How inexact differential shows work is a path function.

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pedradauy · Accepted Answer

A 1-form   F  =      F    i    d      x    i   can be integrated along a path   γ  :  t  ↦  x  (  t  ) giving :      ∫          γ        F  =  ∫      F    i              d              x        i                    d      t        d  tIf   γ is a loop encircling a surface   Σ, by Stokes theorem, we have :      ∫          γ        F  =      ∫          Σ        d  FNow, if   d  F  =  0, the integral of F around any (null homotopic) closed path vanishes. If       γ    1   and       γ    2   are two path with the same end points, then we can build a loop   γ by first following       γ    1   and then       γ    2   in the reverse order and we get (if       γ    1   and       γ    2   are homotopic) :  0  =      ∫          γ        F  =      ∫                  γ        1              F  −      ∫                  γ        2              FTherefore the integral of F is depending only on the path. On the other hand, if   d  F  ≠  0, then this integral is path dependent.Another way, maybe more familiar in physics is the following : using the Poincaré lemma, if   d  F  =  0 then, there is (at least locally) a 0-form V (ie a function) such that dV. In that case :      ∫          γ        F  =      ∫          γ        d  V  =  V  (  b  )  −  V  (  a  )In other words, if   d  F  ≠  0 then F does not derive from a potential.

How inexact differential shows work is a path function.

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