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

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The discussion centers on proving that the covariant gradient of a scalar field is a covariant vector in a general coordinate system. The covariant gradient is expressed as the product of the partial derivative of the scalar field and the dual metric tensor, leading to confusion about its transformation properties under a change of basis. Participants suggest using the chain rule and the Jacobian matrix to clarify the transformation of components. There is a consensus that the demonstration can be simplified, but the exact nature of the mistake in the original proof remains unclear. The conversation emphasizes the importance of understanding tensor transformations in the context of covariant derivatives.
AndersF
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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?
 
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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.
 
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.
 
At first, I derived that: $$\nabla \frac 1{\mu}=-\frac 1{{\mu}^3}\left((1-\beta^2)+\frac{\dot{\vec\beta}\cdot\vec R}c\right)\vec R$$ (dot means differentiation with respect to ##t'##). I assume this result is true because it gives valid result for magnetic field. To find electric field one should also derive partial derivative of ##\vec A## with respect to ##t##. I've used chain rule, substituted ##\vec A## and used derivative of product formula. $$\frac {\partial \vec A}{\partial t}=\frac...