(Now I notice you're the same person who asked this at MathOverflow, where I've previously answered something similar -- if you didn't like the answer then, you won't like it now.)

This is really just expanding on Marek's comment:

How do you compute the stress tensor in a field theory? You vary the action with respect to the metric and see what comes out: $T_{\mu\nu} = 1/\sqrt{-g} \frac{\delta S}{\delta g^{\mu\nu}}$. This makes sense in non-gravitational theories, and the $T^{00}$ component is the energy.

What happens if you do this in a gravitational theory? The metric is dynamical, and varying the whole action with respect to it gives you the equation of motion (i.e., Einstein's equation). So $T^{00}$, defined in this way, where you vary the whole action (including the Einstein-Hilbert term), is just zero: it's the energy of the matter, $T^{00}$, plus the gravitational term, $-\frac{1}{8\pi G} G^{00}$.

This is what "canceled out by the negative energy of the gravitational field" means, but it's kind of a vacuous notion. I wouldn't waste time thinking too hard about the claims people make based on this idea. This isn't a physically useful notion of energy in a gravitational theory.

(Now I notice you're the same person who asked this at MathOverflow, where I've previously answered something similar -- if you didn't like the answer then, you won't like it now.)

This is really just expanding on Marek's comment:

How do you compute the stress tensor in a field theory? You vary the action with respect to the metric and see what comes out: $T_{\mu\nu} = 1/\sqrt{-g} \frac{\delta S}{\delta g^{\mu\nu}}$. This makes sense in non-gravitational theories, and the $T^{00}$ component is the energy.

What happens if you do this in a gravitational theory? The metric is dynamical, and varying the whole action with respect to it gives you the equation of motion (i.e., Einstein's equation). So $T^{00}$, defined in this way, where you vary the whole action (including the Einstein-Hilbert term), is just zero: it's the energy of the matter, $T^{00}$, plus the gravitational term, $-\frac{1}{8\pi G} G^{00}$.

This is what "canceled out by the negative energy of the gravitational field" means, but it's kind of a vacuous notion. I wouldn't waste time thinking too hard about the claims people make based on this idea. This isn't a physically useful notion of energy in a gravitational theory.

This is really just expanding on Marek's comment:

How do you compute the stress tensor in a field theory? You vary the action with respect to the metric and see what comes out: $T_{\mu\nu} = 1/\sqrt{-g} \frac{\delta S}{\delta g^{\mu\nu}}$. This makes sense in non-gravitational theories, and the $T^{00}$ component is the energy.

What happens if you do this in a gravitational theory? The metric is dynamical, and varying the whole action with respect to it gives you the equation of motion (i.e., Einstein's equation). So $T^{00}$, defined in this way, where you vary the whole action (including the Einstein-Hilbert term), is just zero: it's the energy of the matter, $T^{00}$, plus the gravitational term, $-\frac{1}{8\pi G} G^{00}$.

This is what "canceled out by the negative energy of the gravitational field" means, but it's kind of a vacuous notion. I wouldn't waste time thinking too hard about the claims people make based on this idea. This isn't a physically useful notion of energy in a gravitational theory.

(Now I notice you're the same person who asked this at MathOverflow, where I've previously answered something similar -- if you didn't like the answer then, you won't like it now.)

This is really just expanding on Marek's comment:

How do you compute the stress tensor in a field theory? You vary the action with respect to the metric and see what comes out: $T_{\mu\nu} = 1/\sqrt{-g} \frac{\delta S}{\delta g^{\mu\nu}}$. This makes sense in non-gravitational theories, and the $T^{00}$ component is the energy.

What happens if you do this in a gravitational theory? The metric is dynamical, and varying the whole action with respect to it gives you the equation of motion (i.e., Einstein's equation). So $T^{00}$, defined in this way, where you vary the whole action (including the Einstein-Hilbert term), is just zero: it's the energy of the matter, $T^{00}$, plus the gravitational term, $-\frac{1}{8\pi G} G^{00}$.

This is what "canceled out by the negative energy of the gravitational field" means, but it's kind of a vacuous notion. I wouldn't waste time thinking too hard about the claims people make based on this idea. This isn't a physically useful notion of energy in a gravitational theory.