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

• – Qmechanic Jan 8 '17 at 17:31
• @Qmechanic I had already studied the related question, but it is not helping me really. On top of that there are mistakes in the answer. – jac Jan 10 '17 at 19:25
• Why is $\nabla_\mu \left[\dfrac{dx^\mu}{d \tau} \dfrac{dx^\nu}{d \tau} \right]$ not equal to zero since the expression in brackets doesn't depend on $y^\mu$? I think you are doing some trick I don't understand with the "integration by parts". The only thing that depeds on $y^\mu$ is the piece with the delta function. I would have done something like $\dfrac{dx^\mu}{d \tau} \nabla_\mu \to \dfrac{d}{d\tau}$ and gone from there. I am not sure how the details work out though. – Brian Moths Jan 10 '17 at 20:04
• @NowIGet... I am not saying that $\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}$ is zero. A covariant derivative can be written as a sum of a common partial derivative and a number of terms involving christoffel symbols. I just said that the partial derivative is 0. This is because the covariant derivative is a derivative relative to the variable $y^\sigma$. – jac Jan 10 '17 at 20:14
• Ok I think what was wrong with what you did is when you did in the integration by parts, you hit $\dot{x}^\nu$ with $\dfrac{\partial}{\partial \tau}$, but you forgot to hit $\dfrac{1}{\sqrt{-g}}$. This should give you a $\Gamma$ to cancel the other extra $\Gamma$, but I am not sure. – Brian Moths Jan 10 '17 at 23:41

\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) ~-~\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 )) ~-~\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. $$\Box$$
• Clear to me now. Thanks. The identity $\Gamma^\mu_{\mu \sigma}=\frac {1}{\sqrt{-g}}\partial_\sigma \left( \sqrt{-g} \right)$ was the missing link in my reasoning. – jac Jan 15 '17 at 9:48
• $\uparrow$ Ah, Ok. – Qmechanic Jan 15 '17 at 13:58
• @Qmechanic Can you, please, explain where did the term $uv = \dot{x}^{\nu}\delta^{(4)}(y - x(\tau))$ go after integration by parts $\int udv = uv - \int vdu$? – Sergio Jun 2 at 11:16
• @Qmechanic Thank you for reply. If we try to obtain $\nabla_{\mu}T^{\mu\nu}$ for continuous distribution of dust, we get $\nabla_{\mu}T^{\mu\nu} = (\text{continuity eqn}) + (\text{geodesics eqn})$. I think the your "source term" is a kind of continuity eqn for $\delta$-like distribution. – Sergio Jun 4 at 15:09