Conservative force/Proofs

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This mathematics article is devoted entirely to providing mathematical proofs and support for claims and statements made in the article Conservative force. This article is currently an experimental vehicle to see how well we can provide proofs and details for a math article without cluttering up the main article itself. See Wikipedia:WikiProject Mathematics/Proofs for some current discussion. This article is "experimental" in the sense that it is a test of one way we may be able to incorporate more detailed proofs in Wikipedia.

[edit] Properties of conservative forces

The main page states that the following three properties are equivalent for conservative forces:

  1. The curl of F is zero:
    \nabla \times \vec{F} = 0. \,
  2. The work, W, is zero for any simple closed path:
    W = \oint_C \vec{F} \cdot \mathrm{d}\vec r = 0.\,
  3. The force can be written as the gradient of a potential, Φ:
    \vec{F} = -\nabla \Phi. \,

Proof: Let C be any simple closed path and consider a surface S of which C is the boundary. Then Stokes' theorem gives that

\int_S (\nabla \times \vec{F}) \cdot \mathrm{d}\vec{a} = \oint_C \vec{F} \cdot \mathrm{d}\vec{r}

If the curl of F is zero the left hand side is zero - therefore statement 2 is true.

Now assume that statement 2 holds. Let c be a simple curve from the origin to a point $x$ and define a function

\Phi(x) = \int_c \vec{F} \cdot \mathrm{d}\vec{r}.

From the fundamental theorem of calculus now follows that

\vec{F} = \nabla \Phi.

The minus sign is included for convenience in physical calculations. So statement 2 implies statement three.

Finally, assume that the third statement is true. For this we use the following rule:

Lemma For any twice continuously differentiable function f, curl(grad f) = 0.
Proof Writing out the definitions yields
\nabla \times \nabla f = \left( \frac{\partial^2 f}{\partial y \partial z} - \frac{\partial^2 f}{\partial z \partial y} \right) \vec{e}_x + \left( \frac{\partial^2 f}{\partial z \partial x} - \frac{\partial^2 f}{\partial x \partial z} \right) \vec{e}_y + \left( \frac{\partial^2 f}{\partial x \partial y} - \frac{\partial^2 f}{\partial y \partial x} \right) \vec{e}_z
where the ex etc. denote unit vectors. Now use equality of mixed partial derivatives to see that each component vanishes.

The proof now consists only of applying the above theorem to F - which is a gradient by assumption - to see that it's curl is zero.

This shows that statement 1 implies 2, 2 implies 3 and 3 implies 1, therefore the proof is complete.

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