Saturday, August 23, 2014

general relativity - Is it foolish to distinguish between covariant and contravariant vectors?


A vector space is a set whose elements satisfy certain axioms. Now there are physical entities that satisfy these properties, which may not be arrows. A co-ordinate transformation is linear map from a vector to itself with a change of basis. Now the transformation is an abstract concept, it is just a mapping. To calculate it we need basis and matrices and how a transformation ends up looking depends only on the basis we choose, a transformation can look like a diagonal matrix if an eigenbasis is used and so on. It has nothing to do with the vectors it is mapping, only the dimension of the vector spaces is important.


So it is foolish to distinguish vectors on the way how their components change under a co-ordinate transformation, since it depends on the basis you used. So there is actually no difference between a contravariant and covariant vector, there is a difference between a contravariant and covariant basis as is shown in arXiv:1002.3217. An inner product is between elements of the same vector space and not between two vector spaces, it is not how it is defined.


Is this approach correct?


Along with this approach mentioned, we can view covectors as members of the dual space of the contra-vector space. What advantage does this approach over the former mentioned in my post?


Addendum: So now there are contra variant vectors and their duals called covariant vectors. But the duals are defined only once the contravectors are set up because they are the maps from the space of contra vectors to $R$ and thus, it won't make sense of to talk of covectors alone. Then what does it mean that the gradient is a covector ? Now saying because it transforms in a certain way makes no sense.




No comments:

Post a Comment

classical mechanics - Moment of a force about a given axis (Torque) - Scalar or vectorial?

I am studying Statics and saw that: The moment of a force about a given axis (or Torque) is defined by the equation: $M_X = (\vec r \times \...