Following on my previous question, I am interested in understanding a quantity that arises in the derivation of the Einstein equations described there, which appears to be of the nature of a Hamiltonian.

Here's how I see the Hamiltonian's proper definition. Given for example a one-derivative Lagrangian $\mathcal{L}(q, \dot{q})$, we can form its variation with respect to some derivation $\delta q$ of the generalized position:

$$ \delta \mathcal{L}(q, \dot{q}) = \frac{\partial\mathcal{L}}{\partial q} \delta q + \frac{\partial\mathcal{L}}{\partial \dot{q}} \delta \dot{q}, $$

assume that the dot and the $\delta$ commute, and integrate by parts to get

$$ \delta \mathcal{L} = \left( \frac{\partial L}{\partial q} - \frac{d}{dt} \frac{\partial L}{\partial \dot{q}} \right) \delta q + \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}} \delta q \right). $$

If $\delta$ happens to be the time derivative $d/dt$, then we just reorganize to get

$$ \frac{d}{dt} \left( \mathcal{L} - \frac{\partial \mathcal{L}}{\partial \dot{q}} \dot{q} \right) = \left( \frac{\partial L}{\partial q} - \frac{d}{dt} \frac{\partial L}{\partial \dot{q}} \right) \dot{q} $$

The parenthesized expression on the left is minus the Hamiltonian as it is typically defined (sans the substitution of generalized momenta for generalized velocities, so energy). This shows that for a nontrivial ($\dot q \neq 0$) motion, the equations of motion are satisfied if and only if the Hamiltonian is conserved, i.e. conservation of energy.

Now the same thing in GR. According to the derivation here, we have

$$ \delta (\sqrt{-g} R) = \sqrt{-g} G_{ab} \, \delta g^{ab} + \partial_d \sqrt{-g}(g^{ac} \delta\Gamma^d_{\mathstrut ac} - g^{cd} \delta\Gamma^b_{bc}) $$

Taking $\delta = \partial/\partial x^e$, and using the identity $\sqrt{-g} \nabla_e A^e = \partial_e \sqrt{-g} A^e$ referenced in the article, we get

$$ \nabla_d \bigl(R\, \delta^d_e - (g^{ac} \partial_e\Gamma^d_{\mathstrut ac} - g^{cd} \partial_e\Gamma^b_{bc})\bigr) = G_{ab} \, \partial_d g^{ab} $$

From this equation I conclude that the odd quantity inside the derivative on the left should play the role of a Hamiltonian. Based on the fact that the Hamiltonian is the energy, and also that this derivation is just the standard derivation of momentum as the conserved Noether current for spatial translations (as you can see, if the vacuum Einstein equation $G_{ab} = 0$ holds, then the "Hamiltonian" is indeed locally conserved), I feel like the left-hand side should be the divergence of the energy-momentum tensor. Rewriting it with some index raisings made explicit:

$$ \nabla_d g^{df}\bigl(R\,g_{ef} - (g^{ac} g_{df} \partial_e\Gamma^d_{\mathstrut ac} - \partial_e\Gamma^b_{bf})\bigr) = G_{ab} \, \partial_d g^{ab}, $$

I guess my question is:

Is the tensor $T_{ef} = g^{ac} g_{df} \partial_e\Gamma^d_{\mathstrut ac} - \partial_e\Gamma^b_{bf} - R\,g_{ef}$ in any reasonable sense the "energy-momentum of gravity", and likewise is it the Hamiltonian in any reasonable sense? Also, it would be nice to know if it has a more...expressive...form.

  • $\begingroup$ Comments to the post (v1): 1. Technically, you are asking about the energy function, not the Hamiltonian, since no Legendre transformation $v\leftrightarrow p$ was actually performed. Consider to edit the post accordingly in various places to make this clear. 2. For the Hamiltonian formulation of GR, see the ADM formulation. $\endgroup$
    – Qmechanic
    Jun 19, 2017 at 7:01
  • $\begingroup$ @Qmechanic Fixed the title. As for ADM, I did not see any obvious relationship between its structure and this computation, but I would be happy to learn otherwise. $\endgroup$
    – Ryan Reich
    Jun 19, 2017 at 7:05
  • $\begingroup$ 3. There is not a well-defined notion of a gravitational stress-energy-momentum tensor. $\endgroup$
    – Qmechanic
    Jun 19, 2017 at 7:13
  • $\begingroup$ @Qmechanic I have heard this claim before as well. Is, then, the expression I have written the "pseudotensor" (it is, manifestly, not globally defined but rather coordinate-dependent) referred to in this answer? $\endgroup$
    – Ryan Reich
    Jun 19, 2017 at 7:18

1 Answer 1


I'm just going to give a quick pointer and address your first question (Qmechanic already addressed the second in the above comment). Looking at your expression, you can write the scalar curvature as:

$$R=g^{ab}R_{ab}=g^{ab}\left[\partial_{\rho}\Gamma_{ba}^{\rho}-\partial_{b}\Gamma_{\rho a}^{\rho}+\Gamma_{\rho\lambda}^{\rho}\Gamma_{ba}^{\lambda}-\Gamma_{b\lambda}^{\rho}\Gamma_{\rho a}^{\lambda}\right]$$

It would appear (on first glance) that the second derivatives of the metric in your expression may cancel out leaving something like:

$$T_{ef}=g_{ef}g^{ab}\left[\Gamma_{\rho\lambda}^{\rho}\Gamma_{ba}^{\lambda}-\Gamma_{b\lambda}^{\rho}\Gamma_{\rho a}^{\lambda}\right]$$

This does indeed bear a semblence to the stress-energy-pseudotensor in it's various incarnations (it's reminding me of the Moller complex which I suggest you look up) and has the requirement that it only contains first derivatives of the metric (mind you I just glanced, you would have to show the second derivative terms do indeed cancel out). Note there is no unique expression for the pseudotensor of a given system (indeed there are an infinite number of them).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.