world that rules our universe? \( W = \) an intelligent world Now, let's introduce another notion, equally arbitrary: \( W' = \) 17 balls of jello, each with the mind of a baby, but which communicate so as to generate a meta-intelligence (An intelligence that cannot be achieved by any of them, but by their mutual cooperation: Mind you, cells are stupid, but they manage to produce intelligence in a brain, so the analogy is not that far-fetched.) Let's establish an isomorphism \( \varPhi \): \[\varPhi\left(W\right)=W'\] \[\varPhi\left(Q\left(W\right)\right)=\left(Q\left(\varPhi\left(W\right)\right)\right)=Q\left(W'\right)\] IOW, my question is your question disguised under an isomorphism. And I don't know how "stupid" it is. But it's your question. --yawn.

I had not seen this objection of yours. Sorry. First, I'm not proclaiming anything. I don't use exclamation marks, capitals, etc. Or I rarely do. You represented what should be a spherical wave by what looks like a plane monochromatic wave. That was my main objection. Now, it is possible in principle to treat a photon falling in a gravitational field the way you suggest, but it's more involved than what you suggest. It's called the eikonal approximation for optics, and it's only justified in certain cases. It will take me more time to get to it.

It rings a bell, yes. It's been ringing a bell for about 39 years now. I've also produced them in a laboratory, guided by people who knew the stuff backwards. You've set the standards too high. I'm no Mr Spock. Nor are you. You'd have to be a character from a sci-fi script. Your goal is laudable, your means are misled. If you set all units to the proper values, fundamental constants become one and no fundamental equation of physics changes. That must mean something. Think about it. 10-45 in the equations must be explained; 1 doesn't have to be explained. When a number is reduced to 1 by setting the scale, and nothing substantial is affected, something has been explained.

There's no need to explain the carriers, because universal constants don't have to be broadcast. Otherwise, there would be particles/fields out there not fully "aware" that the speed of light is \( c \) or the charge of the electron is \( e \), or Newton's constant is \( G \), using their own local value; and we would be able to tell on account of phenomena at the interfaces where this information hasn't been updated yet. As Markus said, it's not a mechanism; it's a symmetry. And it's hardwired in every piece of physics that we know of. Think of fundamental constants, not as really constants, but as different funny ways of writing the number 1. That's what they are. It's our human yardstick that makes them look funny. Bound states of elementary particles can be said to have volume because they have parameters with dimensions of length with a meaning related to spatial extension. Really elementary particles, like quarks and leptons, don't really have a volume. They do have characteristic length parameters. For example, the Compton wavelength of the electron is, \[\frac{\hbar}{m_{e}c}\] This doesn't mean the electron has a volume. It simply means that if you want to look at the electron at a scale the order of its Compton wavelength (or smaller), your electron no longer looks like a one-particle state, but a many-particle state. You start seeing a cloudy thing made up of many virtual particles.

Your two-photon gravitons would not satisfy the equivalence principle either, as photons couple to electric charge, not to energy. There are many, many issues with your idea. I think it's time you give it a rest.

LOL. I've done my best to lessen the carnage too.

To give you an example of something that would make sense to me, and I'm sure to @swansont too, if you were able to define an entanglement entropy as a function of space time, then you would have an entanglement field. For a system with two degrees of freedom maximally entangled, entanglement entropy is, \[-\frac{1}{2}\log\frac{1}{2}-\frac{1}{2}\log\frac{1}{2}=\log2\] For a more general 2-DOF system with spatial dependence added to the broth, it'd be something like*, \[s\left(x\right)=-\left|c_{1}\left(x\right)\right|^{2}\log\left|c_{1}\left(x\right)\right|^{2}-\left|c_{2}\left(x\right)\right|^{2}\log\left|c_{2}\left(x\right)\right|^{2}\] But you're being very vague about what this entanglement field is. Is something like that what you mean? * The \( c_1 \) and \( c_2 \) would be the amplitudes (equivalent to your A's). Well, to your A's if you corrected the spatial dependence in the wrong place!

Massless bosons cannot be considered point particles, neither can they be considered spherical waves. Localisation for massless particles is a tricky thing. They register their "going through" so to speak, but it's not consistent to consider them at point \( x \). For gravitons, to make things worse, the number that controls how strongly they couple to energy is extremely tiny at the energies presently available. Your gravitons as entangled photons would not work as actual gravitons should, among other things because they would couple with charged particles too strongly --1044 times more strongly than required. You really must go back to the drawing board.

You're confusing Klein bottles with Kaluza-Klein unified model of EM/gravitation.

These are very good questions, and I must say I've been thinking about asking both. One of them I did: "What is this entanglement field?" But I also was kept thinking for a while, how do you trap photons between the energy levels of a semiconductor?

That's why I'd rather work on other, more easy to grasp, aspects of physics. Maybe some day you can explain to me what a graviton is. In the meantime, I'll tell you that the fact that, \[\frac{\alpha_{\textrm{grav}}}{\alpha_{\textrm{em}}}=10^{-44}\] means that it's 10-44 times less likely to scatter a graviton than to scatter an electron off the same scatterer. If you don't know that and don't care for it, don't think has any bearing on your problem, I can't help you. Giving neg reps every which one who offends you by not agreeing with you, won't help your idea along either.

The gravitational fine structure constant is the number that controls coupling of gravitons to energy. https://en.wikipedia.org/wiki/Gravitational_coupling_constant It's the order of 10-45 in international units.* Would you be as kind as to tell me where in your "calculations" is the faintest inkling of how you would evidence this from a purely electromagnetic model? Thank you. Although the non-answer I can only predict. (*) It's a dimensionless number. You calling me, or anybody else here, "troll" won't have much of an effect.

Number of photons is not a conserved quantity. The point is not a single photon; the point is individual photons; namely, multi-photon states for which the number of photons can be counted. They do indeed give you everything. The problem is they give you more than you bargained for. In particular, they give you dilatons, due to the conformal invariance of the theory (you can stretch it and squish it, and squash it, and the theory remains the same.) They also give you ambient space-times that are difficult to deal with, called Calabi-Yau manifolds. They are complexified, and they are Ricci-flat, which is the next best thing after being flat. Technical problems appear when trying to compactify the fields in 26 dimensional Calabi-Yau manifolds, because there are astronomical numbers of ways to compactify. Compactifying means reducing most of those dimensions to little loops that cannot be seen from our perspective, in a cooling universe. I've tried to be as helpful as possible, but you should really start reading what you're told and slowing down on the lecturing everybody else part of your present discourse. Some of my points, and those from others, you've chosen to simply ignore. It's quite irritating. Edit 1: Another problem with string theory is the difficult to get predictions that can be confirmed in the laboratory, as the main features show up at the Planckian scale. Edit 2: I hope that answers some of your questions about superstrings. Now it would be nice to answer some of mine (and others').

You further assume that, \[\omega=\omega\left(x\right)\] This is incorrect and betrays a lot of ignorance on basics of physics. Let's leave aside the fact that single photons cannot be dealt with in terms of wave functions, that there is no momentum operator for photons of the form \( \frac{\hbar}{i}\frac{\partial}{\partial x} \) --or for any other particle in quantum field theory, for that matter--, and other inconsistencies. Photons travelling in the vacuum can be Fourier analised in terms of monochromatic components, \[Ae^{i\frac{\omega}{c}x-i\omega t}\] There is no \( x \) dependence on that Fourier component. There is just a phase velocity, \[\frac{\omega}{k}=v_{f}\] and a group velocity, \[\frac{d\omega}{dk}=v_{g}\] For photons both are constant and equal to \( c \). Now, for massive particles, you have, from the Einstein relation, \[E^{2}=p^{2}c^{2}+m^{2}c^{4}\] the following relation for frequency and \( k \), \[\omega=c\sqrt{k^{2}+\frac{m^{2}c^{2}}{\hbar^{2}}}\] So that you get, \[\frac{\omega}{k}=v_{f}=c\sqrt{1+\frac{m^{2}c^{2}}{k^{2}\hbar^{2}}}\] and, \[\frac{d\omega}{dk}=v_{g}=\frac{ck}{\sqrt{k^{2}+\frac{m^{2}c^{2}}{\hbar^{2}}}}\] Satisfying the constraint, \[v_{f}v_{g}=c^{2}\] Naive application of relativistic waves gives you a superluminal phase velocity and a subluminal group velocity. Those are some basics of wave dynamics for relativistic particles. I saw you were in sorry need of a quick tutorial.

The rigorous theorem requires heavy-duty maths: field operators, states and vacuum, and representations of the Lorentz group. A relatively easy-to-follow discussion is provided by John Baez: https://math.ucr.edu/home/baez/spin_stat.html In a nutshell, what the spin-statistics tells you can be stated as this: The splitting of particle classes into bosons and fermions, as characterized by their exchange properties (statistics): \[\varphi_{1}\left(x\right)\varphi_{2}\left(x'\right)=\varphi_{2}\left(x\right)\varphi_{1}\left(x'\right)\] (bosons) or, \[ \varphi_{1}\left(x\right)\varphi_{2}\left(x'\right)=-\varphi_{2}\left(x\right)\varphi_{1}\left(x'\right) \] (fermions) Can be obtained also from their properties under the Lorentz group (the rotation factor of it) by rotating the whole universe around the midpoint between them. If the vacuum is Lorentz invariant, we won't be doing anything to it, and particles will just be exchanged if the rotation is of angle \( \pi \). The name spin-statistics is because "spin" has to do with properties under rotations, and "statistics" has to do with properties under exchange. "Bosons under rotations are also bosons under exchange; conversely for fermions" is the content of the theorem. I hope that helped.

Bosons do many more things than "overlap" --whatever that means. They scatter charged particles, at very high energies they can even scatter each other, they form Bose condensates, they can form aggregates that travel as solitons through non-linear media, they flip spins, and some of them can even acquire mass. x-posted with MigL.

The fact that spherical waves are not monochromatic has nothing to do with whether they are electromagnetic or not. The waves you put in your picture are inconsistent with the symmetry. Gravitons are spin 2, and photons are spin 1. Gravity cannot be built from spin 1 particles or spin 1/2. That's charted territory, and known not to work. https://www.quora.com/How-do-we-know-that-a-graviton-(if-it-exists)-has-spin-2 MigL has given you the lowdown: The cross section would have to be many orders of magnitude bigger than it actually is. That's why gravitons haven't been detected, not because "physicists don't believe in them". Were gravitons made of photons they would have been detected many decades ago. It's kinda like trying to say that a neutrino is made of electrons. If what you say were right, scattering electrons gravitons would be very easy. You could do it in a garage.

The quantum states of spherical waves (s waves) are not monochromatic plane waves like the ones you're writing. The speed of light in vacuum plays no role in inflation or in present expansion. The cosmological constant is a free parameter of the theory that must be measured independently.

OK. At this point I see you either don't understand what I said, or else, you don't bother to read what I tell you. Any of those is enough for me to ponder that it's no longer worth maintaining a discussion with you.

At least in this world. Star Trek is another matter.

OK. You really must go to a QM book and learn the stuff. Dice have nothing like what elementary particles have. They have infinitely many possible observables, and they're perfectly correlated for every compatible pair. It's as if dice had also colours; and every time you measure a colour, the other die face of the die has the corresponding anti-colour. And also they had density, volume, haziness, etc. And every time you measure that property in the other face of the die die, it has the corresponding anti-property. As long as you measure compatible observables. But when you measure incompatible observables, they become completely uncorrelated. Now, try to do that with dice.

Coming back from death used to be fiction. It still is.