Physical Difference Between Co- and Contravariant Vectors

[deuteron]

TL;DR

What is the physical difference between covariant and contravariant vectors

Hi,

today I have asked a very similar question on the topic, however now my question is more specific and focused, therefore I wanted to ask this again.

From the following thread, Nugatory's answer, I understood that some physical quantities need to be described by contravariant vectors, such as velocity, and others need to be described by covariant vectors, depending on how we want them to behave under a change of basis (in the answer the change of basis was related to stretching or shrinking the basis vectors, and velocity vector components needed to be scaled inversely, therefore it was said that velocity needed to be described by a contravariant, upper index, four vector)

But, using the Minkowski metric, we can lower and raise the index of a vector

So following this, there exists a velocity four vector that is covariant.
What does that correspond to, since we have established that velocity should be describes by a contravariant vector?
I want to think that the covariant vector is physically not the same as the contravariant vector, since the explanation of the velocity vector components scaling inversely with the basis vectors made a lot of sense to me, and it would be confusing for me if the components scaled *with* the basis vectors. However I have also seen on many places that covariant and contravariant vectors describe the same physical quantity, therefore I am very confused about this.

 

[PeterDonis]

From the following thread, Nugatory's answer

I think @Orodruin's response in post #2 is a better one to focus on.

using the Minkowski metric, we can lower and raise the index of a vector

Mathematically, yes. The question is whether there is some physical operation that corresponds to this, or not.

In post #2 of the other thread, @Orodruin described lowering/raising an index with the metric as giving a one-to-one correspondence between vectors ("contravariant vectors") and covectors ("covector" is easier to type than "covariant vector" -- you will also see them called "1-forms" in some sources). But what does this mean? Does this mean that, if I lower the index on the 4-velocity vector, I get a 4-velocity covector? No. If we have a metric, then for any given 4-velocity vector, there is some covector that corresponds to it, but that covector is still a covector; it isn't a tangent vector, it's a gradient, in the terminology @Orodruin used. That's a different thing, physically, from a vector.

 

[robphy]

One can think of the metric-dual of the 4-velocity as the
[metric-induced-]covector that allows the 4-velocity to compute its square-magnitude [as defined by that metric] by counting John A Wheeler's "Bongs of bell" (Misner-Thorne-Wheeler Gravitation, p55)
(the number of intersections of the 4-vector with the covector's family of hyperplanes).

See

as part of my poster http://www.opensourcephysics.org/CPC/posters/salgado-talk.pdf
for the AAPT Topical Conference: Computational Physics for Upper Level Courses (2007)
https://www.opensourcephysics.org/CPC/ .

 

[cianfa72]

Yes, as @robphy highlighted, there is also a beautiful picture about covector as family of hyperplane elements in R. Penrose "The Road to Reality" p.225 - 226.

Let me add that in some cases a covector field (1-form) represents a family of hyperplanes tangent at each point to the level set of some local smooth function $\varphi$.

 

[robphy]

Yes, as @robphy highlighted, there is also a beautiful picture about covector as family of hyperplane elements in R. Penrose "The Road to Reality" p.225 - 226.

Let me add that in some cases a covector field (1-form) represents a family of hyperplanes tangent at each point to the level set of some local smooth function $\phi$.

In the lower-right panel of my poster, I have drawn my renditions
of the magnetic covector field in Ampere-Maxwell Law
and the electric covector field in Faraday's Law
based on the visualizations of Bill Burke (in his Applied Differential Geometry book):

Additional references:
https://physics.stackexchange.com/q...earn-em-with-differential-forms-with-focus-on

I think the earlier sources on these visualizations of differential forms are the works of Jan Schouten (Ricci Calculus and Tensor Analysis for Physicists.)

 

[pervect]

But, using the Minkowski metric, we can lower and raise the index of a vector

So following this, there exists a velocity four vector that is covariant.
What does that correspond to, since we have established that velocity should be describes by a contravariant vector?

I will give an example rather than try and generate an abstract answer.

Suppose we want to create some map from a 4-vector that "picks out" the x component of that vector. Thus, if the components of the vector are (t,x,y,z), it picks out the x component. If it's a 4-velocity (vt, vx, vy, vz), we want to pick out the magnitude of the x component of the velocity.

This process of "picking out" a component of the vector generates a map from a vector to a scalar. So it's by definition a one-form aka co-vector. And as has been mentioned, it can be represtend by a a "stack of plates", the number of stacks pierced by the vector generate the value of the x-component.

[add]
The x-component of the vector is just a scalar. The process of creating a mapping from a vector to it's x component is a one-form. This causes some notational issues, one I'm familiar with use boldvace for the map and non-boldface for the scalar.

 

[ergospherical]

Have you heard of the idea of a "dual basis" in Euclidean space R³? You have a "basis", a list of three vectors $(\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3)$ which define your coordinate axes. Most typically these are taken to be $\mathbf{e}_1 = (1,0,0)$, $\mathbf{e}_2 = (0,1,0)$ and $\mathbf{e}_3 = (0,0,1)$, but you have the freedom to define any basis you want.

Given this basis, you can construct a new set of vectors in the following way. Let $A = \mathbf{e}_1 \cdot (\mathbf{e}_2 \times \mathbf{e}_3)$ be the scalar triple product. Consider:

$\mathbf{E}^1 = A^{-1} \mathbf{e}_2 \times \mathbf{e}_3$
$\mathbf{E}^2 = A^{-1} \mathbf{e}_3 \times \mathbf{e}_1$
$\mathbf{E}^3 = A^{-1} \mathbf{e}_1 \times \mathbf{e}_2$

These have the property that $\mathbf{E}^i \cdot \mathbf{e}_j$ is equal to one if $i=j$ and zero otherwise, e.g. $\mathbf{E}^1 \cdot \mathbf{e}_1 = 1$ whilst $\mathbf{E}^1 \cdot \mathbf{e}_2 = 0$. The list ($\mathbf{E}^1, \mathbf{E}^2, \mathbf{E}^3$) is called the dual basis.

For the standard basis $\mathbf{e}_1 = (1,0,0)$, $\mathbf{e}_2 = (0,1,0)$ and $\mathbf{e}_3 = (0,0,1)$, you find that $\mathbf{E}^1 = \mathbf{e}_1$, etc., so there is no difference between the basis and the dual basis. But for other choices of basis (e.g. those that are not orthonormal), there will be a difference.

The dual basis is useful, because for non-orthonormal bases, it is the dual basis vectors which can be used to pick out the components of a vector. This means to say that if $\mathbf{v} = v^1 \mathbf{e}_1 + v^2 \mathbf{e}_2 + v^3 \mathbf{e}_3$ then $v^1 = \mathbf{E}^1 \cdot \mathbf{v}$ is true (rather than $v^1 = \mathbf{e}_1 \cdot \mathbf{v}$).

This idea can be abstracted a little bit further by defining a completely fresh vector space for dual vectors. In the above example, the "dual basis" were still vectors in the original vector space R³. Whereas, when people refer to the dual basis in differential geometry, they will typically mean a set of maps $\omega^i : V \rightarrow R$. If V = R³, then since we have a built-in dot product, there is a natural way of defining a dual basis given the vectors we have already constructed above: $\omega^i(\mathbf{v}) := \mathbf{E}^i \cdot \mathbf{v}$.