Why does no physical energy-momentum tensor exist for the gravitational field?

Question

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?

Comments

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.

This post imported from StackExchange Physics at 2014-03-12 15:45 (UCT), posted by SE-user Trimok

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See stress-energy momentum pseudotensor (en.wikipedia.org/wiki/…)

This post imported from StackExchange Physics at 2014-03-12 15:45 (UCT), posted by SE-user Trimok

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thinking out loud what if energy-momentum isn't really a tensor quantity. What if it is actually something else, that doesn't transform independently on each point, but needs to transform as a whole in an open set at the same time? Imagine such geometrical things, and take the limit of open sets to individual points, then you recover normal tensors.

This post imported from StackExchange Physics at 2014-03-12 15:45 (UCT), posted by SE-user lurscher

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related: physics.stackexchange.com/q/68382

This post imported from StackExchange Physics at 2014-03-12 15:45 (UCT), posted by SE-user Ben Crowell

Answer 1

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.

This post imported from StackExchange Physics at 2014-03-12 15:45 (UCT), posted by SE-user Arnold Neumaier

Comments

This answer is fine, +1, but seems to me to mystify an idea that should be straightforward. For fields like the electric and magnetic fields -- or the gravitational field in Newtonian mechanics -- we have an energy density that goes like the square of the field. In relativity this clearly isn't going to work: by the equivalence principle, we can always say that the gravitational field is zero at a given point, simply by adopting a free-falling frame of reference.

This post imported from StackExchange Physics at 2014-03-12 15:45 (UCT), posted by SE-user Ben Crowell

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I would guess you can define a non-zero stress-energy tensor for gravity if you consider perturbations about a fixed background spacetime. (Almost certainly if these perturbations are small.) Can anyone with expertise confirm this?

This post imported from StackExchange Physics at 2014-03-12 15:45 (UCT), posted by SE-user Jess Riedel

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@JessRiedel: You can do that, but the object you end up with isn't a tensor. See ned.ipac.caltech.edu/level5/March01/Carroll3/Carroll6.html , at "It is natural at this point to talk about the energy emitted via gravitational radiation."

This post imported from StackExchange Physics at 2014-03-12 15:45 (UCT), posted by SE-user Ben Crowell

Answer 2

The energy-momentum tensor is defined locally, and it's a tensor. In electromagnetism, or in Newtonian gravity, the way we form a local energy density is basically by squaring the field.

The problem with applying this to GR is that the gravitational field $\mathbf{g}$ is zero, locally, in an inertial (i.e., free-falling) frame of reference, so any energy density we form by squaring it is going to be something that can be made to be zero at any given point, simply by a choice of coordinates. But a tensor that's zero for one choice of coordinates is zero for any choice of coordinates, so the whole idea doesn't work for GR.

The other kind of thing you could try would be taking derivatives of the field and using them as ingredients in such a locally defined tensor. This doesn't help, though. There's a discussion of this in Wald, section 11.2. The basic problem is that if you want the result to be a tensor, the derivatives have to be covariant derivatives operating on a tensor. But the only tensor we have available is the metric, and the defining characteristic of the covariant derivative is that it gives zero when you differentiate the metric. (There's a loophole in Wald's argument, however, that bothers me. When forming a tensorial quantity by differentiation, it's sufficient but not necessary that the derivative be a covariant derivative. When we form a curvature tensor from the metric, we do it by taking non-covariant derivatives on the metric in order to form the Christoffel symbols, and then doing further operations involving non-covariant derivatives to get the Riemann curvature tensor -- which is a valid tensor.)

None of this prevents the definition of nonlocal measures of the energy carried by gravitational fields in a certain region. That's why, for example, we can talk about the energy carried by a gravitational wave, but we have to talk about a region that's big compared to a wavelength. However, that won't allow us to define something that can go into the Einstein field equations, which are local because they're a differential equation.

This post imported from StackExchange Physics at 2014-03-12 15:45 (UCT), posted by SE-user Ben Crowell

Answer 3

How about this:

The mathematical expressions for momentum and (kinetic) energy are usually linear and quadratic in the first derivatives of the dynamical variables. E.g. for a classical particle the dynamical variable is just the trajectory variable ${\bf{x}}(t)$, while the momentum is $m{\bf{\dot{x}}}$, and the kinetic energy is $\frac{1}{2}m{\bf{\dot{x}}}$$^{2}$. But in general relativity the first covariant derivative of the gravitational field variable, i.e. of the space-time metric $g_{\mu\nu}(x)$, vanishes: $\nabla_{\lambda}{g}_{\mu\nu}(x)=0$. Accordingly, the energy and momentum of the gravitational field themselves vanish. In special relativistic theories of gravity, on the other hand, the first derivative of the gravitational field variable does not vanish, and neither (therefore) does the field's energy-momentum. This suggests that the gravitational field's lack of energy-momentum in general relativity arises because conserved quantities like energy and momentum are only really definable in terms of space and time, whereas the field in this case is identical to the very geometry of space-time. The following fact bugs me, however: the observed change in orbital periods of binary pulsars is due, we are told, to their orbital energy being carried off in the form gravitational waves. But I don't see how this can be, if the gravitational field has no energy-momentum.

This post imported from StackExchange Physics at 2014-03-12 15:45 (UCT), posted by SE-user David Gunn

Comments

" the observed change in orbital periods of binary pulsars is due, we are told, to their orbital energy being carried off in the form gravitational waves. But I don't see how this can be, if the gravitational field has no energy and momentum." It's not that is has no energy and momentum. It's that there is no way to define a local density of energy and momentum.

This post imported from StackExchange Physics at 2014-03-12 15:45 (UCT), posted by SE-user Ben Crowell

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You're missing the key word, which is "local."

This post imported from StackExchange Physics at 2014-03-12 15:45 (UCT), posted by SE-user Ben Crowell

Answer 4

It is well known in General Relativity (GR) that a tensor describing the energy-momentum of the gravitational field does not exist. However, that does not mean we cannot modify GR and find a tensor that describes gravitational energy-momentum under the premise that GR is not complete. Einstein realized (with Grossmann) in 1913 that gravity gravitates and there must be a tensor that describes that phenomenon. However, he could not produce that entity and introduced a pseudo-tensor instead in 1915. The futility and infinity of pseudo-tensors and other approaches to describe local gravitational energy led to the general statement first mentioned above. However, gravity still gravitates and we must be able to produce a tensor to describe that important fact. \par That was accomplished by extending a classic theorem of differential geometry in a Riemannian spacetime, the Berger-Ebin theorem, to a Lorentzian spacetime called the Orthogonal Decomposition Theorem (ODT) in the paper ``Modified General Relativity" (MGR) published in Gen Relativ Gravit 51, 53 (2019), and updated in the recent paper ``Modified General Relativity and quantum theory in curved spacetime" published in Int J Mod Phys A (2021). The ODT states that an arbitrary symmetric tensor in a Lorentzian spacetime $w_{\alpha\beta}$ can be decomposed into a linear sum of divergenceless tensors $v_{\alpha\beta}$ plus another tensor $ \varPhi_{\alpha\beta} $, which belongs to a subspace orthogonal to that of $v_{\alpha\beta}$: $ w_{\alpha\beta}=v_{\alpha\beta}+\varPhi_{\alpha\beta} $. By returning to Einstein's original postulate of a total e-m tensor $ T_{\alpha\beta} $, which must be divergenceless and locally conserved, the matter e-m tensor $\tilde{T}_{\alpha\beta}$ is no longer divergenceless. A constant multiple of it can be set equal to an arbitrary symmetric tensor and decomposed by the ODT to give: $k\tilde{T}_{\alpha\beta}=v_{\alpha\beta}+\varPhi_{\alpha\beta}$. Lovelock's theorem demands that in 4-D spacetime, the only divergenceless symmetric tensors consisting of a concomitant of the metric and its first two derivatives are the metric and the Einstein tensor. Thus, we arrive at Einstein's equation $ k\tilde{T}_{\alpha\beta}=\Lambda g_{\alpha\beta}+G_{\alpha\beta}+\varPhi_{\alpha\beta} $ with a new tensor that describes the energy-momentum of the gravitational field. Why is that so? $ \varPhi_{\alpha\beta} $ is constructed from the Lie derivative along the line element vector X of both the metric and a product of unit line element covectors u (collinear with X): $ \varPhi_{\alpha\beta}=\frac{1}{2}L_{X}g_{\alpha\beta}+L_{X}(u_{\alpha}u_{\beta}) $. Lie derivatives have the unique property that a tensor constructed from them has the same value when the Lie derivative is expressed with covariant or partial derivatives. Thus, when the connection coefficients (Gamma's) vanish under free-fall, $ \varPhi_{\alpha\beta} $ is invariant. Gravitational energy can be localized. \par It is easy to prove that $ \varPhi_{\alpha\beta} $ vanishes if and only if X is a Killing vector. Of course, in general there are no Killing vectors unless a symmetry is involved.\par
Line element field vectors are virtually never used in the literature except for a few theorems on the evolution of time. It is imperative to understand that a Lorentzian spacetime does not exist without a line element field (X,-X): a non-compact paracompact manifold admits a Lorentzian metric $ g_{\alpha\beta} $ if and only if it admits a line element field (Hawking and Ellis 1973).\par Thus, MGR completes GR by introducing a symmetric tensor that has the necessary properties to describe gravitational energy-momentum. The Einstein equation remains intact without introducing any new vector fields by employing the line element field, which is a fundamental part of a Lorentzian spacetime $(M,g_{\alpha\beta})$.

Answer 5

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 $T_{grav}^{\mu\nu}$, as shown in my own work [1]. As a consequence, you cannot find this tensor in general relativity!

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

This post imported from StackExchange Physics at 2014-03-12 15:45 (UCT), posted by SE-user juanrga

Comments

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 ...

This post imported from StackExchange Physics at 2014-03-12 15:45 (UCT), posted by SE-user Dilaton

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$t_G^{\mu\nu}$ 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.

This post imported from StackExchange Physics at 2014-03-12 15:45 (UCT), posted by SE-user juanrga

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From your username juanrga I'm guessing that you are Juan Ramón González Álvarez, the author of the vixra paper. Physics.SE has a policy that it's OK to cite your own work, but you should disclose the fact that it's your own work.

This post imported from StackExchange Physics at 2014-03-12 15:45 (UCT), posted by SE-user Ben Crowell

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I don't think it's correct to attribute what you call FTG to the list of names you give. The term FTG seems to originate with Baryshev, "Field Theory of Gravitation: Desire and Reality," 1999, arxiv.org/abs/gr-qc/9912003 . Baryshev has a lot of "gee whiz" claims that I don't find plausible. A search of citations to this paper shows only 17 citations, of which 8 are by Baryshev himself.

This post imported from StackExchange Physics at 2014-03-12 15:45 (UCT), posted by SE-user Ben Crowell

Answer 6

It is true that no energy-momentum tensor for the gravitational field exists, however, it is easy to understand why not and then derive a perfectly correct formulation for conservation of energy and momenta of the gravitational field.

The invariance group of special relativity is the Poincare group. Energy and momentum combine in special relativity to form a 4-vector which belongs to a representation of the poincare group. The current of this four vector is the energy-momentum stress tensor whose divergence is zero.

When going from special to general relativity it is often assumed that any tensor quantity can be replaced by a similar one with ordinary derivatives replaced by covariant derivatives, this is not always the case. In general relativity the symmetry is the diffeomorphism group, not the Poincare group or Lorentz group. 4-vectors in GR only exist locally but energy and momentum are not local quantities so they cannot form a 4-vector. Instead they should form an object from a representation of the Diffeomorphism group.

If your spacetime is topologically equivalent (diffeomorphic) to $R^4$ then you can choose any global system of 4 co-ordinates and transform those co-ordinates using Poincare transforms. These are diffeomorphisms so this means you can embed the Poincare group in the diffeomorphism group by choosing such co-ordinates. For this reason it is possible to derive an energy-momentum pseudo-tensor for the gravitational field. It is co-ordinate dependent and not a tensor, but it works.

There is a better approach conceptually that is covariant and works for any topology. This is derived by applying Noether's first theorem directly to the Einstein-Hilbert action using the symmetry generators of the diffeomorphism group which are contravariant vector fields $k^\mu$. The result is a current with a linear dependence on the field $k^\mu$ which simplifies to the Komar Superpotential using the field equations

$J^{\mu} = (k^{\mu;\nu} - k^{\nu;\mu})_{;\nu}$

Using this formulation the energy and momenta belong to the dual of the adjoint representation of the diffeomorphism group.

Edit: I will add one more important point that is often misunderstood.

The matter and radiation part of the energy-momentum-stress tensor can be derived using the formula given in the question applied to the matter+radiation part of the Lagrangian

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

If you use this expression on the full action as suggested in the reference it gives the gravitational equations of motion, which are dynamically zero. It is crucial to understand that this is not how to derive the Noether current which is correctly given by this expression (see Wikipedia for details)

$T_\mu{}^\nu = \left( \frac{\partial L}{\partial \boldsymbol\phi_{,\nu}} \right) \cdot \boldsymbol\phi_{,\mu} - L\,\delta_\mu^\nu$

Some people cunfuse these two things and think that they give the same answer for the full Lagrangian, so that the Noether current must be zero under the field equations. This is certainly not the case. When the Noether current is derived correctly it gives the Komar Superpotential using the field equations and this is not zero. If you take a co-ordinate dependent approach you can alternatively use Noether's theorem to get pseudotensor expressions which again are not equal to zero.

This post imported from StackExchange Physics at 2014-03-12 15:45 (UCT), posted by SE-user Philip Gibbs

Comments

This answer seems like a recap of your personal theories, which I don't think are correct. Even if they were correct, this answer wouldn't address the question, for the reasons you yourself give in the first paragraph. The question is about a local gravitational energy-momentum tensor, not about globally conserved measures of energy-momentum (which, contrary to what you seem to be claiming, can't be defined for a general metric).

This post imported from StackExchange Physics at 2014-03-12 15:45 (UCT), posted by SE-user Ben Crowell

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The Komar Superpotential is not my personal theory. It has been well known to serious relativitists since Komar introduced it in 1959 prola.aps.org/abstract/PR/v113/i3/p934_1 I have said that the energy-momentum tensor for gravity does not exist and have explained why it should not exist. I have not clained that there is a globally conserved energy-momentum. I have claimed that their are globally conserved energy and momenta. If you do not understand why this is you should not be so quick to downvote.

This post imported from StackExchange Physics at 2014-03-12 15:45 (UCT), posted by SE-user Philip Gibbs

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@PhilGibbs I like your answer and I know that you are a serious physicist who knows what he is talking about, so the +1 was mine ;-). Maybe I'll have to come up with an additional question, conserning the very interesting last paragraph ...

This post imported from StackExchange Physics at 2014-03-12 15:45 (UCT), posted by SE-user Dilaton

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@Dilaton, thank you. I am happy to answer further questions because this is something physicists should understand.

This post imported from StackExchange Physics at 2014-03-12 15:45 (UCT), posted by SE-user Philip Gibbs

Answer 7

Any non trivial problem of motion in GR is written in a "non linear dynamic geometry"; this non linearity makes many usual quantities from a flat space-time meaningless if not trivial in GR.