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Starting with the Einstein-Hilbert Lagrangian

$$ L_{EH} = -\frac{1}{2}(R + 2\Lambda)$$

one can formally calculate a gravitational energy-momentum tensor

$$ T_{EH}^{\mu\nu} = -2 \frac{\delta L_{EH}}{\delta g_{\mu\nu}}$$

leading to

$$ T_{EH}^{\mu\nu} = -G_{\mu\nu} + \Lambda g_{\mu\nu} = -(R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R) + \Lambda g_{\mu\nu} $$

But then, in the paragraph below Eq(228) on page 62 of this paper, it is said that this quantity is not a physical quantity and that it is well known that for the gravitational field no (physical) energy-momentum tensor exists.

To me personally, this fact is rather surprising than well known. So can somebody explain to me (mathematically and/or "intuitively") why there is no energy-momentum tensor for the gravitational field?

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A conserved energy-momentum tensor means $\partial_\mu( \sqrt{-g} T^{\mu\nu}) = 0$, but this is not a covariant equation. And the covariant equation $\nabla _\mu( T^{\mu\nu}) = 0$, does not correspond to a conserved energy-momentum tensor. – Trimok Oct 25 '12 at 11:01
See stress-energy momentum pseudotensor (en.wikipedia.org/wiki/…) – Trimok Oct 25 '12 at 11:19

2 Answers

up vote 5 down vote accepted

The canonical energy-momentum tensor is exactly zero, due to the Einstein equation. The same holds for any diffeomorphism invariant theory.

By saying ''it doesnt exist'' one just means that it doesn't contain any useful information.

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The Hilbert tensor $T_{EH}^{\mu\nu}$ represents the stress-energy-momentum of matter plus non-gravitational fields. It is a perfectly physical quantity.

Let me take the cosmological constant as zero for simplicity (it can be also absorbed into the stress-energy-momentum tensor). If you rewrite the Hilbert Einstein equations as

$$T_{EH}^{\mu\nu} + t_{G}^{\mu\nu} = - \left( R_{\mu\nu}^{(1)} - \frac{1}{2}\eta_{\mu\nu}R^{(1)} \right)$$

where the superindex $(1)$ represent the linearised terms in a series expansion over flat background with metric $\eta_{\mu\nu}$, the equations look formally as the equations for a non-linear spin-2 field where $t_{G}^{\mu\nu}$ would represent the energy-momentum tensor for the own gravitational field. The problem is that the sign of $t_{G}^{\mu\nu}$ is incorrect and in fact it is not even a tensor. This problem is specific of general relativity.

The energy-momentum tensor for the gravitational field exists in the field theory of gravity (FTG). This is a true tensor and positive definite. From the perspective of the modern field theory of gravity, it is easy to understand why general relativity lacks an energy-momentum tensor for the gravitational field. In the derivation of general relativity from FTG, it is needed to neglect the field-theoretic energy-momentum tensor for the gravitational field $T_{grav}^{\mu\nu}$ [1]. As a consequence, you cannot find this tensor in general relativity!

[1] General Relativity as Geometrical Approximation to a Field Theory of Gravity

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Hi juanrga, thanks for this answer. But I have to admit that I do not yet understand it fully. How do you obtain $t_G^{\mu\nu}$, I mean from which action, and why is it not a tensor? And what do you exactly mean by field theory of gravity? Do you mean field theory in curved spacetime where coordinate transformations act as gauge transformation and the graviton is the gauge field? Maybe I should look what you say in your paper ... – Dilaton Oct 27 '12 at 17:47
1  
$t_G^{\mu\nu}$ is the difference between the full Einstein tensor and the linearized tensor. It can be alternatively obtained if you write the action in "relaxed form". Then you obtain the Hilbert Einstein equations in "relaxed form", whose source is the matter tensor plus the pseudo-tensor. $t_G^{\mu\nu}$ is not a tensor because does not transform as one. I am referring to FTG, which is the non-geometrical approach to gravity developed by Poincaré, Feynman, Birkhoff, Moshinsky, Thirring, Kalman... The theory is defined in a flat background. – juanrga Oct 28 '12 at 12:09

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