Відео

It's Belgian.

The term Geometric Algebra is due to Clifford himself, see e.g. On the classification of geometric algebras (1882) by W. K. Clifford. The name geometric algebra was brought back by David Hestenes for the same reason that Clifford introduced it: to emphasize the fact that this is the algebra of geometry. Nowadays both terms are in use, although Geometric Algebra tends to be used by those who use an algebra over the real numbers to describe geometrical phenomena, while the term Clifford Algebra tends to be used when referring to more algebraic endeavors, and especially when an algebra is built over fields other than the real numbers.

​@@bivector Clifford couldn't very well call it "Clifford Algebra", now could he? When you use the earlier name, it is an attempt to erase Clifford's legacy, and an attempt to disguise other literature about the same subject. Clifford Algebra is not some mystical "algebra of geometry", there is no unique thing you can call the "algebra of geometry". Clifford Algebra is the algebra of spinor structures defined on top of geometry. This is part of what is maddening about this development, it pretends that GA captures some fundamental mystical geometric properties of the underlying space, rather than the properties of representations of the rotation group which acts on the space. If you can give me a reason to think of this differently, for example, if there is a nontrivial relation that emerges between some analog of geometric algebra defined over Q_p^n and the "p" of the finite field, that would justify using the name for that case of a finite field. But, as far as I know, and I am not an expert, in ALL the cases, the Clifford Algebra is just an additional structure sitting on top of the space, like a tensor section (tensor field in physics language) or a spinor section (spinor field in physics language), and I don't like shoving stuff into the geometry which is already perfectly well defined as a structure sitting on top of the geometry.

@@bivector Also, I would consider the Moebius strip and Klein bottle perfectly well defined geometries, and also CP2, but I don't see how you define a geometric algebra on those. Geometric algebra is an additional structure which only sometimes exists on manifolds.

@@annaclarafenyo8185 study this more before claiming nonsense.

@@ondrejstefik159 I know it better than the inventor.

still watching but e12 is the same as e1*e2. Notation.

@@eldersprig lol I get that ... I thought he was saying that the rhs was the same as something else where that I haven't noticed yet.

I am know that I ask too many questions to early and in hind sight I can see what folk are getting at, it's just that, like in a lecture class it's like hitting a brick wall and you have to ask .. and then you get a reputation for being a pain in the arse because it wasn't all obvious.

Thanks for the detailed feedback! (even if some things were resolved later - always good to know what those blocking moments were). Let me respond to the questions either way. 12:20 true - there's actually two options, either e1*e2 = e2*e1 = 0, or e1*e2=-e2*e1, we explore the second option as the first is quite uneventful. 15:50 we're just working through our first examples there so starting with an all positive metric allows us to show you still have elements that square to -1 etc .. 17:05 'e12' is the same as 'e1*e2' - even though we started from two completely different vectors. (re-illustrating the gauge freedoms inherent to the construction of k-vectors) Give yourself some time to let all of it sink in - its quite a departure from the standard approach - even if the algebra is simple, getting used to the new perspective just takes a while! And please do keep the feedback coming - much appreciated.

@@bivector Thank you for explanations, this is all quite amazing :-)

The fact that most CCs seem to be auto-generated these days is really a shame. I feel bad for what the hearing impaired have to go through. Siggraph should have the budget to pay for manual CCing.

What, you don't want to learn a chocolate-based algebra system?

There's a lot of stuff on STA (spacetime algebra) if you search for it. If you're interested in quantum physics stuff, search for Cohl Furey's work/videos and possibly John Baez. Some Chinese researchers last year published a paper on using the sedenions to build qft.

GR I would also be interested in seeing. Seems doable as GR is just tensor algebra and the electromagnetic field tensor.

@@AkamiChannel HR requires tensor calculus, and in a space with intrinsic curvature. Mathematically messy, and beyond my level of competence and bandwidth. Please publish if you hear of anyone exploring PGA for that.

This is why "SLERP" to interpolate from A to B is usually defined as (B/A)^t * A = exp(t*log(B/A)) * A. You are correct in that this isn't quite the same as exp(t*log(B) + (1-t)*log(A)), though in practice I have doubts that most people would actually notice. They should have the same end-points, but they'll likely take slightly different paths between them.

It's the consequence of angular momentum conservation in Hilbert space. Not sure how that is supposed to become classical. Hilbert space is still a quantum mechanical construct to describe an infinite ensemble. Such an ensemble is simply not necessary for classical systems. The deceptive part about the quantum mechanics of angular momentum is that it is a finite dimensional Hilbert space, which simplifies a lot of things and makes them look almost classical. If you do an actual experiment, however, then you will see that it's not classical at all. A series of Stern-Gerlach magnets turned by 90 degrees should prove to you very quickly that the number of physical degrees of freedom to explain the separation of one beam into two, then four, then eight etc.. can not be finite if we use classical theory. And it isn't. What people aren't explaining to you in beginner's classes for quantum mechanics is where those "infinite degrees of freedom" originate. They are the loss of knowledge about the state of the vacuum on the forward light cone, which is always in our future, i.e. it is fundamentally unknowable.

@@schmetterling4477 I am talking of the stern gerlach, where he shows it can be explained by a classical population of particles that align either parallel or antiparallel to B. But I am sure there is no way to do the same (I hope I were wrong) with a pair of entangled particles, that he did not discuss at all.

@@wolphramjonny7751 A classical population does not explain Stern Gerlach. That's the entire point of these experiments. In a classical population the state would be located in entirety in the individual corpuscle, which would require an infinite phase space per corpuscle. That model is unphysical even in classical mechanics because it neglects the back-reaction of the individual corpuscle on the magnetic field. The localization in angular momentum space is, instead, a collective phenomenon of the magnetic field and the electron (or silver atom if you go with the proper experimental details) field. Whatever quantum of angular momentum "the particle" gains, the field loses and vice versa. Since the em field extends to infinity and propagates at the speed of light, we can only tell that the balance of angular momenta is conserved but not where the missing or gained angular momentum of "the particle" went. This is rarely explained properly at the undergrad and layman level. QM is a good example of a near perfect theory that is being taught extremely poorly (I believe only thermodynamics takes the second rank in that category).

@@wolphramjonny7751 Entanglement is simply a really silly way to think about conservation and symmetry.