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I am little bit dissapointment with action integral in General relativity. The action integral is:

$$ \int Rd^{4}x=\int R_{ij}g^{ij}d^{4}x\tag{1} $$ Where

$$ R_{ij}=\frac{\partial\Gamma^{l}_{ij}}{\partial x^{l}}-\frac{\partial\Gamma^{l}_{li}}{\partial x^{j}}+\Gamma^{l}_{ij}\Gamma^{m}_{lm}-\Gamma^{l}_{im}\Gamma^{m}_{lj}\tag{2} $$ Is the Ricci tensor. The Ricci tensor is in General relativity connected with hamiltonian, and quantity $$ R=R_{ij}g^{ij}\tag{3} $$ is the scalar curvature, which is an invariant quantity, and also total energy of the system. I can't understand, how can I see in Ricci tensor Hamiltonian function.

In every book from General relativity I found something like this:

Action principle...

Lets have an action:

$$ \int\sqrt{-g} R d^4 x=\int\sqrt{-g} R_{ik}g^{ij} d^4 x $$ Where R is the scalar curvature...aaaand, when we do some variation gymnastics like: $$ \delta \sqrt{-g}=-\frac{1}{2}\sqrt{-g}g_{ik}\delta g^{ik} $$ $$ \delta R=R_{ik}\delta g^{ik} $$ and wait when smoke clears, we hopefully arrive to vacuum Einstein field equations: $$ R_{ik}-\frac{1}{2}Rg_{ik}=0 $$ and $$ R_{ik}-\frac{1}{2}Rg_{ik}=\frac{8\pi G}{c^{4}}T_{ik} $$ in the presence of matter. And now we can go further...:)

This is written in every book. But question is: Ok...we arrive to Einstein equations through variation of R. But, why we do variation of R and not some other quantity? I found this identity: $$ R_{ik}=\kappa(T_{ik}-\frac{1}{2}Tg_{ik}) $$ But I can't prove it yet. And its the same, where I started

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    $\begingroup$ Minor comment: What happened to $\sqrt{|g|}$ in eq. (1)? $\endgroup$ – Qmechanic Sep 28 '15 at 14:09
  • $\begingroup$ Nothing happened. This missed $\endgroup$ – marek Sep 28 '15 at 14:18
  • $\begingroup$ This missing $$\sqrt{-g}$$ has no impact on this. I forgot to write it:) $\endgroup$ – marek Sep 28 '15 at 14:20
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    $\begingroup$ It is rather a kind of Lagrangian than the Hamiltonian (obtaining a Hamiltonian formulation for GR is actually pretty difficult). And the quantity you give is the action not the total energy of the system. $\endgroup$ – Sebastian Riese Sep 28 '15 at 14:26
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    $\begingroup$ As was mentioned, R is a Lagrangian (eventially, you apply EL equation on it). The use of R is beause of 3 main reasons - (1) it's scalar (thus invariant of coordinate transformations) (2) it's the simplest scalar that can be built out of Riemann tensor (it's trace) - the main player differential geometry. (3) it's logical - minimising global curvature for a given distribution of energy/momentum. Remember the clasical analog of minimising soap surfaces which carry energy density per area. $\endgroup$ – Alexander Sep 28 '15 at 23:16
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It seems like you've got lost in the subject. To clarify some facts:

  1. The action for General Relativity (Einstein-Hilbert action) is, as usual, an integral of the Lagrangian density over spacetime: $$ S[g] = \frac{1}{16 \pi G} \int d^4 x \sqrt{-g} \cdot R, $$ where $\sqrt{-g}$ is the square root of the determinant of the metric tensor and $R$ is the Ricci scalar curvature of the metric. Why is it so? Because it is a postulate. You can't derive this action from any kind of a fundamental principle (like the equivalence principle). Different actions could also exist. But Einstein-Hilbert action is the simplest of the kind, and therefore gives rise to the simplest geometrical theory of gravity: General Relativity.

  2. The square root of the determinant of the metric tensor $\sqrt{-g}$ is there for a reason: it gives a natural volume element of the Riemannian geometry: $$ d\:\text{Volume} = d^4 x \sqrt{-g} $$ is the invariant spacetime 4-volume element. The square root provides the invariance of the Einstein-Hilbert action under diffeomorphisms (General Coordinate Transformations, GCTs) and therefore the mathematical manifestation of the general principle of relativity.

  3. You have found this identity $$ R_{\mu \nu} = \frac{8 \pi G}{c^4} \left( T_{\mu \nu} - \frac{1}{2} T g_{\mu \nu} \right) $$ and are currently unsure of how it arises in the theory. Actually, this is not an identity but rather a dynamical equation of motion, which is completely equivalent to Einstein's equations. In fact, one could easily derive one from the other. We start from Einstein's equations: $$ R_{\mu \nu} - \frac{1}{2} R g_{\mu \nu} = \frac{8 \pi G}{c^4} T_{\mu \nu} $$ Lets take the trace of this equation (contract it with the contravariant metric): $$ R - \frac{1}{2} R n = \frac{8 \pi G}{c^4} T_{\mu \nu} g^{\mu \nu} $$ $$ \left( 1 - \frac{1}{2} n \right) R = \frac{8 \pi G}{c^4} T $$ where $n$ is the dimensionality of spacetime ($n = 4$). Now I substitute the expression for $R$ in the original equation: $$ R_{\mu \nu} - \frac{8 \pi G}{c^4} \cdot \frac{T g_{\mu \nu}}{(2-n)} = \frac{8 \pi G}{c^4} T_{\mu \nu} $$ $$ R_{\mu \nu} = \frac{8 \pi G}{c^4} \left( T_{\mu \nu} - \frac{1}{n-2} T g_{\mu \nu} \right) $$ For $n = 4$ it reduces to $$ R_{\mu \nu} = \frac{8 \pi G}{c^4} \left( T_{\mu \nu} - \frac{1}{2} T g_{\mu \nu} \right) $$ which means that your equation is completely equivalent to Einstein's equation.

  4. There is a way to construct the Hamiltonian formalism of General Relativity. Take a look at ADM formalism. However my guess would be that you don't need that, and when you were speaking about Hamilton, you were referring to Hamilton's principle which is just a fancy name for the principle of the least action. The principle of the least action lies at the heart of the Lagrangian formalism.

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  • $\begingroup$ @ Hindsight. It's much more clear to me. I didn't know nothing about ADM formalism. I supposed that Energy-momentum tensor is included in that R itself and gets naturally its place on the right side of the equations. But it looks like it is added there to rebalanced Ricci tensor on the left. You can then say, that Ricci tensor is hamiltonian, because it is equal to Energy-momentum tensor. $\endgroup$ – marek Oct 1 '15 at 10:22
  • $\begingroup$ @marek I understood nothing of what you wrote about the relation of the Ricci tensor to the Hamiltonian :) But I am glad it is starting to get more clear to you. $\endgroup$ – Prof. Legolasov Oct 1 '15 at 14:32
  • $\begingroup$ @marek about that relation: theoretical physics is an exact science and therefore we must tend to make precise mathematical statements. In the context of General Relativity we have two different types of things: matter (which is characterized by the energy-momentum tensor) and gravity (which is by definition the curvature of spacetime and is characterized by the Riemann tensor and its contractions). Einstein's equations (and Einstein-Hilbert lagrangian which is used to obtain them) are the equations of motion for gravity. $\endgroup$ – Prof. Legolasov Oct 1 '15 at 14:37
  • $\begingroup$ @marek Equations of motion for matter also exist, and their form depends on the type of matter which you are considering. The easiest case is a probe point particle with negligible mass. The equations of motion of such particle is the well-known geodesic equation. You could also put some fields (scalars / electromagnetism / spinors) on the spacetime and obtain the equations of motion for them. But it will always depend on the geometric properties of spacetime (like the affine connection). $\endgroup$ – Prof. Legolasov Oct 1 '15 at 14:39
  • $\begingroup$ @marek and equations of motion for gravity (Einstein's equations) also depend on the properties related to the specific model of matter. But in this case all these properties can be summarized in the stress-energy tensor, which accounts for the complete information about how your matter interacts with gravity. This is what relativists mean when they say that matter moves in spacetime, and spacetime curvature is given by the matter content. $\endgroup$ – Prof. Legolasov Oct 1 '15 at 14:42

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