View Full Version : outer product of two one forms.
ronblack2003
Nov13-04, 02:01 PM
Given two one forms f = (1,1,0,0,) and g=(-1,0,1,0): what are the components of f(x)g ... would appreciate any help.
ronblack2003
Nov13-04, 02:08 PM
Given two one forms f = (1,1,0,0) and g = (-1,0,1,0):
What would be the Components of f(outer product)g be?
would appreciate any help.
pervect
Nov14-04, 02:54 AM
I'm not quite sure if the "outer product" is the wedge product or the tensor product. Wikipedia seems to think it's the former. See
This Link (http://en.wikipedia.org/wiki/Geometric_algebra)
The wedge product of u ^ v is
u (tensor) v - v (tensor) u
The intent here is to anti-symmetrize the tensor product, The formula above works as written only for rank 1 tensors (but that's what you have).
The tensor product p = u (tensor) v in component notation for rank 1 tensors is:
pij = uivj
Thus (1,2,3) (tensor) (4,5,6) is the second rank tensor (you can think of it as an array)
1*4 ,1*5, 1*6
2*4, 2*5, 2*6
3*4, 3*5, 3*6
Hope this helps
pmb_phy
Nov16-04, 01:16 PM
[QUOTE=pervect]I'm not quite sure if the "outer product" is the wedge product or the tensor product.
Yes. Its the tensor product.
Let "@" be the tensor product and let f = f_u w^v where f_u are components of f and w^u are a coordinate basis 1-forms. Same with g. Then let h = f@g. Then
h = (f_u e^u)@(g_u v e^v) = f_u g_v (w^u@w^v)
Therefore f_u g_v are the components of f@g. w^u@w^v are the covariant basis tensors for the outer product.
Pete
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