Total Potential Energy

We start by reviewing what we actually know about energy at all; It's a well established fact that the total energy of a system i.e the sum of the total potential energies and the total kinetic energies of a system is always constant. We can also theoretically and experimentally verify that the total kinetic energy of a system is the sum of the kinetic energies of individual particles.

Consider a system of 2 particles having masses $m_{1}$ and $m_{2}$ and moving with velocity $v_{1}$ & $v_{2}$ respectively. For simplicity we'll say that they're attracted to each other only by gravitational force and hence they'll have some potential energy. Now,

$TE= U_{tot} + KE_{tot}$

But,

$KE_{tot}= KE_{1} + KE_{2}$

$KE_{tot}= \frac{1}{2}m_{1}v_{1}^{2} + \frac{1}{2}m_{2}v_{2}^{2}$

Now, there is a theorem in physics called the work energy theorem which states that: "Net work done by any conservative force is equal to the change in the kinetic energy of the body" i.e

$\sum W\ =\ \Delta\ KE$

In the case of our system, the work done by gravitational force will increase the $KE$ of the particle and will decrease the potential energy of the same keeping the $TE$ constant. This means that:

$\sum W =\ -\ \Delta\ \Phi$

$\Phi$ should not be confused with $U_{tot}$ . First one is the potential energy of a particle when its kept in the gravitational field of another particle and the latter is the total potential energy of system which we're trying to figure out. In our case $\Phi = \frac{-G m_{1} m_{2}}{x}$ .

Change in kinetic energy means the particles will now have new velocities. Let's say that new velocities are $u_{1}$ and $u_{2}$ respectively.

$\Delta KE_{1}= \frac{1}{2}m_{1}u_{1}^{2} - \frac{1}{2}m_{1}v_{1}^{2}$

Similarly,

$\Delta KE_{2}= \frac{1}{2}m_{2}u_{2}^{2} - \frac{1}{2}m_{2}v_{2}^{2}$

$\Delta KE_{tot}= \Delta KE_{1} + \Delta KE_{2}$

$\Delta KE_{tot}= (\frac{1}{2}m_{1}u_{1}^{2} - \frac{1}{2}m_{1}v_{1}^{2})$ $+$ $(\frac{1}{2}m_{2}u_{2}^{2} - \frac{1}{2}m_{2}v_{2}^{2})$

Now, law of conservation of energy is not only applicable to system of particles but also the individual particles constituting the system itself. If we use the law for single particle then also we will not replace the total potential energy( $U_{tot}$ ) with the potential energy which is present by the virtue of the particle being in another gravitational field ( $\Phi$ ).

For an individual particle

$TE= KE + U$

Now, using the work energy theorem and conservation law for individual particle in the above equation where we left:

$-\Delta \Phi_{tot}= (\frac{1}{2}m_{1}u_{1}^{2} - \frac{1}{2}m_{1}v_{1}^{2})$ $+$ $(\frac{1}{2}m_{2}u_{2}^{2} + \frac{1}{2}m_{2}v_{2}^{2})$

$-\Delta \Phi_{tot}= (TE_{1} - U'_{1}) - (TE_{1} - U_{1})$ $+$ $(TE_{2} - U'_{2}) - (TE_{2} - U_{2})$

$\Delta \Phi_{tot}= (U'_{2} - U_{2}) + (U'_{1} - U_{1})$ $+$ $TE_{1} - TE_{1} + TE_{2} - TE_{2}$

But, when the system was in it's initial state then the particles were not moving at all and the total energy of the system was decided during that time. $TE_{1}$ is that total energy of particle 1 and $TE_{2}$ is that energy of particle 2. Since, they were not moving at all but were still exerting forces on each other, the total energies of the both the particles are equal and that value is equal to the potential energy of that state.

Let's assume that the system was a static system from it's initial state which means that there is no kinetic energy in the system. The above argument still hold true but now since kinetic energy is out of the equation we have $TE_{1} = TE_{2}$ and $U_{1} = U_{2}$ . One loop hole in this argument is that since there is no kinetic energy in the system so it never reached it's final state or it's final state has same energy as it's initial state. If we assume that the particles were moving with constant velocity then we can get rid of this loop hole because then $\Delta KE = 0$ and still $\Delta U$ is non zero. Classical definition of potential energy (gravitational or electrostatic) is the work done in bringing the particle from it's position in the system to infinity and $U_{\inf} = 0$ .

If the final state of the system is infinity then also the above argument will hold true. So, is it safe to assume that: $U_{1}= U_{2} = U$ and $U'_{1}= U'_{2} = 0$ as well as $\Delta \Phi = \Phi_{initial}$

$\Phi_{tot}= 2U$

$\Phi_{1} + \Phi_{2} = 2U$

$\boxed{U = \frac{1}{2}(\Phi_{1} + \Phi_{2})}$

The above equation will form the base of our next argument which states that Potential energy of a system of 2 particles is always mutually shared between the particles .

Now, let's picture a static system of N particles interacting with each other gravitationally. The above equations are only defined for a system of 2 particles.

A system of N particles can be also thought of a collection of sub-systems of 2 particles which are sharing their potential energy. So, we can write the equation for system of N particles as(assuming that particles are point objects and that they're attracted to each other by either gravitational or electrostatic forces):

$U_{tot}$ $=$ $\frac{1}{2} \sum_{i=1}^N [M_{i} \sum_{j = 1}^{N-1} V_{j}(x_{j})]$

where $M_{i}$ is the mass of the individual particles and $V_{j}$ is the potential of $M_{i}$ in presence of the gravitational fields of other N-1 particles except itself. You can try to prove the above result yourself but for now we'll say that it holds true.

Hence its verified that The total potential energy of a system of particles is the sum of the potential energies of the pair of particles