19
$\begingroup$

I've encountered in some books (and even completed an exercise from the Goldstein by using it), a strange notation that seems to work exactly like a gradient, I have tried to look for an explanation but found none yet and I think I can reduce the time spent looking for my answer by posting this question, even if you give me some literature to check I will be greatly thankful:

Why $$\frac{\partial}{\partial\mathbf{r}}=\nabla$$

As I understand it the partial derivative with respect to a vector is like aplying the gradient. I don't know why it seem so odd to me the notion of differentiating something with respect to a vector.

$\endgroup$
24
$\begingroup$

It's purely notation.

Given a real-valued function $f(\mathbf r) = f(x^1, \dots, x^n)$ of $n$ real variables, one defines the derivative with respect to $\mathbf r$ as follows: \begin{align} \frac{\partial f}{\partial \mathbf r}(\mathbf r) = \left(\frac{\partial f}{\partial x^1}(\mathbf r), \dots, \frac{\partial f}{\partial x^n}(\mathbf r)\right) \end{align} so, by definition, $\partial f/\partial \mathbf r$ is a vector of functions that precisely equals $\nabla f$. You may also run into the notation $\nabla_{\mathbf r}f$ which means precisely the same thing.

An advantage of the notations $\partial f/\partial\mathbf r$ and $\nabla_\mathbf r$ is that they make explicit the symbol used to label the argument of the function, and this can sometimes stave off confusion.

Addendum. You may also come across the following notation. Let $\mathbf f(\mathbf r) = \mathbf f(x^1, \dots, x^n)$ be a real, $m$-component vector-valued function of $n$ real variables, then define its derivative with respect to $\mathbf r$ as the following matrix, often referred to as the Jacobian matrix: \begin{align} \frac{\partial\mathbf f}{\partial \mathbf r}(\mathbf r) = \begin{pmatrix} \frac{\partial f^1}{\partial x^1}(\mathbf r) & \cdots & \frac{\partial f^1}{\partial x^n}(\mathbf r) \\ \vdots & \ddots & \vdots \\ \frac{\partial f^m}{\partial x^1}(\mathbf r) & \cdots & \frac{\partial f^m}{\partial x^n}(\mathbf r) \\ \end{pmatrix} \end{align} Consider, for example, the function $\mathbf g(\mathbf r) = \mathbf r$. In this case, you can convince yourself that its derivative $\partial/\partial\mathbf r$ is the identity matrix; \begin{align} \frac{\partial\mathbf g}{\partial \mathbf r}(\mathbf r) = \begin{pmatrix} 1 & \cdots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & 1 \\ \end{pmatrix} \end{align}

$\endgroup$
  • $\begingroup$ So that partial derivative only can make sense when aplying to a function of the n variables, not a vector itself? $\endgroup$ – Mark A. Ruiz Dec 4 '13 at 5:22
  • $\begingroup$ @MarkA.Ruiz Recall that derivatives generally speaking are operations that can be applied to functions. Having said this, perhaps the addendum I included will clear things up a bit more. $\endgroup$ – joshphysics Dec 4 '13 at 7:04
6
$\begingroup$

No, not really.

A gradient is the derivative of a scalar. It is not actually a vector, but a dual vector or 1-form.

http://en.wikipedia.org/wiki/Gradient

Vectors and 1-forms have different transformation properties, and used to be called contra-variant and co-variant vectors, but the language of exterior calculus makes this much cleaner. Intuitively, a 1-form operating on a vector gives you a scalar, and vice versa, and is essentially the definition of the inner product.

In flat space or Cartesian coordinates these are the same thing, but once you go to curved space or curvilinear coordinates they are quite different. So, the relation:

$$\frac{\partial}{\partial \mathbf{r} }=\nabla$$

Really only holds in flat space/Cartesian coordinates.

The partial derivative of a vector is not the gradient! This is because the partial derivative operator does not in fact operate in a coordinate independent way, but scalars, vectors, and tensors are coordinate independent.

Instead, without any further knowledge we can define something called the Lie derivative, which operates on vector and tensor fields and tells you the change of the tensor field along the flow of another tensor field (remember that tensors can be written without reference to coordinates):

http://en.wikipedia.org/wiki/Lie_derivative

Note we have suddenly started talking about vector fields, which are vectors defined at every point on your system.

Parenthetically, Lie derivatives are useful because if you take the Lie derivative of some tensor along a vector and find that it is zero, that vector is called a Killing vector and is a symmetry of the system.

If you have a manifold with a metric tensor defined, you can use it to define the covariant derivative, which is essentially the coordinate-independent partial derivative.

http://en.wikipedia.org/wiki/Manifold

http://en.wikipedia.org/wiki/Covariant_derivative

(And the covariant derivative is useful in defining parallel transport and geodesics.)

http://en.wikipedia.org/wiki/Parallel_transport

http://en.wikipedia.org/wiki/Geodesic

So your notion of oddness is spot on, because you've secretly entered the world of curved space! A very good general overview of these concepts can be found here:

http://preposterousuniverse.com/grnotes/grnotes-two.pdf

http://preposterousuniverse.com/grnotes/grnotes-three.pdf

And now you've gone down the rabbit hole! :-)

$\endgroup$
  • $\begingroup$ I have learned that the Del operator conveyed the vector gradient whilst the "d" operator was the one that produced dual vector fields. $\endgroup$ – Ken Wang Sep 29 '19 at 14:37
1
$\begingroup$

https://en.wikipedia.org/wiki/Matrix_calculus

This wiki article describes the convention in the question as well as a variety of other related conventions.

$\endgroup$
  • $\begingroup$ Yeah, also the notation convention made much more sense to me after reading about the directinal derivative. $\endgroup$ – Mark A. Ruiz Oct 11 '16 at 13:20
  • $\begingroup$ Sure. I figured this might be a useful resource for anyone else confused about this type of stuff. $\endgroup$ – aquirdturtle Oct 12 '16 at 6:32

Not the answer you're looking for? Browse other questions tagged or ask your own question.