# Necessity of the Gibbons-Hawking-York (GHY) boundary term

The fundamental point of my question is whether the GHY-boundary term in general relativity is even necessary at all, and if yes, then why is it so, and what is its physical significance.

Several points:

• Despite the appearance of an integral, the variational principle is a local one. Assume that $\mathcal{D}$ is a regular domain of spacetime, and in the action integral, we integrate over $\mathcal{D}$. It is usually stated that we can demand the variations $\delta g^{\mu\nu}$ vanish outside $\mathcal{D}$, but we cannot make $\nabla_\sigma\delta g^{\mu\nu}$ vanish. However, if $\delta g^{\mu\nu}$ vanishes identically outside $\mathcal{D}$, then its derivatives vanish too, except for maybe specifically on the boundary. But then we extend $\mathcal{D}$ to (symbolically) "$\mathcal{D}+\epsilon$", then now the derivatives vanish too, on the boundary. Is there anything wrong with this reasoning?

• Is there any physical applications of the GHY boundary term? The only one I know is that the Israel junction conditions can be derived from the variational principle, but this derivation utilizes the boundary term (see for example https://arxiv.org/abs/gr-qc/0108048 and https://arxiv.org/abs/1206.1258). However, the junction conditions can be obtained without this too (using a delta function source term, for example).

• The Einstein field equations can be obtained from the Palatini-formalism too, however in that case, $g^{\mu\nu}$ and $\Gamma^\sigma_{\mu\nu}$ are separate configuration variables, as such, the vanishing of both $\delta g$ and $\delta\Gamma$ can be prescribed on the boundary. But then there is no boundary term to append the action with.

With these points in mind, is the GHY term necessary? If yes, then is there any physical significance of it? Why does the Palatini-formalism not require it then?

If not, then why bother with it at all?

• You may be interested in my own question on this subject : physics.stackexchange.com/q/218060 – Cham Apr 27 '17 at 12:30
• @Bence Racsko At physics.stackexchange.com/questions/523361, I've posted a question that may allow you to answer your own. (I hesitate to do that, partly because your technical knowledge greatly exceeds my own, and partly because my question, being only about the "physical significance" part of yours, tries to account for the fact that Einstein accepted the variation of relativity that assigns a spatial extent to fermions--which he had worked out in conversations with Cartan in 1927--without objecting to Pauli's alternative use of fiber bundles as adjuncts to point-like fermions.) – Edouard Jan 6 at 16:21

Regarding the last question, the Palatini formulation doesn't require the boundary term whereas standard GR does because the two formulations are not strictly equivalent. If the action is just the Einstein-Hilbert action, they give the same field equations, true, but for more generic actions (e.g., involving combinations and powers of the Ricci scalar $\mathcal R$), they can give different field equations.