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Frame-Free Vector Derivative

  1. Mar 20, 2006 #1
    In vector analysis, it is possible to express the [tex]\nabla[/tex] operator in terms of a frame[tex]\{\mathbf{e}_1, \mathbf{e}_2, ..., \mathbf{e}_n\}[/tex] and its reciprocal frame [tex]\{\mathbf{e}^1, \mathbf{e}^2, ..., \mathbf{e}^n\}[/tex] by:

    [tex]\nabla = \sum_{k=1}^n \mathbf{e}^k \frac{\partial}{\partial x^k}[/tex]

    where [tex]x^k[/tex] are the vector coordinates for frame [tex]\{\mathbf{e}_k\}[/tex]:

    [tex]\mathbf{x} = \sum_{k=1}^n x^k \mathbf{e}_k[/tex]

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

    [tex](\mathbf{a}\cdot\nabla_{\mathbf{x}})F(\mathbf{x}) = \lim_{\tau\rightarrow 0}\frac{F(\mathbf{x} + \tau \mathbf{a})-F(\mathbf{x})}{\tau}[/tex]

    They then define the differential of [tex]F[/tex] by:

    [tex]\underline{F}(\mathbf{x}, \mathbf{a}) = (\mathbf{a}\cdot\nabla_{\mathbf{x}})F(\mathbf{x})[/tex]

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

    [tex]\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})[/tex]

    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



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

    For orthonormal frame [tex]\{\mathbf{e}_k\}[/tex] we can write:

    [tex]F(\mathbf{x}) = |\mathbf{x}|^r = |(x^1)^2 + (x^2)^2 + ... + (x^n)^2|^{\frac{r}{2}}[/tex]

    where [tex](x^k)^2[/tex] means the square of the kth component and the [tex]k[/tex] 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 [tex]\{\mathbf{e}_k\}[/tex] is an orthonormal frame, for Euclidean spaces we have [tex]\mathbf{e}^k = \mathbf{e}_k[/tex] and so

    [tex]\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[/tex]


    [tex]\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}[/tex]

    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 [tex]\mathbf{x}[/tex]? 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
  2. jcsd
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