# What is the vector field associated with potential energy?

The mere concept of a line integral is defined for a vector field, and I thus thought the following was a rigorous and general definition of potential energy:

Definition: Given a conservative force field $$f:U\to\mathbb{R}^n$$, where $$U$$ is an open subset of $$\mathbb{R}^n$$, we define the potential energy associated to $$f$$ as the function $$V:U\to\mathbb{R}$$ for which $$\int_Cf = V(x_A)-V(x_B)$$ for any piecewise curve $$C\subseteq U$$ that starts at $$x_A$$ and ends at $$x_B$$. Alternatively, our map $$V$$ is that for which $$f = -\nabla V.$$ Observe that $$V$$ is defined up to adding a constant.

In the above definition, potential energy is a property of a force field. However, I don't think this always needs to be the case. In the case of two point-masses $$m$$ and $$M$$, for example, one may talk about the work done in moving the mass $$m$$ through a certain curve even when there does not seem to be a single force field involved (if anything, at any time $$t$$ there is a force field $$f_t$$ created by the point-mass $$M$$) and one may even show -although I'm confused by the proof- that

$$W=\frac{GMm}{r(t_2)}-\frac{GMm}{r(t_1)}$$

is the work done on the point-mass $$m$$ due to gravity as $$m$$ moves from $$\vec{r}(t_1)$$ to $$\vec{r}(t_2)$$, and where $$r(t_i)$$ is the distance between $$m$$ and $$M$$ at time $$t_i$$.

1. What is a general, rigorous definition of potential energy? If we define it by means of the formula

$$W=\int_C\vec{F}\cdot d\vec{r}=U(x_A)-U(x_B)$$

where $$C$$ is a curve that starts at $$x_A$$ and ends at $$x_B$$, then what is the vector field involved in the integral above? How does such vector field relate with the function $$\vec{F}$$?

1. In the case of two (or more) point-masses, what is the vector field involved?

In the general case of $$N$$ point-like masses in $$D$$ dimensions, we have a $$D$$-dimensional position vector $${\vec r}_i$$ for each particle and a corresponding $$D$$-dimensional force vector $${\vec F}_i$$. The work done by the force on the$$i$$-the particle when moving the $$i$$-th particle from position $${\vec r}^A_i$$ to position $${\vec r}^B_i$$ along a path $$C$$ from $$A$$ to $$B$$, keeping fixed all the other particles is $$W_{i,AB}=\int_{C}\vec{F}_i\cdot d\vec{r}_i.$$ If we move all the particles along a path $$C'$$ in the $$DN$$-dimensional space starting from a point $$\{ \vec r_i^A \}$$ and ending at a point $$\{ \vec r_i^B \}$$, we can consider the total work.
$$W_{AB}=\int_{C'}\sum_i\vec{F}_i\cdot d\vec{r}_i.$$ If such a quantity is independent of the path for all the pairs of the initial and final points, the differential form in $$DN$$-dimensional allows introducing a potential energy scalar function as $$W=U(\{ \vec r_i^A \})-U(\{ \vec r_i^B \})$$.
Summarizing, the vector field involved in the work for a system of $$N$$ particles is the $$ND$$ dimensional vector field $$\{ \vec F_i \}$$ ($$i = 1,\dots,N$$).
Thinking the potential energy as a property of the field of force is correct. However, one has not to forget that the connection of the potential energy with observable quantities is through $${\vec F}_i = - \nabla_i U$$, for all the values of $$i$$.
• What is the meaning of $\nabla_i$ at the end of your answer?
• The gradient with respect to the coordinates of the i-th particle. In cartesian coordinates $\nabla_i U = \left(\frac{\partial{U}}{x_i}, \frac{\partial{U}}{y_i}, \frac{\partial{U}}{z_i} \right)$ (in the parenthesis, the three components of the vector). Commented Aug 28, 2022 at 12:06