How Can We Obtain Frame-Free Vector Derivative?

[kryptyk]

In vector analysis, it is possible to express the $$\nabla$$ operator in terms of a frame$$\{\mathbf{e}_1, \mathbf{e}_2, ..., \mathbf{e}_n\}$$ and its reciprocal frame $$\{\mathbf{e}^1, \mathbf{e}^2, ..., \mathbf{e}^n\}$$ by:

$$\nabla = \sum_{k=1}^n \mathbf{e}^k \frac{\partial}{\partial x^k}$$

where $$x^k$$ are the vector coordinates for frame $$\{\mathbf{e}_k\}$$:

$$\mathbf{x} = \sum_{k=1}^n x^k \mathbf{e}_k$$

Hestenes and Sobczyk[84] instead present a frame-free definition of $$\nabla$$ by first defining the directional derivative of some function of a vector via:

$$(\mathbf{a}\cdot\nabla_{\mathbf{x}})F(\mathbf{x}) = \lim_{\tau\rightarrow 0}\frac{F(\mathbf{x} + \tau \mathbf{a})-F(\mathbf{x})}{\tau}$$

They then define the differential of $$F$$ by:

$$\underline{F}(\mathbf{x}, \mathbf{a}) = (\mathbf{a}\cdot\nabla_{\mathbf{x}})F(\mathbf{x})$$

from which they then claim we can obtain the derivative of $$F$$ with respect to the vector $$\mathbf{x}$$ by:

$$\nabla_{\mathbf{x}} F(\mathbf{x}) = \nabla_{\mathbf{a}}(\mathbf{a}\cdot\nabla_{\mathbf{x}})F(\mathbf{x}) = \nabla_{\mathbf{a}} \underline{F}(\mathbf{x}, \mathbf{a})$$

I just don't understand how this last equation allows us to obtain the derivative in a frame-free manner. I still find it necessary to introduce a frame and use partial derivatives. Once the derivative has been evaluated, the frame-dependence can be removed. For example, let

$$F(\mathbf{x})=|\mathbf{x}|^r$$,

then

$$\nabla F = \sum_{k=1}^n \mathbf{e}^k \frac{\partial}{\partial x^k}F(x^1, x^2, ..., x^n)$$

For orthonormal frame $$\{\mathbf{e}_k\}$$ we can write:

$$F(\mathbf{x}) = |\mathbf{x}|^r = |(x^1)^2 + (x^2)^2 + ... + (x^n)^2|^{\frac{r}{2}}$$

where $$(x^k)^2$$ means the square of the kth component and the $$k$$ is not to be interpreted as an exponent. (Yes, this notation is a bit ugly but it seems to be quite engrained in the literature.)

Since $$\{\mathbf{e}_k\}$$ is an orthonormal frame, for Euclidean spaces we have $$\mathbf{e}^k = \mathbf{e}_k$$ and so

$$\mathbf{e}^k \frac{\partial}{\partial x^k}|(x^1)^2 + (x^2)^2 + ... + (x^n)^2|^{\frac{r}{2}} = r |\mathbf{x}|^{r-2} x^k \mathbf{e}^k = r |\mathbf{x}|^{r-2} x^k \mathbf{e}_k$$

thus,

$$\nabla_{\mathbf{x}} |\mathbf{x}|^r = \sum_{k=1}^n r |\mathbf{x}|^{r-2} x^k \mathbf{e}_k = r |\mathbf{x}|^{r-2} \mathbf{x}$$

and so we finally arrive at a frame-free expression for the derivative.

Can this evaluation be carried out in a completely frame-free manner? Would such an approach easily generalize to any frame-independent function of $$\mathbf{x}$$? I'm finding it easier to just evaluate them like I did here - but it seems perhaps by means of the chain rule it can be done more elegantly without mention of frames at all.

 

[fresh_42]

It always depends on what you want to use it for. Of course you need coordinates (a frame) whenever you actually calculate something. The notation with the nabla operator is per se frame-free - and sufficient if we are interested in the transformation it expresses rather than the numbers it spits out at a certain point for a certain function.