With respect to both gravity and electromagnetism, to the best of my understanding potential energy is added or subtracted from a system based on the distance between two objects such as charged or massive objects. Would this then imply that when oppositely charged objects or two massive objects are brought together within a small enough radius that the drop in potential energy would be greater than the combined mass of the two objects making the net mass/energy of the system negative?

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    $\begingroup$ Highly related: physics.stackexchange.com/questions/17082/… $\endgroup$ Jun 21, 2020 at 20:19
  • $\begingroup$ The crux of this question is not why potential energy is subtracted but rather what happens at very small distances where it would seem the energy subtracted would be more than the rest mass! $\endgroup$ Jun 23, 2020 at 13:18
  • $\begingroup$ @DerekSeabrooke I've added some more to my answer, let me know if it answers your question. $\endgroup$
    – DavidH
    Jun 24, 2020 at 10:28

2 Answers 2


Yes, potential energy can be negative: consider Newton’s law of gravitation

$$V = -\frac{GMm}{r}$$

Where $G$ is Newton’s constant, $M$ and $m$ are masses, and $r$ is the distance between them. It can clearly be seen that this is always negative.

The key thing is that the absolute value of potential energy is not observable; there is no measurement that can determine it. The only thing that can be measured is differences in potential energy. So actually there is a redundancy in the equation above: if I add any constant to it, the difference in potential energy for two given separations is the same. The common form of Newton’s law of gravitation is set by the convention that two objects an infinite distance apart have zero gravitational potential energy, but this is purely a convention.

The idea of redundancies in physical descriptions is very important in theoretical physics, and is known as gauge invariance.

EDIT: following some comments by the original poster, I've added some more to this answer to explain the effect on total energy of a system of attracting objects at very short distances.

Let's consider two equal point masses $M$ separated by some distance $r$: the total energy of the system, using the above definition of potential energy, is

$$E = 2Mc^2 - \frac{GM^2}{r}.$$

If the total energy is negative, $E < 0$. We can rearrange this inequality to give a condition on the radius for negative total energy:

$$r < \frac{GM}{2c^2}.$$

Compare this to the Schwarschild radius $r_\mathrm{s} = 2GM/c^2$. The distance at which the Newtonian energy becmes negative is less than the Schwarzschild radius---if two point masses were this close they would be a black hole. In reality we should be using GR to describe this system; the negative energy is a symptom of the breakdown of our theory.

One can do the same calculation with two opposite charges $\pm e$ and find

$$r < \frac{e^2}{8 \pi M c^2 \varepsilon_0}.$$

We can then compare this to the classical electron radius $r_\mathrm{e}$ and similiarly find that $r < r_\mathrm{e}$ for a negative total energy. The classical electron radius is the scale at which quantum fluctuations must be taken into account, so again the negative energy is a symptom of the breakdown of the theory.

  • $\begingroup$ Nice answer. +1 from me. $\endgroup$
    – Gert
    Jun 21, 2020 at 10:12
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    $\begingroup$ How is potential energy unobservable? Is it not real energy? According to relativity energy and mass are equivalent meaning that a change in potential energy should affect the warping of space. $\endgroup$ Jun 21, 2020 at 15:28
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    $\begingroup$ @DerekSeabrook We can only measure differences in energy, not absolute values. Whenever you deal woth potential energy you have to define a reference point (we usually say the potential energy is zero there), but this is arbitrary (the constant from DavidH's) answer. $\endgroup$
    – Sito
    Jun 21, 2020 at 18:16
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    $\begingroup$ @DerekSeabrooke Yes, potential energy is real energy that causes spacetime curvature. For example, the mass-energy of a hydrogen atom includes -27.2 eV of negative PE. Because of it, a hydrogen atom bends spacetime a bit less than a non-interacting proton and electron would. $\endgroup$
    – G. Smith
    Jun 21, 2020 at 19:53
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    $\begingroup$ @DerekSeabrooke That's one thing that often trips people up. General relativity (and many other things) care about the energy gradient. That's why you can have spacetime with ridiculous energy densities (like with the Big Bang/Inflation) without everything collapsing into black holes. You usually don't bother with negative energies in GR because in the end, there aren't any - once you add up all the energies together, you're left with a positive number anyway - e.g. a hydrogen molecule has less energy than two hydrogen atoms (the binding potential energy is negative), but still positive. $\endgroup$
    – Luaan
    Jun 22, 2020 at 10:54

Basically the notion of absolute potential energy is undefined.

Definition: The change in potential energy of the system is defined as the negative of work done by the internal conservative forces of the system.

By this definition we can conclude that we are free to choose reference anywhere in space and define potential energy with respect to it.

For ex: Consider a system of 2 oppositely charged particles which are released from rest. Under the action of their mutual electrostatic forces they move towards each other. The internal electrostatic forces are doing positive work which results into the decrease in potential energy of the system.


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