Assuming the metric convention is $$(-,+,+,+)$$, the action of a relativistic point particle with worldline $$X^\mu(s)$$ is

$$S = -m \int ds \sqrt{-g^{\mu\nu} \dot{X}_\mu \dot{X}_\nu }.$$

To make this an integral over spacetime, we can insert an integration over a delta function

$$S = \int d^4 x \left[ -m \int ds \delta^4(x - X(s) ) \sqrt{-g^{\mu\nu} \dot{X}_\mu \dot{X}_\nu } \right].$$

Now, the Hilbert stress-energy tensor is defined as

$$T_{\mu\nu} = \frac{-2}{\sqrt{-g}} \frac{\delta S}{\delta g^{\mu\nu} }$$

Which in this case becomes

$$T_{\mu\nu} = -m \int ds \frac{\delta^4(x - X(s))}{\sqrt{-g}} \frac{\dot{X}_\mu \dot{X}_\nu }{\sqrt{-g^{\mu\nu} \dot{X}_\mu \dot{X}_\nu }}.$$

This has the wrong sign, however. Choosing for the parameter $$s$$ the coordinate time $$t$$, we write $$X^\mu = (t, \vec{X}(t))$$ thus $$\dot{X}^\mu = v^\mu= (1, \vec{v})$$, with $$\vec{v}$$ the coordinate velocity. This allows one to perform the integration, leaving

$$T_{\mu\nu}(t,\vec{x}) = -m \int dt' \frac{\delta(t - t')\delta^3(\vec{x}-\vec{X}(t'))}{\sqrt{-g}} \frac{\dot{X}_\mu \dot{X}_\nu }{\sqrt{-g^{\mu\nu} \dot{X}_\mu \dot{X}_\nu }} = -m \frac{\delta^3(\vec{x}-\vec{X})}{\sqrt{-g}} \gamma v_\mu v_\nu.$$

This implies $$T_{00} \sim -\gamma m$$ is negative, which cannot be right. What went wrong?

1. A spacetime coordinate $$x^{\mu}$$ has an upper index in physics by convention. Then $$x_{\mu}$$ with a lower index is defined as a composite object $$x_{\mu}:=g_{\mu\nu}x^{\mu}$$, and hence depends implicitly on the metric tensor components $$g_{\mu\nu}$$. OP's error is to not include this implicit dependence when varying wrt. the (inverse) metric tensor.
• Oh, I see. Since I'm varying wrt $\delta g^{\mu\nu}$ I thought I was being clever by lowering the indices on the velocities and raising them on the metric under the square root, but I suppose that's not a consistent way to do it. I applied $g_{\mu\nu} \delta g^{\mu\nu} = - g^{\mu\nu} \delta g_{\mu\nu}$ and then it works out. – Kasper Dec 8 '19 at 14:44