# Field independent definition of "Potential function" (Not Potential Energy)

I know what "Potential Energy" is: A function like $U(x)$ whose negative gradient is equal to the force $F(x)$ generating it: $$F(x)=-\nabla U(x).\tag{1}$$

But the definition of the "Potential Function" itself seems to depend totally on the type of the field.

For example:

1. Electric Potential(Electric Potential Energy per unit charge): $$V_{e}=\frac{U_{e}}{q}.\tag{2}$$

2. Gravitational Potential(Gravitational Potential Energy per unit mass): $$V_{g}=\frac{U_{g}}{m}.\tag{3}$$

As you see every Potential is defined by the corresponding Potential Energy.

My question is:

1. Is there a way to define Potential of a field independent of the "Nature of the Field"? Whether being gravitational, electric, etc? (Again I'm not talking about "Potential Energy" just "Potential".)

2. Is it possible to derive specific "Potentials" like $V_{e}$ and $V_{g}$ as special cases of the first quantity?

• What is wrong with $\vec E(x)=-\nabla V(x)$ where $\vec E$ is the field strength? Commented Nov 3, 2016 at 18:57
• @Farcher I didn't knew this definition exists. But I want a definition which potential of certain fields(e.g. electric) could be extracted from it as special cases of the primary definition. Commented Nov 3, 2016 at 19:02
• Isn't this what you want? I could have written is as $\vec g(x)=-\nabla V(x)$ Commented Nov 3, 2016 at 19:08
• In Physics until relatively recently it was thought to be force (field strength) but now it is thought to be potential. Commented Nov 3, 2016 at 19:25
• @DvijMankad In part my statement was based on this post What is the more fundamental quantity? The electromagnetic field F or the potential A ? and the links therein.but I am on shifting sands here. Commented Nov 19, 2019 at 23:11

Yes. We can define the (scalar) potential of a vector field as the scalar function whose gradient equals that vector field (or its negative, if we so choose). The "nature" of the vector field does not matter, but note that not all vector fields possess a potential. In general we must have a continuosuly differentiable vector field $\vec{V}$ such that $\vec{\nabla}\times \vec{V}=0$. See this Wikipedia article.