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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?

Qmechanic
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Kasper
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1 Answers1

2

  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.

  2. Another common source of sign mistakes is the definition of the Hilbert/metric stress-energy-momentum (SEM) tensor, but OP got that part right.

Qmechanic
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