# Does potential energy in external fields contribute to the mechanical energy?

To my mind, there should only be potential energies internal to a system, since a potential energy is associated with two or more interacting bodies.

However, we often speak of "the potential energy of a charge in an external electric field", for instance. Although I understand this is fine algebraically, since the work done on the source charge is approximated to be zero, how can this be right conceptually?

For instance, if we were to write the Hamiltonian of the test charge, would the system be the test charge and the source? I would say yes, that the mechanical energy of a system only includes potential energies internal to the chosen system, and not potential energies due to external fields (which can't be in the system!).

I was wondering if someone could help to clarify this. Thank you!

• I agree with you first paragraph but not with your third paragraph. You should always consider the entrie system (also the "source"). I refer to my answer.
– user224659
Commented Apr 3, 2020 at 9:27

This is an interesting question. You are right in how you think about potential energy. There are more bodies involved than just the one we are trying to get the trajectorie from. Saying a particle moves in a static external potential is always an approximation. I want to motivate this:

In classical mechanics you often deal with closed systems: You have a single particle which moves in a system subjected to a static external field, described by some potential, e.g. $$\phi \propto x^2$$. Imagine it like this: There is this potential and then you bring in the particle. In all of these exercises it's assumed that the potential is unchanged by introducing this particle and independent of the movement of the particle. In this sense it is static.

But in reality the particle itself can alter this potential. The $$\propto x^2$$ potential is caused by one or multiple external particles and their interactions between each other. Brining in the single particle introduces new interactions and changes the potential. The potential in which the single particle moves now is actually a function of the external generalized coordinates $$\vec{\lambda}$$ and the coordinates of the single particle itself $$\vec{r}$$ $$\phi=\phi(\vec{r},\vec{\lambda})$$ In order to exactly solve for the trajectorie of the single particle you have to solve a very complex Hamiltonian containing all of the external interactions. That means you also have to solve for the movement of the particles causing the potential.

This is something you often can't do, the Hamiltonians are just too complex. What you do is approximate the potential as $$\phi\approx\phi(\vec{r})$$ And assume it's only dependent on some external parameters, that aren't influenced by the single particle. So you assume that brining in the single particle changes nothing. This is a very powerful method and allows us to gain a rough understanding of real life interactions.

But it can have it's downfalls. Let's consider a classical example from CED: You have a large configuration of charges which make up a electromagnetic field. This field is described by some potential, e.g. $$\propto \frac{1}{r}$$. Now you want to consider how a small charge moves in this field. Well then most of the time you assume that the small charge doesn't alter the large configuration and moves in the static $$\propto \frac{1}{r}$$ potential. But now imagine a very fragile but initally stable large configuration which causes this potential. The small test charge would cause a colapse of the configuration. Here the assumption that there is movement in a static external field leads to wrong results.

• I made some edits to make my point more clear.
– user224659
Commented Apr 3, 2020 at 9:22
• This is an amazing answer; so essentially we don't write the Hamiltonian for a single test charge, but rather for the whole system. Except in some cases the Hamiltonian can be approximated as only a function of the position of the test charge - but nonetheless, it is still of the entire system! Commented Apr 3, 2020 at 11:35
• Yes. Just some terminology: What you call the "source" I would call the enviroment and what you call the test charge I would call the system. This generalizes the ideas even further and shows that this approximation is not limited to electrodynamics. Glad this was helpful. I just recently read something very similar in Landaus introduction to statistical physics.
– user224659
Commented Apr 3, 2020 at 11:51