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I am using coordinates $\{t,x,y,z\}$ and a metric $$ds^2=-dt^2+f(t,z)dx^2+f(t,z)dy^2+dz^2$$in which $$\Gamma^\mu_{tt}=0\quad\text{for all }\quad\mu=t,x,y,z.$$

I am then asked to show that a worldline along which $x$, $y$, and $z$ are all constant is a timelike geodesic.

I start with the worldline $x^\mu=(t(\tau),X,Y,Z)$, where $\tau$ is proper time and capital letters stand for constant quantities. The four-velocity is then $u^\mu=\dot{x}^\mu=(\dot{t}(\tau),0,0,0)$ where $\dot{}$ stands for derivative with respect to proper time. I then consider $$u^\mu\nabla_\mu u^\nu$$ My reasoning is that, if I can show that the expression above vanishes, then I've proven $x^\mu$ is a geodesic. I rewrite the above as $$u^\mu(\partial_\mu u^\nu+\Gamma^\nu_{\mu\alpha}u^\alpha)=u^\mu(\partial_\mu u^\nu+\Gamma^\nu_{\mu t}u^t)$$ where I carried out the sum over $\alpha$ by replacing it by $t$, since that is the only non-zero component of $u^\alpha$. Summing over $\mu$, $$u^t(\partial_tu^\nu+\underbrace{\Gamma^\nu_{tt}}_{=0}u^t)+\underbrace{u^i}_{=0}(\partial_iu^\nu+\Gamma^\nu_{ti}u^t)=u^t\partial_tu^\nu.$$ The only non-trivial component is $\nu=t$, which gives me $$u^t\partial_tu^t=\frac{dt}{d\tau}\frac{\partial}{\partial t}\frac{dt}{d\tau}=\frac{d^2t}{d\tau^2}=?$$

This does not seem zero to me. Where did I go wrong? Thanks.

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2 Answers 2

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You have to remember that the RHS of the geodesic equation is zero if and only if you choose affine parameterization. In this case $d\tau^2 = - ds^2 = dt^2$ (because $x,y,z$ are constant). Therefore $t=a\tau+b$ and therefore $\frac{d^2t}{d\tau^2} = 0$.

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$$u^\mu\nabla_\mu u^\nu=u^t\nabla_t u^\nu=u^t(\partial_t u^\nu+\Gamma^\nu_{\mu t}u^\mu)=u^t(\partial_t u^\nu+\Gamma^\nu_{t t}u^t)=u^t\partial_t u^\nu$$

Use the identity $$u^t u_t=-1$$

The derivative of 4-velocity $u^\mu$ wrt proper time $\tau$, is the 4-acceleration $w^\mu$ $$u_t\partial_\tau u^t +u^t \partial_\tau u_t=u_tw^t +u^t w_t=2u^t w_t=0$$

which yields $$u^t \partial_\tau u_t=u^t w_t=0$$ $4$-vector of velocity is orthogonal to $4$-vector of acceleration.

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  • $\begingroup$ I'm not quite sure I get your proof. 4-acceleration is $a^\mu=\dot{x}^\nu\nabla_\mu u^{\nu}$ which in this case is $a^\mu=u^t\partial_tu^\mu$. But the derivative here is with respect to coordinate time, not proper time, so I do not understand your argument above. $\endgroup$
    – martin
    Apr 24, 2020 at 17:00

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