# Frame-Free Vector Derivative

1. Mar 20, 2006

### 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.

Last edited: Mar 20, 2006