Write ∇u with covariant components and contravariant basis

[mishima]

Homework Statement

Write $\nabla u$ in polar coordinates in terms of its physical components and the unit basis vectors $e_i$, and in terms of its covariant components and the contravariant basis vectors $a^i$. What is the relation between the contravariant basis vectors and the unit basis vectors?

Relevant Equations

$$V'^i = \frac {\partial x'^i} {\partial x_j} V^j$$

The first part I'm fairly sure is just the regular gradient in polar coordinates typically encountered:

$$\nabla u= \hat {\mathbf e_r} \frac {\partial u} {\partial r} + \hat {\mathbf e_\theta} \frac 1 r \frac {\partial u} {\partial \theta}$$

or in terms of scale factors:

$$=\sum \hat {\mathbf e_i} \frac 1 h_i \frac {\partial u} {\partial x_i}$$

I don't know how to prove that the $\frac 1 r \frac {\partial u} {\partial \theta}$ for example is the so called "physical" component. Are physical components just the dot product of the gradient with each basis?

Similarly I don't know how to show that any particular part, say $\hat {\mathbf e_\theta} \frac 1 r$ is the contravariant basis vector. Do I just plug it into the transformation equation and verify equality$$V'^i = \frac {\partial x'^i} {\partial x_j} V^j ~~~~?$$

 

[Orodruin]

The physical components are the components of a vector in the normalised basis. Since that basis is orthonormal you indeed get the physical components by taking the inner product with the corresponding normalised unit vector.

The contravariant basis (or, as I prefer, the dual basis) is defined as $\vec E^a = \nabla y^a$, where $y^a$ is the coordinate function. Your text (you should specify which text you are using, it may help people understand where you are coming from) should discuss how it relates to the tangent basis and the normalised basis.

 

[mishima]

I'm using Boas, Mathematical Methods in the Physical Sciences, 3rd ed. The text is light on details in this section, and seems to have left most of the meat to the practice problems. From what I can tell, the relation between the tangent and normalized basis is the scale factor. I am having a hard time understanding how to compute...for example in your equation, what if y is a partial derivative? Can it be? Are you saying if I want the dual basis for the gradient, I should take the gradient of the gradient? Sorry, just confused.

 

[Orodruin]

From what I can tell, the relation between the tangent and normalized basis is the scale factor.

The definition of the tangent basis is that it consists of the tangent vectors of the coordinate lines, they are therefore given by
$$
\vec E_a = \frac{\partial \vec x}{\partial y^a},
$$
the partial derivative of the position vector with respect to the individual coordinates. If you deal with an orthogonal coordinate system, you can introduce the scale factors $h_a$ as the norm of the tangent basis vectors. This means that you can introduce an orthonormal basis by dividing the $\vec E_a$ by their norms, i.e., $\vec e_a = \vec E_a/h_a$ (no sum).

The dual basis $\vec E^a$ is what I was mentioning above. It is defined as the gradient of the coordinate functions $\vec E^a = \nabla y^a$. Here $y^a$ is not a partial derivative, it is a coordinate function, for example $r = \sqrt{x^2 + y^2}$ for the radial polar coordinate. Since the gradient of any function $f$ is orthogonal to the level surfaces of $f$, the dual basis vectors are orthogonal to the level surfaces of the coordinate functions, i.e., the surfaces with that coordinate being constant.

It always holds that $\vec E^a \cdot \vec E_b = \delta^a_b$. However, if your coordinate system is orthogonal, this relation results in $\vec E^a$ being parallel with $\vec E_a$ (and therefore also with the normalised basis vector $\vec e_a$) and that indeed $\vec E^a = \vec E_a/h_a^2$ (no sum).

I believe this should definitely be covered explicitly in textbooks that choose to discuss the tangent and dual bases - not left as exercises. I know at least one book that does ...

what if y is a partial derivative? Can it be?

No, and yes. Once you leave Euclidean space it will turn out that tangent vectors are actually directional derivatives with the partial derivatives as the tangent basis, but that is a story for later (calculus on manifolds). Here however, $y^a$ is a coordinate function.

Are you saying if I want the dual basis for the gradient, I should take the gradient of the gradient?

No, you should take the gradient of the coordinate functions.

 

[mishima]

Alright, so I start with the coordinate functions, not the polar ds element or gradient function. I take the gradient of the 2 coordinate functions. Then...

$$\nabla r(x,y) = cos \theta \mathbf {\vec E_x} +sin \theta \mathbf {\vec E_y}$$

which has magnitude 1. Then from the $\theta$ coordinate function...

$$\nabla \theta(x,y) = \frac {-sin \theta}{r} \mathbf {\vec E_x} +\frac {cos \theta}{r} \mathbf {\vec E_y}$$

which has magnitude 1/r.

Im still not sure from those results what exactly I point to and call the dual (contravariant) basis.

 

[Orodruin]

Those gradients constitute the dual basis:
$$\vec E^r = \nabla r, \qquad \vec E^\theta = \nabla\theta.$$
Note that $\vec E^i = \vec e_i = \vec E_i$ in Cartesian coordinates.

Geometrically, the tangent basis tells you in what direction you go if you change the corresponding coordinate. The dual basis are the surface normals to the surfaces you get when keeping the corresponding cordinate fixed (and change all other coordinates).

 

[mishima]

So, what I just calculated was the dual basis...the gradient of polar 'space' defined by the coordinate functions. Now the problem is asking me to give the gradient of an arbitrary function in this space, right? So I need to find tangent (covariant) components to match this dual (contravariant) basis. Would this just be the dot product again? I project the gradient of my arbitrary function u onto the dual basis vectors just computed?

 

[Orodruin]

So I need to find tangent (covariant) components to match this dual (contravariant) basis.

You just call them covariant components or ”components of the vector in the dual basis”. Unfortunately, some texts focus on the transformation properties of things rather than what they mean.

Would this just be the dot product again? I project the gradient of my arbitrary function u onto the dual basis vectors just computed?

No. The dual (or tangent) basis is not orthonormal, so you cannot do it this way. However, you can dot it with the other basis due to $\vec E^a\cdot \vec E_b = \delta ^a_b$. In reality, this is what you are doing in a Cartesian system too - but there both bases coincide.

For your particular problem, things are much easier than that. You have the function expressed in terms of your coordinates. Just apply the gradient and use the chain rule! You do not even need to know what the basis vectors are!

 

[mishima]

Sorry, I am really confused. You're saying to take the gradient of the components of the gradient? Like for $\theta$

$$\nabla (\frac 1 r \frac {\partial u} {\partial \theta})~~? $$

 

[Orodruin]

No. I am telling you to take the gradient of the function and use the chain rule.

 

[mishima]

My arbitrary u function?

 

[Orodruin]

Yes, which is a function of the coordinates.

 

[mishima]

I get what you are saying, but how do I recognize the "covariant components and the contravariant basis vectors" after? Am I chain ruling between r $\theta$ and x y or r $\theta$ and r' $\theta'$? If the former, why does the partials wrt x y inherently make covariant components? If the latter, why do I need an arbitrary transformation to generate the covariant components?

 

[Orodruin]

I get what you are saying, but how do I recognize the "covariant components and the contravariant basis vectors" after?

I am not sure that you do and it is very difficult to know if you do not show your work.

By definition, the components of a vector in a given basis are the expansion coefficients of the vector in that basis. Thus, if you have an expression for a vector that uses a particular set of basis vectors, whatever multiplies the basis vectors are the corresponding components.

Am I chain ruling between r θθ\theta and x y or r θθ\theta and r' θ′θ′\theta'?

Neither. You are chain-ruling $u$ and whatever coordinates you want to express your gradient in.

 

[mishima]

First off sorry if I'm being frustrating, Ill try to focus in on what is confusing me. It seems like there is some computational aspect I'm missing. I'd like to find out the response to 'if you want to find the covariant components of the gradient in polar coordinates, you...'. You are saying take the gradient of u in polar coordinates and apply the chain rule.

u is a function of r and $theta$, which in turn are functions of x and y.

$$\nabla u(r(x,y),\theta(x,y))$$

I only know one way to do that, which is using

$$=\sum \hat {\mathbf e_i} \frac 1 h_i \frac {\partial u} {\partial x_i} $$

where for polar coordinates the scale factors $h_i$ are 1 and r. So this gives me the regular old polar gradient:

$$\nabla u(r,\theta) = \hat {\mathbf e_r} \frac {\partial u} {\partial r} + \hat {\mathbf e_\theta} \frac 1 r \frac {\partial u} {\partial \theta}$$

Now I am not sure what to chain rule. I thought maybe x and y since r and $\theta$ are functions of x and y.

$$\nabla u(r,\theta) = \hat {\mathbf e_r} \frac {\partial u} {\partial r} (\frac {\partial r} {\partial x}+\frac {\partial r} {\partial y})+ \hat {\mathbf e_\theta} \frac 1 r \frac {\partial u} {\partial \theta}(\frac {\partial \theta} {\partial x}+\frac {\partial \theta} {\partial y})$$

Now, the expressions in parenthesis are almost $\nabla r(x,y)$ and $\nabla \theta(x,y)$ which you pointed out earlier are the contravariant basis. Everything not in parenthesis is therefore a candidate for the covariant components. I suspect the covariant basis e hats need to be dealt with in some way.

 

[Orodruin]

You do not need to care at all about the Cartesian coordinate. I suggest you forget about them for this purpose. All that matters is that you have a function $u(r,\theta)$ and you want to compute $\nabla u(r,\theta)$. You need to apply the chain rule for the gradient to the function $u(r,\theta)$, knowing that both $r$ and $\theta$ are themselves functions. You never need to refer back to the Cartesian coordinate system because you know that $\nabla r = \vec E^r$ and $\nabla \theta = \vec E^\theta$, which are exactly the basis vectors you are interested in having in your expression. You should not use the expression you have given for the gradient, it is not helpful in this case.

Edit: As an example: Note that, if you had a function $f(g(\vec x))$, its gradient would be given by
$$
\nabla (f\circ g) = \frac{df}{dg} \nabla g.
$$
This is the chain rule.
You need to do the corresponding thing for a function $u$ that is a function of two variables.

 

[mishima]

Thanks, I've not encountered that last expression for gradient before. And I don't find it in Boas, I've done every problem in the book up to this section. It seems to be the missing computational ability I'm having. Do you have a good reference where I can study more on that? Multivariable vector calculus?

 

[LCKurtz]

@mishima: You might be interested in this article and the book to which it refers:
https://www.physicsforums.com/insights/the-birth-of-a-textbook/
@Orodruin:

 

[mishima]

Thanks, I see he is an authority on the matter indeed. :) I was able to find the expression in Riley/Hobson/Bence without much commentary...sort of surprised it isn't in Boas.

I see now that the covariant components of $\nabla u$ are just $\frac {\partial u} {\partial r}$ and $\frac {\partial u} {\partial \theta}$ but I couldn't derive it or really explain why.