 # joigus

Senior Members

1539

• #### Days Won

12

joigus had the most liked content!

## Community Reputation

282 Beacon of Hope

• Rank
Primate
• Birthday 05/04/1965

## Profile Information

• Location
(0,0,0)
• Interests
Biology, Chemistry, Physics
• College Major/Degree
Physics
• Favorite Area of Science
Theoretical Physics
• Biography
I was born, then I started learning. I'm still learning.
• Occupation
teacher

## Recent Profile Visitors

6927 profile views

• ### Time Traveler

1. I'm not 100 % sure of what's bothering you, but it's useful to distinguish in these integrals the field point $$x$$ and the source point $$x'$$. Now, the sources $$\boldsymbol{J}$$ do not extend to spatial infinity. Nor do the fields $$\boldsymbol{A}$$ (at the field points: the points at which the field is calculated, $$x$$ ). So, $\int\frac{\partial}{\partial x_{i}}\left[f\left(x\right)g\left(x-x'\right)\right]=\left.fg\right|_{\textrm{boundary}}\rightarrow0$ The boundary is spatial infinity. I hope that helps.
2. 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).
3. 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?
4. ## Can science prove the existence of an intelligent world that rules our universe? Split of Theory of everything

@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.
5. 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.
6. 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.
7. 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.
8. 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.
9. LOL. I've done my best to lessen the carnage too.
10. 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!
11. 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.
12. You're confusing Klein bottles with Kaluza-Klein unified model of EM/gravitation.
13. 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?
14. 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.
×