# Homework Help: Derivatives of basis vectors

1. Oct 5, 2006

### wandering.the.cosmos

In introductory mechanics courses we derive the equations of motion in curvilinear coordinates, especially the $m (d^2/dt^2)x$, by expressing the coordinate basis vectors in terms of their cartesian counterparts, and then differentating them with respect to time.

For example, in 2D polar coordinates we have

$\hat{r} = \cos[\theta] \hat{x} + \sin[\theta] \hat{y}$
$(d/dt) \hat{r} = \frac{d \theta}{dt} (- \sin[\theta] \hat{x} + \cos[\theta] \hat{y}) = \frac{d \theta}{dt} \hat{\theta} = \frac{d \theta}{dt} \hat{\theta}$

My question is, how do we express this in terms of the so-called "modern" notation for basis vectors - the partial derivatives $\{ \partial_\mu \}$ - on a manifold? Do the $\hat{x},\hat{y},\hat{r}$ correspond to $\partial_x, \partial_y, \partial_r$? It doesn't quite make sense to me, to say something like $(d/dt)(\partial_r)$.

Last edited: Oct 5, 2006
2. Oct 6, 2006

### George Jones

Staff Emeritus
Yes, and with the standard dot product, $\mathbb{R}^2$ is a Riemannian manifold.

$\partial_\theta$ means hold $r$ constant, and so is a vector in the $\hat{\theta}[/tex] direction, but it is not normalized. To normalize, use the chain rule to express [itex]\partial_\theta$ as a linear combination of the unit vectors $\partial_x$ and $\partial_y$.

Doing the same for $r[/tex] shows that you were correct for [itex]r$.

Also, the chain rule can be used to express $d/dt$ as a linear combination of $\partial_\theta$ and $\partial_r$.

Last edited: Oct 6, 2006
3. Oct 6, 2006

### HallsofIvy

This may be more information than you want but, strictly speaking, such things as $\hat{x},\hat{y}$ are "vectors" while $\partial_x, \partial_y$ are "covectors". That is, they are "operators" on the vectors and are in the dual space. Of course, there exist a natural correspondence between "vectors" and "covectors" so one can always think of covectors as "being" vectors. In that sense, any inner product, uv, is really "apply the covector corresponding to u to the vector v".

4. Oct 6, 2006

### George Jones

Staff Emeritus
No, $\partial_x, \partial_y$ are tangent vectors (fields), and $dx$ and $dy$ are co(tangent)vectors (fields).

Tangent vectors are derivation operators on functions, and covectors are operators on vectors.

5. Oct 6, 2006

### pervect

Staff Emeritus

I think that what you're probably asking for is the covariant derivative. This is $$\nabla_t$$ rather than $$\frac{\partial}{\partial t}$$.

But your post raises other issues.

In modern notation, basis vectors are usually written ei, where i = 0...3. I.e. basis vectors a boldfaced, and subscripted.

Hatted vectors usually imply a non-coordinate basis. Hatted vectors are usually derived from an orthonormal basis of one-forms, also sometimes called a coframe. They have a very "physical" interpretation.

On the www, the only thing I can think of that might be relevant is http://en.wikipedia.org/wiki/Frame_fields_in_general_relativity, for some guides to notation. It's almost standard notation, except that most textbooks write vectors as boldface, where this article actually put a vector symbol over them, i.e. it writes the basis vectors as $$\vec{e_i}$$ rather than ei, the later being what you'll see in most textbooks.

Most textbooks will talk about orthonormal bases of one-forms, so you might want to find and read that section of your textbook, assuming you have one.

6. Oct 6, 2006

### George Jones

Staff Emeritus
No, wandering.the.cosmos is asking about the relationship between the modern differential view of tangent vectors on manifolds as partial derivatives, and vectors in classical mechanics.

wandering.the.cosmos is working with a differentiable manifold that is a configuration space (in this case, $\mathbb{R}^2$) in classical mechanics, not a differentiable manifold that is a relativisitic spacetime.

Hats over vectors conventionally imply that the vectors are units vectors with respect to a (positive-definite) metric. These vectors could be "coordinate" vectors, or they could be "non-coordinate" vectors.

Hatted vectors are defined quite naturally without reference to one-forms.

7. Oct 10, 2006

### pervect

Staff Emeritus
Let me try again. Suppose we have a metric (?line element?)

ds^2 = dx^2 + dy^2

then dx, dy are an orthonormal basis (ONB) of one forms, because ds^2 = dx^2 + dy^2 and the duals of these one forms are unit vectors. In this case, that means that the coordinate vectors $$\frac{\partial}{\partial x}$$ are unit vectors.

This goes back to your point that dx is a cotangent vector (aka one form), while $$\frac{\partial}{\partial x}$$ is a tangent vector.

We always have $$\frac{\partial}{\partial q} dq = 1$$ for any coordinate q, this represents the combination of a one form dq on a coordinate vector $$\frac{\partial}{\partial q}$$ which results in a scalar, and we can see that this is always a unit scalar.

Now, if we have a metric ds^2 = dr^2 + r^2 d $$\theta$$^2, then dr and r*d$\theta$ are an ONB of one forms, because ds^2 = dr^2 + (r d $\theta$)^2. The line element will always be the sum of the squares of an ONB of one forms (well, except for possible minus signs if we were doing relativity).

The duals of these unit one-forms are again unit vectors, namely $$\frac{\partial}{\partial r}$$ is a unit vector in the r direction, $$\hat{r}$$, and $$\frac{1}{r}\frac{\partial}{\partial \theta}$$ is a unit vector in the theta direction $$\hat{\theta}$$.

While I suppose you may not actually have to use an ONB of one forms, it makes life a lot easier. You can read the ONB of one forms right off the metric.

Last edited: Oct 10, 2006