Doubt in Functional Derivative of Lagrangian

Question

Lecture XXXIII: Lagrangian formulation of GR by Christopher M. Hirata

NON-INTERACTING DUST

Consider a system with a suite of particles {A} each of mass $\mu_{A}$ following some set of trajectories $x^{\mu}(\sigma|A)$ where is a particle index and $\sigma$ is a coordinate on the world line. The action for such particles is:

$$ S = \sum_{A} \mu_{A} \int d\tau = -\sum_{A} \mu_{A} \int\sqrt{-g_{\mu\nu}\dfrac{dx^{\mu}}{d\sigma}\dfrac{dx^{\nu}}{d\sigma}}d\sigma.\tag{28}$$

$$ \delta S = \dfrac{1}{2}\sum_{A}\mu_{A}\int u^{\mu}u^{\nu}\delta g_{\mu\nu}d\tau.\tag{29}$$

Therefore the functional derivative is:

$$ \dfrac{\delta S_{matter}}{\delta g_{\mu \nu}(y^{\alpha})}=\dfrac{1}{2}\sum_{A}\mu_{A} \int u^{\mu}u^{\nu} \delta^{(4)}[y^{\alpha}-x^{\alpha}(\tau|A)]d\tau\tag{30}$$

$\delta^{(4)}$ is the $4$D delta function.

MY QUESTION

I was going through the PDF (attached on top too) on deriving Stress-Energy Tensor from Actions. I came across the equations $(29)$ and $(30)$. Can someone provide help on the intuition or the process of how $(30)$ is derived from $(29)$? I am honestly clueless here.

Above are the equations $(28)$, $(29)$ and $(30)$ from the PDF.

Answer 1

Hints:

$(28)\Rightarrow (29)$: The square root action (28) is invariant under reparametrization of the world-line parameter $\sigma\in[\sigma_i,\sigma_f]$. Formally, we can use the proper time $\tau$ as a world-line parameter. This reparametrization works on-shell for the geodesic/classical path but generically violates the boundary conditions for virtual paths, and hence destroys the stationary action principle. Nevertheless (29) can still be viewed as an on-shell equation (where we have identified the geodesic/classical path by other means).

$(29)\Rightarrow (30)$: Use $$\frac{\delta g_{\mu\nu}(x)}{\delta g_{\alpha\beta}(y)} ~=~\frac{1}{2}\left( \delta_{\mu}^{\alpha}\delta_{\nu}^{\beta} + \delta_{\nu}^{\alpha}\delta_{\mu}^{\beta}\right)\delta^4(x\!-\!y),$$ cf. e.g. this Phys.SE post.