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Cross product part 2

Anonymous Tue 26 May 18:51:30 2020 No.11714589 View Reply Original Report

Quoted By: >>11714621 >>11714672 >>11714708 >>11714718 >>11714739 >>11714771 >>11715188 >>11716464

Cross product results in a third vector that's perpendicular to the other two.

Why tho? In a 2-dimensional space this happens because you take the perpendicular component of the other vector. But I can't see the mental leap from this to the 3-dimensional interpretation. You got two vectors on a 2d plane and somehow the result of a*b*sin(theta) is a vector on a whole another plane? And it just happens to be perpendicular to the other two? Did humans just force it to be that way for convenience or does this actually come from something?

My second, much simpler question (pic related): why isn't w = v x r?

Anonymous

Anonymous Tue 26 May 2020 18:59:12 No.11714621 Report

Quoted By: >>11715356

>>11714589
>In a 2-dimensional space this happens because you take the perpendicular component of the other vector. But I can't see the mental leap from this to the 3-dimensional interpretation.

in 3D, your initial two vectors form a plane. The third vector's direction is picked such that it is perpendicular to both of the first two vectors. the reason that it faces one way rather than the other is the answer to:
>why isn't w = v x r
because we have to establish some sort of direction with the cross product. we have picked it such that you follow the right-hand rule in establishing the direction of the third vector.

don't get confused about the $|a||b|sin\theta$ notation, that is just for determining the magnitude of the vector, where theta is the angle between them. This does not give you the direction of the third vector, you either need to calculate it yourself by finding what direction is perpendicular to both vectors, or simply by doing the cross product in matrix notation.

Anonymous

Anonymous Tue 26 May 2020 19:22:08 No.11714672 Report

Quoted By:

>>11714589
>why isn't w = v x r
>implying there is only one perpendicular vector to v and r

Anonymous

Anonymous Tue 26 May 2020 19:32:21 No.11714708 Report

Quoted By:

>>11714589
>Why tho?
>Did humans just force it to be that way for convenience or does this actually come from something?

Whatever way you visualize and/or interpret anything in math, it's all a sentimental tool to analyze reality. Is using a stick as a weapon something humans forced or did it actually come from the something (the physics allowing momentum)? Depends on perspective.

Questions like that won't really help your understanding of math, from my experience it mostly leads to circular reasoning.

Anonymous

Anonymous Tue 26 May 2020 20:15:55 No.11714718 Report

Quoted By:

>>11714589
>why isn't w = v x r?
Dimensionally this doesn't make sense. It would imply |w| = |v||r|sin(theta), giving |w| units of m^2/s

Anonymous

Anonymous Tue 26 May 2020 20:24:31 No.11714739 Report

Quoted By:

>>11714589
for three vectors $u,v,v \in \mathbb{R}^3$ the determinant $\det(u,v,w)$ is the signed volume of the parallelepiped spanned by u,v,w.
the cross product $u \times v$ is defined to be the unique vector which satisfies
$\langle u\times v,w\rangle = \det(u,v,w)$
for all $w \in \mathbb{R}^3$ . plug in $w = u$ to get
$\langle u\times v,u\rangle = \det(u,v,u) = 0$
the determinant being zero because the parallelepiped with two same edges degenerates into zero volume. therefore the cross product u × v is perpendicular to u. put w = v to get the same thing for v.

so no, it doesn't "just happen". there's a reason for it.

Anonymous

Anonymous Tue 26 May 2020 20:28:30 No.11714750 Report

Quoted By: >>11714761 >>11715371

Btw guys why the fuck is angular momentum and angular velocity pointing upwards away from the plane anyway?
Where's the intuition?

Anonymous

Anonymous Tue 26 May 2020 20:30:57 No.11714761 Report

Quoted By: >>11714766

>>11714750
it's the best way to unambiguously define a "direction" of a rotation--by using the axis of rotation. otherwise, if you try to do it "in the direction" of rotation, where would the vector point?

Anonymous

Anonymous Tue 26 May 2020 20:32:58 No.11714766 Report

Quoted By: >>11714805

>>11714761
>where would the vector point?
The direction of v of course. It changes all the time but you can always see the direction from it

Anonymous

Anonymous Tue 26 May 2020 20:33:34 No.11714771 Report

Quoted By: >>11714776

>>11714589
Let i,j,k be the standard basis for R^3. Then i x j = k by the definition of the cross product. Now let a,b,c be an orthonormal basis for R^3, write in terms of the standard basis and follow the computations to get a x b = +- c. Since any two non-colinear vectors can be scaled and then completed uniquely to form an ortholinear basis for R^3, one finds that a x b is perpendicular to both a and b.

Anonymous

Anonymous Tue 26 May 2020 20:34:36 No.11714776 Report

Quoted By:

>>11714771
orthonormal* not ortholinear

Anonymous

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Anonymous Tue 26 May 2020 20:41:58 No.11714805 Report

Quoted By: >>11714824 >>11714843 >>11717645

>>11714766
>it changes all the time
no, it would be pointing tangentially to the circle at every point. the tangential velocity is only defined for a single point, not for the entire rotating object
>pic related
at any given moment, all of these are valid velocity vectors. there are an infinite number of these. why not define it using a single, unambiguous vector?

Anonymous

Anonymous Tue 26 May 2020 20:44:52 No.11714824 Report

Quoted By: >>11714843

>>11714805
To add on, it doesn't make much sense to speak of "constant angular velocity/momentum" with a vector that changes over time.

Anonymous

Anonymous Tue 26 May 2020 20:46:55 No.11714843 Report

Quoted By:

>>11714805
>>11714824

Thank you anon.

Anonymous

Anonymous Tue 26 May 2020 22:25:06 No.11715188 Report

Quoted By:

>>11714589
>My second, much simpler question (pic related): why isn't w = v x r?
Not doing your phys 1 homework for you.

Anonymous

Anonymous Tue 26 May 2020 23:27:53 No.11715356 Report

Quoted By:

>>11714621
OP, this guy speaks the truth.

Anonymous

Anonymous Tue 26 May 2020 23:37:14 No.11715371 Report

Quoted By:

>>11714750
not everything is intuitive anon. if you're astonished now then you would freak if you studied quantum mechanics

https://www.youtube.com/watch?v=NeXIV-wMVUk

Anonymous

Anonymous Wed 27 May 2020 06:40:32 No.11716464 Report

Quoted By: >>11716555

>>11714589
>Why tho?
Vectors (in Euclidean space, at least) have two aspects - length (induced by the Riemannian structure which just identifies the space of tangent vectors with points in the initial space) and direction (from the structure of a vector space).

So this
> a*b*sin(theta)
only tells us about the length of the vector. You can pick any direction you want but it should be consistent wrt product properties. Two vectors define a plane. In 3 dimensions, a plane defines a vector (the normal vector, up to length and two directions). Thus we can choose between two vector directions in 3D (this is possible due to an additional structure in the Euclidean space).
This breaks in 4D and higher, because two vectors still define a plane but no single line is defined by this plane. It turns out that the natural generalisation is the so-called wedge product which, without an additional structure (inner product/Hodge star/volume form), cannot be defined for vectors. Instead, it is defined for linear functionals on vectors.

To sum up, this is how the vector product is viewed nowadays in mathematics.

1. Have a vector space
2. Automatically have a dual space of linear functionals on the initial vector space (which itself is also a vector space)
3. Introduce a structure of graded algebra on the dual space by defining the wedge product
4. Step 3 can be motivated by the desire to have a tool to calculate the volume of a n-dimensional parallelotope spanned by n vectors
5. With an additional structure we can define the wedge product for vectors instead.
6. It turns out that the result of this wedge product of two vectors can be identified with a new vector only in 3D and 7D.
7. Call it 'vector product' in 3D.

Obviously, this is slightly above what is usually taught to engineers, so feel free to skip this.
Also, you may be interested in the history of cross product. It arose naturally when Hamilton was developing his theory of quaternions.

Anonymous

Anonymous Wed 27 May 2020 07:26:12 No.11716555 Report

Quoted By:

>>11716464
>(induced by the Riemannian structure which just identifies the space of tangent vectors with points in the initial space)
nice flexing

Anonymous

Anonymous Wed 27 May 2020 08:19:42 No.11716678 Report

Quoted By:

Guys, the cross product produces an axial or pseudo-vector, which is actually a tensor of rank 2 (its Hodge dual). So many weird properties disappear when you think about them as tensors.

Anonymous

Anonymous Wed 27 May 2020 14:23:06 No.11717645 Report

Quoted By:

>>11714805
what is the tangent velocity if you would have it to explain to a brainlet?

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