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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pedradauy

Answered 2022-09-27
Author has **8** answers

A 1-form $F={F}_{i}\text{d}{x}^{i}$ can be integrated along a path $\gamma :t\mapsto x(t)$ giving :

${\int}_{\gamma}F=\int {F}_{i}\frac{\text{d}{x}^{i}}{\text{d}t}\text{d}t$

If $\gamma $ is a loop encircling a surface $\mathrm{\Sigma}$, by Stokes theorem, we have :

${\int}_{\gamma}F={\int}_{\mathrm{\Sigma}}\text{d}F$

Now, if $\text{d}F=0$, the integral of F around any (null homotopic) closed path vanishes. If ${\gamma}_{1}$ and ${\gamma}_{2}$ are two path with the same end points, then we can build a loop $\gamma $ by first following ${\gamma}_{1}$ and then ${\gamma}_{2}$ in the reverse order and we get (if ${\gamma}_{1}$ and ${\gamma}_{2}$ are homotopic) :

$0={\int}_{\gamma}F={\int}_{{\gamma}_{1}}F-{\int}_{{\gamma}_{2}}F$

Therefore the integral of F is depending only on the path. On the other hand, if $\text{d}F\ne 0$, then this integral is path dependent.

Another way, maybe more familiar in physics is the following : using the Poincaré lemma, if $\text{d}F=0$ then, there is (at least locally) a 0-form V (ie a function) such that dV. In that case :

${\int}_{\gamma}F={\int}_{\gamma}\text{d}V=V(b)-V(a)$

In other words, if $\text{d}F\ne 0$ then F does not derive from a potential.

${\int}_{\gamma}F=\int {F}_{i}\frac{\text{d}{x}^{i}}{\text{d}t}\text{d}t$

If $\gamma $ is a loop encircling a surface $\mathrm{\Sigma}$, by Stokes theorem, we have :

${\int}_{\gamma}F={\int}_{\mathrm{\Sigma}}\text{d}F$

Now, if $\text{d}F=0$, the integral of F around any (null homotopic) closed path vanishes. If ${\gamma}_{1}$ and ${\gamma}_{2}$ are two path with the same end points, then we can build a loop $\gamma $ by first following ${\gamma}_{1}$ and then ${\gamma}_{2}$ in the reverse order and we get (if ${\gamma}_{1}$ and ${\gamma}_{2}$ are homotopic) :

$0={\int}_{\gamma}F={\int}_{{\gamma}_{1}}F-{\int}_{{\gamma}_{2}}F$

Therefore the integral of F is depending only on the path. On the other hand, if $\text{d}F\ne 0$, then this integral is path dependent.

Another way, maybe more familiar in physics is the following : using the Poincaré lemma, if $\text{d}F=0$ then, there is (at least locally) a 0-form V (ie a function) such that dV. In that case :

${\int}_{\gamma}F={\int}_{\gamma}\text{d}V=V(b)-V(a)$

In other words, if $\text{d}F\ne 0$ then F does not derive from a potential.

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