# Boundary conditions in variational principles

In classical mechanics, the condition to fix the variation of the trajectory at the endpoints has a clear-cut meaning. We want the system to propagate from $x\in\mathcal{C}$ to $y\in\mathcal{C}$, therefore, we only consider curves that have these two points as endpoints, so the variations also have to respect this. It is clear.

Moving on to relativistic field theory and the (covariant) Lagrangian approach of it, these conditions however become much less transparent.

The action is usually defined as $$S[\phi]=\int_{\mathcal{D}}\mathcal{L}(\phi,\nabla\phi...\nabla^{(k)}\phi)\ d^nx,$$ and we consider only variations $\delta\phi$ for which $\delta\phi|_{\partial\mathcal{D}}=0$.

In particular, it is accepted, that the derivatives of the variation $\nabla_\mu\delta\phi$ cannot be set to zero on the boundary.

Sources I have seen, however did not explain any of this, they just postulated.

Questions:

• What is the motivation to postulate the vanishing of the variation at the boundary?

• What is the reason one cannot also postulate the vanishing of the derivative of the variation at the boundary?

• Is there any significance to the domain $\mathcal{D}$ in the action? In particular, I have seen some books demand the integral to be extended to the entire manifold, while some books explicitly said that $\mathcal{D}$ should be a compact or precompact domain.

Further notes:

• I have seen a throwaway note in Wald's General Relativity that, when defined the functional derivative $\delta S/\delta\phi$ as $$\delta S[\phi]=\int_{\mathcal{D}}\frac{\delta S}{\delta\phi}\delta\phi\ d^nx,$$ said that more generally, $S$ is functionally differentiable if there exists a tensor distribution $\frac{\delta S}{\delta\phi}$ such that $$\delta S[\phi]=\langle \frac{\delta S}{\delta\phi},\delta\phi\rangle.$$ This made me think that maybe one should take the variations $\delta\phi$ as test functions. But then, if they are usual test functions and have compact support, then taking $\mathcal{D}=\text{supp}(\delta\phi)$ will reproduce the usual boundary conditions, but if we extend the domain of integration outside $\mathcal{D}$, then because the test function is identically zero outside its support, then we can also set the derivatives zero, right? But if we take the variations to be tempered test functions, then the domain always have to be the entire manifold.

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• It is possible to derive the Israel junction conditions from variational principles, but in this case, the Gibbons-Hawking-York boundary term must be appended to the action. This provides a direct application of the GHY term, which itself is a consequence of not being able to set the derivatives of the variation to zero, so here I see this claim validated by practical results, yet the underlying reason still eludes me.