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In order to derive the geodesic equation of motion from the covariant conservation of the energy-momentum tensor we have to do the following procedure:

$$ T^{\mu\nu}_{\space\space\space\space;\mu}= \partial_\mu T^{\mu\nu}+\Gamma^{\mu}_{\space\space\sigma\mu}T^{\sigma\nu}+\Gamma^{\nu}_{\space\space\sigma\mu}T^{\mu\sigma} $$

$$ T^{\mu\nu}_{\space\space\space\space;\mu}= \frac{1}{\sqrt{-g}} \partial_\mu (\sqrt{-g} T^{\mu\nu})+ \Gamma^{\nu}_{\space\space\sigma\mu}T^{\mu\sigma}.$$

The first term is zero since the energy momentum tensor is divergenceless. Using the definition for the energy-momentum tensor:

$$ T^{\mu\nu}(x)= \frac{m}{\sqrt{-g}} \int u^{\mu}u^{\nu} \delta^4(x-z(\tau))d\tau $$

where: $ u^{\mu} = \frac{dz^\mu(\tau)}{d\tau}$.

  1. When taking the derivative $ \partial_\mu (\sqrt{-g} T^{\mu\nu})$ it doesn't act in the $u^{\mu}u^{\nu}$ terms, only on the delta function and I dont understand why.

  2. Another thing is when we integrate by parts there is a boundary term of the form: $$ -\frac{m}{\sqrt{-g}} u^{\nu} \delta^4 (x-z(\tau)) \ \Big|_a^b $$ where it is being evaluated at the proper time limits that I labeled as $a$ and $b$. The other two terms combine in order to give the geodesic equation. Why do we ignore the boundary term?

Qmechanic
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ContravariantMind
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1 Answers1

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  1. The particle velocity $u^{\nu}(\tau)=\dot{z}^{\nu}(\tau)$ is independent of the spacetime coordinate coordinate $x^{\mu}$.

  2. The boundary terms mentioned by OP signify the creation and annihilation of a particle, and do indeed modify the continuity equation with source terms, cf. e.g. my Phys.SE answer here.

Qmechanic
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