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Potential energy is usually defined using a field and a particle that experiences the field force, as the work down in moving a unit particle from infinity to a position in that field.

But some physics text books describe the particle placed there as possessing potential energy, others that the potential energy is "stored" in the field itself, which appear to conflict with one another. So what is the modern meaning of potential energy for a particle in a field?

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Contrary to most introductory and intermediate textbooks, a single, non-interacting particle cannot possess potential energy. Potential energy is a property of a system of interacting particles and/or fields. A minimum of two entities is required. It is probably more useful to think of potential energy as interaction energy.

The concept of potential energy (or interaction energy) follows nicely from the concept of system. Suppose you have several interacting particles and/or fields (protons in an electric field for example) in your system. Further suppose there are other charged particles outside your system in the surroundings. The system's potential energy is merely a way of accounting for the mutual pairwise interactions within the system. More precisely, the change in the system's potential energy is the opposite of the work done by these internal interactions. See chapter 6 of Matter & Interactions by Chabay and Sherwood (Third edition, Wiley, 2011)

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yes, I would tend to agree with your view and summarise it as work-done on system + potential energy of system + kinetic energy of interacting particles = const. But does the idea of a point in the system having a potential make sense then? –  Physiks lover Oct 18 '12 at 20:51
    
@Physikslover No. Potential energy is a property of the entire system, not of any one particle. Be careful when you say "potential" as opposed to "potential energy." The are potentially confusing but are not the same thing. "Potential" can be "gravitational potential" which is "gravitational potential energy per unit mass" or "electric potential" which is "electric potential energy per unit charge." Textbooks that assign potential energy to a particle are blatantly wrong. –  user11266 Oct 18 '12 at 21:47
    
When textbooks suggest that a particle in a gravitational field has a potential energy mgh, how would you express it? –  Physiks lover Oct 19 '12 at 15:43
    
@Physikslover Very simply, it should be stated as "the system consisting of particle plus gravitational field has potential energy." And by the way, it should be $mg\Delta h$ rather than just $mgh$ because the CHANGE in height is what's relevant. Yes, yes, I know one can arbitrarily put the zero anywhere, but the notation should more accurately reflect that. –  user11266 Oct 22 '12 at 0:08
    
@Physikslover One could also say that the gravitational field did work on the particle in the amount $mgh$. But note now that the system is just the particle, and not particle plus field. –  user11266 Oct 22 '12 at 0:18
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You have some space $\mathcal{M}$ (e.g. $\mathbb{R}^3$), you have time which are values in $\mathbb{R}$, you have a particle characterized in terms of coordinates $\{q_{\ i}(t)\}$ for it, which is a map from time into space. Lastly, the theory is essentially given by a mathematical relation $$D(\{q_{\ i}(t)\})=0,$$ which gives the dynamics. For example $$D(\{q_{\ i}(t)\})=\vec\nabla\Phi(\vec q(t))-m\vec q''(t).$$

Noethers theorem is a mathematical insight on specific such dynamical systems which gives an energy function $H_D$, such that

$$\forall t:D(\{q_{\ i}(t)\})=0\ \ \Longrightarrow\ \ \forall t:H_D(\{q_{\ i}(t)\})=E,$$ where $E$ is some real depending on the initial conditions.

For the above example, you have $$H_D(\{q_{\ i}(t)\})=\frac{m}{2} (\vec q''(t))^2+\Phi(\vec q(t))=E.$$ The Hamiltonian gives the total energy and as its value here is seperated in the two summands $\frac{m}{2} (\vec q''(t))^2$ and $\Phi(\vec q(t))$, the energy-distribution can vary between the two. One can speak of "storing" the energy in the kinetic or the potential energy part. At the same time, the value of $H_D$, and also of $\Phi$ alone, critically depends on the value of $\{q_{\ i}(t)\}$ and so it makes sense to talk of the particles potential energy, i.e. the potential energy associated with the particle. In conclusion, the dynamics determines the energy function and the value of all energy expression necessarily depend on the functions $(\{q_{\ i}(t)\})$ (together with its derivatives). As $H_D=E$ is the actual unchanging energy of the particle, you can speak of some energy being stored in the potential in the above sense, but only if you're aware of the fact that this only makes sense in the context of a particle given by $(\{q_{\ i}(t)\})$ which fulfills the dynamics of the system.

As a side note, what is said above essentially also translates to field theory. I.e. things work out similarly, if instead of particle coordinates $q_{\ i}(t)$, you speak of some fields $\psi_i(\vec x,t)$ and dynamics $D(\{\psi_{\ i}(\vec x,t)\})=0$, like $$D(\psi(\vec x,t))=(\Box+m^2)\psi(\vec x,t),$$ in the case of the Klein-Gordon equation. In (quantum) field theory, it is actually more common to directly express the dynamics in terms of a Lagrangian or Hamiltonian.

Now if you have a theory with more dynamical objects than a particle (for example if you consider interactions between charged particles together with a changing electrical field), then the total conserved energy function will depend on both of them. If, in this case, you still make sense of associating two energy quantities with both objects (like if you can make out term which only depend on one of the field), then it's suggestive to speak of energy exchange between the two. In that case "energy stored in the other field" is a an intuitive notion. To keep track of thing, you want to make yourself clear what the quantity is, which is actually conserved.

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I think there are two sources of confusion.

First of all, there is an energy associated with a field regardless of whether there are particles moving in it. Think for instance of an inductor that produces a current when you remove the external bias. The energy for this comes from the magnetic field.

Secondly, it is a matter of book-keeping. A particle moving from one place to another, in an external field, is associated with an energy. If you find this energy by calculating the work done on the particle along the path, you tend to think of this energy as stored in the particle. But you might as well calculate the difference in the total field (particle + external) between the two configurations and calculate the energy from this field. It is a matter of convenience how you look at it.

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