First video of a trilogy about minimal surfaces - surfaces that minimise surface area given boundary conditions. Please LIKE, SUBSCRIBE, COMMENT as usual. The next video is how soap films take the shape they do (rather than why). There are different conventions for the Gauss map: do you take the outward normal or inward normal? And what does outward or inward mean if the surface is not compact? There are also different conventions for the shape operator / second fundamental form, where you might see something like S(phi_u) = -n_u instead. Regardless, all those conventions differ by negative signs, so it doesn't matter too much. (Except for the people who define mean curvatures in such a way that the mean curvature of a sphere is negative. Or people who define shape opeartor such that on the unit sphere it is not the identity map. What are you thinking?)
@moonshine77535 ай бұрын
I remember my professor specifying that there are two different choices for the normal vector and you could choose whatever. (As long as it remains consistent of course)
@charlievane5 ай бұрын
Thanks
@mathemaniac5 ай бұрын
Thank you so much!!!
@melontusk73583 ай бұрын
Thank you for your contributions to this great channel.
@TTuCbKoTp9Ic5 ай бұрын
5:07 for those who like me didn't understand the reason why the normal vector can only be assosiated with a point of the circle and not the whole sphere, just imagine moving it across the cylinder along v. As it stays perpedicular to the surface, it can only be directed toward a point on the circle
@adriencances1345 ай бұрын
Astonishingly clear and well written, with great animations, keep up this amazing work man!
@krzysiekczajkowski4275 ай бұрын
Please do more on differential geometry. Great video!
@what-is-math5 ай бұрын
Mine are not as cool, but I have a full course. Come take a look ;)
@moonshine77535 ай бұрын
Love this. I just finished studying these things this semester and you nailed it! I'm surprised you managed to not mention eigenvalues, but I think it's good, they weren't really necessary to the video.
@mathemaniac5 ай бұрын
The omission of eigenvalues is a deliberate choice. To justify eigenvalues being real takes a bit more time (and it requires something about real symmetric matrices), and I am not entirely sure why that theorem is intuitively true.
@benhsu425 ай бұрын
Thanks!
@darkshoxx5 ай бұрын
Great video and easy to follow explanations. Quick Tangent 😉: I went through all of this at university in our second year, and we only ever got two visual demonstrations/examples about this, and I was surprised it didn't come up here. It was about how Gaussian curvature is a measure of internal geometry and therefore independent of external transformations. Meaning if you roll a piece of paper into a cylinder, they will both have the same Gaussian curvature (namely zero). And for the same reason, a slice of pizza that is curved by gripping the crust and bending it will remain straight towards the tip and not bend downwards, because that shape would have positive Gaussian curvature, but pizza has a Gaussian curvature of zero, so that can't happen. In other words, you can't bend a flat surface in two directions at the same time.
@mathemaniac5 ай бұрын
Well I have made this exact video before! However that video signficantly underperformed, and I have already mentioned it in this video in hopes of bringing it to attention again (though it doesn't seem to help that this video is also significantly underperforming).
@darkshoxx5 ай бұрын
@@mathemaniac oh dang, sorry to hear that. Best of luck
@smotala115 ай бұрын
Can't find that video. Do add a link, sounds exciting!
@chalkchalkson56395 ай бұрын
Btw this is a very important notion in physics! for example, even though the minkowsky metric is not trivial and on first glance things might look curvilinear, the space has 0 Gaussian curvature. Turns out - coordinates with 0 Gaussian curvature are exactly those frame where special relativity is sufficient to describe everything (if you are careful) and coordinates with non-zero curvature are exactly those where you get """gravity""". Both these notions must obviously be intrinsic properties of the space, rather than being representation dependent
@robertwilsoniii204816 күн бұрын
Which makes perfect sense because only gravity contains acceleration. That being said, special relativity is also incomplete because accrleration is the only way to *change* speed. So reality has to be described by general relativity to be complete. This also cements "time" as illusory imo, the time variable judt represents relative differences in the speed of simultanious motion due to differences in the relative accelerations in other directions to that motion slowing down those relative motions with respect to each other. @@chalkchalkson5639
@padraiggluck29805 ай бұрын
Thanks!
@Iovemath5 ай бұрын
8:56 Gaussian Curvature in a sense is the measurement of how the surface can reshape to a flat plane without distortion. A paper can easily curve into a cylinder by just roll it up. So the curvature of a cylinder is 0.
@_spartan117965 ай бұрын
Fantastic video!
@Phanimations5 ай бұрын
TO EVERYBODY WHO READS THIS. Please upvote this video, and maybe let it play in the background multiple times, it's legitimately some of the best content on youtube, and it's not getting the views it deserves. Since I've begun to work on videos of a similar genre and style, I know how much time and effort goes into not just creating, but VERIFYING everything, along with making sure its intelligible. It's wayyyy too easy to use math jargon and confuse the heck out of everyone watching. That he manages to break things down so well along with having stellar production value is not something the internet should take for granted. Tl;dr I'm glazing mathemaniac but it's well deserved
@mathemaniac5 ай бұрын
Not sure about the playing in the background part, but thanks for your recognition of efforts.
@Phanimations5 ай бұрын
@@mathemaniac Well... I understand that it's not exactly the best possible solution. But unfortunately this video isn't performing the greatest, and if someone, after they watch the video, lets it play in the background, then the watch time goes way up. Of course, this is your channel, but I am genuinely dismayed by seeing this incredible video not get the recognition it deserves
@chalkchalkson56395 ай бұрын
In case you are somehow looking for subjects to cover in this series, I think it would be really cool to discuss how this framework generalises to Riemann manifolds and metric tensors. Not only because it's really neat to relate curvature and different notions of distance (like "how different are our longitudes and latitudes" vs "how many meters are we apart"), but also because it's always fun when geoemtry and fancy physics intersect :)
@prashantkumarsingh61805 ай бұрын
Thank you for this wonderful explanation.
@ayte15495 ай бұрын
Surprised how few comments this video has, so I'm leaving my own. Great content!
@channalbert5 ай бұрын
Something about all this treatment is screaming "use geometric algebra" to me.
@mrervinnemeth28 күн бұрын
Well, the objects which measure the rate of change and live in the tangent plane are differential k-forms. The shape operator is in fact a geometric product, with the trace and and determinant being the two components of the geometric product, the inner and exterior products. I was also thinking about GA during the video. Topology will remain stuck at surfaces unless they start using k-vectors.
@jomilariola4435 ай бұрын
Great Video, Never expected Curvature being measured via linear algebra, but makes sense in retrospect Q: btw whats the background music? (ex: 1:50), it feels familiar…
@mathemaniac5 ай бұрын
I think the background music is automatically tagged by KZbin - just pull down the description to see. (These are copyright-free music provided by KZbin itself)
@jomilariola4435 ай бұрын
@@mathemaniac Thanks! Never knew that the auto tag feature existed (I should’ve scrolled a little further when checking the description)
@akshatrai90075 ай бұрын
The legend is back
@hansisbrucker8135 ай бұрын
Really nice. I learned a lot 🤓👍
@AwNaw1205 ай бұрын
Just curious based on your chosen topic: are you a fan of the works of Pavel Grinfeld (aka the mathisbeautiful channel guy)? Reading his texts really helped me wrap my mind around tensors, curvature, and minimal surfaces, not unlike your videos. And thanks for the channel, great stuff as always!
@mathemaniac5 ай бұрын
I have never heard of that person actually. Although for this video (or this coming series of videos) I haven't used it, the book Visual Differential Geometry and Forms by Tristan Needham is a gem.
@geraltofrivia94245 ай бұрын
Great content.
@kovanovsky22335 ай бұрын
Great video! Since I majored in physics and studied GR, I'm wondering if you're gonna cover Riemann tensor, Ricci tensor, and Ricci scalar. Would be great to see how they are connected to these two types of curvature (or how perhaps they are the same as these two) 😃
@mathemaniac5 ай бұрын
Very unlikely that I will cover those because there are already many videos out there talking about this. What you have said is basically intrinsic curvature, while in this video, they are extrinsic curvatures. However in 3D, the Theorema Egregium says that the Gaussian curvature is actually intrinsic.
@EarlJohn615 ай бұрын
"It's bendy..." "It has the same shape as my girlfriend... (hubba hubba)" I'm sure there are others that are nearly as descriptive. 😅😅😅
@Unknown-mf4of3 ай бұрын
Who is, no doubt, Canadian, so no one else knows her.
@ringoffire05 ай бұрын
I just learned all about this yesterday, what a crazy coincidence. Too bad I didn’t see this vid until now
@alexboche13495 ай бұрын
Nice Video! Can you clarify how/if this applies to higher dimensional manifolds?
@mathemaniac5 ай бұрын
Usually it wouldn't be defined this way for higher-dimensional manifolds. You can search for second fundamental form (or extrinsic curvature) for how it applies to higher-dimensional manifolds.
@nafizabdoulcarime50825 ай бұрын
Great video
@posqeak5 ай бұрын
How do we describe all spirals and vortices?
@alexboche13495 ай бұрын
Gaussian curvature discussed in other vid: kzbin.info/www/bejne/bXvYaplvpMl-jpY
@mathemaniac5 ай бұрын
I have already linked it in the description and the cards that are at the bottom of the description.
@TheJara1235 ай бұрын
Changing topic from Lie algebra...great... whatever topic you take we are with you!!! Thanks man...
@apteropith5 ай бұрын
i've been trying to work out something about the curvature of hyperboloids clearly, a hyperboloid embedded in ordinary 3D space has a varying positive gaussian curvature but, a hyperboloid generated by rotations in a minkowski space is a surface of uniform _negative_ curvature this makes some sense given that it's effectively an anti-sphere, but how one would _calculate_ that curvature as negative has never been clear to me but i'd previously been given oversimplified accounts of how gaussian curvature is calculated, so now i'm guessing that in this matrix-formulation the metric tensor has to be inserted somewhere somewhen (i've never liked the matrix formulations for minkowski spaces, because that "somewhere" and "somewhen" is deeply unintuitive to me)
@lucashou49205 ай бұрын
Doesn't this definition of curvature with normal vectors require the existence of some ambient space?
@graysoncroom5 ай бұрын
yeah this is the extrinsic view of curvature (in the style of Oneill's Elemententary Differential Geometry)
@mastershooter645 ай бұрын
Extrinsic and Intrinsic differential geometry are equivalent due to Nash Embedding Theorem
@Czeckie5 ай бұрын
does the choice of the normal matter? it seems to me it could change the sign of curvatures
@mathemaniac5 ай бұрын
That's addressed in the pinned comments. It could change the sign of curvatures, but as long as you are consistent in your choice of normal, it doesn't matter.
@dmitriycherniakin95524 ай бұрын
Can somebody explain why w lies on tangent plane?
@Mayo-r3v5 ай бұрын
يجب السرد التاريخي ويجب تعريف الحاجه لهذه الرياضيات الجميله والأسباب الحقيقيه لوجودها
@LinkenCV5 ай бұрын
Q: topologically similar shapes(for example, torus and tea cup) have the same overall curvature? Is this question even have meaning?(not matematician)
@mathemaniac5 ай бұрын
Well - I do plan on making a video on "total curvature" at some point, but not this time. (Probably the end of the year???) If you are a bit impatient, search for Gauss-Bonnet theorem.
@fariesz67864 ай бұрын
my maths brain hears: _B ϕ⃗ᵤ_ my linguistics brain hears: _bi faiyu_
@Svuem5 ай бұрын
Can someone explain to me (physics student) why they made us learn differential forms and pullbacks? And why you would ever need complex differential forms dz=dx+idy and dz*=dx-idy. I know how to compute stuff but I have zero idea how this relates to curvature
@nickharrison37485 ай бұрын
is it Reimann Geometry that define Curvature
@coloradoing91725 ай бұрын
Riemannian
@alephNull_9 күн бұрын
Im here trying to find out why a donut has zero curvature 😅
@Yoseph-ph7hh5 ай бұрын
new video!!
@jamesaston62155 ай бұрын
A fold is infinitely curvy?
@mathemaniac5 ай бұрын
Assuming the fold is a "limit" in the sense that the "turning radius" is getting smaller and smaller, then the Gaussian curvature is still 0, but the mean curvature diverges.
@fuuryuuSKK4 ай бұрын
even just an edge between two faces has that property of det=0, tr/2=div
@as-qh1qq5 ай бұрын
well made
@as-qh1qq5 ай бұрын
In the sphere map, the normal vector isn't on the surface of the sphere but at the origin...the sphere is the locus of the tip of all possible unit normals.
@Galinaceo05 ай бұрын
Isnt there a torus that isnt curved? (the "pacman torus" or something)
@mathemaniac5 ай бұрын
Yes, but this video is only for surfaces in 3D. I think the flat torus cannot be embedded in 3D, and can only be embedded in 4D.
@tetraedri_18345 ай бұрын
@@mathemaniacNo smooth embeddings exist, but C^1-isometric embedding of a flat torus in R^3 does!
@z0ru4_5 ай бұрын
Rien compris
@robertwilsoniii204816 күн бұрын
How ironic that you need to use straight lines to measure curvature. That just seems wrong. But it's apparant to me that gaussian curvature was the basis for gravitational probe b's measure of geodetics and frame dragging in earth's orbit. One thing to keep in mind is thst straight lines are necessary for the existence of irrational lengths and numbers. The proof of the existence of irrationals *requires* straight lines, therefore if space is curved then it *must* also be discrete. And if it is curved and discrete, *then* the use of straight lines (via differentials) to *measure* curvature is unjustified and *wrong.* There must be a way of expressing curvature without using straight lines of any kind.
@lateefahmadwanilaw8948Ай бұрын
❤❤❤
@Jonathan-ij2ky5 ай бұрын
g
@DrDeuteron5 ай бұрын
In cartography, there are a bunch of curvatures, but the two most popular are based on the radius of curvature of contour lines and the rate of change of slope along the steepest descent. They are available for the surface of the 🌏 at NASADEM….but they’re relative to the geoid, not the ellipsoid, and definitely not the tangent plane…so basically it’s a flat earth model because, ironically, that’s what geologists and hydrologists want.
@sallylauper82225 ай бұрын
What is the curvature of an actual real tree? A shape that's like genus 10,000+ with insane topological complexity. Does the term "curvature" even apply to a shape of this complexity? Is there a tricky mathematical way to estimate the curvature despite it's complexity? And thirdly, what the heck is Poincare's conjecture? Oh yeah, and does the measure of curvature have anything to do with the measure of angles in 3 dimensions?
@fuuryuuSKK4 ай бұрын
curvature in the sense it is used here is defined independently for each specific point on the surface, because different parts of an object can be curved differently. for example, the trunk of a tree, a branch thereof, and stems of its leaves all might be approximated as cylinders, at different rotations and scales, or perhaps as cones, which would just shift the circle the local normals describe a bit away from being in a plane with the origin in the direction the cone narrows to. I should note that the definition was given in a very concise, but perhaos unintuitive way. Curvature, as used here, is a function, which belongs to an object, takes a point on the surface of that object, then looks at the transformation matrix describing what happens to the normal vector (aka "away" direction) when the point gets jiggled around slightly, and then returns one of two base-independent properties of that matrix. This means that curvature can be, and usually is, different depending on which point you select. Two points could have the same curvature, but in that case these points' immediate surroundings would usually look very similar.
@MattHudsonAtx2 ай бұрын
Lindenmayer and Prusiencowics showed the geometry of plants and trees (really, nearly all biological growth) has fractal dimension.
@kovacsattila8993Ай бұрын
You can't describe curvature on point, only on a surface. But surfaces are made of points. Curvature can't exist, proved.
@bbryant4605 ай бұрын
0:00 zero seconds in, wrong. n-punctured sphere, torus and Klein bottle all have constant Gaussian curvature of 0. So no, I can readily think of surfaces that are flat.
@mathemaniac5 ай бұрын
I should have said "embedded" surfaces. The surfaces you mentioned still has curvature (extrinsic) when embedded.
@bbryant4605 ай бұрын
@@mathemaniacit depends entirely on the embedding. Take for example the embedding of the Klein bottle in E^4: (cos(v)cos(u), cos(v)sin(u), 2sin(v)cos(u/2), 2sin(v)sin(u/2)). This is an isometric embedding, hence the induced metric is the Euclidean metric.
@mathemaniac5 ай бұрын
Which has curvature, just extrinsic in 4D. The cylinder in 3D also has curvature because it has extrinsic curvature, even though of course its intrinsic curvature is 0.
@bbryant4605 ай бұрын
@@mathemaniac no, the extrinsic curvature of the above embedding is 0. (Historically, mathematicians were unsure if one could even do this for the Klein bottle) Your initial point was the viewer couldn’t think of other flat surfaces. You then changed the point to finding an embedding where the extrinsic curvature is zero. You’re now wrong on two counts. You can embed the an annulus in E^3 so that the extrinsic curvature is zero as well. I’ll leave that as an exercise as it is significantly simpler than the Klein bottle.
@mathemaniac5 ай бұрын
I can be wrong in this, but I'm not sure why the extrinsic curvature of the embedding is 0 in your parametrisation (or is that a typo?). I don't see how the normal vector can remain unchanged on the surface. Is there any reference for this? Equivalently, can you provide the normal vector that doesn't change across the surface? The annulus has extrinsic curvature 0 because it is a part of a plane. Because its closure has a boundary, it isn't exactly comparable to the Klein bottle. As far as I understand, if you want to have both intrinsic and extrinsic curvature 0 across all the surface, it has to be a part of a plane. The reason why I changed the point was my admission of omission of the condition that I wanted to specify in the first place, which should be clear from the context given in the entire rest of the video.