B A covariant vs contravariant vector?

[etotheipi]

We have a basis {$\mathbf{e}_1$, $\mathbf{e}_2$, $\dots$} and the corresponding dual basis {$\mathbf{e}^1$, $\mathbf{e}^2$, $\dots$}. I learned that a vector $\vec{V}$ can be expressed in either basis, and the components in each basis are called the contravariant and covariant components respectively. So if $$\vec{V} = v^i \mathbf{e}_i = v_j \mathbf{e}^j$$ then the contravariant components are $(v^1, v^2, \dots)$ and the covariant components are $(v_1, v_2, \dots)$, and you have to transform these differently if you want to express $\vec{V}$ in terms of another basis.

But I wondered why I often see the terms "covariant vector" and "contravariant vector"? I thought any vector $\vec{V}$ has covariant and contravariant components. As an example, Wikipedia says that velocity is a contravariant vector, but can't velocity have covariant components?

So I hoped someone could explain the distinction here; sorry if this is confused! Thanks!

 

[fresh_42]

I guess you better ask this in a physics forum. Physicists made use of these adjectives independent from their mathematical usage, and as far as I could understand, also switched them.

The mathematical meaning is:
If we have a class of objects and functions between them (=category), e.g. vector spaces and linear transformations, and we apply a machinery (=functor) which transforms them into a different class of objects (= different category), e.g. their sets and any function between sets, then two things can happen:
$$
[f: V \longrightarrow W ]\stackrel{\mathcal{F}}{\longrightarrow} [\mathcal{F}: \mathcal{F}(V)=\{V\} \longrightarrow \mathcal{F}(W)=\{W\} ]
$$
Our machinery here is the forgetful functor $\mathcal{F}$: it simply forgets the linear structures and turns everything in a mapping of only sets. In this case the direction of the mappings $f$ and $\mathcal{F}(f)$ are the same, from $V$ to $W$. Those functors are called covariant, as they preserve the direction.

If the direction is swapped by a funktor $\mathcal{L}$, from $\mathcal{L}(W)$ to $\mathcal{L}(V)$ then it is called contravariant. An example are the linear functionals of a vector space. This means $\mathcal{L}$ turns a vector space $V$ into the vector space $\mathcal{L}(V)=V^*$ of all linear functions $V \longrightarrow \mathbb{R}$. Now what happens to $f: V \longrightarrow W$?
Here we define $\mathcal{L}(f): W^* \longrightarrow V^*$ by $(\mathcal{L}(f)(\omega))(v)=\omega(f(v))$, where $v\in V$ and $\omega \in W^* = \{W \stackrel{linear}{\longrightarrow} \mathbb{R}\}.$ In this case $\mathcal{L}(f)$ changed the mapping direction.

This is the mathematical meaning. The physical is a different one.

 

[etotheipi]

I'm still pretty new to the theory of vector spaces so I don't understand some of the terminology, but I think I can gauge the general idea.

I had no idea the Physics definition was different to the Maths one! That should help with further research. Thanks!

 

[PeroK]

I'm still pretty new to the theory of vector spaces so I don't understand some of the terminology, but I think I can gauge the general idea.

I had no idea the Physics definition was different to the Maths one! That should help with further research. Thanks!

It tends to be the way things are defined. If we start with the four-position, $x^{\mu}$, that is by definition a contravariant vector. In the sense that it is defined using the upper indices (i.e. the contravariant form). And it has the corresponding dual or co-vector $x_{\mu}$.

The four-velocity is then defined as $u^{\mu} = \frac{dx^{\mu}}{d\tau}$ and likewise is a contravariant vector.

The gradient on the other hand, which is defined as $\partial_{\mu} = \frac{\partial}{\partial x^{\mu}}$ is a covariant vector.

I think this is more conventional than, say, defining the four-velocity as neither one nor the other and then talking about its covariant and contravariant forms. In any case, it should amount to the same thing.

 

[etotheipi]

It tends to be the way things are defined. If we start with the four-position, $x^{\mu}$, that is by definition a contravariant vector. In the sense that it is defined using the upper indices (i.e. the contravariant form). And it has the corresponding dual or co-vector $x_{\mu}$.

Yes this was the context. The author was saying that the components of one 4-vector transformed contravariantly, whilst the components of another 4-vector of gradients of a scalar field transformed covariantly, to justify the different placement of the $\mu$'s.

I think this is more conventional than, say, defining the four-velocity as neither one nor the other and then talking about its covariant and contravariant forms. In any case, it should amount to the same thing.

I see; that makes things clearer. Thanks!

 

[troglodyte]

This is the mathematical meaning. The physical is a different one.

Not in all cases.In the language of mathematical physics one argue in strict mathematical language.Some theoretical physics profs prefer this language to describe physics.For example the canonical quantization for Boson particles without gauge potentials uses this formalism.I have actually one Prof who uses this stuff to explain quantum field theory.In some cases it can be painful to understand but in some cases you see structures and connections on a very deep level.For example the connection between ket vector to the bra dual vector has its mathematical roots in Frechet Riesz isomorphism and is the key to define the hermitian conjugation .