# 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? – Farcher Nov 3 '16 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. – Hamed Begloo Nov 3 '16 at 19:02
• Isn't this what you want? I could have written is as $\vec g(x)=-\nabla V(x)$ – Farcher Nov 3 '16 at 19:08
• @Farcher So that means potential is a quantity which its dimension/unit is variable. Since gravitational potential has different dimension than electric potential. Is it right? Now another question which quantity is more fundamental/primitive? "Field strength" or "Potential"? – Hamed Begloo Nov 3 '16 at 19:16
• In Physics until relatively recently it was thought to be force (field strength) but now it is thought to be potential. – Farcher Nov 3 '16 at 19:25

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.