Covariance & Contravariance of Vectors

[som]

I don't know whether my question is write or not. Is there any way to obtain the covariant component of the same vector $$\vec{V}$$?
or is it just the components when written in terms of spherical coordinate unit vectors?

 

[Chestermiller]

I don't know whether my question is write or not. Is there any way to obtain the covariant component of the same vector $\vec{V}$?
or is it just the components when written in terms of spherical coordinate unit vectors?

You need to write it in terms of the coordinate basis vectors.

 

[som]

Oh ok. I missed the 2nd page. This is the first time I get a fully satisfactory and clear answer. Another thing I want to make clear. As I understand the coordinate basis vectors are covariant basis vectors and the reciprocal basis vectors are contravariant. So a vector can be represented either by a set of contravariant components and covariant basis or by a set of covariant components and contravariant basis. So is it meaningful if one say that a vector is contravariant?

 

[Chestermiller]

Oh ok. I missed the 2nd page. This is the first time I get a fully satisfactory and clear answer. Another thing I want to make clear. As I understand the coordinate basis vectors are covariant basis vectors and the reciprocal basis vectors are contravariant. So a vector can be represented either by a set of contravariant components and covariant basis or by a set of covariant components and contravariant basis. So is it meaningful if one say that a vector is contravariant?

No. Like you said, only the components. But some people call the contravarisnt components a “contravarisnt vector.”

 

[som]

In some sources I found the reason behind this co-contra nomenclature is the way they transform under a basis transformation. I found the examples of contravariant vectors as position vector, velocity, acceleration etc. and that of covariant ones as grad(f). For contravariant vectors (or more correctly vectors with contravariant components), the matrix that transforms the vector components must be the inverse of the matrix that transforms the basis vectors. On the other hand for vectors with covariant components, the components must be transformed by the same matrix as the change of basis matrix. I doubt this property to hold in general.

For instance, consider the velocity vector in cartesian system. It turns out both the covariant and contravariant components are same. Now they appear indistinguishable in terms of the way they transform under a basis transformation. Am I thinking write or something I missed?

 

[Chestermiller]

In some sources I found the reason behind this co-contra nomenclature is the way they transform under a basis transformation. I found the examples of contravariant vectors as position vector, velocity, acceleration etc. and that of covariant ones as grad(f). For contravariant vectors (or more correctly vectors with contravariant components), the matrix that transforms the vector components must be the inverse of the matrix that transforms the basis vectors. On the other hand for vectors with covariant components, the components must be transformed by the same matrix as the change of basis matrix. I doubt this property to hold in general.

For instance, consider the velocity vector in cartesian system. It turns out both the covariant and contravariant components are same. Now they appear indistinguishable in terms of the way they transform under a basis transformation. Am I thinking write or something I missed?

Isn't the transformation matrix between covariant and contravariant for Cartesian coordinates the identity matrix? I'm sure the property carries over.

What book are you using. Back in the day, I used Vector Analysis With an Introduction to Tensor Analysis by Wills.

 

[som]

Yes, you are right. It is unit matrix. But I meant to say something else. I found in wikipedia that the very name 'covariant' and 'contravariant' has been given purposefully according to their behaviour under basis transformation. Here is the link.
https://en.m.wikipedia.org/wiki/Covariance_and_contravariance_of_vectors
I say I doubt this description to hold in general. Because if the covariant and contravariant components of a vector become equal as in Cartesian system then how one could say they behave differently.

 

[Orodruin]

To have any difference between contravariant and covariant components, you need to go away from Cartesian coordinate system. The relation between the co- and contravariant components is given by the components of the metric tensor, which in Cartesian coordinate systems is the identity. In a general curvilinear coordinate system, this will no longer be the case.

It is also worth pointing out that when one goes away from a Euclidean space to a general manifold, the coordinate bases are bases of different vector spaces. While the tangent vectors are spanned by the partial derivatives $\partial_\alpha$ and naturally have contravariant components, its dual space is spanned by the differentials $dx^\alpha$ with naturally covariant components. It is not generally the case that a metric, and thus a relation between tangent vectors and dual vectors, exists.

 

[stevendaryl]

I would say that there are subtle differences between contravariant and covariant components, even in Cartesian coordinates.

Let's look at an example: You take a map of your town, and put a Cartesian coordinate system on it, by marking parallel lines north-south and east-west. Every location on the map is characterized by a pair of numbers $(x,y)$. A convenient set of basis vectors is $e_1$, representing a unit displacement in the x-direction, and $e_2$ representing a unit displacement in the y-direction. So an arbitrary vector $\vec{v}$ can be written as: $v^1 e_1 + v^2 e_2$. Now, you want to come up with a dual basis $e^1, e^2$ defined implicitly by the requirement:

$e_i \cdot e^j = \delta^j_i$

Clearly, you can just let $e^1 = e_1$ and $e^2 = e_2$, right? Sort-of...

The problem is that we chose $e_1$ and $e_2$ to be unit vectors in the x- and y- directions, respectively. But what's a unit? We can choose whatever we like to be a unit. For example, we can measure $x$ and $y$ in feet, in which case $e_1$ represents a displacement of one-foot in the x-direction. So if we have the equation $e_1 \cdot e^1 = 1$, that means that $e^1$ has a length of 1 (feet)^-1. But suppose we switch over to the metric system. In this case, $e_1$ has a length of 0.3077 meters, and $e^1$ has a length of 3.25 (meter)^{-1}.

So it's not literally true that $e^1 = e_1$. They only have the same length in one particular system of measurements. But they transform oppositely under changes of scale.

 

[Orodruin]

I actually have an issue with this. Your approach would give the unit vectors physical units, which is what leads to them changing when you change units. Instead, the unit vectors may be thought of as dimensionless and squaring to one (dimensionless). I think things become much clearer with this convention as all components of any vector expressed in terms of unit vectors would have the appropriate physical dimension. If you choose a dimensional basis, it cannot square to one in arbitrary units and the components of, for example, a velocity vector would have units of 1/time.

 

[stevendaryl]

I actually have an issue with this. Your approach would give the unit vectors physical units, which is what leads to them changing when you change units. Instead, the unit vectors may be thought of as dimensionless and squaring to one (dimensionless).

It's hard for me to get a grasp of what a dimensionless basis vector means.

The way I thought of it was that the basis vector $e_x$ is the tangent to the parametrized curve $x(s), y(s), z(s)$ such that $\frac{dx}{ds} = 1$, $\frac{dy}{ds} = 0$, $\frac{dz}{ds} = 0$. If it's not that, then what is it? But if it is that, then it seems to me that $e_x$ would carry along the same units as $x$ does. Or maybe you're saying that the coordinates are dimensionless, also?

 

[Orodruin]

If both coordinates and basis vectors carry dimension of length then the dimension of the position vector would be length^2 as $[x^i \vec e_i] = [x^i][\vec e_i] = \mathsf{L}^2$.

The more elegant (my opinion, it is subjective) way of defining the basis vectors is by having the coordinates carry the dimension of the position vector.

Note that if you parametrise with the curve length, then $dx/ds = 1$ is independent of the units you pick as $s$ carries the same units as $x$. The unit vectors define the directions and essentially have dimension $\mathsf{L/L}=1$, i.e., you go along the tangent of the line for $\ell = 1$ m in direction $\vec e$ and you end up $\ell\vec e$ relative to where you started.

 

[stevendaryl]

Anyway, regardless of the conventions for whether basis vectors have dimensions, or not, there is a difference between tangent vectors and covectors when it comes to changing units.

A typical tangent vector might be a particle's velocity, which could measured in units (feet) (seconds)^-1. A typical covector might be a gradient. For example, the change in temperature as you travel North might be measured in (degrees K) (feet)^-1. Suppose you have a velocity vector $v$ with magnitude say 15 feet/second and a temperature gradient G of 10^{-4} degrees/foot. Now convert from feet to meters. The magnitude of $v$ would go from 15 to 4.62 (multiply by .3077). The magnitude of G would go from 10^-4 to 3.25 x 10^-4. So you multiply by 3.25, the reciprocal of .3077.

So dimensional vectors and covectors scale differently under changes of units. Whether you want the basis vectors to reflect this, or not, is a matter of convention, I guess.

 

[Orodruin]

Anyway, regardless of the conventions for whether basis vectors have dimensions, or not, there is a difference between tangent vectors and covectors when it comes to changing units.

I fundamentally disagree with this. Changing units does not change the vectors. Only the numerical values we assign to their components. How those values change depends only on the dimensions of those components.

A typical tangent vector might be a particle's velocity, which could measured in units (feet) (seconds)-1. A typical covector might be a gradient. For example, the change in temperature as you travel North might be measured in (degrees K) (feet)-1. Suppose you have a velocity vector vvv with magnitude say 15 feet/second and a temperature gradient G of 10^{-4} degrees/foot. Now convert from feet to meters. The magnitude of vvv would go from 15 to 4.62 (multiply by .3077). The magnitude of G would go from 10-4 to 3.25 x 10-4. So you multiply by 3.25, the reciprocal of .3077.

All of this is a direct result of the units of the vector components. If you change the time units instead, only one of them will change because only one of them has non-trivial time dimension.

Also, the magnitudes of the vectors do not change. You just expressed them in different units and the magnitude is a dimensional quantity.

Of course, the length dimensions of a vector is generally indicative of how that vector will naturally generalise when going away from a nomalised basis.

 

[som]

The coordinate basis vectors defined as ${\hat a}_q=\frac{\partial{\vec r}}{\partial q}$ can indeed have a dimension. If I make a simple step out from Cartesian to Plane Polar coordinate system, then the coordinate basis vectors are
${\hat a}_r={\hat r}$, ${\hat a}_\theta={r\hat \theta}$
So the contravariant components of velocity are $v^1=\dot r$, $v^2=\dot \theta$
The reciprocal basis vectors are
${\hat a}^r={\hat r}$, ${\hat a}^\theta={\frac{1}{r}\hat \theta}$
So the covariant components of velocity are $v^1=\dot r$, $v^2=r^2\dot \theta$. Hence all the basis vectors as well as the components may have different dimensions.

Upto this I do not have any confusion. We have some rules to find the covariant and contravariant components of a vector, we have followed those and worked them out. This is also consistent with the interconversion with the help of metric tensor because here $g_{11}=1$, $g_{22}=r^2$, $g_{12}=g_{21}=0$.

My point is that when we say the covariant components are {${\dot r}, r^2{\dot \theta}$} and the contravariant components are {${\dot r}, {\dot \theta}$}, is there any framework or methodology or definition by which we can show that the 1st one actually co-varies with something and the 2nd contra-varies with that same thing?

 

[stevendaryl]

I fundamentally disagree with this. Changing units does not change the vectors. Only the numerical values we assign to their components.

Yes, but the way that the numerical values associated with velocity vectors changes under a change of units is the opposite of the way that a gradient changes under a change of units. That has to be the case, because the product $\vec{v} \cdot \nabla T$ is a number of dimensions degrees/second, which obviously can't depend on the units we're using for distance.

 

[Orodruin]

Yes, but the way that the numerical values associated with velocity vectors changes under a change of units is the opposite of the way that a gradient changes under a change of units. That has to be the case, because the product $\vec{v} \cdot \nabla T$ is a number of dimensions degrees/second, which obviously can't depend on the units we're using for distance.

Yes, but I do not find this particular to scalar products like this and therefore not something that fundamentally has to do with co-/contravariance. Take $\vec v t$ for example, this is a displacement vector and the value does not depend on the units we use for time.

My point is that when we say the covariant components are {${\dot r}, r^2{\dot \theta}$} and the contravariant components are {${\dot r}, {\dot \theta}$}, is there any framework or methodology or definition by which we can show that the 1st one actually co-varies with something and the 2nd contra-varies with that same thing?

If you only have the components of a vector in one coordinate system you cannot get far. You need to know which basis they are written in for them to hold meaning. If you know the components in two systems you can check explicitly if the transformation rules are satisfied.

 

[som]

Yes, but I do not find this particular to scalar products like this and therefore not something that fundamentally has to do with co-/contravariance. Take $\vec v t$ for example, this is a displacement vector and the value does not depend on the units we use for time.If you only have the components of a vector in one coordinate system you cannot get far. You need to know which basis they are written in for them to hold meaning. If you know the components in two systems you can check explicitly if the transformation rules are satisfied.

Can you give an explicit example for clarification?

 

[Chump]

I would say that there are subtle differences between contravariant and covariant components, even in Cartesian coordinates.

Let's look at an example: You take a map of your town, and put a Cartesian coordinate system on it, by marking parallel lines north-south and east-west. Every location on the map is characterized by a pair of numbers $(x,y)$. A convenient set of basis vectors is $e_1$, representing a unit displacement in the x-direction, and $e_2$ representing a unit displacement in the y-direction. So an arbitrary vector $\vec{v}$ can be written as: $v^1 e_1 + v^2 e_2$. Now, you want to come up with a dual basis $e^1, e^2$ defined implicitly by the requirement:

$e_i \cdot e^j = \delta^j_i$

Clearly, you can just let $e^1 = e_1$ and $e^2 = e_2$, right? Sort-of...

The problem is that we chose $e_1$ and $e_2$ to be unit vectors in the x- and y- directions, respectively. But what's a unit? We can choose whatever we like to be a unit. For example, we can measure $x$ and $y$ in feet, in which case $e_1$ represents a displacement of one-foot in the x-direction. So if we have the equation $e_1 \cdot e^1 = 1$, that means that $e^1$ has a length of 1 (feet)^-1. But suppose we switch over to the metric system. In this case, $e_1$ has a length of 0.3077 meters, and $e^1$ has a length of 3.25 (meter)^{-1}.

So it's not literally true that $e^1 = e_1$. They only have the same length in one particular system of measurements. But they transform oppositely under changes of scale.

$v^1 e_1 + v^2 e_2$ Why not have everything in a lower index, here? Why is one raised and one lowered?

 

[stevendaryl]

$v^1 e_1 + v^2 e_2$ Why not have everything in a lower index, here? Why is one raised and one lowered?

It's just a convention, but whether the index is raised or lowered tells how the object transforms under coordinate changes. Here's a simple illustration:

At the top, we have a two-vector basis, $e_1$ and $e_2$. We have a vector $V$ that can be written in terms of that basis as:

$V = 2 e_1 + 6 e_2$

In terms of the basis $e_1, e_2$, we can write the components of $V$ as:

$V^1 = 2$, $V^2 = 6$

At the bottom, we have a different basis, $e_a = 2 e_1$ and $e_b = 3 e_2$. In terms of this basis:

$V^a = 1$, $V^b = 2$

So when the basis vectors get bigger, the components get smaller, so that the combination $V^\mu e_\mu$ remains constant.

Whether the coordinate is raised or lowered tells how the object changes. (Assuming that we're dealing with vectors or tensors or their components)

 

[Chump]

It's just a convention, but whether the index is raised or lowered tells how the object transforms under coordinate changes. Here's a simple illustration:

View attachment 224059

At the top, we have a two-vector basis, $e_1$ and $e_2$. We have a vector $V$ that can be written in terms of that basis as:

$V = 2 e_1 + 6 e_2$

In terms of the basis $e_1, e_2$, we can write the components of $V$ as:

$V^1 = 2$, $V^2 = 6$

At the bottom, we have a different basis, $e_a = 2 e_1$ and $e_b = 3 e_2$. In terms of this basis:

$V^a = 1$, $V^b = 2$

So when the basis vectors get bigger, the components get smaller, so that the combination $V^\mu e_\mu$ remains constant.

Whether the coordinate is raised or lowered tells how the object changes. (Assuming that we're dealing with vectors or tensors or their components)

Thanks for the reply. I have a few more questions. If this isn't the proper forum, please tell me; and I'll create a new thread.

So, I'd like to be sure I understand: The second representation of V that you have there is a covariant representation because the basis, let's call it basis two, is a covariant one, right? However, it just so happens that the component, which has another basis, let's call it basis one, incorporated into it, has a raised index, as opposed to a lower one because the component contravaries to basis two?

2nd question: I've seen the components of the metric tensor represented by $g^{jk} = e_{(j)} \cdot e_{(k)}$ and $g_{jk} = e^{(j)} \cdot e^{(k)}$ and $e_i \cdot e^j = \delta^i_j$. Are these non-standard notations? Because they are confusing, especially the mixed tensor, I would think that they're all like this: $g_{jk} = e_{(j)} \cdot e_{(k)}$ and $g^{jk} = e^{(j)} \cdot e^{(k)}$ and $e_i \cdot e^j = \delta^j_i$.

3rd question: How are differentials and partial derivatives incorporated into these representations of vectors? By that I mean, I understand that bases can be reprsented in terms of each other; how can the differentials and partial derivatives establish that?