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


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?

1 Answer 1


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 $i$.

  • $\begingroup$ What is the meaning of $\nabla_i$ at the end of your answer? $\endgroup$
    – Sam
    Commented Aug 28, 2022 at 10:51
  • $\begingroup$ 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). $\endgroup$ Commented Aug 28, 2022 at 12:06

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