[ 3 / biz / cgl / ck / diy / fa / ic / jp / lit / sci / vr / vt ] [ index / top / reports ] [ become a patron ] [ status ]
2023-11: Warosu is now out of extended maintenance.

/sci/ - Science & Math


View post   

File: 340 KB, 768x1024, 1536537160443.jpg [View same] [iqdb] [saucenao] [google]
10102402 No.10102402 [Reply] [Original]

In one dimension, static equillibrium is defined by one equation: ΣF=0
In two dimensions, there are three: ΣFx=0, ΣFy=0, ΣM=0.
In three dimensions, there are six: ΣF=0, ΣM=0, where each of these is a vector equation 2ith three components.
So
1 => 1
2 => 3
3 => 6

How many equations in four dimensions?

>> No.10102406

>>10102402
>M
What the fuck is M?

>> No.10102407

>>10102406
Moment=r x F

>> No.10102419

>>10102402

in 3 dimensions you have to balance both normal forces and shear forces, in short you have to be sure that there's no torque and no translational elements.

long short story yes, it's 6 components but only if the forces are symmetrical

>> No.10102423

>>10102419
Right. I understand why there are 6 equations for a 3D body. But what about a hypothetical 4D body, or a collection of point masses in 4D space?

>> No.10102440

>>10102423

4d body has 10 equations, always if conditions are symmetrical

collection of point masses is a different story, you'd need to define the interactions between the elements and them deform them.

>> No.10102444
File: 78 KB, 700x734, 1535197110919.jpg [View same] [iqdb] [saucenao] [google]
10102444

>>10102440
>4d body has 10 equations
How come? And what are they?

>> No.10102454

>>10102444

Before we think at 4d, are you familiar with continuum mechanics and tensors?

because the thing you've described in 3D is called Cauchy stress tensor

>> No.10102460
File: 306 KB, 552x510, test (8).png [View same] [iqdb] [saucenao] [google]
10102460

>>10102402
Let the symplectic manifold [math](M,\omega)[/math] acquire a symmetry [math]G[/math] such that symplectic reduction gives rise to an symplectic submanifold [math]C \cong \mathfrak{g}^*[/math] isomorphic to the coadjoint orbits of [math]G[/math], where [math]\mathfrak{g} = \operatorname{Lie}G[/math] is the Lie algebra of [math]G[/math], then all motion are parametrerized by the generators of [math]G[/math].
For a free rigid body in dimension [math]n[/math], the symmetry group is [math]G = \mathbb{E}(n)[/math] the Euclidean group, which has generators [math]\{p_i,m_i\}_{i=1}^{n}[/math], where [math]p_i[/math] are the momenta generating translations and [math]m_i[/math] are the moments genrating rotations, satisfying commutation relations [math][p_i,m_j] = \epsilon_{ijk}p_k[/math], etc.. In dimension [math]n = 3[/math] you precisely get the Lie algebra of [math]SU(2)[/math], which is a universal double cover of [math]SO(3)[/math]. Hence in general the number of generators of motion are the genrators of the Lie algebra of the universal central extension [math]\hat{G}[/math] that fits into the diagram
[eqn]0\rightarrow H\rightarrow \hat{G} \rightarrow G\rightarrow 0,[/eqn]
where [math]H \subset G[/math] is a central subgroup of integrals of motion.

>> No.10102463

>>10102454
Uhh im really not familiar, no. I guess i should shut up and read more first.

>> No.10102472

>>10102463

I didn't mean to use fancy terms, it's just their name.

Are you in college?

>> No.10102477

>>10102472
I am. Im a lowly mech engineering undergrad and i was thinking about this in class. I have taken any continuum/fluid mechanics yet

>> No.10102491

>>10102477

you will understand what's going on in the future, be patient and remember to study linear algebra properly.

I hope you have a brillant career OP, keep the curiosity going

>> No.10102499

>>10102491
ty~

>> No.10102547

Each dimension adds another row/column to a square matrix. 1D is a 1x1 matrix. 2D is a 2x2 matrix. 3D is 3x3 matrix. The number of equations you need is equal to the number of terms on the diagonal, plus the number of terms above the diagonal.

So for 3D, 3x3 matrix, you have 3 terms on the diagonal, and 3 above the diagonal.

Then, for 4D, and thus a 4x4 matrix, you get 4 terms on the diagonal, and 6 above it. So 10 equations

>> No.10102564

>>10102547
>10 equations
It's one matrix equation

>> No.10102568

>>10102564
Same shit

>> No.10102585

>>10102568
Not unless they are independent.

>> No.10102588

>>10102585

for a rigid body they must be independant, you can't collapse a 3d body into 2d, that is what would happen if the equations were linearly dependant.

>> No.10102592

>>10102585
Sure, bur they are unless you're doing some theoretical math nerd shit. For all intents and purposes you can just call it 10 equations 99% of the time

>> No.10102933

>>10102547
This exactly answers my question, thank you!