Covariant coordinates don't co-vary

[George Keeling]

Homework Statement

I am studying co- and contra- variant vectors and I found the video at youtube.com/watch?v=8vBfTyBPu-4 very useful. It discusses the slanted coordinate system above where the X, Y axes are at an angle of α. One can get the components of v either by dropping perpendiculars to the axes (v_i) or by dropping a line parallel to the other axis (vⁱ). These give correct results for the norm vⁱv_i and the dot product vⁱ w_i. (I have not shown w). So the vⁱ are called contravariant and the v_i are called covariant. According to the video Dirac thought this was a great example.

But both v_i and vⁱ contra-vary with a change of scale of the basis vectors. This contradicts some definitions of contravariant and covariant components, e.g. this one on Wikipedia. These definitions say that covariant components co-vary with a change of scale.

Is there a simple resolution to this apparent contradiction?

Homework Equations

Norm and dot product were calculated by expressing v_i and vⁱ in Cartesian coordinates. Quite a lot of equations! Along the way we found the metric for the slanted coordinate system.

The Attempt at a Solution

We can demonstrate the problem by drawing basis vectors on the diagram and then another set double their size.

 

[Orodruin]

The covariant and contravariant components belong to different basis vectors.

 

[Cryo]

Why not think about vectors in terms of partial derivatives? You have space, which you can label with some coordinates $\left\{x^{(1)}, x^{(2)}, \dots\right\}$. Imagine there is a differentiable, scalar function $f=f\left(x^{(1)}, x^{(2)}, \dots\right)$ defined everywhere in space. Consider the difference of the value of function between two closely-spaces points

$df=f\left(x^{(1)}+a^{1},\dots\right)-f\left(x^{(1)},\dots\right)$, where $a^{\dots}\to 0$

Keeping only first order terms the result is:

$df=a^i \partial_i f + \mathcal{O}\left(a^2\right)$

Clearly the difference $df$ will not depend on the coorinate system. You can compute how the partial derivatives $\partial_i f=\frac{\partial f}{\partial x^{(i)}}$ change when you change the coordinates. It follows that $a^i$ must change in the 'opposite' way, to ensure invariance of $df$: that's your contra-variant vector. The rules for changing the co-variant vectors, i.e. $w_i$, follow from the fact that $w_i a^i = scalar$, since $w_i$ is a linear functional over the vector space of contra-variant vectors, i.e. by definition $w_i a^i = scalar$

 

[George Keeling]

Thank you Orodruin. Cryo is a bit harder. Following Orodruin I immediately saw that I needed smaller covariant basis vectors which I call e_x, e_y. I realize that the up/down position of the indices is debatable. Now I have a new diagram showing a few of the basis vectors.

To get v with the covariant coordinates we must have

v = v_xe_x + v_ye_y . . . . . . . (1)​

I have drawn them in on the diagram and e_x is clearly smaller than e^x, which I have conveniently drawn so that v^x = 1. We must also have

v_xe_x = v^xe_x . . . . . . . (2)​

and similarly

v_ye_y = v^ye^y . . . . . . . (3)​

which give us

v = v^xe^x + v^ye^y . . . . . . . (4)​

which is as it should be.
(3) gives us

v_y = e^y(e_y)^-1v^y . . . . . . . (5)​

where (e_y)^-1 is the inverse of e_y: or e_y(e_y)^-1 = 1.
(5) shows us that v_y does indeed vary with e^y with a strange constant of proportionality (e_y)^-1v^y.

PS Is there a way to do hard spaces in BB Code? I couldn't see anything in https://www.physicsforums.com/help/bb-codes.

 

[Orodruin]

Your diagram is not correct. Generally the dual basis is not parallel to the tangent basis (it is orthogonal to the coordinate surfaces, not tangent to the coordinate lines).

 

[George Keeling]

Orodruin:
Is it true, like contravariants, that for covariant bases / coordinates

v = v_xe_x + v_ye_y

 

[Orodruin]

A vector can be written in terms of any basis with the corresponding components. When you have a coordinate system with coordinate functions $x^a$ you can introduce the tangent basis $\vec e_a = \partial \vec x/\partial x^a$ and the dual basis $\vec e^a = \nabla x^a$, respectively. The contravariant components by definition are those components that express a vector in the tangent basis and the covariant components by definition are those that express a vector in the dual basis.

You can write $\vec v = v^a \vec e_a$ or $\vec v = v_a \vec e^a$, but generally this is not the same as $v_a\vec e_a$, which does not make sense because you are then creating a vector that has the same contravariant components as the covariant components of $\vec v$ in this particular coordinate system.

 

[George Keeling]

I did say

covariant basis vectors which I call e_x, e_y. I realize that the up/down position of the indices is debatable.

. In that context the answer to my question

v = v_xe_x + v_ye_y

is yes because (positioning indices less debatably)

$\vec v = v_a \vec e^a$

.
Now that I have learned that $\vec e^a = \nabla x^a$, it should be easy to make progress. (Famous last words)

 

[Cryo]

I think the other contributors already explained the significance of derivatives when talking about vectors on manifolds, which is what you are doing. I will just add one more thing about the dual space.

As I said in my earlier post one can introduce the space of contra-variant vectors $v^i$ through partial derivatives. What was confusing for me, when I was learning, is how to think about the dual space, i.e. $v_i$. Sometimes people say that if $v^i$ is a vector then $v_i$ is like a surface through which this vector (field) $v^i$ flows. I do not quite like this (e.g. see Bachman "Geometric approach to differential forms" -> Sec. 5.7 "How not to visualize a differential 1-from"). What I have come to realize is that the best way to think about dual space is as functionals - which is the normal way do define it (e.g. Halmos "Finite-dimensional vector spaces").

I.e. you define your contra-variant vector space $V$ whichever way you like (as long as it makes sense). Then the dual space is a space $V^*$ of linear functions $\mathbf{f}: V \to \mathbb{R}$ (or $\mathbb{C}$), i. e. if $\mathbf{v}, \mathbf{w} \in V$, $\alpha, \beta \in \mathbb{R}$ (or $\mathbb{C}$) and $\mathbf{f} \in V^*$ then:

$\mathbf{f}\left(\alpha\mathbf{v} + \beta\mathbf{w}\right) = \alpha \mathbf{f}\left(\mathbf{v} \right) + \beta \mathbf{f}\left(\mathbf{w} \right) = real\, number$ (or $complex\, number$)

This may be a weird way of looking at it, but it makes it plainly obvious that the dual space $V^*$ is different from $V$. How do you square this with the conventional notions? Let's say your basis for (finite-dimensional) $V$ is $\mathbf{e}_i$, i.e. $\mathbf{v} \in V \implies \mathbf{v}=v^i\mathbf{e}_i$. Then you could introduce a basis-set of functionals $\mathbf{q}^i: V\to \mathbb{R}$ such that:

$\mathbf{q}^i \left(\mathbf{e}_j\right) = \delta^i_j = \begin{cases} 1, & \text{if i=j}.\\ 0, & \text{otherwise}. \end{cases}$

Then any member of dual space $\mathbf{f} \in V^*$ is given by $\mathbf{f}=f_i \mathbf{q}^i$, and:

$f_i v^i = \mathbf{f}\left(\mathbf{v}\right) = f_i v^j \mathbf{q}^i \left(\mathbf{e}_j\right) = f_i v^j \delta^i_j$

 

[Cryo]

Thinking of dual space as of space of linear functions also explains why co-variant vectors must co-vary, since $\mathbf{f}\left(\mathbf{v}\right)=scalar$ is a statement that does not depend on coordinates, so that $scalar$ has to be the same irrespective of the coordinte system and basis sets used to span the vector and dual vector spaces.

 

[George Keeling]

Following Orodruin's great tip

$ê^j = \nabla x^j$

(I am using ^ instead of → over basis vectors, after Sean Carroll.) Since I already knew the inverse metric of the skewed system, it was easy to calculate that
$$ ê^1=\hat x - \frac { \hat y } {\tan \alpha} \\
{\rm and} \phantom {10000000000000000000000000000000000000000000000000000000000}\\
ê^2=\frac { \hat y } {\sin \alpha} $$ where $\hat x, \hat y$ are the Cartesian basis vectors. In can now redraw my diagram ($x=1,y=2$)

$ê^x$ is $\pi / 2 - \alpha$ below the Cartesian X axis. Therefore the angle between $ê_y$ and $ê^x$ is a right angle, and the projected line from $v_y$ is parallel to $ê^x$ as I should have known and Orodruin intimated when he wrote in #5 "the dual basis is ... orthogonal to the coordinate surfaces, not tangent to the coordinate lines". Likewise $ê^y$ is parallel to the projected line from $v_x$ and always on the Cartesian Y axis. In addition $|ê^i |>|ê_i |$.

All this did not answer my original question that "both $v^i$ and $v_i$ contra-vary with a change of scale of the basis vectors". I had to use the pullback operator to find the metric of the primed system and then showed that if $ê'_x=aê_x$ and $ê'_y=bê_y$ ($a=2$ in the diagram above) then
$$v'_x=a^2v'^x+abv'^y \cos \alpha\\
{\rm and} \phantom {10000000000000000000000000000000000000000000000000000000000}\\
v'_y=b^2v'^y+abv'^x \cos \alpha$$ Which does answer the question - the transformed covariant coordinates do covary with the contravariant basis vectors.

I have obviously left out quite a bit of detail. It is elsewhere. Thanks Dirac and PF and please like this post if I am correct.

 

[aradii]

Homework Statement

View attachment 233849
I am studying co- and contra- variant vectors and I found the video at youtube.com/watch?v=8vBfTyBPu-4 very useful. It discusses the slanted coordinate system above where the X, Y axes are at an angle of α. One can get the components of v either by dropping perpendiculars to the axes (v_i) or by dropping a line parallel to the other axis (vⁱ). These give correct results for the norm vⁱv_i and the dot product vⁱ w_i. (I have not shown w). So the vⁱ are called contravariant and the v_i are called covariant. According laser to the video Dirac thought this was a great example.

But both v_i and vⁱ contra-vary with a change of scale of the basis vectors. This contradicts some definitions of contravariant and covariant components, e.g. this one on Wikipedia. These definitions say that covariant components co-vary with a change of scale.

Is there a simple resolution to this apparent contradiction?

Homework Equations

Norm and dot product were calculated by expressing v_i and vⁱ in Cartesian coordinates. Quite a lot of equations! Along the way we found the metric for the slanted coordinate system.

The Attempt at a Solution

We can demonstrate the problem by drawing basis vectors on the diagram and then another set double their size.

About the video you posted i have a question guys wouldn't the metric tensor need to be multiplied by 3 differentials since we're in 3-space? It seems like it would need dx dy dz. Can the metric tensor manipulate only dx and dy to move through 3-space? Not sure I understand.

 

[George Keeling]

Welcome to Physics Forums aradii!
I re-watched that video but only part 1. The first five minutes are about any number of dimensions and we saw things like the metric equation $ds^2=g_{ij}dx^idy^j$ which is using the Einstein summation convention and $i,j$ run from $0...(n-1)$ or $1...n$ where $n$ is the number of dimensions. In the second half it is only talking about two dimensions, (Euclidean or oblique). In that case in Euclidean $dx^1=dx , dx^2=dy$ and there is no $dx^3$ which would have been $dz$. The metric equation in full is$$
ds^2=g_{11}dxdx+g_{12}dxdy +g_{21}dydx +g_{22}dydy
$$Also the metric is always symmetric so $g_{12}=g_{21}$
Hope that helps!

 

[Orodruin]

About the video you posted i have a question guys wouldn't the metric tensor need to be multiplied by 3 differentials since we're in 3-space? It seems like it would need dx dy dz. Can the metric tensor manipulate only dx and dy to move through 3-space? Not sure I understand.

As described in #13, the metric is a type (0,2) tensor. It takes two tangent vectors and map them to a number (that we typically call the "inner product" between those vectors). In essence, this is why there are two $dx^\mu$ in the definition of the metric (although really it should be $g = g_{\mu\nu} dx^\mu \otimes dx^\nu$, we physicists are often a bit sloppy with notation. What you seem to be thinking about is more akin to the volume form, which in an $N$-dimensional metric space is an $N$-form generally given by $\eta = \sqrt{|g|} dx^1 \wedge \ldots \wedge dx^N$, where $|g|$ is the metric determinant. The metric $g$ describes the geometry of the space (in terms of lengths and angles between two vectors, which it has two $dx$s) while the volume form $\eta$ describes volume elements and the volume spanned by $N$ vectors, which is why it has $N$ $dx$s.