Equivalence of the geodesic equation and the continuity equation for the energy-momentum tensor

Question

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. \begin{equation} \nabla_\mu T^{\mu\nu}=0\tag{1} \end{equation} 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 \begin{equation} 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} \end{equation} 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 \begin{equation} -\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} \end{equation} 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?

Answer 1

Just be careful with what quantity depends on what argument, cf. above comment by user NowIGetToLearnWhatAHeadIs. Then it works like a charm:

$$\begin{align} \nabla^{(y)}_{\mu} T^{\mu\nu}(y) ~=~& \partial^{(y)}_{\mu} T^{\mu\nu}(y) ~+~\Gamma^{\mu}_{\mu\lambda}(y) T^{\lambda\nu}(y) ~+~\Gamma^{\nu}_{\mu\lambda}(y) T^{\mu\lambda}(y) \cr ~=~& \frac{1}{\sqrt{-g(y)}}\partial^{(y)}_{\mu} \left(\sqrt{-g(y)}T^{\mu\nu}(y)\right) +\Gamma^{\nu}_{\mu\lambda}(y) T^{\mu\lambda}(y) \cr ~\stackrel{(2)}{=}~&\frac{m}{\sqrt{-g(y)}} \int_{\tau_i}^{\tau_f} \!\mathrm{d}\tau ~\dot{x}^{\nu}\dot{x}^{\mu}\partial^{(y)}_{\mu}\delta^4(y\!-\!x(\tau )) ~+~\Gamma^{\nu}_{\mu\lambda}(y) T^{\mu\lambda}(y) \cr ~=~&-\frac{m}{\sqrt{-g(y)}} \int_{\tau_i}^{\tau_f} \!\mathrm{d}\tau ~\dot{x}^{\nu}\dot{x}^{\mu}\partial^{(x)}_{\mu}\delta^4(y\!-\!x(\tau )) ~+~\Gamma^{\nu}_{\mu\lambda}(y) T^{\mu\lambda}(y) \cr ~=~&-\frac{m}{\sqrt{-g(y)}} \int_{\tau_i}^{\tau_f} \!\mathrm{d}\tau ~\dot{x}^{\nu} \frac{d}{d\tau}\delta^4(y\!-\!x(\tau )) ~+~\Gamma^{\nu}_{\mu\lambda}(y) T^{\mu\lambda}(y) \cr \stackrel{\text{int. by parts}}{=}&~\frac{m}{\sqrt{-g(y)}} \int_{\tau_i}^{\tau_f} \!\mathrm{d}\tau ~\ddot{x}^{\nu} \delta^4(y\!-\!x(\tau)) ~+~\Gamma^{\nu}_{\mu\lambda}(y) T^{\mu\lambda}(y)\cr &~-~\frac{m}{\sqrt{-g(y)}}\left[\dot{x}^{\nu}\delta^4(y\!-\!x(\tau))\right]_{\tau=\tau_i}^{\tau=\tau_f} \cr ~~~~~~\stackrel{(2)}{=}~&\frac{m}{\sqrt{-g(y)}} \int_{\tau_i}^{\tau_f} \!\mathrm{d}\tau\underbrace{\left\{\ddot{x}^{\nu}+ \Gamma^{\nu}_{\mu\lambda}(x(\tau))\dot{x}^{\mu}\dot{x}^{\lambda} \right\}}_{\text{geodesic eq.}}\delta^4(y\!-\!x(\tau ))\cr &~-~\frac{m}{\sqrt{-g(y)}}\left[\dot{x}^{\nu}\delta^4(y\!-\!x(\tau))\right]_{\tau=\tau_i}^{\tau=\tau_f}\cr \stackrel{\text{geodesic eq.}}{=}&~~-~\frac{m}{\sqrt{-g(y)}}\underbrace{\left[\dot{x}^{\nu}\delta^4(y\!-\!x(\tau))\right]_{\tau=\tau_i}^{\tau=\tau_f}}_{\text{source terms}}. \end{align}$$ 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. $\Box$

Answer 2

There is much easier route to the geodesic equation: Consider a cloud of non-interacting dust particles with proper mass density $\rho_0$ and common four velocity $u^\mu$. They have $$ T^{\mu\nu}= \rho_0 u^\mu u^\nu. $$ Energy-momentum conservation says that
$$ 0 = \nabla_\mu T^{\mu\nu} = \rho_0 u^\mu\nabla_\mu u^\nu + u^\nu \nabla_\mu(\rho_0 u^\mu). $$ The second term is zero by particle conservation, and the first one is the geodesic equation.

Carroll's argument is equivalent to this one, but complicated by his need to introduce delta functions in order to isolate a single particle.

Aah -- just saw Sergio's comment.