# [Physics] Equivalence of the geodesic equation and the continuity equation for the energy-momentum tensor

conservation-lawsgeneral-relativitygeodesicshomework-and-exercisesstress-energy-momentum-tensor

I am stuck with an exercise in Sean Carroll's Spacetime and Geometry (Chapter 4, Exercise 3).
The goal is to show that the continuity of the energy-momentum tensor, i.e. $$\nabla_\mu T^{\mu\nu}=0\tag{1}$$
is equivalent to the geodesic equation in the case of a free particle.
The energy-momentum tensor of a free particle with mass $$m$$ moving along its worldline $$x^\mu (\tau )$$ is
$$T^{\mu\nu}(y^\sigma)=m\int d \tau \frac{\delta^{(4) }(y^\sigma-x^\sigma(\tau ))}{\sqrt{-g}}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}.\tag{2}$$
Taking the covariant derivative of this tensor gives
\begin{align} \nabla_\mu T^{\mu\nu}=&m\int d \tau \nabla_\mu\left[ \frac{\delta^{(4)}(y^\sigma-x^\sigma(\tau ))}{\sqrt{-g}}\right]\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}\cr &+m\int d \tau \frac{\delta^{(4)}(y^\sigma-x^\sigma(\tau ))}{\sqrt{-g}}\nabla_\mu\left[\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}\right].\tag{3} \end{align}
The first covariant derivative of the right-hand side of the above equation reduces to an ordinary partial derivative, as the argument is a scalar. This allows us to apply partial integration to this term. The second covariant derivative has an argument that is not explicitly dependent on $$y^\sigma$$, so the covariant derivative can be written as a multiplication of this tensor with the appropriate Christoffel symbols. This finally leads us to
\begin{align} &-m\int d \tau \frac{\delta^{(4)}(y^\sigma-x^\sigma(\tau ))}{\sqrt{-g}}\frac{d^2x^\nu}{d\tau^2} \cr &+ m\int d \tau \frac{\delta^{(4)}(y^\sigma-x^\sigma(\tau ))}{\sqrt{-g}}\left[ \Gamma^\mu_{\mu\sigma}\frac{dx^\sigma}{d\tau}\frac{dx^\nu}{d\tau} + \Gamma^\nu_{\mu\sigma}\frac{dx^\mu}{d\tau}\frac{dx^\sigma}{d\tau} \right].\tag{4} \end{align}
The continuity equation requires
$$-\frac{d^2x^\nu}{d\tau^2} + \Gamma^\mu_{\mu\sigma}\frac{dx^\sigma}{d\tau}\frac{dx^\nu}{d\tau} + \Gamma^\nu_{\mu\sigma}\frac{dx^\mu}{d\tau}\frac{dx^\sigma}{d\tau}=0.\tag{5}$$
This is the geodesic equation with an extra term, i.e. the term in the middle and with an incorrect sign for the first term.
Can I get rid of this term in the middle by changing the parameter $$\tau$$ of the worldline? What about the incorrect sign? What did I do wrong?

\begin{align} \nabla^{(y)}_{\mu} T^{\mu\nu}(y) ~=~& \partial^{(y)}_{\mu} T^{\mu\nu}(y) ~+~\Gamma^{\mu}_{\mu\lambda}(y) T^{\lambda\nu}(y) ~+~\Gamma^{\nu}_{\mu\lambda}(y) T^{\mu\lambda}(y) \cr ~=~& \frac{1}{\sqrt{-g(y)}}\partial^{(y)}_{\mu} \left(\sqrt{-g(y)}T^{\mu\nu}(y)\right) +\Gamma^{\nu}_{\mu\lambda}(y) T^{\mu\lambda}(y) \cr ~\stackrel{(2)}{=}~&\frac{m}{\sqrt{-g(y)}} \int_{\tau_i}^{\tau_f} \!\mathrm{d}\tau ~\dot{x}^{\nu}\dot{x}^{\mu}\partial^{(y)}_{\mu}\delta^4(y\!-\!x(\tau )) ~+~\Gamma^{\nu}_{\mu\lambda}(y) T^{\mu\lambda}(y) \cr ~=~&-\frac{m}{\sqrt{-g(y)}} \int_{\tau_i}^{\tau_f} \!\mathrm{d}\tau ~\dot{x}^{\nu}\dot{x}^{\mu}\partial^{(x)}_{\mu}\delta^4(y\!-\!x(\tau )) ~+~\Gamma^{\nu}_{\mu\lambda}(y) T^{\mu\lambda}(y) \cr ~=~&-\frac{m}{\sqrt{-g(y)}} \int_{\tau_i}^{\tau_f} \!\mathrm{d}\tau ~\dot{x}^{\nu} \frac{d}{d\tau}\delta^4(y\!-\!x(\tau )) ~+~\Gamma^{\nu}_{\mu\lambda}(y) T^{\mu\lambda}(y) \cr \stackrel{\text{int. by parts}}{=}&~\frac{m}{\sqrt{-g(y)}} \int_{\tau_i}^{\tau_f} \!\mathrm{d}\tau ~\ddot{x}^{\nu} \delta^4(y\!-\!x(\tau)) ~+~\Gamma^{\nu}_{\mu\lambda}(y) T^{\mu\lambda}(y)\cr &~-~\frac{m}{\sqrt{-g(y)}}\left[\dot{x}^{\nu}\delta^4(y\!-\!x(\tau))\right]_{\tau=\tau_i}^{\tau=\tau_f} \cr ~~~~~~\stackrel{(2)}{=}~&\frac{m}{\sqrt{-g(y)}} \int_{\tau_i}^{\tau_f} \!\mathrm{d}\tau\underbrace{\left\{\ddot{x}^{\nu}+ \Gamma^{\nu}_{\mu\lambda}(x(\tau))\dot{x}^{\mu}\dot{x}^{\lambda} \right\}}_{\text{geodesic eq.}}\delta^4(y\!-\!x(\tau ))\cr &~-~\frac{m}{\sqrt{-g(y)}}\left[\dot{x}^{\nu}\delta^4(y\!-\!x(\tau))\right]_{\tau=\tau_i}^{\tau=\tau_f}\cr \stackrel{\text{geodesic eq.}}{=}&~~-~\frac{m}{\sqrt{-g(y)}}\underbrace{\left[\dot{x}^{\nu}\delta^4(y\!-\!x(\tau))\right]_{\tau=\tau_i}^{\tau=\tau_f}}_{\text{source terms}}. \end{align} The source terms naturally break the continuity equation (1) because they correspond to the creation & annihilation of energy-momentum of a particle. Away from creation & annihilation source terms, the continuity equation (1) should be satisfied, which then enforces the geodesic equation. $$\Box$$