I cannot support you financially but I can only support you through my gratefulness for continuing this series. You are doing math's work. Thank you.
@richardchapman1592
6 ай бұрын
Know the feeling about not being ableto pay the wages of intellectuals.
@OelHardrada
10 ай бұрын
Happy for the new episode of this series. It has inspired me to share with fellow classmates.
@dr_drw
10 ай бұрын
Love this series
@andrewpeuny
10 ай бұрын
Nice job, looking forward to the following chapters
@blacklistnr1
10 ай бұрын
Math drinking game: Take 1 drink for every "basis" Take 2 drinks for every "thus"
@_anbh0143
9 ай бұрын
thanks for this amazing series !! its really exciting !!! keep up the good work !!
@UliTroyo
10 ай бұрын
Great as always, thank you!
@minamur
29 күн бұрын
imagine there's a basis for a space with fewer vectors than the standard basis for that space. that would mean that the the standard basis must have some extra vectors, but that would make the standard basis linearly dependent, i.e. not a basis. "the standard basis is not a basis" is a contradiction, so the premise, "there is a basis for a space with fewer vectors than the standard basis" must be false.
@Alexander-oh8ry
10 ай бұрын
I find the explanation of basis size = dimension too drawn out, i suggest keeping the non-geometric algebra shorter
@bubbacat9940
10 ай бұрын
You say that lines are represented as ax+by+c=0, but couldn't you represent lines in some other form likr y=mx+b or r=a*sec(θ+o) where there are only 2 dimensions?
@sudgylacmoe
10 ай бұрын
You could represent it some other way, but they aren't as useful as the form I showed here.
@APaleDot
10 ай бұрын
One problem with y = mx + b is that vertical lines are not representable because infinity is not a real number, therefore it doesn't represent the entire space of 2D lines.
@frsrblch
10 ай бұрын
The form y=mx+b has no orientation or magnitude, it is merely the set of all points where that equation is true. A line in geometric algebra has orientation and magnitude. Two lines could be overlapping, but could have different orientations (pointing in different directions), or have different magnitudes.
@Qebafhzn
10 ай бұрын
Is there such a thing as an infinite-dimensonal geometric algebra?
@sudgylacmoe
10 ай бұрын
Yes there is, as long as you can find a symmetric bilinear form there. However, some annoying things happen there, such as the geometric product not being continuous.
@frsrblch
10 ай бұрын
David Hestenes starts from this in Clifford Algebra to Geometric Calculus. Build the infinite-dimensional algebra first, and then every finite algebra is a subalgebra of that.
@2fifty533
9 ай бұрын
@@sunnythebridger7529what??
@sunnythebridger7529
9 ай бұрын
@@2fifty533 nvm, I just watched video while sleepy.
@PaulMurrayCanberra
10 ай бұрын
Numbers have an infinite dimension if they are considered to be products of prime numbers. Not sure about numbers mod n.
@TechJolt3d
10 ай бұрын
I'm keeping notes on these videos in a personal notebook, though i'm at 1.5, so gonna get here maybe in december if I get more free time
@kuretaxyz
10 ай бұрын
Hey! I'm first. Love your videos.
@Manisphesto
10 ай бұрын
Hmmm... What if we visualize 4D vectors as two 2D vectors in separate planes, just like how 2D vectors can be visualize as 2 scalars. Think about it like this (greek letters for scalars, and latin letters as vectors) αx + βy + γz + δw = u + v. Where w to z are basis vectors (pretend they have hats), u and v are vectors that are perpendicular to each other, and α to δ are scalars. I wonder what geometric algebra with 4D vectors be like...
@sudgylacmoe
10 ай бұрын
I actually do use this at times, but it's still not a way to visualize the 4D space itself.
@user-fl5nv7oh3z
Ай бұрын
Most people enter math world from a real world. As you can enter, there has to exist a transformation from the math base MB to the real world base RWB. A simple number c in the real world can be seen as a vector c*{1}, so c is just the length of the vector {c}. Let c be 1, so this vector can act as a base vector. I now embed this vector in a n-dimensional space, such, that the vector is represented in MB by { 1/sqrt(n), 1/sqrt(n), 1/sqrt(n), ...}. I now can expand the RWB to 3 dimensions and the 2 additional dimensions span a plane normal to c. How will the projection of the MB to this plan look like?
@rtg_onefourtwoeightfiveseven
10 ай бұрын
11:31 Could a basis for (say) all functions R -> R not be, for example, the set of all functions on the real numbers that equal 1 at one point and 0 everywhere else? This set is linearly-independent, and every function can be expressed as a combination of these functions by simply representing the function pointwise. Is there some quirk to do with the fact that the basis is uncountable that keeps us from calling it a fully-described basis of functions R->R? Or is it not valid for some other reason?
@sudgylacmoe
10 ай бұрын
It's not valid because to be a basis, every vector must be able to be written as a *finite* linear combination of the basis vectors. Even though there is a generalization of bases called Schauder bases that allows for infinite sums, it still only works for countable infinity.
@rtg_onefourtwoeightfiveseven
10 ай бұрын
@@sudgylacmoe I see. I wasn't aware of the finite restriction, thanks. Does the hypothetical basis for the function space describe all functions using finitely many basis elements, or does that require Schauder bases?
@sudgylacmoe
10 ай бұрын
It manages to use finitely many basis elements. And I wouldn't say it's hypothetical, it does exist, it's just that we can't describe it.
@ApriiSR
10 ай бұрын
@@sudgylacmoe huh, why doesn't it work for uncountable infinity
@sudgylacmoe
10 ай бұрын
The very definition of a sequence is a function from the naturals to some other set. All notions of series, limits, etc. relies on this fundamental idea, which is only defined for the natural numbers. Now that I think of it, you could maybe use ordinals, but that gets more hairy because there's no natural way to well-order the real numbers.
@evandrofilipe1526
10 ай бұрын
We are so back edit: 12:20 chapter 7 confirmed??
@cmilkau
9 ай бұрын
Numbers are an interesting case of linear spaces in the context of field extensions. For instance, the complex numbers can be considered a 2D real vector space, and the real numbers can be considered an infinitely-dimensional rational vector space.
@AMADEOSAM
10 ай бұрын
Thank you! Great and Top
@cmilkau
9 ай бұрын
5:30 Before you can assign the number of basis vectors to the dimension, you have to prove that all bases (of that space) have the same number of elements.
@gabitheancient7664
10 ай бұрын
exercise 1: I'm not sure what do you mean my dimension here, so I'll assume it's geometric dimension (which tbh I don't know how to describe) my argument: in an n-dimensional space, you have n "directions" to project your vector to to decompose it in, if you first decompose one vector and draw lines through each decomposition vector, you can, with any other vector, just project it on these lines to also decompose it in respect with those vectors, because again, you have n vectors pointing to n different "directions"
@davidhand9721
4 ай бұрын
I don't understand why people find it so difficult to illustrate 4D. Use an animation. Each 3D frame is a slice. You can use some gradients to give an impression of 4D locality if you wish. You can use color as a 4th coordinate, too. It's not so hard.
@kristoferkrus
6 ай бұрын
11:00 But don't you have a redundancy, since for example 1x + 1y - 2 = 0 is the same line as 2x + 2y - 4 = 0? I'm not sure I agree that the number of dimensions of lines in 2D is 3; the parameters a, b and c that you use to represent the lines forms a three-dimensional space, but I don't think the lines themselves do. To see this, we could also imagine that we choose to add the restriction a^2 + b^2 + c^2 = 1. We will still be able to represent the same set of lines as before, but the equation a^2 + b^2 + c^2 = 1 is the equation for a sphere, which is a two-dimensional object, so now a, b, and c form a two-dimensional space, not a three-dimensional one. Also, an alternative but equally valid representation of the lines are with just an angle α and a distance d from the origin, such that cos(α)x + sin(α)y = d, and that representation is just two-dimensional. *Edit:* I guess that the word "dimension" _is_ multidimensional and can be defined in different ways, as you said, so maybe you can define it as the number of parameters used to describe an object as well, and I guess that by that definition a line in 2D space could be three-dimensional, but only if it is parameterized by three parameters. But this would require the parameterization of the line to be a property of the line itself, which feels a bit odd to me.
@sudgylacmoe
6 ай бұрын
A lot of people have said this before. A more accurate way of saying it would be "This way of representing lines is 3D". You can't just ignore the scalar factor because even if the equations x + y - 2 = 0 and 2x + 2y - 4 = 0 represent the same line, when adding, they can produce different results. See kzitem.info/news/bejne/tWaYnmGEg3pobKg for a video I made exploring this linear space in more detail.
@lucassiccardi8764
9 ай бұрын
Very interesting video and very interesting series: thanks for your work! One question: the "absurd" definition you mention at the end of the video, that "1D objects in 3D space can be described as the 6D subspace of 2D elements in the 16D algebra created by the 4D space of 2D objects in 3D space" ...where does it come from? Did you make it up or is it something that comes up in a proof? The fact that 6D subspaces and 16D algebra are connected by the definition interests me very much, and I'm curious, however theoretical this may be, what was the original context for such a definition. Cheers!!
@sudgylacmoe
9 ай бұрын
It comes from PGA, and I have a video on it here: kzitem.info/news/bejne/kZ9p1ZeCoZSpfWk. In particular, it's talking about how lines in 3D space are represented by the bivectors in 3D PGA.
@lucassiccardi8764
9 ай бұрын
@@sudgylacmoe Thank you very much! Great channel!
@Alan-zf2tt
10 ай бұрын
Hmmm ... @ 3:55 or thereabouts ... (A) proposed third basis will either intersect 2-d plane completely (or) exclusive-OR intersect at one and only one single point. Proposed third basis is either a linear combination of two earlier bases whether these are orthogonal or not OR an acceptable basis orthogonal to 2-d plane accordingly? Constructs a third basis independent of two earlier bases. (B) If a point exists in n-space and is not reachable by a vector from some arbitrary origin in n-space then that spanning system lacks completeness regardless of how many vectors are used in its proposed spanning set? So construct a proposed additional basis that satisfies (A). Iterate this operation making allowances for increases in dimension of event space if that is unknown until the point becomes accessible from such an arbitrary origin. (C) Test proposed spanning set to see if it has bases numbering more or less than dimension of space it is proposed to span. Adjust members of spanning set until it has n-components in n-space. This should ensure independence and spanning of n-space of each additional basis even if the original two were spanning but not orthogonal in first place. The basis in original 2-space is allowed to be non-orthogonal. (D) any new proposed basis can be tested against each existing 2-plane spanned by any two bases discarding any proposed basis if it does not satisfy condition in (A) that it intersect subsequent 2-planes at one and only one single point. Allowances are made for any dodginess in original 2d spanning set. The only exception to this limit of intersection at one and only one point in any enclosed 2-plane is trivially accepting that all spanning sets will intersect at some arbitrary origin and also at infinity. As the test for infinity is beyond scope of this comment it is not dealt with here (should keep analytics and calculus bods happy?)
@TheNerd484
10 ай бұрын
I think if you imagine dimensions as degrees of freedom, the lines in 2d example makes perfect sense
@sudgylacmoe
10 ай бұрын
Yes, but the whole point is that there are different ways to think of dimension.
@TheNerd484
10 ай бұрын
Of course, I was just thinking of that particular example@@sudgylacmoe
@angeldude101
10 ай бұрын
But there's the matter of tilting your perspective slightly in such a way that 2D lines really do appear to only have 2 degrees of freedom: distance from the origin, and _angle._ In the 3D linear space, the angle is split into a linear combination of two orthogonal lines, but it's possible to change the coefficients of said basis lines without changing the actual line being represented. There's a redundant degree of freedom despite still having 3 linearly independent algebraic dimensions. Just to add more confusion to what "dimension" means.
@05degrees
10 ай бұрын
@@angeldude101 well actually 2D lines may be placed to live in a 2D projective space (with only one extra line there, so it shouldn’t change the dimension, intuitively) so yeah. But also an nD projective space is most helpfully defined as an (n+1)D vector space (with zero vector excluded) where scalings of a vector are all treated as the same thing. So in this sense there are these two dimensions differing by 1. Though I still would say that the space of 2D lines is 2D, even if for calculations with it we would use 3D vectors.
@PaulMurrayCanberra
10 ай бұрын
No it doesn't. Lines have two degrees of freedom, not three. Ax+By+C=0 describes the same line as 2Ax + 2By + 2C = 0.
@philipoakley5498
10 ай бұрын
And then you have the real physics dimensions of: Length, Time, Mass, Temperature, electric Current, Molecular quantity, Light intensity, and the horrific angular units (uncounted cancellations;-). It's a nightmare!
@Nick_Tag
8 ай бұрын
silly question... don't fractals allow non-integer dimensionality? But then i guess basis set would be non-integer.
@sudgylacmoe
8 ай бұрын
There are many different definitions of dimension used in different parts of mathematics. With fractals, the definition used is Hausdorff dimension, which is different than any of the uses of the word "dimension" in this video.
@Nick_Tag
8 ай бұрын
wow thanks. For context I mainly asked because Wolfram's physics project mentioned it is (currently) plausible for our universe, and I see GA, WPP (& DeBroglie-Bohm) as similarly powerful; as an alternative / intuitive frameworks to re-interpret known expressions, and to possibly derive new predictions (as WPP did for an upper speed limit on entanglement)
@guidosalescalvano9862
10 ай бұрын
An interesting question is what the basis of the dimension vector is. Bivectors seem like different beasts than univectors in terms of the dimensionality vector…
@TheGildedMackerel0
10 ай бұрын
woot!
@Pluralist
9 ай бұрын
@HoSza1
10 ай бұрын
[a1, a2, ..., an, ...] ↦2^a1 * 3^a2 * 5^a3 * ... * pn^an is a bijection between an infinite dimensional space (vectors of infinite length) and a one dimensional space (nonnegative integers) where pn is the n-th prime numbers. But wait, shouldn't bijections be possible only between same dimensional spaces?
@sudgylacmoe
10 ай бұрын
The thing that's required to be between same-dimensional spaces is bijective linear transformations, not just bijective functions. This function is not a linear transformation.
@HoSza1
10 ай бұрын
Surely the restriction to linear functions makes it easy to prove a theorem about it, and I wonder if it's possible to require something less strong additional condition, that still guarantees the same dimensionality.
@YouTube_username_not_found
10 ай бұрын
Does the concept of dimension even make sense in this example you propose🤔?? How are you defining "dimension" here? Surely it isn't "cardinality of basis". This wasn't the initial reply I was going to make, but after re-evaluating the example, I realized a much more crucial detail in it that deems it flawed. I will not reveal it now, maybe you would 🙂 like to try find it yourself. Can you find it?
@HoSza1
10 ай бұрын
@@KZitem_username_not_found I'm no psychic so I can't know which of the missing details confused you. I admit I omitted the descriptions of the base fields, but that's not that hard to figure out. Of course the elements of the vectors above are not simply natural numbers but we need to define a field over the naturals, in order the example to be correct and complete. The dimension is the cardinality of the base as usual, which is infinite in this case for the vectors.
@YouTube_username_not_found
10 ай бұрын
@@HoSza1 "I'm no psychic so I can't know which of the missing details confused you." I feel like you didn't read my reply 😐. There is something in the example that makes it flawed and that's what I asked you to find. Or maybe what you imply by this sentence is that there is nothing wrong and that I am merely confused. I am not 😑confused. "I admit I omitted the descriptions of the base fields, but that's not that hard to figure out." You omitted more than that. This is the issue I was going to adress in my original reply but didn't after noticing the bigger issue. And indeed it is not hard to figure them out, this isn't my problem here. "Of course the elements of the vectors above are not simply natural numbers" You mean the components of vectors? Using the word _element_ is a little confusing since it usually refer to _element of a set_ . At first I thought you were referring to the elements of the ending set/codomain which are definitely the nonzero natural numbers. But here I should stop you. *You* are the one who said _"(nonnegative integers) where pn is the n-th prime numbers"_ , *now* you say *they are not simply natural numbers* 🤨?? the example as you have described it is about a bijection that maps between each positive integer and the infinite vector containing the exponents "an" of its prime factorisation, those can be *only* natual numbers. If not, then what can they be ?? There is no other option, otherwise you will *modify entirely* the sets being considered. Also, notice how only finitely many "an" are nonzero, so we are not working with the whole set of infinite length vectors (let's call them sequences from now on). "we need to define a field over the naturals" OK, *now* I am confused 😕, what is _"field over the naturals"_ supposed to mean?? Note that I do know what a field is (otherwise I wouldn't watch a video or comment about something I still haven't fully grasped). We say "a vector space over a field". I am not sure what you are talking about here. The issue is, the sets of natural sequences and positive integers aren't vector spaces, both of them lack the existence of inverses in them, this is without mentioning that there is no way to make them closed under saclar multiplication since scalars form a field which requires us to leave the integers and go to at least the rationals. "The dimension is the cardinality of the base" And since we no longer have a vector space structure, the notion of _basis_ is no longer well-defned. At least one should refine the definition or try to generalize it, if there is any chance to make the notion makes sense in this setting.
@emjizone
8 ай бұрын
8:32 Although n*1=n this is true for numbers, this relation is *NOT specific to numbers.* In fact, *this relation is true for ANY vector!* So *I don't see how this proves that numbers can't be be linearized in more dimensions.* I mean I really thing this is *not* a proof. For example, octonions are numbers, right? Well, can't you linearize the space of octonions in 8 (infinite and independent) dimensions?
@sudgylacmoe
8 ай бұрын
At that part of the video I am specifically talking about real numbers. I am writing this with high school students in mind who might not even know about anything past the real numbers.
@Alan-zf2tt
9 ай бұрын
Hmmm number 2? In examples of polynomials and geometric spaces and maybe others is an important theme: "nested" space? With either of above two examples it is easy to imagine and accept 1D space is nested within 2D space is nested in 3D spaces is nested in ... and so on. I saw a KZitem video linking physics to linear algebra in which notion of a vector is not too important. Maybe this was a case of treating number of exclusive variables as a separate dimension - I cannot find that video now - shame - Vectors and linear algebra were used with 5 and 7 bases. 5-based one used Planck scale variables and 7-based one used mass, distance, ... type of things. I wish i had saved it
@sudgylacmoe
9 ай бұрын
That nested space idea is subspaces. I talk about them partway through this video: kzitem.info/news/bejne/sGpmzq6ocGN_Y6w
@Alan-zf2tt
9 ай бұрын
@@sudgylacmoethank you for reply and link. They are appreciated. I wondered if isomorphisms worked between similar subspaces all separated from each other but contained in a larger subspace. It seemed a profound thought at the time
@Alan-zf2tt
9 ай бұрын
@@sudgylacmoe thank you for your reply. Were I attending your lectures the next question I would ask would be along lines of: does the math on subspaces force a universal origin on all nested subspaces or could any number of subspaces exist even tho they do not share a common origin? Maybe I am also asking is common zero (origin) common to all subspaces? I think it would be pleasant if subspaces in a space could all have different zeros/origins
@sudgylacmoe
8 ай бұрын
@@Alan-zf2tt Sorry, I forgot to reply to this. The standard theory of subspaces requires subspaces to have the same origin as the original space.
@Alan-zf2tt
8 ай бұрын
@@sudgylacmoe thank you
@GamingKing-jo9py
7 ай бұрын
lines in 3D space can be described as the bivector subspace in the 4D complex space in 3D space. conplex referring to either the complex plane or maybe just a regular plane did i get this part right? it still doesn't make sense
@sudgylacmoe
7 ай бұрын
It's referring to PGA. Lines in 3D space can be described using bivectors in 3D PGA, which is built from the algebraically-four-dimensional space of planes in 3D space.
@GamingKing-jo9py
7 ай бұрын
just rewatched the swift addendum on pga, and now it makes sense! very cool!
@tablettorrensabellan
3 күн бұрын
Sorry, I don't accept that the minimum number of parameters to define a line in 2D is 3. You just need the slope and intercept. And you can just obtain those from the 3 parameters by dividing the whole equation by b. Then the parameter for y is always 1 since b/b =1.
@sudgylacmoe
3 күн бұрын
I never said that the minimum number of parameters to define a line in 2D is 3. I just was saying that in this particular representation (which is surprisingly useful), the number of parameters is 3. Also your representation doesn't work because it can't represent vertical lines. One way to truly represent 2D lines with two components is to use an angle and a distance from the origin.
@tablettorrensabellan
2 күн бұрын
@@sudgylacmoe Thanks a lot for your answer!! I really think that you are among the top ten best expaliners in KZitem!!! Your videos are crystal clear, and not boring at all... Congratulations!!!
@tedsheridan8725
10 ай бұрын
6:50 I don't understand this argument that you can't truly visualize 4D space. We have 2D vision, and perceive 3D objects one 2D perspective at a time. You can do the same thing with 4D objects, as you show. So if we can't "truly visualize" 4D space, then you can't "truly visualize" 3D space. And yet we do every day. The trick is to learn how to think in 4D so that animations of 4D objects make sense. It is by no means easy but certainly not impossible.
@kovanovsky2233
10 ай бұрын
I don't think 2D lines are actually 3 dimensional since we can set a=1 (in ax+by+c=0) and the equation can still describe all possible 2D lines. Therefore, 2D lines are 2 dimensional.
@sudgylacmoe
10 ай бұрын
That won't be able to describe horizontal lines, because there a = 0. But even still, that's not a linear space anymore because adding two lines with a = 1 would produce a line with a = 2. Even if it seems like there's an extra scalar factor hanging around in there, it has ramifications in how you add vectors, so it can't be ignored.
@kovanovsky2233
7 ай бұрын
@@sudgylacmoe Oops, I think I meant b=1 (so it can be arranged as y = mx+c line formula). But I see your point, if b=1, I wouldn't be able to describe vertical line instead. Thanks for your answer.
@ayouliyouli
7 ай бұрын
I am under the impression that you voluntarily confuse the dimension of the object, the dimension of the space containing the object, and the dimension of the set of similar objects in everything you say. Here I use the word "dimension" always in the sense "size of the basis". The fact that they are all called dimension should not be confusing. Yes, the set of two-dimensional lines is of dimension three, even though we are talking about 1 dimensional objects living in a 2 dimensional space. Instead of using a confusing vocabulary, you just have to be precise in your wording. So in the short about quaternions, when you say "quaternions are not 4D objects" and conclude that they should be considered 3D objects geometrically, what exactly do you mean ? Do you mean that a single quaternion can be seen as a 3dimensional object in some space ? Do you mean that a quaternion can be seen as an object that lives in 3D space ? Do you mean that the set of quaternions is 3 dimensional ? (I know the last one is not what you mean, but I had to ask for symmetry). What I'm actually understanding is that you can associate to each quaternion some combination of 2D planes living in a 3D space. Is that what you mean when you say that quaternions are 3dimensional objects ?
@sudgylacmoe
7 ай бұрын
About quaternions, I mean that they live in three-dimensional space. They're a combination of a scalar and a plane in 3D space, and they naturally represent three-dimensional rotations. Geometrically, everything about quaternions is three-dimensional. I don't know if you read my whole response on that short, but it was partially in response to other videos I've seen where they try to use four-dimensional geometry to describe quaternions, which is a huge hindrance and seriously misses their point.
@ricardodelzealandia6290
5 ай бұрын
As an interesting aside, you say "thus" using an A#.
@sudgylacmoe
5 ай бұрын
...I have semi-perfect pitch, and now I will never be able to not hear this. I can hear it in my mind already.
@ricardodelzealandia6290
5 ай бұрын
@@sudgylacmoe I'm sorry. If it's any consolation, you're in A in the subsequent videos. 😁
@NotNecessarily-ip4vc
9 ай бұрын
For well over 300 years (ever since Newton vs Leibniz) we have defined 0 and 1 (and their geometric counterparts) as follows: 0 = not-necessary 0D = not-necessary 1 = necessary 1D = necessary (Newton won so above are his definitions. Newton conflated "natural" with "necessary" and was largely ignorant of Geometry.) A year ago quantum physics proved that Leibniz was actually correct (the universe is "not locally real") which looks like this: 0 = necessary 0D = necessary 1= not-necessary 1D = not-necessary Since Mathematics > Physics > Chemistry > Biology... the implications of the definitions of 0 and 1 changing are world altering. "Only the zero-of yourself is necessary" is now a true statement. That's neat to think about. A little over a year ago the zero-of yourself was not-necessary. See how the facts change over time? Newton really set humanity back with his conflated definitions. Zero is the most important number in mathematics and is both a real and an imaginary number with a horizon through it. It's geometric counterpart zero-dimensional space is the most important dimension in physics and is both a real and an imaginary dimension with an event horizon through it. Quarks are zero-dimensional color-charged electricity and the Monad is the zero-dimensional space binding our quarks together with the strong force; the hue-monad (or soul).
@NotNecessarily-ip4vc
9 ай бұрын
Monad (from Greek μονάς monas, "singularity" in turn from μόνος monos, "alone") refers, in cosmogony, to the Supreme Being, divinity or the totality of all things. The concept was reportedly conceived by the Pythagoreans and may refer variously to a single source acting alone, or to an indivisible origin, or to both. The concept was later adopted by other philosophers, such as Gottfried Wilhelm Leibniz, who referred to the Monad as an *elementary particle.* It had a *geometric counterpart,* which was debated and discussed contemporaneously by the same groups of people. [In this speculative scenario, let's consider Leibniz's *Monad,* from the philosophical work "The Monadology", as an abstract representation of *the zero-dimensional space that binds quarks together* using the strong nuclear force]: 1) Indivisibility and Unity: Monads, as indivisible entities, mirror the nature of quarks, which are deemed elementary and indivisible particles in our theoretical context. Just as monads possess unity and indivisibility, quarks are unified in their interactions through the strong force. 2) Interconnectedness: Leibniz's monads are interconnected, each reflecting the entire universe from its own perspective. In a parallel manner, the interconnectedness of quarks through the strong force could be metaphorically represented by the interplay of monads, forming a web that holds particles together. 3) Inherent Properties: Just as monads possess inherent perceptions and appetitions, quarks could be thought of as having intrinsic properties like color charge, reflecting the inherent qualities of monads and influencing their interactions. 4) Harmony: The concept of monads contributing to universal harmony resonates with the idea that the strong nuclear force maintains harmony within atomic nuclei by counteracting the electromagnetic repulsion between protons, allowing for the stability of matter. 5) Pre-established Harmony: Monads' pre-established harmony aligns with the idea that the strong force was pre-designed to ensure stable interactions among quarks, orchestrating their behavior in a way that parallels the harmony envisaged by Leibniz. 6) Non-Mechanical Interaction: Monads interact non-mechanically, mirroring the non-mechanical interactions of quarks through gluon exchange. This connection might be seen as a metaphorical reflection of the intricacies of quark-gluon dynamics. 7) Holism: The holistic perspective of monads could symbolize how quarks, like the monads' interconnections, contribute holistically to the structure and behavior of particles through the strong force interactions.
@NotNecessarily-ip4vc
9 ай бұрын
[2D is not the center of the universe, 0D is the center of the mirror universe]: The mirror universe theory is based on the concept of parity violation, which was discovered in the 1950s. Parity violation refers to the observation that certain processes in particle physics don't behave the same way when their coordinates are reversed. This discovery led to the idea that there might be a mirror image of our universe where particles and their properties are flipped. In this mirror universe, the fundamental particles that make up matter, such as electrons, protons, and neutrinos, would have their charges reversed. For example, in our universe, electrons have a negative charge, but in the mirror universe, they might have a positive charge. Furthermore, another aspect of the mirror universe theory involves chirality, which refers to the property of particles behaving differently from their mirror images. In our universe, particles have a certain handedness or chirality, but in the mirror universe, this chirality could be reversed. Leibniz or Newton: Quantum mechanics is more compatible with Leibniz's relational view of the universe than Newton's absolute view of the universe. In Newton's absolute view, space and time are absolute and independent entities that exist on their own, independent of the objects and events that take place within them. This view implies that there is a privileged observer who can observe the universe from a neutral and objective perspective. On the other hand, Leibniz's relational view holds that space and time are not absolute, but are instead relational concepts that are defined by the relationships between objects and events in the universe. This view implies that there is no privileged observer and that observations are always made from a particular point of view. Quantum mechanics is more compatible with the relational view because it emphasizes the role of observers and the context of measurement in determining the properties of particles. In quantum mechanics, the properties of particles are not absolute, but are instead defined by their relationships with other particles and the measuring apparatus. This means that observations are always made from a particular point of view and that there is no neutral and objective perspective. Overall, quantum mechanics suggests that the universe is fundamentally relational rather than absolute, and is therefore more compatible with Leibniz's relational view than Newton's absolute view. What are the two kinds of truth according to Leibniz? There are two kinds of truths, those of reasoning and those of fact. Truths of fact are contingent and their opposite is possible. Truths of reasoning are necessary and their opposite is impossible. What is the difference between Newton and Leibniz calculus? Newton's calculus is about functions. Leibniz's calculus is about relations defined by constraints. In Newton's calculus, there is (what would now be called) a limit built into every operation. In Leibniz's calculus, the limit is a separate operation. What are the arguments against Leibniz? Critics of Leibniz argue that the world contains an amount of suffering too great to permit belief in philosophical optimism. The claim that we live in the best of all possible worlds drew scorn most notably from Voltaire, who lampooned it in his comic novella Candide.
@tomholroyd7519
10 ай бұрын
I get so tired of watching Hershey Fonts drawn in slow motion. We know what they are. Just draw them in one frame. Damn.
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