Potential associated with a conservative force field F

[AntonioJ]

Homework Statement

Given the potential energy, the force is obtained as F = -∇U(r). A conservative force field F is associated with a potential f by F = ∇f.

Relevant Equations

Does the first expression arise from this last one? If so, with -∇U(r), would one obtain the electric field E instead of the force F?

Given the potential energy, the force is obtained as F = -∇U(r). A conservative force field F is associated with a potential f by F = ∇f. Does the first expression arise from this last one? If so, with -∇U(r), would one obtain the electric field E instead of the force F?

 

[anuttarasammyak]

F=-\nabla U
is enough. I feel no necessity to introduce f of f=-U + const.

 

[bamboum]

Force, field and potential are 3 different things. But can be correlated each other. Field like E is a space deformation (can be due to an extra charge for example) then some field like the electrostatic can be associated to potential V, E= -nabla V is correct. Then when comes another charge q in the field Coulomb law acts and F=qE. So you have U(r)= qV(r).
V is generally determined with a constant. For electrical field V=0 when r is infinite.
Mathematically this constant disappears in calculation (derivation or integration)

 

[vanhees71]

A force is a vector field, $\vec{F}(\vec{x})$. If it's conservative, there exists a scalar potential, $U$, then by definition
$$\vec{F}(\vec{x})=-\vec{\nabla} U(\vec{x}).$$
If $\vec{\nabla} \times \vec{F}=0$ in an open singly-connected neighborhood of a point, then there exists a potential (at least) in this neighborhood (Poincare's Lemma).

The potential is determined only up to an arbitrary additive constant. Indeed it's convenient to define it to go to 0 at infinity (if possible for the given force).