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?

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.

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.

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.

Answer 4

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.