Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The gradient is usually written as the product of the unit vectors times the derivative with respect to that coordinate. In Einstein summation convention:

$\hat e_i \partial_i$

I've seen it written as so in some places.

Is this wrong and is one of them supposed to be a contravariant vector, because otherwise it won't transform as a tensor between coordinate system?

share|cite|improve this question
up vote 1 down vote accepted

I would say that "this definition depends on the orthonormality of $(\hat{e}_i)$", rather than it is "wrong", but that's it. More generally, if $\{e_1 \ldots e_n\}$ is a basis of $\mathbb{R}^n$ and $\{e^1\ldots e^n\}$ is the dual basis (that is, the only set of vectors with property $e^i \cdot e_j = \delta^i{ }_j$) then

$$\nabla f =\frac{\partial f}{\partial x^i}e^i.$$

P.S.: You can read more on Itskov's book Tensor Algebra and Tensor Analysis for Engineers:

share|cite|improve this answer
What I meant mostly is if the expression $\nabla f =\frac{\partial f}{\partial x^i}e^i$ is always correct in any curvilinear coordinates. So it is, right? – fiftyeight Jun 26 '12 at 16:59
@fiftyeight: Of course. Link:… – Giuseppe Negro Jun 26 '12 at 17:00

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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