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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?

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  • $\begingroup$ Related: physics.stackexchange.com/q/39526/2451 $\endgroup$
    – Qmechanic
    Jan 8 '17 at 17:31
  • $\begingroup$ @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. $\endgroup$
    – jac
    Jan 10 '17 at 19:25
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    $\begingroup$ 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. $\endgroup$ Jan 10 '17 at 20:04
  • $\begingroup$ @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 $. $\endgroup$
    – jac
    Jan 10 '17 at 20:14
  • $\begingroup$ 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. $\endgroup$ Jan 10 '17 at 23:41
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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. $\Box$

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  • $\begingroup$ 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. $\endgroup$
    – jac
    Jan 15 '17 at 9:48
  • $\begingroup$ $\uparrow$ Ah, Ok. $\endgroup$
    – Qmechanic
    Jan 15 '17 at 13:58
  • $\begingroup$ @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. $\endgroup$
    – Sergio
    Jun 4 '19 at 15:09
  • $\begingroup$ @Qmechanic 1. In the third line/equality did you just cancel out the $\sqrt{-g(y)}$ with the $\sqrt{-g}$ inside the $T^{\mu\nu}(y)$ integral? 2. Isn't the $\sqrt{-g}$ inside the integral dependent on $x$? $\endgroup$
    – Nothingham
    Aug 16 '20 at 19:37
  • $\begingroup$ 1. Yes. 2. Yes, but there is also a delta function. $\endgroup$
    – Qmechanic
    Aug 16 '20 at 19:59
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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.

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