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In this answer it was stated that potential energy is a property of a system and not an individual particle.

If we have two particles (1 and 2) interacting via a conservative force, we can write an equation regarding the total energy of the system and its conservation:

KE1 + KE2 + PE(sys) = KE1' + KE2' + PE(sys)'

Where the ' indicates final.

I know this equation is incorrect, but my question is why is it incorrect?

I most definitely am interpreting this answer incorrectly.

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  • $\begingroup$ When one part of the system is effectively invariant---as in the case where human sized things are moving around near the surface of the Earth---it is expedient to ascribe the potential energy to the small thing. It's just that if someone took the planet away it would affect the behavior of the experiment. On the other hand, if you are talking about the elastic energy of a spring of natural length $l_0$ with a mobile mass on each end, then the potential should be written $U(\vec{x}_1,\vec{x}_2) = \frac{1}{2}k(|\vec{x}_1-\vec{x_2}|-l_0)^2$ and is obviously a function of both positions. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Oct 9, 2015 at 2:20
  • $\begingroup$ Your equation is correct. Why do you think that its incorrect? $\endgroup$
    – garyp
    Commented Oct 9, 2015 at 12:56
  • $\begingroup$ @garyp So when finding the total energy of the system it would be correct to count the potential energy once (instead of twice for both particles)? $\endgroup$
    – andrew
    Commented Oct 9, 2015 at 21:16
  • $\begingroup$ Potential energy is the energy associated with the interaction between pairs of objects, and the positions of those objects. So you calculate it once for every pair of objects. Two objects: once. Three objects: three times (and add them together). Four particles: six times (and add them together). One particle: not defined! If you throw a ball, there are two objects interacting by gravity: the ball and the earth. $\endgroup$
    – garyp
    Commented Oct 9, 2015 at 22:45
  • $\begingroup$ @garyp You can derive KE1+PE(sys)=KE1'+PE(sys)' from the work-energy theorem and the definition of potential energy. Doesn't this contradict KE1 + KE2 + PE(sys) = KE1' + KE2' + PE(sys)'? I'm still a bit unclear on this. Could you provide any references on this subject? Thanks. $\endgroup$
    – andrew
    Commented Oct 10, 2015 at 4:50

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To offer insight on the comment about potential energy being defined for a system, rather than just the body itself.

This stems from the definition of potential energy. For a body in a gravitational field, the source of the field had to do work in order to bring this body to its current position. Generally, potential energy is defined to be zero at an infinite distance away from the source of the field. Without identifying the source, there is no sense of where this reference point is. You can't measure a quantity that relies on a reference point without a reference point! Hence, there is no concept of what the potential energy of the body is. Potential energy is defined for a system. Without both a source and the body, there is no true sense of potential energy.

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