# How can space warp if it is a non-thing?

Discussion in 'Physics & Math' started by nicholas1M7, May 27, 2011.

1. ### Farsight

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You've jumped to the wrong conclusion.

Start from scratch and think about the electromagnetic field is curved space. The electron has an electromagnetic field, which is isotropic. It also has magnetic dipole moment, which should signal rotation to you. So think in terms of frame-dragging, and picture the particle as an extended entity like this:

This is a flat two-dimensional depiction, but it ought to suffice. Now imagine another particle just like it, and think of them in terms of dynamical vortices of stress-energy. If they have no initial relative motion, two similar vortices will move linearly apart. We draw radial "electric field lines" to depict this linear motion. However if they have some initial lateral motion, the vortices will rotate around one another. We draw concentric "magnetic field lines" to depict this rotational motion.

Equation 1 is inappropriate because there are no "point charges". A charge is a region of space where the curvature is total. Equation 2 isn't appropriate because it isn't some big bland large-scale curvature.

No. Nobody has told me that.

It's true because this is what Maxwell was saying, only he didn't get it quite right. He talked about molecular vortices, he didn't know about electrons. He had the vortices down as being in the space the particles moved through rather than the particles themselves. Yes, it isn't an accurate description of quantum electrodynamics, but I think it's a fairly good description of the underlying reality. A start at least.

It's classical.

The photon and neutrino aren't static, set them aside. Most of the particle zoo are ephemera. Look at the lifetimes.

OK.

Yes, I've read The Refractive Index in Electron Optics and the Principles of Dynamics by Ehrenberg and Siday.

3. ### Magneto_1Super PrincipiaRegistered Senior Member

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This is because, when studying General Relativity (GR), and studying the works of: Gauss, Ricci, and Riemannian, these works are for the "physicists" and "Geometers." of the world.

When studying General Relativity (GR), and studying the works of: Christoffel Bianchi, and Kruskal, these works are for the "physicists" and "Advanced Mathematicians" of the world.

When you want to build a foundation for (GR), study the works of Gauss, Ricci, and Riemannian.

When you become more proficient in the subject and study of (GR) then I would suggest studying the works of Christoffel, Bianchi, and Kruskal.

Last edited: Jun 14, 2011

5. ### AlphaNumericFully ionizedRegistered Senior Member

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You're the guy who didn't (and likely still doesn't) know what a tensor was or the difference between rank and components. You're hardly in a position to talk about being proficient at this stuff. I am absolutely certain you couldn't pass a university level exam on general relativity.

7. ### przyksquishyValued Senior Member

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What's wrong with it? On the one hand, we have the equation of motion experiments show charges follow in an electromagnetic field. On the other hand, we have the sort of equation of motion they'd follow in curved space. They don't match. That kills the idea stone dead right there.

That's not "scratch". "Scratch" is how charges are observed to behave in an electromagnetic field, which ample experiments have shown is accurately modelled by the Lorentz force equation. That's what you need to be able to explain.

Maybe not, but there are charges that get close enough, and they obey that equation. So you don't get off the hook. That equation is a summary of behaviour even highschool students get to test. If you can't recover that equation one way or another, you literally can't explain highschool physics.

Then what is it? How do you model it mathematically, and how do you recover the Lorentz force?

Well now you know it and you should revise what you are saying accordingly.

No, even if you have perfectly understood Maxwell, that by itself doesn't make anything true.

Based on what? You'd just like it to be that simple, so it is? :bugeye:

In what way that distinguishes them from the electron?

Why do you think I didn't mention any of those? :bugeye:

8. ### leentjeRegistered Member

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Who on earth is Riemannian: a soul brother of Ramanujan?

Personally I think it would be better to start your studies with Euclid; later you could continue with the works of Cevian, Laplacian, Lagrangian, Abelian, Hamiltonian and Lebesgian, but of course it all depends on where you fit into the Gaussian.

9. ### AlexGLike nailing Jello to a treeValued Senior Member

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Are you serious? Bernard Rieman developed Riemannian geometry.

If you're not familiar with it, at least you should recognize it.

10. ### TachBannedBanned

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You mean BernHard RiemanN. He's just being funny.

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12. ### CptBorkValued Senior Member

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Oh, now you decided to throw Kruskal in there too? Why? Is the name mentioned in the Wikipedia entry on black holes, or something? Without Bianchi's identity, you wouldn't even be able to set up the Einstein field equations, let alone study GR. Without the Christoffel symbols, you couldn't even get started on GR, there'd be no way of expressing the geodesic equation. Kruskal you don't need unless you're studying the hypothetical physics inside a black hole event horizon.

Stupid advice. As someone who has actually studied GR and passed a graduate course in it, I can advise you that you don't need to spend years "studying the works of Gauss, Ricci and Riemann", you only need the specific results that apply to curved spacetime. And stop mentioning Ricci please, it's a dead giveaway that you're just patronizing us by throwing famous names around as if you actually understood the substance of their work. Besides, Ricci hardly did a thing for the theory in comparison to folks like Hilbert and Minkowski.

My suggestion to you: I know you might think it makes you sound more knowledgeable when you throw around famous names, but it doesn't make you look smarter in the eyes of people who are actually familiar with the discoveries that made those names famous. You're not fooling anyone about your lack of knowledge in this area, and you're hardly the first one to have ever tried.

13. ### Magneto_1Super PrincipiaRegistered Senior Member

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Kruskal, only, because Kruskal has convinced physicists that there is a technique that will allow you to keep the calculations alive at the "Event Horizon" at the Schwarzschild Radius and surface of the "Black Hole." Kruskal was able so resolve the infinity problem that arises when your measuring distance (r) is equal to the Schwarzschild radius (rs); when making measurements relative to the "Mean Center" of the "Black Hole" Event Horizon.

I personally, am still not comfortable with that Kruskal solution. I think that there is still a simpler more classical way of resolving the infinity problem that occurs at the surface of the Black Hole Event Horizon.

I have not been able to derive this simple solution yet. So in the meantime let's accept the Kruskal solution; many "Advanced Abstract Mathematicians" have completely accepted it. And I am also ok with it.

CptBork stick with the physics, and try not follow in the footsteps of your brother, the "Circus Act" known as: AlphaNumeric = Tashja = Tach = Guest254 = Przyk = George Weatherill

14. ### AlphaNumericFully ionizedRegistered Senior Member

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Kruskal coordinates are not the only ones valid across the event horizon. Eddington-Finklestein coordinates are too. In fact there's infinitely many different coordinates which work across the event horizon, though many are just trivially related to one another.

I'll inform the physics community of your discomfort.

Perhaps you should learn how to do basic coordinate transformations first? You couldn't even do the correct Cartesians -> Polar transform when I discussed it with you.

It's not a matter of choosing to accept the solution, it is a perfectly valid solution to the EFEs. Nothing more needs to be considered in regards to 'accepting' it. Whether or not they are the best to examine particular properties of the space-time is a different issue.

That's the spirit, when you can't retort several peoples' criticism accuse them of being the same person with different accounts.

I don't have any sock accounts. I don't need sock accounts, I've slapped your arguments down with just this account, what benefit would hiding behind other accounts give me? Many times I've even argued with Tach.

15. ### AlphaNumericFully ionizedRegistered Senior Member

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And just to illustrate przyk and I are not entirely in lock step, I'm not above saying things which Farsight might end up running with (though he won't understand it) and that I can contribute high level physics/maths, not just insults, consider the following :

Actually it is possible to put electromagnetism and GR on very similar footings, though not in the way Farsight is claiming.

In differential geometry you can construct the Riemann curvature tensor via covariant derivatives on vectors, $[\nabla_{a},\nabla_{b}]\xi^{c} = R^{c}_{dab}\xi^{d}$ (see here) where $\nabla \sim \partial + \Gamma$ where $\Gamma$ is the Christofell connection term. This is obviously a map on the tangent space and by looking at the indices the c,d indices are associated to that map, while the a,b select which map, ie $R^{c}_{dab} \sim (R^{c}_{d})_{ab}$ where $R^{c}_{d}$ is a matrix acting on the vectors and there's a bunch of them, as selected by the a,b indices. More formally $R \in \textrm{End}(M) \times \Lambda^{2}(M)$ where M is the manifold in question. This is constructed in terms of a tangent bundle's properties.

In gauge theory you have a G-bundle, where the fibres are copies of the gauge group. In this case you now have a covariant derivative $D \sim \partial + A$ where A is a Lie algebra valued 2-form. From this you can once again consider the commutator and you get a very familiar object, $[D_{a},D_{b}] = \partial_{a}A_{b} - \partial_{b}A_{b} + [A_{a},A_{b}] = F_{ab}$, the gauge field strength! But this has only 2 indices so how does it match? Remember that the connection is now Lie algebra valued and so we can expand it in terms of the Lie algebra generators, $A_{a} = A^{i}_{a}T^{i}$. The $T^{i}$ are themselves matrices, so we can make the matrix indices explicit via $A_{a} \to (A_{a})^{\mu}_{\nu}$. Thus we have $F_{ab}$ is a set of matrices, where the choice of a,b decides which matrix we're considering.

As a result Maxwell's tensor (or the generalised non-Abelian Yang-Mills tensor) is to gauge theory what Riemann's curvature tensor is the general relativity. Both of them satisfy Bianchi constraints of the form $[\nabla_{a},[\nabla_{b},\nabla_{c}]] + \textrm{permutations} = 0$ (for those of you who might have noticed, yes, the Bianchi identity is indeed expressible as the Jacobi identity!). Both of them can be used to construct actions.

However, this doesn't mean the electromagnetic field is curved space, it can be viewed in terms of the curvature of an abstract mathematical space. Since Farsight is always whining about how people like myself (supposedly) take mathematics too literally I'm sure he wouldn't want to be a hypocrite and accept a mathematical abstraction as literal physical reality, right?

Actually, I tell a little lie. It is possible to view electromagnetism in terms of curved space, you just have to allow for extra compact dimensions. But Farsight has also whined many times about how silly string theorists are to consider extra compact dimensions so that open is out for him too. Such a shame he didn't dismiss everything he doesn't understand, he might almost have something to justify, in some vague way, some of the noise which exits his mouth.

16. ### CptBorkValued Senior Member

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Well ok, why don't we stick with the physics, then? I have refrained from insulting you as a person and chosen to stick with the physical ideas you present. You, on the other hand, have consistently misrepresented your knowledge and abilities, while taking baseless shots at me for the fact that I'm still a grad student. Throwing famous names around doesn't constitute a physics argument, either.

17. ### przyksquishyValued Senior Member

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Actually, not as much as you seem to think, and...
...unfortunately I think I'm already responsible for that. What originally got Farsight all excited about electromagnetism being curved space was this comment I directed at RJ:
By "equivalence postulate" of sorts, I meant that the replacement $\partial_{\mu} \rightarrow D_{\mu}$ in field theory was analogous to the principle of general covariance. The remaining blanks are, of course, exactly what you go on to fill in the rest of your post.

The reason for the parenthetical remark, by the way, is simply that guage field theory isn't the area of physics I'm most familiar with. You made a few interesting points in your post that helped solidify the analogy a little for me - eg. the point about the extra Lie matrix indices. I also found it a little odd that in both GR and guage field theory all the derivatives were in spacetime and none were in the more abstract space associated with guage transformations - ie. all the indices on the curvature tensor are of the same "type" in GR but not in Yang-Mills. But I suppose thats simply due to the fact the tangent vector space in GR happens to be locally homeomorphic to the spacetime manifold, right?

Kaluza-Klein, right? Though I think the point I was making to Farsight still holds, since I was just talking about curvature in space, and not in spacetime. If you include a time-like dimension and consider velocities much less than c, I can see you might be able to get the geodesic equation to reduce to something like
$\ddot{x}^{i} \,+\, \Gamma^{i}_{00} \,+\, \Gamma^{i}_{0j} \dot{x}^{j} \,+\, \Gamma^{i}_{j0} \dot{x}^{j} \,=\, 0 \,,$​
which starts to look a bit more like the Lorentz force. Incidentally I was making a similar point in the Inflation and Curvature thread about the geodesic equation needing the time-like components in order to be able to reproduce Newtonian gravity.

18. ### TachBannedBanned

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You are getting infinitely more hilarious when you pretend to know GR than you pretend to know SR, did you know that?

I am delighted to be part of the circus act. Especially when we benefit from the participation of an international group of famous clowns: John Duffield, R. Kemp, etc.

19. ### Farsight

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I've been talking about this for quite some time now. It isn't my original idea, have a look around the internet for this kind of thing. You'll see plenty of mention of curved spacetime too, but remember what I've said previously about the distinction between curved spacetime and curved space. Also take a look at topological charge. Electrons aren't point particles, we can diffract 'em.

20. ### Farsight

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No it doesn't.

"Scratch" is pair production, and spin angular momentum and magnetic dipole moment and electron diffraction. Scratch is electromagnetic four-potential and electron models.

I can explain it better than most, but you aren't listening.

I don't know. The Lorentz force in terms of potentials is given as $\mathbf{F} = q(-\nabla \phi- \frac{\partial \mathbf{A}}{\partial \mathbf{t}}+\mathbf{v}\times(\nabla\times\mathbf{A}))$. It goes part of the way but it talks in terms of an electric potential φ and a magnetic potential A. IMHO that's misleading, because it's the electromagnetic field and there is only one potential. Moreover the expression doesn't address charge q. So all the explanatory meaning of Maxwell's theory of "molecular" vortices and the linear and rotational motion of electrons is lost.

You can make an electron (and a positron) out of a photon via pair production. Think of a cubic lattice. You're standing in one cell. Grab the horizontal above you, and push upwards. As it moves up, that's the displacement current D. Look off to one side while you're doing this and the rotation rate of the adjacent horizontal is the magnetic field B, whilst the slope is the electric field E. Look further sideways and both reduce with distance. Look the other way sideways and the curvature is opposite, so there's no overall curvature so no overall charge q. Also note that where you're pushing upwards there is no rotation and no slope, so there's no B or E, but the pressure you've exerted is Aμ. This is sketching out a "static" photon. It isn't actually static, it's moving at c, you're not there, and nor is the lattice. In pair production this pulse displaces its own path and ends up in a closed path. You can only create a pair because there's nothing to brace against. Conservation of angular momentum applies.

21. ### przyksquishyValued Senior Member

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You've listed a bunch of things that mainstream theories can already model quantitatively.

What's wrong with the models we already have? Where do you, or anyone else, show that your own model of the electron actually works quantitatively? The problem is that you are saying a lot of things that explicitly contradict mainstream theories like QED which actually have quantitative experimental support. For example, in QED, photons don't form bound states and even if they could, they couldn't form a spin-1/2 bound state.

No, there are specific things you are saying that are fundamentally incompatible with electromagnetism. You haven't addressed the fact that curved space immediately breaks Lorentz invariance (which is a property of electromagnetism) for example.

You can pack things up into one electromagnetic four-potential, which is a four-vector in space time. In that formalism $\phi$ becomes the time-like component $A^{0}$. In the more compact relativistic notation the Lorentz force equation is $f^{\mu} = q F^{\mu\nu}v_{\nu}$, where the electromagnetic field $F_{\mu\nu}$ is given in terms of the four-vector potential $A_{\mu}$ by $F_{\mu\nu} = \partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu}$.

But so what? This is all just mainstream electrodynamics that you could find in any good textbook on the subject or even on Wikipedia. You do not need to try to support things that are already well known. What you need to do, and haven't done, is link those equations to everything you say about electromagnetism that is not mainstream.

You can also make photons by annihilating an electron-positron pair. What's your point? You can also use a nonlinear crystal to generate photon pairs out of individual photons. And you can go further than that: electromagnetic calorimeters in particle accelerator experiments work by converting the energy of a single electron or photon into a whole cascade of lower energy electrons and photons.

And because you say all this, it's automatically all true...?

How do I use any of this to make quantitative predictions anyway?

22. ### Farsight

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Yes, but people forget that it's the Schrödinger wave equation, and can’t seem to appreciate that pair production converts a wave moving linearly at c into two waves moving rotationally at c. I've had people insist to me that the electron is a point particle.

It isn't my model. Various people have attempted to provide a model, but they tend not to get much attention. See for example Is the electron a photon with toroidal topology?, The nature of the electron, and Rotating Hopf-links: a realistic particle model by E Unz.

You’re misreading QED there. The experimental evidence of say Two-photon physics says that photons can interact to form fermions, see this report. In QED you’ll read that this is via “higher order processes” or you may read that a photon “can fluctuate into a fermion-antifermion pair”. But again, take the hard scientific evidence at face value. You start with two photons, you end up with an electron and a positron, and nothing else was involved. Those two photons interacted all right, and they formed two spin ½ bound states.

Because it doesn’t break it. See http://www.classicalmatter.org/ClassicalTheory/OtherRelativity.doc again. If you are made of waves, you always measure wave speed to be the same. Everything is made of waves, the apparent wave speed is unchanged, so Lorentz invariance holds.

Thanks. It still doesn’t address q mind. Ever thought about what a four-vector actually is?

I’ll try harder.

You can also make photons by annihilating a proton-antiproton pair. Photons are the lowest common denominator, electromagnetic waves where Aμ is more fundamental than E or B.

No problem.

No. It's not an issue of whether what I or anybody says is true. It's whether it fits with the scientific evidence, or it doesn't. For example photons don't form bound states and even if they could, they couldn't form a spin-1/2 bound state doesn't.

This is postdiction, but some interesting things come out of it. I know a guy who’s got the mass ratios worked out. He started from a different place, but it’s arguably “the same elephant” in that the electron is a quasi-spherical photon standing wave. The proton too.

23. ### AlphaNumericFully ionizedRegistered Senior Member

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Just to further illustrate how Mr Kemp's paranoia is unfounded here is myself, Guest and Tach engaging in somewhat of a 3 way (light) argument.

Perhaps you'll just weave that into your twisted view of the world in order to avoid facing up to your short comings Mr Kemp?

Just as I love how Mr Kemp tries to insult people for working towards or having doctorates, something he doesn't have, I love how you tell people they have misread quantum electrodynamics when your only experience with it is reading pop science books, while przyk (and many others who have pointed out your errors) can actually do some QED and other field theory. You don't read textbooks (by your own admittance), you can't do the mathematics and you have no knowledge of current research beyond pop science summaries you read in magazines or newspapers yet you presume to try to tell it to those who not only have read textbooks, completed lecture courses and read current research but also engage in peer reviewed research.

QED does not say a bound state of a photon is the electron or positron. They are entirely different quantum fields. The interaction means the energy and momentum is shifted from one field to another. Yes, it is possible the electron is a bound state of something else, something more fundamental but that is not what QED says, as it treats electrons as fundamental. So it is you who has misread QED. Well, I say 'misread' but you cannot misread something you never read in the first place.

Spend a little less time feeding your delusions of grandeur and a little more time reading (unless you are reading your own drivel).