Surface area vector

[mikeeey]

Hello every one .
What is the Surface Area vector form in exterior algebra ,I mean by that the Surface Area vector as an exterior form in 3D , just like the volume form .


THANKS

 

[Matterwave]

Wouldn't that just be a 2-form? A n-form gives a definition of an n-dimensional volume, so an area should be associated with a 2-form.

 

[mikeeey]

yes its a 2-forms , but 2-form is a co-variant second order tensor , but here the surface area is a vector ,
this is why i want to know ,
in mechanics the stress distribution formula is $F^i = \sigma^\ij dA_j$
where F is the force vector and (Sigma ) is the mechanical second order stree tensor and A is the Area vector
while in exterior algebra it's written like this $F^i = T^i_jk dx^j\wedgedx^k$
where T is a third order tensor , when using calculus e.g. co-variant derivative , sigma with give 2 christoffel symbols while the T will give 3 christoffel symbols

 

[mikeeey]

$F^i =\sigma^ij dA_j$
$F^i= B^i_j_k dx^j \wedge dx^k$

 

[robphy]

The "surface area element" can be thought of as a "[3D-]vector" only in 3-D.
In the cross-product, the oriented parallelogram formed from the factors is more fundamental than the vector perpendicular to that parallelogram.
From a tensor algebra viewpoint, to get a vector from the oriented parallelogram,
one has to use the Hodge-dual (often symbolized by *), which involves the $\epsilon_{ijk}$ symbol.

 

[mikeeey]

you mean $dA_i = \epsilon_ijk dx^j \wedge dx^k$

 

[mikeeey]

$dA^i = \epsilon _{ijk} dx^j \wedge dx^k$

 

[Matterwave]

$dA^i = \epsilon _{ijk} dx^j \wedge dx^k$

Yes, it's basically this, but you might have some normalization factors in there, I'm not quite sure.

EDIT: Oh, and in your formula the i has moved from lower index to upper index, so you have to raise the index in there somewhere. :)