How to derive the volume of an n-dimensional hypersphere (the long version)

[@aidanhennessey5586]

I’m still so amazed by the fact that the 5th demential hypersphere has the greatest area. It seems so arbitrary and it hurts my head to try to visualize. Great vid

[@heinzhuberti3583]

As stated in the Wikipedia article of the n-sphere, it is not really reasonable to compare volumes of different dimension. You probably wouldn't compare a length of a line with the surface area of a square.

[@vynneve]

That observation is meaningless. As it makes no sense to compare values of different units. like, what is 1 meter to 1 meter^2 or to a volume? It's exactly like comparing 1 meter to 1 second. They are just totally different units. What you can do! Is check out the ratio that the hypersphere takes up, compared to a n-cube. That trend steadily decreases. You start with 79% (pi/4), 52%, (5D) 16%.... The more dimensions you add the less space the sphere "takes up" because of the extra degrees of freedom. This is a much more cool and meaningful observation imo.

[@ananay010]

I visit this video every year because honestly, this is beautiful.

[@fbg00]

Great video. Thanks for posting this. One point about terminology. At least in mathematics, we call the surface of the unit ball in 3 dimensions "the 2-sphere". Here the "2" indicates the dimension of the manifold which is the surface of the ball. So the circle is the 1-sphere. In n dimensions, one has the unit n-ball ith boundary the (n-1)-sphere.

[@NoNameAtAll2]

isn't 2-sphere its surface? not the insides

[@dylf14]

What an unexpected conclusion. That was some fantastic wizardry right there.

[@DeclanMBrennan]

Very neat and an unexpected link between hypervolumes and the Gausian distribution at least for me. We could also get rid of the Pi and do the volume entirely in terms of factorials. Starting with your equation for volume: Vn= Pi^(n/2) / Gamma(n/2+1) Looking at the demonimator and remembering that Pi^(1/2)= Gamma(1/2) which means Pi= Gamma(1/2)^2 So: Pi^(n/2)= (Gamma(1/2)^2)^(n/2) = Gamma(1/2)^n Which gives: Vn= Gamma(1/2)^n / Gamma(n/2+1) It's naughty of me but I like non-integer factorials. Gamma(n)= (n+1)! So finally we have: Vn= (3/2)! / (n/2)!

[@KaiHenningsen]

I remember once deriving hypersphere volumes (and, I think, surfaces) from first principles (mostly for fun), based on integrating the space under a circle, in increasing dimensions. Then some years later, I reconstructed the results (at least the volume, don't remember if also the surface) based on what I remembered of the results, where the main point was you get an additional factor of pi every two dimensions. Was many years ago, I don't remember much else. But I do remember that I never once needed the gamma function. The result was no more complicated that the well-known formulas for two and three dimensions.

[@jakenbaked87]

That result is interesting to think about. At higher dimensions, a unit radius seemingly doesn't provide enough "space" for the n-sphere to grow. We see that if you include the radius term, it's contribution grows faster than the gamma function, hence a sphere with higher dimensions AND larger radius grows as expected.

[@quasistarsupernova]

hi other jake :)

[@oscarmeyer516006]

very interesting! I am hoping to do Physics and Maths next year in university and I found this fascinating so thanks for uploading with less knowledgeable people in mind :)

[@yugeshkeluskar]

The Best video that exits on KZbin

[@FranFerioli]

This was enlightening on many levels, but I will never forgive you the joke about the gamma function and the hang man. NEVER…

[@chen-yunchang7105]

You saved my statistical mechanics homework! Thank you!

[@thomasmayer8590]

So the basic concept is integrating n times?

[@sam111880]

Yes, i like this derivation i have seen it before as well have done it before. I also derived it using the Jacobin determinant but its a mess doing coordinate transforms way. My only issue is the justification of saying that Vn(r) = integral (Sn * r^n-1 dr ). As you have to kind of assume this relation to be true for you to derive the Volume from the surface area.

[@evgeniiakorchevskaia3801]

Hi! Could you please explain the transition from \int S_n(r) dr to \int Sn \cdot r^{n-1}dr? Can't see where r^{n-1} came from. it's at 11:46.

[@AbhishekSachans]

Surface area of radius r in n dimensional space = r^(n-1) * surface area of unit sphere in the same space. (Just verify it quickly for 2 & 3 dimensions)

[@lychex]

Any one know what type of input device Dr. Quark uses? Looks like a very good device to take notes during lectures.

[@rebloyola5048]

Tku very much Is there any Physical reality attachable to a Vn tending to Zero?

[@luciasevillarodriguez9032]

very useful video and a very entertaining explanation, thank you very much!!

[@1lucasgrange]

I wonder why it peaks at n=5? Also, as your graph at the end is continuous, does that mean that this holds for fractional hyperspheres too?

[@toxxikanshul]

Q) why are we using this gaussian type curve for each dimension/cordinate? it may also happen that the curve for each ‘x’ value may not be gaussian type? Basically i want to know from where does this trick originates?

[@Kuribohdudalala]

Your videos are really good. Beside learning a lot, you got me to laugh at your hangman joke

[@sam111880]

As well as the r^n-1 term is a little wishy wash in it. As sure it make since by seeing the pattern in the 4 pi*r^3 /3 derivative = 4pi*r^2 . But that's just a special case. Sorry if my typing /spelling is a little misunderstood. It is cool to note that the sphere volume gets smaller to 0 in higher dimensions but the hyper cube/rectangles get infinitely big. It just shows weird things happen in higher dimensions. :)

[@shashankjoshi3863]

11:39 sir how do you write S(r)=S_n*r^(n-1)

[@5ty717]

Hence if volume/surface area is at a premium which is not at all established … we might gravitate toward an informational existence in a N=6 (spacial) hypersphere rather than N~5 or N~7 Nsphere?

[@leif1075]

Wait aminute! Pulling the e^x^2 term through the integral sign only works if the two functions are the same or if each function is defined only in terms of one of the two variables right? Because generally speaking the product of two intrgrals does not equal the integral of the product or a double integral of two functions

[@Ka97A]

Actually I was thinking about a pie just before you put the pie image in the screen. I really laughed :v pretty cool! Thanks!

[@Ray-Z-j4w]

Thanks for sharing!! It's sooo interesting.

[@harshitaapandey1402]

Hi, could you please explain to me how you got the dx1dx2 element to equal rdrd(theta)? How were you able to draw that graph?

[@sayanjitb]

i am assuming you are not familiar with coordinates and plots in different coordinates. dx1dx2 does imply area in let's say 2 -D cartesian coordinate system, so when you try to resolve area in plane polar system it becomes r dr d(theta). increment in r is dr and arc length is r*d(theta) [wish you know], so it boils down to area r*dr*d(theta)

[@quosswimblik4489]

Yes but how would you find it's log equivalent if you were using this N-Sphere exponent instead of a standard N-Cubed exponent.I imagine it would be a pretty curve but will it ever be seen.

[@cronos586]

What (hardware/software) do you use for drawing your workbooks?

[@AlbertArmea]

This looks like OneNote.

[@WilSnipeForFood]

Albert Armea I was just about to ask the same thing. thanks!

[@cronos586]

Looks like it was also answered on reddit: www.reddit.com/r/math/comments/2pv93i/how_to_derive_the_volume_of_an_ndimensional/ "Answer from Dr. Quark himself (2 years ago). "Xournal for the inking, Camtasia for screen capture; hardware is a regular tablet PC." Xournal is free (Linux and Win; OSX requires MacPorts + GTK extension). Camtasia is proprietary ($99). A Wacom Bamboo tablet may be a good substitute for a tablet PC."

[@LeoAlekseyev]

The software is a Windows port of Xournal on a regular tablet PC. OneNote or Windows Journal would probably also work.

[@p-swayang]

Thank you sir.... Really it is helpful. Keep uploading more and more such videos.

[@lostinsocorro]

Can I store the same amount of "stuff" in an n = 1 sphere in an n = 11 sphere? I'm trying to understand the concept of volume in n-space.

[@jeremylane3139]

yes, there are lots of copies of S_1 in S_11. For example, setting the last 9 coordinates to be 0, you get (x_1,x_2,0,...,0), and if x_1^2+x_2^2 =1 this describes points in S_11, but it also describes a copy of S_1, since there is a bijective (diffeomorphism) map from S_1 to S_11 taking (x_1,x_2) to (x_1,x_2,0,...,0). The thing to keep in mind though is that S_1 and S_11's volumes are measured with different units (i.e. meters versus meters^11) so this doesn't contradict the computations (because the two quantities aren't directly comparable).

[@seijurouhiko]

When he does the integration in polar coordinates, he is integrating r over [0,2pi] and theta over [0,infinity] which is wrong.

[@LittlePeng9]

Can I ask what type of software you used in this and other videos? Great explantation by the way

[@JannetSaif]

why you used guassian function for it?

[@euyin77]

I think that a recursion argument also could be used. Am I wrong?

[@casa1420]

In 1:48, hard or impossible?

[@mattacarregui3934]

I’m so confused. When u integrate a 1D line of length pi along a radius in 2D, u get 2*pi*r; with a 2D surface area(4pi*r^2) in 3D space you get the volume of a sphere. So why can’t you integrate the “surface volume” of the 4D sphere (8pi*r^3) to get the 4D hyper volume? (2pi*r^4)

[@rafaellisboa8493]

what is the name of the program you are using to write things down? looks very practical

[@JuliAn-qj2ld]

Probably Xournal

[@aderinsolajoshua1186]

which software are you using to write there

[@xxnotmuchxx]

Does that also mean the area and n-volume also decrease to zero as you increase dimensions?

[@LittlePeng9]

Oh, I forgot to ask: at one point you derived that 2r*dr=r^2. Doesn't dr mean infinitesimal value?

[@jjepsuomi9]

Cool, exactly what I was looking for! Thank you very much =) I was searching for this for the exact same reason as you were x) I'm doing a problem involving "curse of dimensionality" in Bishop's book "Neural Networks for Pattern Recognition" :)

[@dnquark]

+jjepsuomi9 The discussion in Bishop's book is very nice. I discovered it long after making the video.

[@adriantupper105]

Love this derivation. Still not sure how you jump from dx1dx2 to rdrd(theta). I know they are both elements of surface area but they are different shapes!

[@skyecassio1313]

Not sure if this ist still relevant to you or if you've figured it out by now... But since there might be others with the same question I'll try and answer it anyway. Essentially what you're doing is rewriting your original coordinates into others - for example, if you're going into polar coordinates, your original coordinates are (x1, x2, x3) (those are just the normal, euclidian ones that are perpendicular to each other). However, coordinates have the sole purpose of describing where something is, and as such you are free to build your own description using your own coordinates. If you're working with a cylinder it makes sense to use the fact that you have a set radius - this is going to be your first new coordinate (r). Next, you can look at the symmetry of a cylinder - in euclidian coordinates your symmetry axis would be z, so we are going to keep that as it simplifies some things. Now you have a radius and a hight, the only thing that's left to fully describe your position is where on the "floor plane" you are. Since you're dealing with a cylinder (with a circle as its cross-section) we are going to use an angle (theta) for that. Now if you compare your original coordinates with your new ones, you'll see that you can describe points on each of your original axes (x1, x2, x3) using your new coordinates (r, z, theta). You'll be left with x1=r*cos(theta), x2=r*sin(theta) and x3=z. Now, if you're dealing with a "rewritten" volume integral you have to consider the fact that it's not just the equation itself that needs integrating, but also your new coordinates. This is where the "r" in your r*drd(theta) comes into play. It's called the Jacobian and is the determinant of the matrix you get when you derive each of your original coordinates by each of your new ones. You also have to factor in where you are integrating to and from, as in it won't necessarily be from -1 to 1 (just an example) but rather from 0 to 2pi (if you're integrating over theta). The Jacobian will be different for each set of your new coordinates, like when you're using spherical ones it's r^2*sin(theta). Hope this helps.

[@escapiststupor]

love this a trillions times!

[@lucasmedina5097]

Wonderful proof :). What program did you use for the writing in this video?

[@WhyIsTheArray]

Thank you for this video. It was very helpful to understand.

[@dnquark]

What I'm writing at 9:42 is r^(n-1) becomes r^(n-2) (1/2) (2 r). There's an extra r that appears.

[@sayanjitb]

dear sir , i had known that sphere in 3-dimension sphere is denoted by S^2 ,but you showed that it was being S^3. i'am bit confused. please help ..

[@CheckeeAintAmused]

Why? Is multiplication of infinitesimals no longer commutative?

[@chimetimepaprika]

Thanks, my dude.

[@fmcamargo]

Great post, very explanatory.

[@Schlaousilein67]

Thank you for the video !

[@mcp866]

Can someone explain to me why we use the area under a Gaussian?

[@brentoids]

M CP if you watch it to the end you realize it is a hack to figure out the volume of an n-sphere. the integral of the Guassian to the n-th power can be written as the surface of an n-sphere times the half of the gamma function of n/2, aka 0.5*gamma(n/2). And the integral of gamma is sqrt(pi) which leads us to the solution that the surface area is (sqrt(pi))^n / (0.5*gamma(n/2)) .

[@sam111880]

The issue is if the relationship between volume and area doesn't hold in all dimensions then this weird counter-intuitive behavior may not exist so are math hinges on believing this relation. As such for higher dim stuff recently with dim 8 and 24 aka the leech lattice one can solve packing problems. I don't fully understand it yet. But there is some big areas that are not so clear to me. But like the parallel postulate in geometry we may just have to except some relations and show that we may have different geometries and volumes in them. So its highly problematic to generalize packing problems to higher dimensions without proper understanding /alot of choices to define stuff. But also beautiful math of polytopes volume formulas come out if we generalize in a natural manner using determinants of parallelepiped...etc. Ok enough rambling good video.

[@Jocularious]

Amazing video. I love it.

[@thomaschansler754]

The boundary of the disk is a 1-sphere, the 2-sphere is the boundary of a ball. Disk is standard enough terminology alone.

[@jonathandixon91]

Amazing! Thanks, helped me a ton

[@axdmlm]

And yet the volume of an n dimensional unit cube is always 1. Or is it?

[@owenscott573]

Great video

[@neurophilosophers994]

So if our universe is inside of a fractal holographic black hole universe/multiverse Riemannian Sphere Packing I’ve called it from Penrose’s Twistor Theory ideas, I think his Riemann spheres are virtual Kugelblitz black hole’s on event horizon, then you iterate this infinitely you find that at any arbitrary supra Planck dimension reality has 0 volume. Awesome

[@neurophilosophers994]

Which would make sense from a VSL cardinality spectrum viewpoint

[@seijurouhiko]

Change of Variables, its the jacobian

[@elislam158]

شكرا 🤍

[@Iohannis42]

I have been trying to define a wormhole.

[@annaisabanana6848]

wow... thank you so much!!

[@annaisabanana6848]

im currently doing a level maths and i dont think ive actually audibly been like "ahhhhhh" when i realised where something was going. Thank you again

[@rashmiranjansahu8271]

best video.

[@tengfeicao]

awesome!

[@dnquark]

@fbg00 D'oh! That's what happens when you let a physicist produce math videos :) I'm going to go yell at my math major friends for not spotting this.

[@noodle9754]

The part where the gaussian came into play was so syck

[@seijurouhiko]

but should be "d theta dr", not "dr d theta".

[@immanuelpanjaitan]

like this full