# Why conservative field implies a potential function such that: $\vec{F}=\nabla \phi$?

I understand intuitively why a conservative field should be path independent. But, I can't figure out why the conservative field must has a potential function such that:

$$\vec{F}=\nabla \phi$$

I want a proof or intuitive explanation please.

Thanks.

• The (general) reason is Stokes theorem. – ಠ_ಠ Aug 12 '16 at 0:37
• Stokes theorem is in the last chapter in my textbook, shall I continue with the book sequence or can I understand it independently ? – Mohamed Mostafa Aug 12 '16 at 0:40

Because line integrals of $F$ around closed loops are zero, one can construct $\phi$ by deciding $\phi(x_0)=0$ for some $x_0$ and then defining $\phi(x)=\int_{C(x)} F \cdot dr$ where $C(x)$ is any path from $x_0$ to $x$. (Here I am assuming the domain of $F$ is path-connected. Notably I am not assuming that it is simply connected.)

This is well-defined because of the fact that integrals around closed loops vanish. Now the direction of maximal increase of $\phi$ is a direction $v$ which is aligned with $F$, so that $F \cdot v$ is as large as possible. In other words, $\nabla \phi$ is proportional to $F$. One can refine this a bit further to find that the proportionality constant is $1$.

• You first assumed that there is a potential function of the conservative field then showed how to find it. My question is why to assume that there is a potential function when the field is conservative ? – Mohamed Mostafa Aug 12 '16 at 0:25
• @MohamedMostafa You can construct $\phi$ using the path integral formula and then verify that it is a potential function for $F$. – arctic tern Aug 12 '16 at 0:47

This is the converse of the gradient theorem, a proof can be found here.

Check out the Helmholtz decomposition.

It says that any vector field $\vec{F}$ can be written uniquely as $\vec{F} = \nabla \varphi + (\nabla \times \vec{A})$.

The gradient part is, of course, conservative, so it is the conservative part of $\vec{F}$.

The curl part vanishes for a conservative field because: you can integrate along any loop $$\int_\gamma \vec{F} \cdot \vec{ds} = \int_\gamma (\nabla \phi + (\nabla \times \vec{A})) \cdot \vec{ds} = \int_\gamma (\nabla \times \vec{A}) \cdot \vec{ds}$$ and that is zero (since $\vec{F}$ is conservative); on the other hand, by the Stokes theorem that last integral is also the flux of $(\nabla \times \vec{A})$ through any surface delimited by the loop $\gamma$; so $(\nabla \times \vec{A})$ is zero when integrated over any surface, however tiny, so it must be zero.