How inexact differential shows work is a path function.

Makaila Simon 2022-09-26 Answered
How inexact differential shows work is a path function.
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Answers (1)

pedradauy
Answered 2022-09-27 Author has 8 answers
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 t
If γ is a loop encircling a surface Σ, by Stokes theorem, we have :
γ F = Σ d F
Now, 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 F
Therefore 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.
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