Suppose you have $n$ particles with masses $m_1, \cdots, m_n$ with position vectors $\vec{r_1}, \cdots, \vec{r_n}$. Now without any forces except gravity acting on them, conservation of mechanical energy for multiple particle states that the quantity $$\displaystyle \sum_{0 \leq i \neq j \leq n} \frac{-Gm_im_j}{||r_i - r_j||} + \sum_{0 \leq i \leq n} \frac{m_i}{2} ||\frac{dr_i}{dt}||^2 = \text{constant}$$

I understand the mathematical proof (see 13-3 here) of proving this (derivative w.r.t time is zero), but I don't understand how Feynman intuively reasons why this should be true:

We can understand why it should be the energy of every pair this way: Suppose that we want to find the total amount of work that must be done to bring the objects to certain distances from each other. We may do this in several steps, bringing them in from infinity where there is no force, one by one. First we bring in number one, which requires no work, since no other objects are yet present to exert force on it. Next we bring in number two, which does take some work, namely $W_{12}=−\frac{Gm_1m_2}{r_{12}}$. Now, and this is an important point, suppose we bring in the next object to position three. At any moment the force on number 3 can be written as the sum of two forces—the force exerted by number 1 and that exerted by number 2. Therefore the work done is the sum of the works done by each, because if $F_3$ can be resolved into the sum of two forces, $F_3=F_{13}+F_{23}$ , then the work is $\int F3⋅ds= \int F13⋅ds+ \int F23⋅ds=W_{13}+W_{23}$. That is, the work done is the sum of the work done against the first force and the second force, as if each acted independently. Proceeding in this way, we see that the total work required to assemble the given configuration of objects is precisely the value given in Eq. (13.14) as the potential energy. It is because gravity obeys the principle of superposition of forces that we can write the potential energy as a sum over each pair of particles.

The parts which I don't understand are:

  1. Why the total potential energy doesn't depends on the order in which the objects are placed ?
  2. When you're bringing the third object from "infinity", this arguement assumes that the first and second object stays still. But won't the third object, while coming from "infinity", pull the first two objects already placed, so ain't the calculations wrong ?
  3. Why should this type of construction of bringing particles from infinity give the correct formula for potential energy ?
  4. Now if I agree on the fact that the potential energy for such configuration on particle $i$ is indeed $-m_i \displaystyle \sum_{0 \leq j \leq n, j \neq i} \frac{-Gm_j}{||r_i - r_j||} $, are there any intuive explanation for which I should expect that the total sum of the mechanical energies of all the particles to be constant even when they interact with each other? I'm sorry if this is obvious to everybody else, but I do find it bit counterintuive to imagine that even for more than two particles, under mutual conservative force the total mechanical energy should be constant.

Exact same confusion happens when gravitational field is replaced by electric force, so you can use that to answer too.


1 Answer 1

  1. Consider a mass $M_1$ fixed in space. Now when we bring another mass $M_2$ towards it at a constant velocity, the potential energy can be seen as the work done to move the body $M_2$ towards $M_1$. But consider an observer moving with $M_2$: what this observer sees is that $M_1$ is moving towards $M_2$, and by equivalence the total potential energy of both is equal; hence the order of mass placement is irrelevant.
  2. Yes, the third mass $M_3$ attracts both previous masses; however, the calculations are not wrong. To illustrate this, consider the first two masses held at a fixed distance apart, and so when we move the third mass in, all the bodies will be accelerating towards each other. However, the total potential energy at each point in time is a function of only the position, hence the equations are correct; the kinetic energy of the bodies derives from the decrease in the potential energy of the system.
  3. When our system of masses interacts with each other, the system is closed; there is no work being done by any external force.Thus, by energy conservation, if no energy is supplied to or dissipated from the system, the total energy must remain constant. (I'm afraid I don't fully understand this point, would you mind explaining your confusion about the third point in a bit more detail?)
  • $\begingroup$ Re #1 Well I find it obvious for $n = 2$ too. But how do you explain the irrelevance of order in which they're put for $n \geq 3$ ? Also why should this type of construction give the correct potential energy ? For #2, I don't understand what you mean by "However, the total potential (...) energy of the system." potential energy respect to what ? I have close to zero understanding of fields, but the gravitational field is changing then when the mases are moving, right ? So then how it's a function of position ? $\endgroup$
    – katana_0
    May 17, 2018 at 14:35
  • $\begingroup$ Also I have update the ordering of the questions in the original question. $\endgroup$
    – katana_0
    May 17, 2018 at 14:39

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