Geometric Algebra in 3D - Bivector Addition

[@jackrubin6303]

Thank you. I am 67 years old and have learnt something new!

[@---om4wh]

Your videos are great man! I am so happy to see a channel that goes deeper than the conceptual understanding. Pop math can only take you so far! Keep up the amazing quality!

[@Math_oma]

+Erik Imathiu-Jones Thanks!

[@AtheistBulldozer]

Mathoma Apologies for contacting you on this video but I am not sure how else I can ask you this question... which will be an ironic question. You offered to chat to me about metaethics from that thread I encountered you on (Harmful Opinion video on UPrager). Would you be cool if I added you on google hangouts? Thanks

[@Math_oma]

+AtheistBulldozer Sure, you can add me on Google hangouts.

[@brianwilson3656]

Thank you so much for these videos. Watching this GA series was my covid sanity project. I was originally motivated by GA applied to multibody dynamics. Can't wait to derive the "spatial vector algebra" (see Roy Featherstone) from these fundamentals.

[@jamesmarlar4231]

Thanks, excellent presentation. Please continue these all the way through Geometric Calculus.

[@rajendramisir3530]

Wow! I find the bivector concept very interesting. I like the explanation and discussion you give here of bivectors. The proof and torque example are well done.

[@osamahasan3288]

Excellent presentation for all the previous lectures in the list. and I hope that I will watch all the list soon. Thank you. nice drawing of the diagrams.

[@Ferret440]

The example at the end was excellent! Thank you for this series :D

[@wyboo2019]

kind of proud of myself but i think i figured out why theres bivectors in 4d that aren't 2-blades. in 3d you have 3 basis vectors and 3 choose 2 = 3 ways of arranging them to get basis bivectors, but in 4d, you have 4 basis vectors and 4 choose 2 = 6 ways to couple them to get basis bivectors, so you can make two bivectors that do not share a common basis vector (eg e1e2 and e3e4)

[@Holobrine]

Any physics applications of trivectors?

[@Michallote]

Volume isn't it?

[@Michallote]

I think that intuitively density is a scalar mass divided by a trivector volume

[@Michallote]

So for example we can have a fluid with vector velocity v going through an Bivector area A if the fluid is incompressible i.e. density is constant then we just have v A = Q Only the perpendicular component of v to A will amount to flow so in order for this to be true Q has to be a trivector

[@Holobrine]

@@Michallote Are there any instances of volume being multiplied by a vector to get a bivector?

[@jaredburton6164]

Thank you! I love this video!

[@pronounjow]

So a bivector is more like a general area of effect/magnitude on the plane it's part of, instead of a specific area of a specific shape? Does the extent of the bivector's effect depend on the basis vectors?

[@Math_oma]

+Jo Reven Yeah, it could be a _blob_ of area if you want. However, it's easier the see the plane being represented if I choose to draw it using parallelograms. As for the second question, let me ask you (and everyone reading): what do you think?

[@pronounjow]

Mathoma I'm thinking: Yes, the basis vectors are the extent of the full effect, but the effect continues to radiate outside the extent of the basis vectors to the rest of the plane in smaller and smaller amounts, kinda like the gravitational field.

[@MultivectorAnalysis]

As with vectors, the apparent extension into space of a bivector area is for illustrative purposes, while many applications have the bivector treated as occurring at a single point (cf. the tangent space of a point on a manifold). First lets imagine a vector field, say an electric field. You illustrate the field by drawing arrows extending from selected points in space. The direction a particular arrow is pointing tells you the direction of the force a positively charged particle would experience at that point, while the length of the arrow is proportional to the magnitude that force. That particular vector is for a single point in space, and does not extend out through other points in space (each of which having their own associated electric field vector). The same applies to bivector fields, representing say a magnetic field. Each point in space has an associated bivector. The magnitude of the bivector tells you the intensity of the magnetic field at that point, while its planar attitude and rotational orientation relate to the direction of the force a moving charged particle would experience at that point. Again, no actual extension here.

[@pronounjow]

Nick Okamoto I think I was going for something like that... maybe.

[@JimSmithInChiapas]

I answer this question by saying that bivectors are a concept and mathematical object defined by human beings, for the purpose of capturing certain aspects of the geometrical content of problems in an algebraically manipulable form. The people who originated the concept chose to define bivectors as possessing the characteristics "area" (magnitude) and "orientation", but not "shape" or "position". Therefore, two bivectors are equal iff they are of equal area, lie in the same or parallel planes, and have the same orientation. The choice of basis vectors is immaterial to determining whether they're equal.

[@clariellechang6668]

what would be the interpretation of G(1)? it isn't quite the same as just scalar math because of that single component vector which is also the pseudoscalar

[@Math_oma]

Just one vector, whose span is a line, two elements in its basis, and you have the vector and pseudoscalar being the same.

[@porglezomp7235]

Synthesizing the facts in this video made me think about the fact that while all torques in 3d are equivalent to applying a single force on some lever (the net torque), in 4d you might require multiple distinct levers and you don't get a simple net torque equivalent to one vector!

[@MrRyanroberson1]

in 4d if you want an axle (what you're calling 'lever') then the axle will be a sheet instead of a line (car axles and door hinges are lines with thickness; in 4d you would need planes with thickness)

[@MrRyanroberson1]

I misinterpreted what you meant - in any number of dimensions, the lever is defined as the line connecting the point of force to the center of mass, or the leverage point, depending on the system. The rotation is always uniquely the plane traced out by the force and the lever. in 3d, every lever with a leverage point is given an axle which is primarily only (3-2 = ) 1 dimension, but with added thickness. in 4d, axles would be sheets, in order to constrain all torques to the desired plane.

[@lumipakkanen3510]

Does this also mean that in 4d you can get an object spinning in one plane and still freely manipulate it along another plane without having to take that spin into account?

[@johnhare8208]

This is amazing

[@sdmartens22]

Extremely impressive video as always. I owe you a video series; if you enjoy tensors try XylyXylyX. I was hoping for a correspondence between planes and bivectors; my guess would be given a plane with an area you project down onto the xy,yz,zx planes to get the components of the bivector.

[@Math_oma]

+Shannon Martens Yes, that's a nice video series you're referring to. Tensor algebra is another topic that has a lot of obfuscation surrounding it and he does a nice job getting to the point of what they are.

[@MrRyanroberson1]

so i've been watching your series and i don't think you ever covered the product of two bivectors?

[@MrRyanroberson1]

i've got it: any bivector can be expressed as e1e2e3 times some vector. by factoring out e1e2e3 on the left and right sides, the vector product can be used (so BC -> e1e2e3bce1e2e3; the effect of e1e2e3 on both sides is to negate everything)

[@Amit1994-g9i]

Make a video on spinors

[@SenSeisetakor]

Isn't this bivector simply a vector cross product?

[@Math_oma]

+Vedran Mamić Pocrnja Ah, but isn't the cross product simply a bivector (actually the dual)?

[@MultivectorAnalysis]

The most commonly cited difference is that bivectors are not restricted to three dimensional spaces. Another significant advantage is that the outer product is associative while the cross product is not.

[@gundamlh]

do you mean, (let * denote cross (outer) product) a*b*c != a * (b*c) ?

[@gguevaramu]

Dear Mathoma, I have learned a lot from your videos, and your exposition is that good that usually I don't have questions. But I came across this problem and maybe you can help me The plane of rotation of a vector is given by B = e1e2 + e2e3 + e1e3 to make it unitary it is necessary to divide by Sqrt(3) this factor provides part of the scalar that informs the angle of rotation. The problem occurs when I add these 3 bivectors, the result is this B = e1e2 - e3e2 + e1e3 = (e1 - e3)e2 + e1e3 = (e1 - e3)e2 + (e1 - e3)e3 B= (e1 -e3)(e2 +e3) To make B unitary it is necessary to divide by 2, not by sqrt(3), and this is a problem because I don't get the same angle of rotation, but it is giving me the right plane of rotation. Tiago Rodrigues was very helpful, and the key was that B is not formed with orthogonal vectors. So, the area has to be calculated with the module of the vectorial product. In advance thank you for any comment

[@tiagorodrigues3730]

Hey, I'm not Mathoma, but I can see that when you turn $e_1 \wedge e_3$ (which is a bivector with magnitude 1) into $(e_1 - e_3) \wedge e_3$ (which is a bivector with magnitude $\sqrt{2}$) you need to put up a scalar factor to compensate. So: $$ B = e_1 \wedge e_2 - e_3 \wedge e_2 + e_1 \wedge e_3 = (e_1 - e_3) \wedge e_2 + e_1 \wedge e_3 = (e_1 - e_3) \wedge e_2 + {1 \over \sqrt{2}} (e_1 - e_3) \wedge e_3 = (e_1 - e_3) \wedge (e_2 + {1 \over \sqrt{2}} e_3) $$ has the magnitude you want. Note: if you have difficulty reading the TeX code, try pasting the content between dollar signs into the live demo in mathjax.org or katex.org. Cheers!

[@tiagorodrigues3730]

@@gguevaramu No problem, I must actually apologise. Reviewing your argument and mine, I found that *I actually made a mistake* and was *completely off track.* Your original calculation was correct, and the actual answer is really $B = (e_1 - e_3) \wedge (e_2 + e_3)$. The only thing is that $B$'s magnitude is actually $\sqrt{3}$, not 2: while both $e_1 - e_3$ and $e_2 + e_3$ have length $\sqrt{2}$, they are not orthogonal to one another, so the magnitude of their outer product *isn't* $\sqrt{\sqrt{2}^2 + \sqrt{2}^2} = 2$. It's easier to check with the squared norm $|B|^2 = B B^\dagger= (e_1 - e_3) \wedge (e_2 + e_3) (e2 + e_3) \wedge (e_1 - e_3)$: $$ (e_1 - e_3) \wedge (e_2 + e_3) (e_2 + e_3) \wedge (e_1 - e_3) = = (e_1 - e_3) \wedge (e_2 + e_3) \cdot (e_2 + e_3) \wedge (e_1 - e_3) = (e_1 - e_3) \cdot [ (e_2 + e_3) \cdot (e_2 + e_3) \wedge (e_1 - e_3) ] = (e_1 - e_3) \cdot [ ((e_2 + e_3) \cdot (e_2 + e_3)) (e_1 - e_3) - ((e_2 + e_3) \cdot (e_1 - e_3)) (e_2 + e_3) ] = (e_1 - e_3) \cdot [ 2 (e_1 - e_3) - (-1) (e_2 + e_3) ] = (e_1 - e_3) \cdot [ 2 e_1 + e_2 - e_3 ] = 3$$ Which shows that the magnitude of $B$ is $\sqrt{3}$. Note that an important formula for reducing bivectors is $(u \wedge v) \cdot (a \wedge b) = u \cdot (v \cdot (a \wedge b))$, as well as $u \cdot (a \wedge b) = (u \cdot a) b - (u \cdot b) a$. Also, because $B$ and $B^\dagger$ are coplanar (indeed, $B = -B^\dagger$), their geometric product is equal to their inner product. I hope that clears up the confusion, I'm very sorry once again! P.S.: I guess I went about doing it the hard way, it would be easier to distribute $B$ to get $B = e_2e_3 - e_3e_1 + e_1e_2$, then the product $B B^\dagger = (e_2e_3 - e_3e_1 + e_1e_2) \cdot (-e_2e_3 + e_3e_1 - e_1e_2) = 3$ is easy to compute.

[@gguevaramu]

@@tiagorodrigues3730 I apologize to you and I don't have words to say how much I appreciate your time. Thanks a lot

[@gguevaramu]

@@tiagorodrigues3730 You know, there is a very simple way to get the area magnitude of any bivector, and it is the magnitude of vectorial product. It is out of geometric algebra but is very simple. But I found very interesting the way you did the same, just one thing bothers me, I been looking for a book showing the formulas to reduce bivectors especially the inner product of two bivectors. I have seen several books but none of them show how to get these formulas. I know vector algebra but does not help much. Do you know a book to study this topic?

[@tiagorodrigues3730]

@@gguevaramu I have used David Hestenes's _New Foundations for Classic Mechanics,_ where it is supposed to be exercise 1.1 (a) in chapter 2. It relies on a reduction formula given as equation (1.18) in that same chapter ($(A_r \wedge b) \cdot C_r = A_r \cdot (b \cdot C_s)$, which only holds if $0 < r < s$), but *that* one's derivation is a bit handwavy. Dr Hestenes gives the definition of the inner product of two blades as the $|r-s|$-graded part of the geometric product of those blades, where $r$ and $s$ are the grades of each blade. By linearity you get the definition for full multivectors. If you represent both blades as outer products of mutually-orthogonal vectors and then carry out the multiplication, you should be able to arrive at this formula, though I confess not having done this myself (maybe I should). Regardless, I think that at least braving the exercises of chapter 2 of that book should give you a reasonable handle on working with multivectors.