Thank you for making my medical licensing exam seem easy. This is the motivation I needed!
@PaPa-kr5yt Жыл бұрын
Almost same here. The animation is an art, and the lecture was a constructive (not really self-deceived lol ) procrastination and made my current work easy.
@alexmitchell53393 жыл бұрын
Here from Grand Illusions! Thanks Tim!
@tombarry97365 жыл бұрын
This is blowing my mind right now. The geometry of how 3 dimensions can be inverted without intersecting itself.
@2beJT5 жыл бұрын
What a great teacher
@andyeverett195712 жыл бұрын
What a privilege to view this, thanks to all that made this possible and to Professor Johnson for its clarity!
@epsilonkleiner011 жыл бұрын
I wish I'd known about this video earlier...it makes fibrations so much easier to understand! Thanks a lot Mr. Johnson!
@juliocesarmoreno59612 жыл бұрын
That is a long Johnson. No pun intended.
@kcflitter28797 жыл бұрын
This is the craziest coolest stuff, thanks for making these videos!!!
@seansmith54685 жыл бұрын
JRE brought me some good fibrations
@remusilie69585 жыл бұрын
me2
@HR-yd5ib5 жыл бұрын
Good old Eric weinstein and Lawrance Krauss should take a lesson from this guy on how to explain things clearly.
@Ramin23404 жыл бұрын
It's so weird I was half paying attention when it was brought up on JRE then a yearish later it's the topic of my final year Physics dissertation wtf is going on!!
@nickacelvn4 жыл бұрын
Ha ha you should write a song.
@mazumazu88733 жыл бұрын
Geometry is difficult for me.
@Icingde4th11 жыл бұрын
This was remarkably easy to understand and quite enjoyable, thanks!
@nickacelvn4 жыл бұрын
Explain in back to me in a nutshell!
@skyclaw4 жыл бұрын
This has got me revisiting all the knot theory I did for my MSc.
@Cognitiveleaper5 жыл бұрын
This fellow enjoys his occupation.
@MelodyAnnHM9 жыл бұрын
Very clear and helpful talk! Thank you for uploading!
@nateschmidt58354 жыл бұрын
Great visualization for those who can understand the math
@imathaddict11 жыл бұрын
Someone, somewhere, has passed the great opportunity to call Brougham Bridge as The Quaternion Bridge. Just saying. Neat lecture by the way. Congrats.
@rockapedra11302 жыл бұрын
I thought it went a little fast. Probably from trying to squeeze the whole thing into 50 minutes. I’ll prob just watch it again and pause every so often. Thanks! Loved the animation at the end!
@mtb4u2 жыл бұрын
This is fantastic. Very well explained!
@SpotterVideo9 ай бұрын
Conservation of Spatial Curvature: Both Matter and Energy described as "Quanta" of Spatial Curvature. (A string is revealed to be a twisted cord when viewed up close.) Is there an alternative interpretation of "Asymptotic Freedom"? What if Quarks are actually made up of twisted tubes which become physically entangled with two other twisted tubes to produce a proton? Instead of the Strong Force being mediated by the constant exchange of gluons, it would be mediated by the physical entanglement of these twisted tubes. When only two twisted tubules are entangled, a meson is produced which is unstable and rapidly unwinds (decays) into something else. A proton would be analogous to three twisted rubber bands becoming entangled and the "Quarks" would be the places where the tubes are tangled together. The behavior would be the same as rubber balls (representing the Quarks) connected with twisted rubber bands being separated from each other or placed closer together producing the exact same phenomenon as "Asymptotic Freedom" in protons and neutrons. The force would become greater as the balls are separated, but the force would become less if the balls were placed closer together. Therefore, the gluon is a synthetic particle (zero mass, zero charge) invented to explain the Strong Force. An artificial Christmas tree can hold the ornaments in place, but it is not a real tree. String Theory was not a waste of time, because Geometry is the key to Math and Physics. However, can we describe Standard Model interactions using only one extra spatial dimension? What did some of the old clockmakers use to store the energy to power the clock? Was it a string or was it a spring? What if we describe subatomic particles as spatial curvature, instead of trying to describe General Relativity as being mediated by particles? Fixing the Standard Model with more particles is like trying to mend a torn fishing net with small rubber balls, instead of a piece of twisted twine. Quantum Entangled Twisted Tubules: “We are all agreed that your theory is crazy. The question which divides us is whether it is crazy enough to have a chance of being correct.” Neils Bohr (lecture on a theory of elementary particles given by Wolfgang Pauli in New York, c. 1957-8, in Scientific American vol. 199, no. 3, 1958) The following is meant to be a generalized framework for an extension of Kaluza-Klein Theory. Does it agree with some aspects of the “Twistor Theory” of Roger Penrose, and the work of Eric Weinstein on “Geometric Unity”, and the work of Dr. Lisa Randall on the possibility of one extra spatial dimension? During the early history of mankind, the twisting of fibers was used to produce thread, and this thread was used to produce fabrics. The twist of the thread is locked up within these fabrics. Is matter made up of twisted 3D-4D structures which store spatial curvature that we describe as “particles"? Are the twist cycles the "quanta" of Quantum Mechanics? When we draw a sine wave on a blackboard, we are representing spatial curvature. Does a photon transfer spatial curvature from one location to another? Wrap a piece of wire around a pencil and it can produce a 3D coil of wire, much like a spring. When viewed from the side it can look like a two-dimensional sine wave. You could coil the wire with either a right-hand twist, or with a left-hand twist. Could Planck's Constant be proportional to the twist cycles. A photon with a higher frequency has more energy. ( E=hf, More spatial curvature as the frequency increases = more Energy ). What if Quark/Gluons are actually made up of these twisted tubes which become entangled with other tubes to produce quarks where the tubes are entangled? (In the same way twisted electrical extension cords can become entangled.) Therefore, the gluons are a part of the quarks. Quarks cannot exist without gluons, and vice-versa. Mesons are made up of two entangled tubes (Quarks/Gluons), while protons and neutrons would be made up of three entangled tubes. (Quarks/Gluons) The "Color Charge" would be related to the XYZ coordinates (orientation) of entanglement. "Asymptotic Freedom", and "flux tubes" are logically based on this concept. The Dirac “belt trick” also reveals the concept of twist in the ½ spin of subatomic particles. If each twist cycle is proportional to h, we have identified the source of Quantum Mechanics as a consequence twist cycle geometry. Modern physicists say the Strong Force is mediated by a constant exchange of Gluons. The diagrams produced by some modern physicists actually represent the Strong Force like a spring connecting the two quarks. Asymptotic Freedom acts like real springs. Their drawing is actually more correct than their theory and matches perfectly to what I am saying in this model. You cannot separate the Gluons from the Quarks because they are a part of the same thing. The Quarks are the places where the Gluons are entangled with each other. Neutrinos would be made up of a twisted torus (like a twisted donut) within this model. The twist in the torus can either be Right-Hand or Left-Hand. Some twisted donuts can be larger than others, which can produce three different types of neutrinos. If a twisted tube winds up on one end and unwinds on the other end as it moves through space, this would help explain the “spin” of normal particles, and perhaps also the “Higgs Field”. However, if the end of the twisted tube joins to the other end of the twisted tube forming a twisted torus (neutrino), would this help explain “Parity Symmetry” violation in Beta Decay? Could the conversion of twist cycles to writhe cycles through the process of supercoiling help explain “neutrino oscillations”? Spatial curvature (mass) would be conserved, but the structure could change. ===================== Gravity is a result of a very small curvature imbalance within atoms. (This is why the force of gravity is so small.) Instead of attempting to explain matter as "particles", this concept attempts to explain matter more in the manner of our current understanding of the space-time curvature of gravity. If an electron has qualities of both a particle and a wave, it cannot be either one. It must be something else. Therefore, a "particle" is actually a structure which stores spatial curvature. Can an electron-positron pair (which are made up of opposite directions of twist) annihilate each other by unwinding into each other producing Gamma Ray photons? Does an electron travel through space like a threaded nut traveling down a threaded rod, with each twist cycle proportional to Planck’s Constant? Does it wind up on one end, while unwinding on the other end? Is this related to the Higgs field? Does this help explain the strange ½ spin of many subatomic particles? Does the 720 degree rotation of a 1/2 spin particle require at least one extra dimension? Alpha decay occurs when the two protons and two neutrons (which are bound together by entangled tubes), become un-entangled from the rest of the nucleons . Beta decay occurs when the tube of a down quark/gluon in a neutron becomes overtwisted and breaks producing a twisted torus (neutrino) and an up quark, and the ejected electron. The production of the torus may help explain the “Symmetry Violation” in Beta Decay, because one end of the broken tube section is connected to the other end of the tube produced, like a snake eating its tail. The phenomenon of Supercoiling involving twist and writhe cycles may reveal how overtwisted quarks can produce these new particles. The conversion of twists into writhes, and vice-versa, is an interesting process, which is also found in DNA molecules. Could the production of multiple writhe cycles help explain the three generations of quarks and neutrinos? If the twist cycles increase, the writhe cycles would also have a tendency to increase. Gamma photons are produced when a tube unwinds producing electromagnetic waves. ( Mass=1/Length ) The “Electric Charge” of electrons or positrons would be the result of one twist cycle being displayed at the 3D-4D surface interface of the particle. The physical entanglement of twisted tubes in quarks within protons and neutrons and mesons displays an overall external surface charge of an integer number. Because the neutrinos do not have open tube ends, (They are a twisted torus.) they have no overall electric charge. Within this model a black hole could represent a quantum of gravity, because it is one cycle of spatial gravitational curvature. Therefore, instead of a graviton being a subatomic particle it could be considered to be a black hole. The overall gravitational attraction would be caused by a very tiny curvature imbalance within atoms. In this model Alpha equals the compactification ratio within the twistor cone, which is approximately 1/137. 1= Hypertubule diameter at 4D interface 137= Cone’s larger end diameter at 3D interface where the photons are absorbed or emitted. The 4D twisted Hypertubule gets longer or shorter as twisting or untwisting occurs. (720 degrees per twist cycle.) How many neutrinos are left over from the Big Bang? They have a small mass, but they could be very large in number. Could this help explain Dark Matter? Why did Paul Dirac use the twist in a belt to help explain particle spin? Is Dirac’s belt trick related to this model? Is the “Quantum” unit based on twist cycles? I started out imagining a subatomic Einstein-Rosen Bridge whose internal surface is twisted with either a Right-Hand twist, or a Left-Hand twist producing a twisted 3D/4D membrane. This topological Soliton model grew out of that simple idea. I was also trying to imagine a way to stuff the curvature of a 3 D sine wave into subatomic particles. .-----
@ivanbarbosa815 жыл бұрын
Great respect for teachers.
@flashinglightstf2 жыл бұрын
I absolutely love it and I have no idea why.
@andyeverett19575 жыл бұрын
It made more sense the 2nd or 3rd time through. Clear as it can get, the rest is my fault.
@CarnifaxMachine5 жыл бұрын
Well, it made sense the first 7 minutes or so. It's clear that to comprehend the whole thing I need to watch more Rick and Morty.
@jamma24612 жыл бұрын
Really nice clear talk. One tiny nitpicky comment is that when you are talking about fibrations, really you are just talking about fibre bundles - fibrations are a little more general.
@geometerfpv2804 Жыл бұрын
They are really not the same thing...a fiber bundle has homeomorphic fibers, a fibration has only homotopic fibers. That's a big difference. Oh what I would give for the path space fibration to be a fiber bundle...😂
@donrichards514 Жыл бұрын
the difficulties in expressing irrational coordinates
@gewinnste4 жыл бұрын
Isn't an easier way to envision S3 to think of the 4th dimension as time and then S3 "starts" as a small version of S2 (actually a point) and this sphere grows to actual S2, then gets smaller again to a point? Well the "growing-velocity" is cosinoid, is what needs to be added.
@geometerfpv28043 жыл бұрын
The point isn't necessarily to "envision" S3. The Hopf fibration can be used to get useful topological information about S3. Having a fibration (so, not just a parametrization or graph of S3, but a way of mapping it into a base space such that the fibers are homotopic to one another) gives you information about the homotopy groups of S3 (it gives an "exact sequence"). Fibrations are an important part of homotopy theory in general.
@NilesJohnson12 жыл бұрын
@jamma246 Ah, of course you're right! I should have made that comment during the talk :)
@chaoshengzhe12 жыл бұрын
this is damn better than the "dimension" video!!!
@dlbattle1008 жыл бұрын
Hamilton's original carvings are gone, but there's a plaque.
@sorrysonofa12 жыл бұрын
very good talk!
@nicholaswilliams45072 жыл бұрын
Outstqanding lesson. Thanks for posting.
@kinbolluck476 Жыл бұрын
No thank *you*
@spam20805 жыл бұрын
I don't speak this language. I need to learn mathematics.
@patrickleahey45745 жыл бұрын
but the visual art of this is accessible and beautiful
@kinbolluck476 Жыл бұрын
i like balls too
@honeybunny82226 жыл бұрын
Can you please make a video on topology verify the hopf bundles are in fact bundles
@nahkaimurrao4966 Жыл бұрын
I thought I would never understand what s vibration was but this is very simple
@kinbolluck476 Жыл бұрын
Feet plz
@jaimeafarah7445 Жыл бұрын
The quaternion q = W+iX+jY+kW can be expressed as a 2×2 matrix Q₂ₓ₂ = [ Q₁₁ , Q₁₂ ; Q₂₁ , Q₂₂ ] = [ W + iX , Y + iZ ; −Y + iZ , W − iX ] The determinant of Q is 1 e.g. det Q = W² + X² + Y² + Z² ≡ 1 hence Q⁻¹ = [ W − iX , −Y − iZ ; Y − iZ , W + iX ] The projection η = qkq⁻¹ in matrix form is : QKQ⁻¹ ≡ H where K is considered a quaternion in matrix form as: K = [ 0 , i ; i , 0 ] since η = qkq⁻¹ = q' = ( W' , X' , Y' , Z' ) similarly H = QKQ⁻¹ = Q' = [ W' + iX' , Y' + iZ' ; −Y' + iZ' , W' − iX' ] Then, H₁₁ = (QKQ⁻¹)₁₁ = 2i(WY + XZ) = W' + iX' which means W' = 0 and X' = 2WY + 2XZ. Also, H₁₂ = (QKQ⁻¹)₁₂ = 2YZ − 2WX + i(W² − X² − Y² + Z²) = Y' + iZ' thus Y' = 2YZ − 2WX and Z' = W² − X² − Y² + Z² The determinant of H = det (QKQ⁻¹) = det Q ⋅ det K ⋅ det Q⁻¹ = (1)(1)(1) = 1 = ( 2WY + 2XZ )² + ( 2YZ − 2WX )² + ( W² − X² − Y² + Z² )² which is a unit 2-sphere.
@Illuminateur10 жыл бұрын
These might be the same question, 1. Where in the 3-ball is the 2-sphere? 2. Why are the projected circles linked?
@IzzyisDizzyandFizzy3 жыл бұрын
This is so helpful, thank you!
@darkanderson87615 жыл бұрын
thank you, much better than in physics lesson on medington high
@donrichards514 Жыл бұрын
Thank You Well said.
@ryder112113 жыл бұрын
great teacher. thanks
@alexgil46234 жыл бұрын
Muchas gracias...
@sergeyten16834 жыл бұрын
In most important part of lecture you mostly see only lecturer's back. He is writing something on the board but it's almost impossible to see what he is writing
@GeonGeorge5 жыл бұрын
Where does one learn stuff like these?
@zachariahhanson17928 жыл бұрын
Can you trace out a shape on the 3-sphere that would not leave a circle in the stereographic projection? I'm guessing so. If this is correct then why don't we see any different shapes than circles in the (amazing) video at the end?... I'm also wondering about the line (the one circle that doesn't resemble a circle), what point on the 3-sphere is that projected from? -Surely if it's the point "at the top" (where the projecting light rays come from) then it should be out of sight infinitely far away? I'm hoping I'm making sense and that I don't sound horribly confused. Thanks!
@gastropuce95373 жыл бұрын
That line is actually a circle of infinite radius, that's why it looks like that.
@geometerfpv28043 жыл бұрын
The circles aren't an effect of the projection, they are the fibers of the Hopf fibration. The Hopf map from S3->S2 has fibers (inverse images of single points) each homeomorphic to the circle.
@rhaldryn75113 жыл бұрын
He has a roll of toilet paper handy if some student shits themselves trying to understand this.
@brennaboucher9995 жыл бұрын
Why wouldn't R x S1 be an infinite flat circle, like R x R with rounded corners? And why wouldn't S1 x S1 be an infinite sphere? Wouldn't a torus be better described by S2 x a hollow S1? Or couldn't that infinite cylinder be described by R x S2? Is there an assumption I missed?
@zakkatz91445 жыл бұрын
You have to think about the definition of the cartisian (cross) product. It takes every pair of points from S1 (a circle) and paires them to a point in R (the real line). So S1×R is the circle with strait lines going off to infinity. Im omitting some technicalities that would be covered in a rigorious math class.
@zakkatz91445 жыл бұрын
Similarily for the other cartisian products in your comment. S1×S1 is every point in S1 paired with a point of the same circle. I assume when you say infinite, you mean points of the sphere go on "forever". The teo circle, though, are "finite", so you will get a finite result, as you could verifying by googling the definition of cartisian product and testing it yourself
@brennaboucher9995 жыл бұрын
@@zakkatz9144 It sounds like you mean that a cartesian product is specific result of two finite matrices multiplied (describing a specific points, shapes or objects), and not the product of a platonic "line" or "circle" or "sphere" per se. Am I understanding you correctly? Thanks!
@zakkatz91445 жыл бұрын
@@brennaboucher999 I'd suggest you google the definition of cartisian product. It doesnt invovle matrices, but sets. If you're unfamiliar with the notion of sets, khan academy has some great videos on the matter. If you are, then A×B takes every a in A and b in B, creates an tuple, or ordered pair, (a,b), and returns a new set with every such ordered pair. So if A= {1,2}, B={a,b}. Then A×B= {(1,a), (1,b), (2,a), (2,b)}. Note that order matters: (1,a) is not equal to (a,1). That means that A×B is not necessarily equal to B×A.
@zakkatz91445 жыл бұрын
Also note sets need nod be finite. The real line is not a collection of finite points. The "finite" part was of S1×S1. Both have finite length, so the cartisian product will be "finite" in the sense that it will also have finite length
@mihail2638 жыл бұрын
Thank you for your wonderful lecture. I have a question about difference between Hopf band and (S^1)x(I). Obviously(?) they are homeomorphic. But do they have different fiber structure?
@eddiebATL8 жыл бұрын
+Mihail Arabadji I'd suggest posting this question in a place like reddit.com/r/learnmath or even reddit.com/r/math if /r/learnmath doesn't work. This video is approaching 10,000 views in the course of five years, (which I am, of course, really proud of). However, those reddits can get 10,000 views in a day and from professional mathematicians. Great place to ask for weirder math help.
@sebastianschulz15838 жыл бұрын
+Mihail Arabadji Well, they are indeed homeomorphic and (depending on what exactly you mean by that) have the same structure as fiber bundles. However, you might be missing an important point here, namely that you really want to look at the Hopf band as embedded in $R^3$ (or $S^3$, it doesn't matter for the point to come). Hence, you would really ask for a homeomorphism $R^3 ightarrow R^3$, that takes the Hopf band to $(S^1)x(I)$. That is impossible, as you would unlink the boundary spheres! Put differently, while the Hopf band seems trivial on its own, it becomes interesting once it is nested in $R^3$.
@redaabakhti7684 жыл бұрын
@@sebastianschulz1583 however the boundary circles can be unlinked in R^4 without crossing
@ricci17297 жыл бұрын
excellent...............................
@Czeckie3 жыл бұрын
9:00 isn't that a definition of a fibre bundle and not fibration?
@nickacelvn4 жыл бұрын
3.35 separated by what? Planck length?
@Crasshopperrr11 жыл бұрын
22:10 Brougham Bridge in Dublin!
@OscarMaris5 жыл бұрын
I'm surprised they were able to stop drinking long enough to make a bridge
@TheAgentJesus2 жыл бұрын
Are you fucking serious. THAT'S ALL A FIBRE IS?? WHY the HELL could NO ONE explain that before in terms that were anything less than absolutely mystifying?? Thank you so much for this lol
@geometerfpv2804 Жыл бұрын
Any textbook on topology defines a fiber as the inverse image of a point, it's not a secret 😅. If you are talking about youtube videos, most people talking about things like the Hopf fibration don't actually understand it. They are math/physics communicators, not mathematicians/physicists, and there is a big difference.
@kavishkhamesra3582 Жыл бұрын
#hopffibration
@kritikakhanwal6146 жыл бұрын
What is the meaning of having an inverse of q?
@NilesJohnson6 жыл бұрын
q^{-1} means the multiplicative inverse, using quaternion multiplication. This is called the "reciprocal" on the wikipedia page about quaternions.
@humilityconsulting74996 жыл бұрын
28:47 so where did the S¹ come from in this fibration?
@Scigatt6 жыл бұрын
Multiplying q by k^r(r real mod 4) from the right.
@bencopeland45145 жыл бұрын
Got here because of the JRE. I found this explanation far more comprehensible than Eric Weinstein’s.
@stefan92295 жыл бұрын
Weinstein likes the jargon too much. Half of not understanding him is not knowing the jargon. I think he does it because he likes to sound smart which is unnecessary because he is clearly incredibly smart
@Adam_Boy5 жыл бұрын
on the other hand this is a 45 min video in a particular format to just teach you this topic and nothing else. The JRE talk is just that - a talk, where they brush over this topic for about a measly 15 min or what? So you really can't compare apples with oranges here. Weinsteins himself also says that Joe is not a meathead that won't ever get what he is saying but they rather lack the time for him to understand the topic in that talk.
@georgewashington12004 жыл бұрын
jay maybe he meant geodude evolution
@cg76223 жыл бұрын
I got lost at the Cartesian product stuff oop-
@kritikakhanwal6146 жыл бұрын
how is the result obtained at 28.22, why is w=0
@NilesJohnson6 жыл бұрын
Ah -- I used w to mean "the first coordinate". I should have said "this point lies in the 3-dimensional subspace where the first coordinate is 0".
@robinnilsson83175 жыл бұрын
@@NilesJohnson Seriously? Not a single like from the lecturer answering a youtube question(s) with an direct and good answer in a channel that isn't his? Well Niles, just sit back - Joe Rogan has brought the views - and you shall get the recognition you deserve!
@NilesJohnson5 жыл бұрын
@@robinnilsson8317 Ha! Not sure I would want to get what I deserve, but I'm glad people are discovering this content :)
@SpontaneityJD5 жыл бұрын
Damn, I feel stupid now
@shootingwithguns33082 жыл бұрын
Erases the board. *eraser marks*... Looks about right.
@Hambxne4 жыл бұрын
oh eric, what have you done to by brain
@kpw84u2 Жыл бұрын
H is for Hamiltonians -- which are a whole other sidebar 😂😂😂
@McGroppy5 жыл бұрын
JRE brought me here...
@FerrariMekhari3 жыл бұрын
A chalk board lil
@gert-janscharstuhl44464 жыл бұрын
hmmmmmm donuts
@dlbattle1008 жыл бұрын
Wake up camerman.
@kasparasvisockas42895 жыл бұрын
This teacher is toochaotic, comes out as insecure of himself.