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For simplicity without loss of generalization, consider a free particle.

When using the Principle of Least Action, I imagine all variations of the true path between $t_1, t_2$ regardless of whether they're possible or not. But whereas some possible paths can be realized by varying the initial and corresponding final velocities, others are going to require an additional virtual potential $\delta V(q)$ to give the particle the corresponding force required. Yet the potential is always assumed to be zero in the Lagrangian when evaluating the action in this case.

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No they don't, that is precisely the point of the principle of least action. You need to look at all paths, regardless of whether or not they obey the equations of motion. Then, the paths with the least action are the ones that do obey the equations of motion.

In short, the job of the principle of least action is to tell us which paths are "legal". If we already knew which paths were legal, then there would be no point in using the principle.

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  • $\begingroup$ But when implementing a least-action principle, don't we in practice often consider only a subset of paths, and isn't the choice of this subset often closely related to which paths are more physical? For example, if I want to prove Snell's law from the principle of least time, it's going to be more practical if I consider only paths that are straight except at the interface between the two media. In QED, we're not going to consider Feynman diagrams where an electron decays into two photons, because the spins can't add up. I don't think it's possible to define what "all" paths would mean. $\endgroup$ – user4552 Oct 6 '18 at 2:18
  • $\begingroup$ @BenCrowell For the Snell's law example, that's just an example of doing the action minimization in two steps: first you minimize over the pre-reflection and post-reflection paths (giving straight lines), then you minimize over the reflection point. You don't get the straight lines for free -- those also come from action minimization. $\endgroup$ – knzhou Oct 6 '18 at 10:14
  • $\begingroup$ @BenCrowell The situation in QED is different, because it's quantum, but still we integrate over all field configurations in the path integral, including those that do not conserve energy or angular momentum, etc. The fact that an electron can't decay into two photons is then derived. For instance, you can infer the Feynman rules from the Schwinger-Dyson equations that come out of the path integral. $\endgroup$ – knzhou Oct 6 '18 at 10:16
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Given an action $$S[q]~=~\int_{t_i}^{t_f} \! dt~L,$$ it seems OP is asking the following.

Q: Is it a requirement to the variational principle that each virtual trajectory $[t_i,t_f]\to \mathbb{R}^n$ can be realized (as a solution to Newton's 2nd law) by applying an additional appropriate external force $F(t)$? In other words, that each virtual trajectory is a classical solution to a modified action $$\tilde{S}[q]~=~\int_{t_i}^{t_f} \! dt~\tilde{L},\qquad \tilde{L} ~=~L+F_i(t)q^i,$$
with a source term?

A: No, that is not a requirement.

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