Variation of the Einstein-Hilbert action in D dimensions without the Gibbons-Hawking-York term

Question

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

Comments

It was not clear to me from your question why d=4 is special. The normalization of the total derivative term seems irrelevant to your argument. Anyway, the question is flawed. In trying to establish a good variational principle for an action, you need to look at all of the terms involving small variations of the metric. You have left out the terms that give the boundary stress-tensor, $T^{ab} \delta g_{ab}$. If you impose your naive Neumann boundary conditions, this stress tensor contribution will no longer vanish.

This post imported from StackExchange Physics at 2015-11-13 08:03 (UTC), posted by SE-user user2309840

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It's anyway better to start with a scalar field to see what's going on. The GH term is the analog of the boundary term you need for a scalar field if you start with the non-traditional kinetic term $\phi \Box \phi$.

This post imported from StackExchange Physics at 2015-11-13 08:03 (UTC), posted by SE-user user2309840

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I don't see your point. There's no problem with the matter lagrangian terms and the stress-tensor contribution. The surface term only comes from the EH lagrangian in the usual way. The $\frac{2}{D - 2}$ is NOT a normalization constant. It comes from the contraction of the metric on itself ; $g_{\mu \nu} \, g^{\mu \nu} = D$. The gravitationnal lagrangian remarkable identity (with that factor) was already pointed out by Padmanabhan before.

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

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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 $\partial \, \Omega$) ?

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

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In varying the Einstein-Hilbert action with respect to the metric, you need to integrate by parts twice. In your question, you have only considered the first integration by parts. The second will produce additional terms of the form $T^{ab} \delta g_{ab}$ that will not vanish with your boundary condition. Again, it's easier to see this if you start with the analog problem -- a scalar field with kinetic term $\phi \Box \phi$.

This post imported from StackExchange Physics at 2015-11-13 08:03 (UTC), posted by SE-user user2309840

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I don't see how a scalar field could be usefull here. What I'm discussing is intrinsic to the EH lagrangian. In the variation above, I gave the final variational result from the EH action. The stress-tensor you are discussing is coming from the matter sector. I don't see where's the problem. If we include matter into the total action, variation gives something like $\delta S = \int(G_{\mu \nu} + \Lambda \, g_{\mu \nu} + \kappa \, T_{\mu \nu}) \, \delta g^{\mu \nu} \sqrt{- g} \; d^D x + \int g^{\mu \nu} \; \delta R_{\mu \nu} \; \sqrt{- g} \; d^D x$. The last is the surface I'm talking.

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

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Massless scalar field : Pose $\mathscr{L} = (\partial_a \, \phi)(\partial^a \, \phi)$ (whatever an arbitrary constant in front) and $\mathscr{L}' = -\, \phi \; \partial_a \, \partial^a \, \phi$. Then $\mathscr{L}' = \mathscr{L} - \partial_a (\phi \; \partial^a \, \phi) \ne \mathscr{L} - \partial_a \Big( \phi \; \frac{\partial \mathscr{L}}{\partial(\partial_a \, \phi)} \Big)$. It's not the same as for gravity, in case of $D = 4$.

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

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The stress tensor I am talking about comes from varying your ${\mathcal L}_H$ with respect to the metric and integrating by parts. There is a boundary term of the form $\int_{\partial \Omega} T^{ab} \delta g_{ab}$ which I believe you have dropped.

This post imported from StackExchange Physics at 2015-11-13 08:03 (UTC), posted by SE-user user2309840

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I don't see any problem here. Look at my second equation above. It includes everything. Now, according to the last equation above, the second integral (in the second equation) contains the divergence of $g_{\lambda \kappa} \; \delta P^{\mu \lambda \kappa}$. To cancel that surface term, we don't need to ask that $\delta g_{\mu \nu} = 0$ and $\partial_{\lambda} \, \delta g_{\mu \nu} = 0$ on $\partial \, \Omega$. We just need to impose $\delta P^{\mu \lambda \kappa} = 0$. This is the trick I'm talking about, and it works only if $D = 4$.

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

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Let us continue this discussion in chat.

This post imported from StackExchange Physics at 2015-11-13 08:03 (UTC), posted by SE-user user2309840