# Showing that the gradient of a scalar field is a covariant vector

• AndersF
By the chain rule, the covariant derivative of the Jacobian is the covariant derivative of the field: ##\frac{\partial / \partial x'^i}{\partial / \partial x^j}=\frac{\partial}{\partial x'^i}\frac{\partial}{\partial x^j}##. So the covariant derivative of the field is a covariant vector!

#### AndersF

Homework Statement
Prove that the covariant gradient of a scalar field is a covariant vector
Relevant Equations
##\nabla f=\frac{\partial f}{\partial x^{i}} g^{i j} \mathbf{e}_{j}##
In a general coordinate system ##\{x^1,..., x^n\}##, the Covariant Gradient of a scalar field ##f:\mathbb{R}^n \rightarrow \mathbb{R}## is given by (using Einstein's notation)

##
\nabla f=\frac{\partial f}{\partial x^{i}} g^{i j} \mathbf{e}_{j}
##

I'm trying to prove that this covariant gradient ##\nabla f## is indeed a covariant vector. To do so, I'm trying to show that it transforms as a 1-covariant tensor under a change of basis.

Let ##C## be the transition matrix from a basis ##\{\mathbf e_i\}## to a basis ##\{\tilde {\mathbf e}_i\}##, that is, ##\tilde {\mathbf e}_i= \mathbf e_iC^i_j##.

The covariant derivative increases the contravariant tensor order of the tensor by one unit. Since the partial derivative of a scalar field is indeed a covariant derivative, the object ##\frac{\partial f}{\partial x^{i}}## will therefore be a 1-covariant tensor which I will call ##F_i##.

On the other hand, the contraction between the dual metric tensor ##g^{ij}## and ##F_i## will raise the subscript ##i## of ##F_i##, and the resulting object will be a 1-contravariant tensor: ##g^{ij}F_i\equiv H^j##.

But then, ##\frac{\partial f}{\partial x^{i}} g^{i j}=H^j## will transform as the contravariant components of a contravariant vector ##\mathbf{v}=H^j\mathbf{e}_{j}##: ##\tilde H^j=(C^{-1})^j_kH^k##, which is just the opposite of what I have to prove...

Where is my mistake? How could this be proved?

Last edited:
I don't recognise your notation. I'm more familiar with things like:
$$g^{ij} = g^{k'l'}\frac{\partial x^i}{\partial x^{k'}}\frac{\partial x^j}{\partial x^{l'}}$$If you do the same for the basis vector and partial derivatives, you can express the components of ##\nabla' f## in terms of the components of ##\nabla f## and derive/confirm the relevant transformation rule.

PS the result should drop out in a couple of lines.

• AndersF
PeroK said:
PS the result should drop out in a couple of lines.
I too think that the demonstration should be shorter than what I have tried, but I don't find the problem...

By the way, the components of the transition matrix ##(C^i_j)## of my notation are the terms ## \frac{\partial x^i}{\partial x'^j}## of your notation, aren't they?

AndersF said:
I too think that the demonstration should be shorter than what I have tried, but I don't find the problem...
If I understood what you were doing, I'd try to help!

AndersF said:
Homework Statement:: Prove that the covariant gradient of a scalar field is a covariant vector
Examine the chain rule: ##\partial / \partial x^j = (\partial x'^i/\partial x^j) \partial / \partial x'^i##. The Jacobian ##\partial x'^i/\partial x^j## is your transition matrix.