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I understand that, in curvilinear coordinates, one can define a covariant basis and a contravariant basis. It seems to me that any vector can be decomposed in either of those basis, thus one can have covariant components and contravariant components of the same vector, depending on the chosen basis. What is confusing to me, however, is when people talk about covariant and contravariant vectors. Do they just mean the covariant/contravariant components of vectors or there are indeed two distinct types/classes of vectors? If the latter, covariant vectors can only be decomposed into covariant bases and contravariant vectors only in contravariant bases?

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We don't talk of covariant and contravariant bases. Start with the basis $\{\mathbf e_i\}$. Then a general vector can be written $$\mathbf v = v^i \mathbf e_i$$ Now if you double the length of a basis vector, you must halve the length of the componant. The components are said to be contravariant, because they change opposite to the basis. In index notation this vector is simply written $v^i$, and we call it a contravariant vector meaning that the components are contravariant.

The inner product

$$ \mathbf u \cdot \mathbf v = g_{ij}u^iv^j $$ prompts the definition $$ u_j = g_{ij}u^i $$ The $u_j$ are components of a vector in the dual space. Because the inner product is invariant, the components $u_j$ change opposite to contravariant components, which means they change in the same way as the basis vectors. They are called covariant components, and we refer to them as covariant vectors.

Technically contravariant vectors are in one vector space, and covariant vectors are in a different space, the dual space. But there is a clear 1-1 correspondence between the space and its dual, and we tend to think of the contravariant and covariant vectors as different descriptions of the same vector.

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  • $\begingroup$ what about the components $u^i$? Aren't those the components of the same vector $\mathbf{u}$? $\endgroup$ – Botond Apr 6 at 11:24
  • $\begingroup$ I have added a para to clarify. $\endgroup$ – Charles Francis Apr 6 at 12:10
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    $\begingroup$ God I've been waiting to read something like this for years! Thanks. $\endgroup$ – G.Clavier Apr 6 at 12:32
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You have a basis ${\bf e}_i$ in some vector space.

The contravariant components of a vector ${\bf v}$ are given by ${\bf v}=v^i{\bf e_i}$, as Charles Francis says.

The covariant components of a vector ${\bf v}$ are given by $v_i=\mathbf v\cdot\mathbf e_i$

I think that's a more basic way of thinking about them than going in to their transformation properties - though that is of course true.

Incidentally it's then obvious that $\mathbf u\cdot\mathbf v=\sum u_i v^i$ (or $\sum u^iv_i$)

I would say (though mathematicians would disagree and will probably downvote this answer as heretical) that a 'physics' vector is neither covariant nor contravariant. It's a pointing arrow. If you want to do anything useful with it you have to write down its components, which can be either covariant of contravariant.

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  • $\begingroup$ The problem with the arrow arises, surely, when we try to apply it to grad $\phi$ when working in a non-orthogonal co-ordinates basis, {$\mathbf{e}_i$}. The components of grad $\phi$ on this basis exist but do not give us what we'd like to have, namely {$\frac{\partial \phi}{\partial x_i}$}. [These derivatives are components on the dual basis to {$\mathbf{e}_i$}!] So I'd argue that it's not always $useful$ to think of vectors as arrows. $\endgroup$ – Philip Wood Apr 7 at 12:05
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    $\begingroup$ If you are standing on a hillside (height $\equiv \phi$) then you know the direction and the magnitude of grad $\phi$ just by releasing a marble and seeing which direction it goes in and how fast it accelerates. You can draw that as an arrow. Without writing down any components and hence without involving a basis. But I agree that this is tricky stuff. $\endgroup$ – RogerJBarlow Apr 7 at 12:20
  • $\begingroup$ RogerJBarlow Thanks for replying. It's not that I'm saying that a gradient can't be thought of as an arrow, so much as the arrow thing not being so $suggestive\ of\ useful\ ways\ to\ proceed$ in the case of a gradient, if we're working on a non-orthogonal basis. I don't think we're really at odds. $\endgroup$ – Philip Wood Apr 7 at 15:42

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