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kryptyk
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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.
\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.
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