A system of particles interacting via conservative forces, will also respect $U_2+K_2=U_1+K_1$

Question

Studying vector calculus you learn to prove that a particle moving in a gravitational field will, in that field respect that $dU=-dW$. From this you can conclude $U_2+K_2=U_1+K_1$.

This is easy to prove in here but I fail to see how to prove it for, suppose, $n$ charged particles or massive particles.

How can I prove such a thing? Namely, prove that $U_2+K_2=U_1+K_1$ is true if the nature of all the forces in your system are conservative (irrotational). Obviously here $U_j=\sum_i U_i$ and $K_j=\sum_i K_i$, i.e., at snapshot (2) $U_2$ is the sum of all the potential energies in the system, and the same for kinetic energy.

Answer 1

A force is conservative iff there exists a potential $\Phi$ such that ${\bf F} = -\nabla\Phi$. The Lagrangian for a system of $n$ particles acting under (any number of) conservative forces can be written on the general form

$$L = \sum_{i=1}^n \frac{1}{2}m_i{\bf \dot{r_i}}^2 - \Phi({\bf r_1},{\bf r_2},\ldots,{\bf r_n})$$

which leads to the equation of motion

$$m_i{\bf \ddot{r_i}} = -\nabla_{{\bf r_i}}\Phi$$

Multiplying with ${\bf \dot{r_i}}$ and summing over all particles gives us the desired result

$$\frac{d}{dt}\sum_{i=1}^n\frac{1}{2}m_i{\bf \dot{r_i}}^2 = -\sum_{i=1}^n{\bf \dot{r_i}}\cdot\nabla_{\bf r_i}\Phi \equiv -\frac{d\Phi}{dt} \implies K+U={\rm const.}$$

where $K=\sum_{i=1}^n\frac{1}{2}m_i{\bf \dot{r_i}}^2$ and $U = \Phi$ are the total kinetic and potential energy respectively. This argument covers the case where we have more than one force in play for which we can write $\Phi=\Phi_{\rm force~1}+\Phi_{\rm force~2}+\ldots$.

Answer 2

Hint: You wish to prove $\sum_i { \tfrac{1}{2}mv_i^2}- \sum \dfrac{GM_im_j}{r_{ij}}=\text{constant}$, where $r_{ij}=\sqrt{(x_i-x_j)^2+(y_i-y_j)^2+(z_i-z_j)^2}$. This is the case of gravitation(or electrostatic force).

So if you prove $\dfrac{dK}{dt}=-\dfrac{dU}{dt}$, then this implies $K+U=\text{constant}$.