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In paper Topological Gravity as the Early Phase of Our Universe there's statement:

Hamiltonian of gravity would vanish by time reparameterization invariance.

How to derive such result?

Qmechanic
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Nikita
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    You might be interested in this question of mine. – Javier Oct 01 '20 at 21:36
  • @Javier , thank you! But my question not about particle, question is about gravity! – Nikita Oct 01 '20 at 21:43
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    @Nikita -- The answer for gravity is essentially a generalization to 4 dimensions of the answer for the point particle. It is well worth understanding the point particle, which contains many of the most important concepts, but with many fewer additional technical complications. – Andrew Oct 01 '20 at 22:03
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    @Nikita If you look at the answers, you can see that the same logic applies. Reparametrization invariance is all that is needed, the specific system doesn't matter. If the connection isn't clear (which is understandable), you should edit that into your question, so people can give better answers. – Javier Oct 01 '20 at 22:15
  • @Andrew, but gravity is described by gravity field, not by relativistic particle – Nikita Oct 01 '20 at 23:17
  • @Javier, in any system with reparametrization invariance Hamiltonian =0? – Nikita Oct 01 '20 at 23:18
  • @Javier, what about free scalar field on gravity background? As I understand, $H\neq 0$, but we have reparametrization invariance – Nikita Oct 01 '20 at 23:34
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    @Nikita Yes I understand :) But a crucial symmetry in gravity is diffeomorphism invariance, also known as coordinate invariance. A relativistic particle has the same symmetry but in a much simpler setting, called reparameterization invariance. This reparameterization invariance / coordinate invariance / diffeomorphism invariance is the deep reason why the Hamiltonian is zero. It's a very good idea to understand how this works for the relativistic point particle, before going to gravity -- the same basic story applies to gravity but there are a number of additional technical complications. – Andrew Oct 02 '20 at 02:34
  • From the ADM formalism, this result may be deduced, when taking variation with the respect to the lapse. As well, the variation with respect to the shift will give : $P_i = 0$. However, seems that there must be mire intuitive argument – spiridon_the_sun_rotator Oct 02 '20 at 06:38
  • @Andrew, yes, I understand example with particle. To fully understand this, could you please explain, how this idea work with free scalar field on gravity background. – Nikita Oct 02 '20 at 07:56
  • @Nikita Well, I'd start with pure gravity (a scalar field on a fixed curved background will have a non-vanishing hamilontian so you need to look at a scalar field coupled to gravity, which is more complicated than just gravity). The basic idea is that if you start with the Lagrangian of GR and use Dirac's quantization procedure, you find the Hamiltonian is equal to sum of four first class constraints, related to diffeomorphisms for four spacetime dimensions. There are many places this is worked out in detail. For instance, see Eq 2.5 of Nelson and Teitelboim, 1978. – Andrew Oct 02 '20 at 13:30
  • @Andrew, so we need some additional terms in Lagrangian, not only covariances free action? – Nikita Oct 02 '20 at 19:36
  • @Nikita I wrote an answer finally. You don't need additional terms in the Einstein-Hilbert action (beyond the Gibbons-Hawing-York boundary terms). The main point of your question actually follows just from the diffeomorphism invariance, and is not specific to the form of the Lagrangian. (Ie, if you added other diff invariant terms like $R^2$ or $R_{\mu\nu}R^{\mu\nu}$, you would still get a Hamiltonian that vanished on the constraint surface). – Andrew Oct 03 '20 at 02:40

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I won't give a full derivation -- for that I will point to the literature (or really I'll point to my comment where I gave a ref to the literature), where this calculation has been done many times.

Instead I'll give an outline.

We start with the Einstein-Hilbert action$^\star$

\begin{equation} S = \frac{1}{16\pi G} \int {\rm d}^4 x \sqrt{-g} R \end{equation}

We then define the lapse function $N=(-g^{00})^{-1/2}$, shift vector, $N_i = g_{i0}$, and spatial metric $h_{ij} = g_{ij}$. (The index heights are crucially important here, and please note I am choosing the fastest way through the algebra intentionally so that I do not take it on myself to explain every step).

We also can derive a canonical momentum $\pi^{ij}$, which comes from differentiating the action wrt time. (If you work through the Dirac procedure, you will see the lapse and shift do not have an associated momentum, and instead are Lagrange multipliers for first class constraints).

Then the Lagranian can be written in the canonical form

\begin{equation} \mathcal{L} = - g_{ij} \dot{\pi}^{ij} - N H - N_i P^i + ({\rm total\ derivative}) \end{equation}

where the Hamiltonian constraint $H$ associated with the lapse $N$ is \begin{equation} H = \sqrt{h} \left(^{(3)}R+h^{-1} (\frac{1}{2}\pi^2 - \pi^{ij} \pi_{ij})\right) \end{equation} and the momentum constraints $P^i$ associated with the shifts $N_i$ are \begin{equation} P^i = -2 \nabla_j \pi^{ij} \end{equation}

where $^{(3)}R$ and $\nabla_j$ are the Ricci scalar and covariant derivative associated with the spatial metric $h_{ij}$, respectively.

As I said in the comments, the thing you specifically asked about, namely the fact that the Hamiltonian is zero on the constraint surface, is not special to gravity and instead is related to diffeomorphism invariance, which is also exhibited by the free relativistic particle (but easier to understand in that context). However, this is the canonical formulation of gravity.

$^\star$ strictly speaking one needs to supplement the action with the Gibbons-Hawking-York boundary terms.

Andrew
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  • It's very helpful! But what with scalar field with diffeomorphism invariance? For me, this is most unclear example, which contradict general statement... – Nikita Oct 03 '20 at 06:29
  • So if you have a scalar field coupled to gravity, $S=\int \sqrt{-g} [R - \frac{1}{2} (\partial \phi^2) - V(\phi)]$, then you will find the action can still be written in a canonical form, schematically like $S = \dot{g} \pi + \dot{\phi} p - N H - N_i P^i$, (where $p$ is the momentum of the scalar field), but now the Hamiltonian and momentum constraints will have contributions from the stress energy of the scalar field. Does that help? – Andrew Oct 03 '20 at 10:46
  • Yes, but I wanna to see how Hamiltonian for scalar field vanishes? – Nikita Oct 03 '20 at 11:03
  • If you vary the Einstein-Hilbert action with respect to $N$, you get the constraint equation $H=0$. The only effect of adding a scalar field is to add additional terms to $H$, but it does not change the overall structure. A crucial point is that $H$ is not identically zero, but rather $H=0$ is a constraint that the metric (and if present any matter fields) have to satisfy. I don't know if this helps but I think there's less here than you might think, if you understand what happens for pure gravity then adding matter fields is a relatively minor change. – Andrew Oct 03 '20 at 20:45