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Matrix analog of del operator?

  1. Dec 28, 2011 #1
    The del operator is often informally written as (d/dx, d/dy, d/dz) or [itex]\hat{x}[/itex][itex]\frac{d}{dx}[/itex]+[itex]\hat{y}[/itex][itex]\frac{d}{dy}[/itex]+[itex]\hat{z}[/itex][itex]\frac{d}{dz}[/itex], a pseudo-vector consisting of differentiation operators. Could there be a pseudo-matrix operator like it? What would one be differentiating with respect to- that is, the physical or geometric interpretation (i.e., the x, y, z above are the coordinates in three-space). Would the operator be of any use?
     
  2. jcsd
  3. Dec 30, 2011 #2
    I am not sure how much use it would be, but you could always just put it into a column vector:

    D = {d/dx, d/dy,d/dz}
    (Where {} denote column vector despite being written left to right)

    Then to apply to a function, you would write Df(x,y,z). The function is a scalar, so it just gets multiplied through by everyone. The result is {df/dx, df/dy, df/dz}

    Remember though, with operators, that fD is not generally Df.

    Also, to be totally accurate, if you treat f as a 1x1 "matrix", then in order to make it commute properly with D in the case of fD, you would do fD^T, which is a row vector (while Df is a column vector). The result is [fd/dx,fd/dy,fd/dz]

    Bear in mind, this can get tricky if you are using a non-Cartesian coordinate system, because if you apply this differentation to an object with basis vector components, you will need to differentiate the basis vectors too - in Cartesian coordinates, it doesn't matter, but in cylindrical, spherical, toroidal, etc., it will matter.
    I hope that answers your query!
     
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