# Rigorous definition of potential energy of a system

In the mechanics textbook by Kleppner and Kolenkow, the concept of potential energy is introduced by considering the behaviour of particle under the action of a force $$\vec{F}(\vec{r})$$ that depends on its position. To the best of my understanding, the potential energy is defined as follows:

Given a conservative force $$\vec{F}(\vec{r})$$, a function $$U:\mathbb{R^3}\to\mathbb{R}$$ defined by $$U(\vec{r})-U(\vec{r_0})=\int_{\vec{r}_0}^{\vec{r}}\vec{F}_c\cdot d\vec{r}$$ is called a potential energy function of $$\vec{F}$$

The textbook then goes on to give the expressions of various force laws such as $$\vec{F}(r)=\frac{A}{r^2}\hat{r}$$.

My confusion arises from when I try to extend this definition to a system consisting of multiple particles, and the interaction force(s) are a function of the postitions of both particles, say the gravitational interaction given by $$\vec{F}_{12}=-\frac{Gm_1m_2}{|\vec{r_1}-\vec{r_2}|}\hat{r}_{12}\qquad \vec{F}_{21}=-\frac{Gm_1m_2}{|\vec{r_1}-\vec{r_2}|}\hat{r}_{21}$$

In this case,

1. Is there a separate potential energy associated with each of the interaction pairs or are they considered as one when defining the potential energy function?
2. What does it mean for such a force (or a pair of forces) to be conservative? Should the work done be function of the initial and final positions of both particles or their relative initial and final positions?
3. How is the potential energy function defined and what is it a function of?

## 2 Answers

First of all the gravitational force is $$\mathbf{F}_{2}=-\frac{Gm_1m_2}{|\mathbf{r}_1-\mathbf{r}_2|^2}\hat{\mathbf{r}}_{12}$$ and vice versa, where $$\mathbf{F}_2$$ is the force on particle 2 and $$\hat{\mathbf{r}}_{12} = \frac{\mathbf{r}_2-\mathbf{r}_1}{|\mathbf{r}_2-\mathbf{r}_1|}$$, you forgot to square ;)

We define
$$$$V(|\mathbf{r}_1 - \mathbf{r}_2|) = -\frac{Gm_1m_2}{|\mathbf{r}_1-\mathbf{r}_2|}$$$$ and then we can write \begin{align} \mathbf{F}_{2}&=-\frac{Gm_1m_2}{|\mathbf{r}_1-\mathbf{r}_2|^2}\hat{\mathbf{r}}_{12}=-\nabla_2 \bigg(-\frac{Gm_1m_2}{|\mathbf{r}_1-\mathbf{r}_2|} \bigg) =-\nabla_2 V(|\mathbf{r}_1 - \mathbf{r}_2|) \\ \mathbf{F}_{1}&=-\frac{Gm_1m_2}{|\mathbf{r}_1-\mathbf{r}_2|^2}\hat{\mathbf{r}}_{21}=-\nabla_1 \bigg(-\frac{Gm_1m_2}{|\mathbf{r}_1-\mathbf{r}_2|} \bigg) = -\nabla_1 V(|\mathbf{r}_1 - \mathbf{r}_2|) \\ \end{align} where $$\nabla_i$$ acts on particle $$i$$. Therefore we can can interpret $$V(|\mathbf{r}_1 - \mathbf{r}_2|)$$ as the potential energy. This answers your first and third question. There is one potential which describes the interaction between the two particles. It yields both the force on particle 1 and on particle 2. Depending on which force you want to know, you calculate the derivative with respect to the corresponding coordinates. Now the fact that there exists a potential already answers your second question. By definition a force is conservative if it can be described as the negative gradient of a potential, which is the case here.

If you want to calculate the work, it really depends on which work you're talking about. Usually we talk about the work that a force does. So for example we can consider the work done on particle 1 by the force exerted by particle 2. Or we could talk about the work done on particle 2 by the force exerted by particle 1. In the latter case, we would calculate $$$$W_2 = \int_{\mathbf{ r}_{2,i}}^{\mathbf{r}_{2,f}} \mathbf{F}_2 \cdot \mathrm{d}\mathbf{r}_2.$$$$ Notice that $$W_2$$ depends on the trajectories of both particles, because the force depends on the positions of both particles.

Finally, let me note, that we don't have to make life too complicated. In the case of the two body problem we can introduce a relative coordinate $$\mathbf{r} = \mathbf{r}_2 - \mathbf{r}_1$$ and a center of mass coordinate $$\mathbf{R} = \frac{1}{m_1 + m_2} \big( m_1 \mathbf{r}_1 + m_2 \mathbf{r}_2 \big)$$. We can treat them independently and it turns out that because of momentum conservation, the center of mass moves with constant velocity. Thus, we need only consider the relative coordinate, which means that we have reduced our two-body problem to a one-body problem which makes things a lot easier.

• Thankyou. That clears up a few things. I had a few follow up questions. 1) Would it be correct to say that this only works if the force of interaction depends on the relative position vector specifically, rather than the absolute positions of each of the particles? 2) Is the work done in question i.e. $W=-\Delta U$, then, the work done by the interaction on both the particles as their separation changes from $\vec{R}_0$ to $\vec{R}$ ? Commented Jun 11 at 3:17
• 1) No, the force does not have to depend only on the relative position in order to be conservative (and therefore allow a potential to be defined). However, usually in physics when we consider the force between two particles, it only depends on their relative position vector, as in this example. 2) You're refering to the difference in potential energy after the distance changes, right? Correct, then this difference is equal to the work done on particle 1 by the force exerted by particle 2 plus the work done on particle 2 by the force exerted by particle 1. We can see that this holds by... Commented Jun 11 at 8:11
• ...energy conservation. The kinetic energy of particle 1 plus the kinetic energy of particle 2 plus the potential energy stays constant. Thus a change in potential energy corresponds to minus the sum of changes of the kinetic energies. The change of the kinetic energy of particle $i$ is equal to the work done on it. Commented Jun 11 at 8:19

In the mechanics textbook by Kleppner and Kolenkow, the concept of potential energy is introduced by considering the behaviour of particle under the action of a force $$\vec{F}(\vec{r})$$ that depends on its position. To the best of my understanding, the potential energy is defined as follows:

Given a conservative force $$\vec{F}(\vec{r})$$, a function $$U:\mathbb{R^3}\to\mathbb{R}$$ defined by $$U(\vec{r})-U(\vec{r_0})=\int_{\vec{r}_0}^{\vec{r}}\vec{F}_c\cdot d\vec{r}$$ is called a potential energy function of $$\vec{F}$$

... My confusion arises from when I try to extend this definition to a system consisting of multiple particles, and the interaction force(s) are a function of the postitions of both particles, say the gravitational interaction given by $$\vec{F}_{12}=-\frac{Gm_1m_2}{|\vec{r_1}-\vec{r_2}|}\hat{r}_{12}\qquad \vec{F}_{21}=-\frac{Gm_1m_2}{|\vec{r_1}-\vec{r_2}|}\hat{r}_{21}$$

You are right to be on alert here. The concept of a force acting between two particles, or forces acting between three particles, etc., is being generalized to a force field. The term "field," when used in physics, just means a function of space (or spacetime).

Let us see how the field concept arises...

The way we do it is usually by introducing the concept of a "test charge" (or test mass, or whatever). Call the test charge $$q_0$$ at location $$\vec r_0$$

We write the force on the test charge (not yet a "force field") as: $$\vec F_{on\;q_0\;at\;\vec r_0} = \sum_{i=1}^N k\frac{q_0 q_i(\vec r_0 - \vec r_i)}{|\vec r_0 - \vec r_i|^3}$$

But then we notice, hey, that test charge really could be anywhere, so could write a field of force that just depends on the general coordinate $$\vec r$$ like: $$\vec F(\vec r)= \sum_{i=1}^N k\frac{q_0 q_i(\vec r - \vec r_i)}{|\vec r - \vec r_i|^3}\;.$$ This field is a vector function of space that will tell us the force that would be exerted on a charge of $$q_0$$ were it to be placed at $$\vec r$$.

And it is this function of all space that we can obtain by taking the derivative of a different function of all space, the potential energy: $$\vec F(\vec r)= \sum_{i=1}^N k\frac{q_0 q_i(\vec r - \vec r_i)}{|\vec r - \vec r_i|^3} = -\vec \nabla U(\vec r)\;,$$ where $$U(\vec r) = \sum_{i=1}^N k\frac{q_0 q_i}{|\vec r - \vec r_i|}\;.$$

In electrostatics, in order to get rid of the traces of the "test charge" we also usually define the electric field and the electric potential as: $$\vec E(\vec r)= \sum_{i=1}^N k\frac{q_i(\vec r - \vec r_i)}{|\vec r - \vec r_i|^3} = -\vec \nabla \phi(\vec r)\;,$$ and $$\phi(\vec r) = \sum_{i=1}^N k\frac{q_i}{|\vec r - \vec r_i|}\;.$$

• I see. But is it always possible to express the force on each particle arising from an interaction as a function of a single vector $\vec{F}(\vec{R})$ to define a force field? It works in this example because the force depends only on the relative position vector, but is that always the case? Commented Jun 11 at 3:05
• It's not always the case, but it is the case in many common situations. Examples where it is not the case include: (1) non-conservative forces; and (2) velocity-dependent potentials like the potential for the lorentz force $U=q\phi -q\vec v\cdot \vec A$, there the force if given by $-\vec\nabla U + \frac{d}{dt} \frac{\partial U}{\partial \vec v}$
– hft
Commented Jun 11 at 3:31
• Thank you. That helps a lot Commented Jun 11 at 3:48