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Correct form for d²r?

  1. Feb 4, 2014 #1
    If dr = (dx, dy, dz) so, which is the correct form for d²r? Is (d²x, d²y, d²z) or (dy^dz, dz^dx, dx^dy) ?
     
  2. jcsd
  3. Feb 4, 2014 #2

    tiny-tim

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    what isr ? :confused:
     
  4. Feb 4, 2014 #3
    you must be joking
     
  5. Feb 4, 2014 #4

    Mark44

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    Seems like a reasonable question to me.
     
  6. Feb 5, 2014 #5
    I imagine he's referring to the "numerator" of a second derivative such as ##\frac{d^2 \vec{r}}{dt^2}## in analogy with his identification of the differential form ##d\vec{r}## with the "numerator" of a first derivative.
     
  7. Feb 5, 2014 #6
    Is this even. No doubts.
     
  8. Feb 5, 2014 #7

    PeroK

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    I've seen d²r used as an alternative to dS in surface integrals.
     
  9. Feb 6, 2014 #8
    The correct form of ##\mathrm{d}^2r=\mathrm{d}(\mathrm{d}r)## is 0. Your question is incredibly ambiguous, and I wouldn't expect anyone to understand what you're asking without clarification.
     
  10. Feb 6, 2014 #9
    This is theoretical identity that in the practice is useless.

    I just asking which are the components of the vector d²r.

    The components of the vector r is (x, y, z);

    of dr is (dx, dy, dz)

    But, I have doubt if the components of vector d²r is (d²x, d²y, d²z) or (dydz, dzdx, dxdy) or other...
     
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