Consider the standard Einstein-Hilbert action in $D \ne 2$ dimensions spacetimes :
\begin{equation}
S_{EH} = \frac{1}{2 \kappa} \int_{\Omega} R \; \sqrt{- g} \; d^D x,
\end{equation}
where $\Omega$ is an arbitrary region of the spacetime manifold. Arbitrary variations of the metric components give two terms :
\begin{equation}
\delta S_{EH} = \frac{1}{2 \kappa} \int_{\Omega} G_{\mu \nu} \; \delta g^{\mu \nu} \, \sqrt{- g} \; d^D x + \frac{1}{2 \kappa} \int_{\Omega} \partial_{\lambda} \big( \sqrt{- g} \; (g^{\mu \nu} \, g^{\lambda \kappa} - g^{\mu \kappa} \, g^{\nu \lambda}) \, \nabla_{\kappa} \; \delta g_{\mu \nu} \big) \, d^D x.
\end{equation}
The last term can be transformed into a surface integral, by Gauss theorem, and lead to an apparent problem. If we ask that the metric variations vanish on the boundary ; $\delta g_{\mu \nu} = 0$ on $\partial \, \Omega$, the surface integral **does not vanish and the variational principle cannot be applied**. We can't ask that the partial derivatives $\partial_{\kappa} \; \delta g_{\mu \nu}$ vanishes too on the boundary surface. Usually, this forces us to introduce the **Gibbons-Hawking-York** surface integral into the gravitationnal action to remove that variational issue. I don't like that. I think that the GHY term is very "unatural" and artificial (feel like a patching work), and would prefer another way. Apparently, there's one other way but it seems to work only when $D = 4$ (see below).

It is well known that the EH lagrangian can be decomposed into two terms : a bulk term and a surface (divergence) term :
\begin{equation}
R = g^{\mu \nu} (\Gamma_{\lambda \kappa}^{\lambda} \; \Gamma_{\mu \nu}^{\kappa} - \Gamma_{\mu \kappa}^{\lambda} \; \Gamma_{\lambda \nu}^{\kappa}) + \frac{1}{\sqrt{- g}} \; \partial_{\mu} \big( \sqrt{- g} \; (g^{\mu \nu} \, g^{\lambda \kappa} - g^{\mu \lambda} \, g^{\nu \kappa}) \, \partial_{\nu} \; g_{\lambda \kappa} \big).
\end{equation}
Lets call the first term $\mathscr{L}_{H}$ (quadratic in first partial derivatives of $g_{\mu \nu}$). The second term involves second derivatives of the metric. Using this decomposition, we could verify this **remarkable identity** :
\begin{equation}
\frac{1}{2 \kappa} \; R = \mathscr{L}_H - \frac{2}{D - 2} \, \frac{1}{\sqrt{- g}} \; \partial_{\mu} \Big( \sqrt{- g} \; g_{\lambda \kappa} \; \frac{\partial \mathscr{L}_H}{\partial (\partial_{\mu} \, g_{\lambda \kappa})} \Big).
\end{equation}
Now, this is very similar to a classical system with the following lagrangian (but only if $D = 4$) :
\begin{equation}
L' = L(q, \dot{q}) - \frac{d}{dt} \Big( q^i \, \frac{\partial L}{\partial \dot{q}^i} \Big).
\end{equation}
It is easy to prove that this classical lagrangian gives exactly the same equations as $L(q, \dot{q})$ under arbitrary variations $\delta q^i$, if we just impose the canonical momentum $p_i = \partial L / \partial \dot{q}^i$ to be fixed at the boundaries ; $\delta p_i = 0$ at $t_1$ and $t_2$ (while $\delta q^i$ stay arbitrary) :
\begin{align}
\delta S &= \int_{t_1}^{t_2} \Big( \frac{\partial L}{\partial q^i} \; \delta q^i + \frac{\partial L}{\partial \dot{q}^i} \; \delta \dot{q}^i \Big) \, dt - \delta (q^i \, p_i) \Big|_{t_1}^{t_2} \\[12pt]
&= \int_{t_1}^{t_2} \Big[ \frac{\partial L}{\partial q^i} - \frac{d}{dt} \Big( \frac{\partial L}{\partial \dot{q}^i} \Big) \Big] \, \delta q^i \; dt + \int_{t_1}^{t_2} \frac{d}{dt} \Big( \frac{\partial L}{\partial \dot{q}^i} \; \delta q^i \Big) \, dt - \delta (q^i \, p_i) \Big|_{t_1}^{t_2} \\[12pt]
&= (\text{usual Euler-Lagrange variation of } S) + \big(p_i \; \delta q^i - \delta (q^i \; p_i)\big)\Big|_{t_1}^{t_2} \\[12pt]
&= (\ldots) - q^i \; \delta p_i \Big|_{t_1}^{t_2}.
\end{align}
The last term cancels if we ask that $\delta p_i(t_1) = \delta p_i(t_2) = 0$, and we get the usual Euler-Lagrange equation for an arbitrary variation $\delta q^i$. For the gravitationnal action above, we can ask the same ; $\delta P^{\mu \lambda \kappa} = 0$ on $\partial \, \Omega$, where
\begin{equation}
P^{\mu \lambda \kappa} = \sqrt{- g} \; \frac{\partial \mathscr{L}_H}{\partial (\partial_{\mu} \, g_{\lambda \kappa})},
\end{equation}
while $\delta g_{\mu \nu}$ is still arbitrary on the boundary. Apparently, this works but **ONLY WHEN** $D = 4$ (so $\frac{2}{D - 2} = 1$ in the lagrangian identity above) !

**Now my questions are the following :**

Is this procedure well defined and rigorous ? **Is it really possible to completely throw away the GHY counter-term with this procedure ?**

Does it make sense to fix the spacetime region $\Omega$ and its boundary $\partial \, \Omega$, while the metric is still varied on it ($\delta g_{\mu \nu} \ne 0$ on $\partial \, \Omega$) ? (I suspect some issues here.)

If everything is fine, does that mean that we really can't derive the Einstein equation from the EH action (+ matter terms and without the GHY counter-term), when $D \ne 4$ ?

**I find all this very surprising !** I firmly believed that the usual Einstein equation $G_{\mu \nu} + \Lambda \, g_{\mu \nu} = -\, \kappa \, T_{\mu \nu}$ was valid for any D dimensions. I'm not so sure anymore.

**More details :**

Take note that the "canonical momentum" $P^{\mu \lambda \kappa}$ defined above can be seen as a complicated function of $g_{\mu \nu}$ and $\partial_{\lambda} \, g_{\mu \nu}$ (like $p_i$ is a function of $q^i$ and $\dot{q}^i$). The elements $P^{\mu \lambda \kappa}$ do not form a tensor. Explicitely, they have the following shape :
\begin{equation}
P^{\mu \lambda \kappa} = \frac{1}{4} \; \sqrt{- g} \; \mathcal{M}^{\mu \lambda \kappa \nu \rho \sigma} \; \partial_{\nu} \, g_{\rho \sigma},
\end{equation}
where $\mathcal{M}^{\mu \lambda \kappa \nu \rho \sigma}$ is a complicated tensor defined from the metric inverse :
\begin{equation}
\mathcal{M}^{\mu \lambda \kappa \nu \rho \sigma} = \frac{1}{4 \kappa} \big( g^{\mu \nu} \, g^{\lambda \rho} \, g^{\kappa \sigma} + \ldots \big),
\end{equation}
with the symetry properties $\mathcal{M}^{\mu \kappa \lambda \nu \rho \sigma} = \mathcal{M}^{\mu \lambda \kappa \nu \sigma \rho} = \mathcal{M}^{\nu \rho \sigma \mu \lambda \kappa} = \mathcal{M}^{\mu \lambda \kappa \nu \rho \sigma}$. We can write the quadratic bulk lagrangian like this :
\begin{equation}
\mathscr{L}_H = \frac{1}{4} \; \mathcal{M}^{\mu \lambda \kappa \nu \rho \sigma} (\partial_{\mu} \, g_{\lambda \kappa})(\partial_{\nu} \, g_{\rho \sigma}),
\end{equation}
which looks a bit like the standard lagrangian for a massless scalar field :
\begin{equation}
\mathscr{L}_{\text{scalar field}} = \frac{1}{2} \; g^{\mu \nu} (\partial_{\mu} \, \phi)(\partial_{\nu} \, \phi).
\end{equation}
The lagrangian of the electromagnetic field also have a similar structure. We can prove with some labor the following relation to the pesky surface term, which appears under a general variation of the metric, **but only if** $D = 4$ (remember that $g_{\mu \nu} \, g^{\mu \nu} \equiv D$) :
\begin{equation}
\frac{1}{2 \kappa} \; \sqrt{- g} \; (g^{\mu \lambda} \, g^{\nu \kappa} - g^{\mu \nu} \, g^{\lambda \kappa}) \, \nabla_{\nu} \, \delta g_{\lambda \kappa} \equiv g_{\lambda \kappa} \; \delta P^{\mu \lambda \kappa},
\end{equation}
which can be canceled if we ask $\delta P^{\mu \lambda \kappa} = 0$ on $\partial \, \Omega$, instead of $\delta g_{\mu \nu} = 0$. **This is really remarkable, and almost miraculous !**

The only author I know who showed some parts of the previous exposition (without discussing the case $D \ne 4$) is *T. Padmanabhan*. See for example these papers :

http://arxiv.org/abs/1303.1535

http://arxiv.org/abs/gr-qc/0209088

If the answer to all the questions above is really "yes", then I feel
that this procedure should be teached in all courses on classical
general relativity ! The GHY counter-term could be trashed. **But
then what about that $D = 4$ restriction ?** This is puzzling.

Any opinion on this ?

This post imported from StackExchange Physics at 2015-11-13 08:03 (UTC), posted by SE-user Cham