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Vector Product Rule Deviation

  1. Aug 22, 2009 #1
  2. jcsd
  3. Aug 23, 2009 #2
    I found a link that might be helpful:

  4. Aug 24, 2009 #3
    I don't think there is anything wrong with proving these identities in cartesian coordinates. If you have an identity such as [tex]\nabla (fg)=f\nabla g+g\nabla f[/tex], this just tells you that one vector is equal to another vector. If this is true in one coordinate system, it is true in all coordinate systems.
  5. Sep 10, 2009 #4
    If is a lot less efficient and less intuitive to prove vector relationships by resorting to expressing the vectors as individual components. People are much more likely to be able to see simplifications in vector algebra if they have some higher level algebra tools available to them.
  6. Sep 11, 2009 #5


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    Homework Helper

    The identities of vector calculus are easily derived and proven by algebraic mean given a few lemmas. It would be bad to prove them in cartesian coordinates because it would be messy, lengthly, inelegant, and would exagerate the importance of coordinates.
    Anyone who disagrees should post a cartesian coordinates proof of
    [tex]\mathbf{(a\times\nabla)\times b+a\nabla\cdot b=a\times(\nabla\times b)+(a\cdot\nabla)b}[/tex]
    from another thread
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