# Covariant and contravariant tensors

• I
dsaun777
Is there a purpose of using covariant or contravariant tensors other than convenience or ease in a particular coordinate system? Is it possible to just use one and stick to one? Also what is the meaning of mixed components used in physics , is there a physical significance in choosing one over the other? For instance when breaking down a vector into its component form you can use covariant basis and contravariant components and vice versa but why? Sorry if this question seems redundant but I'm just puzzled.

[Moderator's note: Moved from a mathematical forum. This is about physics. A mathematical answer would probably be a different one and might increase confusion rather than clarify it.]

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Here is a mathematical answer: https://www.physicsforums.com/insights/what-is-a-tensor/
but as mentioned, the physical answer is a different one, because in physics tensors are tools, whereas in mathematics they are just certain objects. As the covariant parts are isomorphic to a contravariant version of it, there is little difference mathematically, whereas it is important to distinguish them in physics.

dsaun777
Here is a mathematical answer: https://www.physicsforums.com/insights/what-is-a-tensor/
but as mentioned, the physical answer is a different one, because in physics tensors are tools, whereas in mathematics they are just certain objects. As the covariant parts are isomorphic to a contravariant version of it, there is little difference mathematically, whereas it is important to distinguish them in physics.
That is a nice brief generalization of what tensor can represent but it gives definitions based on undefined terms. Why even use contravariant and covariant tensors why not just stick to one, why are they mixed? There are no definitions of covariant and contravariant in your 'mathematical answer". I see that they are tools for physics but why have both. It seems like you show up to a job with two tape measures, one measuring meters and the other measuring feet. Both do the same operation. In terms of strictly theoretical physics can not we just disregard one?

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There are no definitions of covariant and contravariant in your 'mathematical answer".
This is because it makes no mathematical sense. One part are ordinary vector spaces (contravariant), the other dual vector spaces (covariant). Physicists use it this way, the terms contravariant and covariant in mathematics are defined for functors and not for tensors. Furthermore a contravariant functor refers to the dual category, which is exactly the opposite of how physicists use it. So I avoided the terms as they belong mathematically in a different context, and physicists use them conversely than mathematicians do. This is a mess, and the reason why I said "mathematical answer". IIRC then large parts of the discussion beyond the article is about these contradictions.

Technically it makes a difference whether we consider a vector space ##V## or its dual space ##V^*##. They might be isomorphic, but one consists of directional arrows, the other one of functions which eat directional arrows. Hence physicists have to distinguish them carefully. E.g. a derivative can be both, the vector "slope at point" or the linear function "multiply by the slope factor", i.e.
$$f(x)=x^2 \Longrightarrow f'(x) = \{\,(x,x^2);(1,2x)\,\}$$
where we have a pair of point and attached direction, or
$$f(x)=x^2 \Longrightarrow f' \, : \,x \longmapsto 2x$$
where we have the function multiply by two.

It is still the same thing, the derivative of ##x^2##, but one is a vector at a point, the other the instruction how to apply the derivative. That's why co- and contravariant are distinguished in physics: do we mean the slope or how to apply the slope? Mathematically there is not much difference between ##2## and ##\text{ multiply by }2##.

• dsaun777
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It seems like you show up to a job with two tape measures, one measuring meters and the other measuring feet.
One practical difference is that in applications contravariant components are typically measured in metres per something (e.g. m/s) but covariant components are typically measured in somethings per metre (e.g. J/m).

• dsaun777
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One practical difference is that in applications contravariant components are typically measured in metres per something (e.g. m/s) but covariant components are typically measured in somethings per metre (e.g. J/m).
Careful though, this depends on the dimensions of your coordinates. If coordinates have different dinensions (such as for polar coordinates) so will the components.

• DrGreg
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Careful though, this depends on the dimensions of your coordinates. If coordinates have different dinensions (such as for polar coordinates) so will the components.
That's a good point. The example I gave in post #5 was not a good one.

If you are using Cartesian coordinates with a Euclidean metric, it's difficult to notice the difference between contravariant and covariant (mathematically, there is a difference but the components look the same). But, for example, in 2D polar coordinates ##(r, \theta)## measured in ##(\text{metres}, \text{ radians})##, a typical contravariant vector's components might have units ##(\text{m/s}, \text{rad/s})##, and a typical covariant vector's components might have units ##(\text{J/m}, \text{J/rad})##.

dsaun777
One practical difference is that in applications contravariant components are typically measured in metres per something (e.g. m/s) but covariant components are typically measured in somethings per metre (e.g. J/m).
So the units of the covariant components usually contain mass or energy and the contravariant components contain velocities and their derivatives... Why?

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So the units of the covariant components usually contain mass or energy and the contravariant components contain velocities and their derivatives... Why?
Careful though, this depends on the dimensions of your coordinates. If coordinates have different dimensions (such as for polar coordinates) so will the components.
The covariant components are multilinear functions. What does a matrix entry for, say ##\varphi \, : \,\mathbb{R}\otimes \mathbb{R}\otimes \mathbb{R} \longrightarrow \mathbb{R}## has as units?