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.


3 Answers 3


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}

  • $\begingroup$ So that partial derivative only can make sense when aplying to a function of the n variables, not a vector itself? $\endgroup$ Commented Dec 4, 2013 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. $\endgroup$ Commented Dec 4, 2013 at 7:04
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    $\begingroup$ Doesn't it contradict Wikipedia which says (en.wikipedia.org/wiki/Directional_derivative) that $\partial f / \partial \mathbf{r}$ is the same as $\nabla_{\mathbf{r}}$ and $D_\mathbf{r} f$ which are directional derivatives, i.e. operator that gives a scalar, as opposed to a gradient which gives a vector? In other words, $\partial f / \partial \mathbf{r}$ is not a "vector of functions" but rather $\mathbf{r} \cdot \nabla f(\mathbf{r})$ - a scalar. $\endgroup$
    – grjj3
    Commented Jun 27, 2020 at 20:20
  • $\begingroup$ @grjj3 No. Notice that on that page, the notation for directional derivative uses $\mathbf x$ for the argument of the function being differentiated and $\mathbf v$ for the vector in whose direction the derivative is being taken. In particular, the notation on that page implies that $\partial f(\mathbf x)/\partial \mathbf v = \mathbf v\cdot \partial f(\mathbf x)/\partial \mathbf x$. All of this is consistent with what's written in the answer above. I will note, however, that the notations are overloaded -- there is enough abuse that their meanings can be ambiguous. $\endgroup$ Commented Jun 28, 2020 at 6:29
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    $\begingroup$ @grjj3 Yes definitely. Unless the symbol for the argument of the function is explicitly included, there is an ambiguity. I'm not a huge fan of these notations for this very reason, but sometimes they can be convenient when the context resolves the ambiguity. Notation that's unambiguous regardless of context can sometimes be cumbersome. $\endgroup$ Commented Jun 28, 2020 at 18:48

No, not really.

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


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):


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.



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



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:



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

  • $\begingroup$ 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
    Commented Sep 29, 2019 at 14:37


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

  • $\begingroup$ Yeah, also the notation convention made much more sense to me after reading about the directinal derivative. $\endgroup$ Commented Oct 11, 2016 at 13:20
  • $\begingroup$ Sure. I figured this might be a useful resource for anyone else confused about this type of stuff. $\endgroup$ Commented Oct 12, 2016 at 6:32

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