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Why is a conserved vector field a gradient of a certain func

  1. Dec 9, 2015 #1
    I know that if a vector field is conserved then there exits a function such that the gradient of this function is equal to the vector field but am just curious to know the reason of it.
  2. jcsd
  3. Dec 9, 2015 #2


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    Hi ZA,

    You of course googled conservative vector field but there must be something that isn't clear to you. What specifically ?
  4. Dec 9, 2015 #3
    Hi! I've figured this out, by the stocks theorem and that the work done by a conservative force on a closed curve is zero but why it isn't true for the non conservative force?
  5. Dec 9, 2015 #4


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    That comes from the "Fundamental Theorem of Calculus", that if F is a differentiable function such that F'= f then [itex]\int_a^b= f(b)- f(a)[/itex]. A "conservative force" is the derivative of some "energy function". The integral depends only on the end points, not the path between the end points. The integral around a closed path, since the "end points" are the same point, is 0.
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