Understanding 2D Tensors: Covariant, Contravariant, and Physical Components

[Hrvoje Zoric]

Explain, plase to me: 2D covariant, conravariant and physical components of vector. I can not find this thematic in my official study book.

 

[techmologist]

Explain, plase to me: 2D covariant, conravariant and physical components of vector. I can not find this thematic in my official study book.

Say you have a particular coordinate system for the 2D plane, such as polar coordinates. In this case the position vector as a function of the polar coordinates is

$$ \textbf{x} {(r, \theta)} \quad = \quad r \cos \theta \textbf{e}_1 + r \sin \theta \textbf{e}_2$$

Then your new basis vectors at any point in space are defined by

$$ \textbf{g}_r = \frac{\partial \textbf{x}}{\partial r} \quad = \quad \cos \theta \textbf{e}_1 + \sin \theta \textbf{e}_2$$

$$\textbf{g}_{\theta} = \frac{\partial \textbf{x}}{\partial \theta} \quad = \quad -r\sin \theta \textbf{e}_1 + r\cos \theta \textbf{e}_2$$

Notice that g_r is already a unit vector, but g_theta is not. Typically, these generalized basis vectors will not be unit vectors. These also happen to be orthogonal because the polar coordinate system is an orthogonal coordinate system. But for more general coordinate system, the generalized basis vectors are not typically orthogonal, either.

Now suppose you have a vector field $\textbf{F}$ which is a function of position. At any position in 2D space, you can write F as a linear combination of the local basis vectors:

$$ \textbf{F} = F^r \textbf{g}_r + F^{\theta} \textbf{g}_{\theta}$$

If I remember the terminology right, $F^r$ and $F^{\theta}$ are the covariant components of the vector field.

In addition to the local basis $\textbf{g}_r$ and $\textbf{g}_{\theta}$, it is useful to define the dual basis at that same point in space, $\textbf{g}^r$ and $\textbf{g}^{\theta}$ in such a way that

$$ \textbf{g}_r \cdot \textbf{g}^r \quad = \quad \textbf{g}_{\theta} \cdot \textbf{g}^{\theta} = 1 $$

and

$$ \textbf{g}_r \cdot \textbf{g}^{\theta} \quad = \quad \textbf{g}^r \cdot \textbf{g}_{\theta} = 0 $$

You can also express the vector field $\textbf{F}$ in terms of this dual coordinate basis:

$$ \textbf{F} = F_r \textbf{g}^r + F_{\theta} \textbf{g}^{\theta}$$

Assuming I don't have it backwards, $F_r$ and $F_{\theta}$ are the contravariant components of the vector field $\textbf{F}$ at that point.

Since the generalized basis vectors are typically not dimensionless unit vectors, the dimensions of the covariant components of a field will typically be different from the Cartesian components of that same field. To correct for this, you can define the physical components

$$ \overline{F}^r = F^r |\textbf{g}_r| $$

and

$$ \overline{F}^{\theta} = F^r |\textbf{g}_{\theta}| $$

These would be the components you would get if you normalized the local basis vectors to unit length.

 

[Noctisdark]

First of all you must know why these are used, they are called dual co-ordinates, they are used because sometimes we need to change system of co-ordinates, as you may know in GR, co-ordinates are very wacky, they are nothing like cartesian(etc...) systems, so one has to come up with ideas to make life easier, I'm going to introduce them in a very special case so you get a feel for them and what di they represent, i want you to take a paper and a pencil and bare with me, you will draw a co-ordinate system with only 2 bases (2 lines) and name the angle between $\theta$, (if $\theta = \frac{\pi}{2}$ then you have good old orthogonal cartesian co-ordinates so avoid this case), now highlight a point A in your co-ordinate system and draw a vector for the origin to A, the problem is how to determine the components of that vector, let's call it $\vec u$, there are two ways, you can project your vector on the x-axis and on the y axis, you'll get 2 component $A_1$ and $A_1$, then your vector $\vec u = A_1 \hat e^1 + A_2 \hat e^2$, here $\hat e^1, \hat e^2$ reprensent the unit vector on the x,y axis respectiv, $A_1, A_2$ are called the covariant components of $\vec u$ and they represent the parralel projection of $\vec u$ in such a way $A_1 = \vec u \cdot \hat e^1 , A_2 = \vec u \cdot \hat e^2$, when someone tells you that the parralel project of vector on axis yield a good representation of a vector, you might argue and wonder if will can perform and orthogonal projection of that vector and yes we can, we get a different set of component called contravariant components of $\vec u$, and we write $\vec u = A^1 \cdot \hat e_1 + A^2 \cdot \hat e_2$ where ( $A^1, A^2$) is the set of component, but what are the weird unit vectors $\hat e_1, \hat e_2$?, draw on your paper 2 orthogonal lines to the x and y axis, and call x* the one orthogonal to the y-axis and the other y*, $\hat e_1$ a vector on x* and $\hat e_2$ is the unit vector on y* such that $\hat e^1 \cdot \hat e_1 = 1$ and $\hat e^2 \cdot \hat e_2 = 1$, you can see that $\hat e^1 \cdot \hat e_2 = 0$ and so on,the component are the projection on these axis, if you have drawn a pic you would be aware of what's going on here, we are just being picky in our choice of co-ordinates, Next post I'll make an example

 

[Noctisdark]

Now since we know how we can represent a vector in a fairly simple co-ordinates, can we know it's magnitude, or it's dot product with other vectors ?, yes we can and for that purpose I shall introduce the great and wonderful (I allways liked it's name) the metric tensor, it written as $g_{\mu \nu}$ and defined as $g_{\mu \nu} = \hat e_{\mu} \cdot \hat e_{\nu}$ (a lot more on the internet I'm being basic here), $\hat e^{\mu} \cdot g_{\mu \nu} = e^{\mu} \cdot\hat e_{\mu} \hat e_{\nu} = \hat e_{\nu}$ so we can go from contravariant to covariant just by the use of the metric tensor, there is a summation of $\mu$, Einstein summation convention applied when 2 terms (dummy indices repeat more than once, that's what repeat means), $A^{\mu}A_{\mu} = A_{\mu}A^{\nu}g_{\mu \nu}$, geometrically you can see that $|\vec u|^2 = A^{\mu}A_{\mu}$, but $\vec u \cdot \vec u = (A_{\mu}\hat e^{\mu}) \cdot (A^{\nu} \hat e_{\nu})$ and it is exactly what we though, so as an example i want you to set $\theta = \frac{\pi}{4}$ and set A = (2,3) and work out it length, and come up with a formula to calculate the scalar.product if two vectors !,