Re: resolve to perpendicular components, because they are independent

Mati and Timo are probably right, but...

mmeron@xxxxxxxxxxxxxxxxxx wrote:
> In article <1137745432.546723.23260@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>, "Ken S. Tucker" <dynamics@xxxxxxxxxxxx> writes:
> >
> >kenneth.bull@xxxxxxxxx wrote:
> >> Hi Physicers,
> >> Any help will be greatly appreciated.
> >> It is my understanding that we often resolve vectors (like force,
> >> velocity) into perpendicular components because they are so called
> >> "independent." If one component changes value, it doesn't affect the
> >> value of the other pependicular components (I guess this is where
> >> "Independent comes from").
> >> Yet I have been recently shown how to resolve vectors (force) into
> >> components that aren't perpendicular using a reverse-parallelogram
> >> rule. Say a force is acting on a structure like < at the left point, I
> >> have to resolve it into two forces alone the two branches (not
> >> orthogonal). I encountered this in my self study of certain questions,
> >> so I don't really have anyone reliable to ask why this is viable in
> >> light of my previous knowledge of "independent vectors need to be
> >> perpendicular."
> >> Can someone shed light on any of this?
> >
> >I think you want to study the *Kronecker Delta*. I accept that
> >a number of axes x^1, x^2...x^n are *independant* if
> >((& is partial))
> >
> >&x^a / &x^b = 1 if a=b, and 0 if a=/=b.
>
> Everybody (well, nearly everybody) is confusing the poor guy. Let me
> try to set the record straight.
>
> Mathematically, a set of vectors is independent if non of them can be
> expressed as a linear combination of the other ones. For the 3D case
> you care about it simply means that 3 vectors are independent if
> they're not lying in the same plane. Translating to axes it means the
> same, you cannot have 3 axes lying in the same plane, otherwise the
> angles between them are arbitrary. So, no, independent vectors do not
> need to be orthogonal.
>
> But, but... non orthogonal axes are a pain. Many of the formulas you
> use implicitly assume orthogonal vector base. Once it is not
> orthogonal then, to begin with the scalar product isn't given by the
> simple formula (a1,a2,a3) /dot ((b1,b2,b3) = a1*b1 + a2*b2 + a3*b3.
> You've a more complex formula instead, one that is taking the angles
> betwee the axes into account. Vector product is even worse. And when
> you get to vector operators like grad or curl, all hell breaks out.

I find nonorthogonal axes easier than orthogonal, indeed a Curl
becomes A_u,v - A_v,u (== &A_u/&x^v - &A_v/&x^u), because
manipulating equations in tensors is streamlined by notation.

For me, orthogonal metrics are challenging, specifically defining
a metric that can be transformed to

X_uv = 1 or 0 (u=v or u=/=v)

in a more general metric like,

g_uv = X_uv + A_u B_v .

Ken




















> So, I'll second Timo's advice, i.e. unless you've a good enough reson
> to go non-orthogonal, don't. But, if you choose to do, it is legit
> (though painful).

.