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

In summary: 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!
  • #1
AndersF
27
4
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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  • #2
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.
 
  • #3
PS the result should drop out in a couple of lines.
 
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  • #4
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?
 
  • #5
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!
 
  • #6
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.
 

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

What is a scalar field?

A scalar field is a mathematical function that assigns a scalar value (such as temperature or pressure) to every point in a space.

What is a gradient?

The gradient of a scalar field is a vector that represents the direction and magnitude of the steepest increase of the scalar field at a given point.

What is a covariant vector?

A covariant vector is a type of vector that transforms in a specific way under coordinate transformations. In the context of scalar fields, the gradient is considered a covariant vector because it transforms in a specific way under changes in the coordinate system.

How do you show that the gradient of a scalar field is a covariant vector?

To show that the gradient of a scalar field is a covariant vector, we can use the chain rule to express the gradient in terms of partial derivatives with respect to each coordinate. Then, by applying the transformation rules for partial derivatives, we can show that the gradient transforms in the same way as a covariant vector under coordinate transformations.

Why is it important to show that the gradient of a scalar field is a covariant vector?

It is important to show that the gradient of a scalar field is a covariant vector because it allows us to use vector calculus techniques to analyze and solve problems involving scalar fields. This enables us to better understand physical phenomena and make predictions about them.

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