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We defined the action as: $$\mathcal{S}(t)=\int_{t_1}^{t_2}\mathcal{L}(q_i,\dot{q_i},t) dt$$ where $q_i(t_1)$ and $q_i(t_2)$ are known and fixed. Hamilton's principle states that the path that is followed has minimum action. Suppose we know just the initial coordinates of a system i.e. $q_i(t_1)$ and not its final coordinates. How can we find out the path followed by the system using the least-action principle (Hamilton's Principle)? As it seems to me that it can only be used when both end points are known.

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This is a subtle point. In principle, $q(t_2)$ (as well as $q(t_1)$) is indeed held fixed at its unknown final value. However, in practice, to a apply Hamilton's principle you don't actually directly fix those path endpoints and calculate any actions, as you might expect you would.

In practice (for classical systems - things are different in quantum mechanics), you always solve the general problem first, by ignoring your initial condition and requiring that the action be stationary for arbitrary fixed initial and final positions. (By "arbitrary fixed", which may sound contradictory, I mean that you are requiring the action to be stationary over the space of trajectories with fixed endpoints, but you're separately solving the problem for all possible choices of fixed endpoints, rather than just solving it for a single specific choice of endpoints.)

This gives you the Euler-Lagrange equation, when turns out to be independent of those endpoints. Then you forget about the action entirely and just work directly with the Euler-Lagrange equation, which is a differential equation that can be tackled as an initial-value problem. In practice, this second step ends up being completely independent from the initial step of varying the action - the action formalism gets you the E-L equation, but then you can forget about it and jump directly to the E-L equation.

So it's this weird conceptual two-step, where you hold the trajectory endpoint fixed to an unknown variable (literally just the dummy variable "$q(t_2)$", with no fixed value assigned). Since the value of $q(t_2)$ ends up not entering into the differential equation that pops out, you don't actually need to know it ahead of time.

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OP is correct: The stationary action principle (SAP)/Hamilton's principle(HP) needs$^1$ boundary conditions (BCs), i.e. both initial and final conditions. This is because we need the $$\text{boundary-terms}~=~\left[\sum_{j=1}^np_j\delta q^j \right]_{t=t_i}^{t=t_f}~=~0\tag{1}$$ to vanish when we vary the action $\delta S$ to find stationary paths.

In conclusion: The SAP/HP can not be applied directly to solve an initial value problem (IVP).

(Of course the SAP/HP can be used indirectly in a certain sense: 1. First use SAP/HP with pertinent BCs to establish the EOMs in the first place. 2. Next use the EOMs to solve an IVP.)

See also e.g. this related Phys.SE post.

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$^1$ Note that if the Lagrangian $L(q,t)$ does not depend on velocities $\dot{q}$, i.e. the system is static, then we don't need any BCs.

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  • $\begingroup$ Respectfully, this is nonsense. The stationary action principle is almost exclusively used to solve initial value problems (in classical mechanics). You just need to be a little cleverer in how you apply it. The fact that it's not directly formulated as an IVP doesn't meet that you can't indirectly use it to solve IVPs. $\endgroup$ – tparker Feb 27 at 14:02
  • $\begingroup$ I updated the answer. $\endgroup$ – Qmechanic Feb 27 at 14:15
  • $\begingroup$ I don't mean to beat you over the head with this, but the indirect usage isn't just some parenthetical curiosity; it is in fact essentially the only way that Hamilton's principle is ever applied in classical mechanics. The direct usage is pretty much completely useless because you rarely know both path endpoints. $\endgroup$ – tparker Feb 28 at 0:09

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