# Proving worldline is geodesic

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.

$$u^\mu\nabla_\mu u^\nu=u^t\nabla_t u^\nu=u^t(\partial_t u^\nu+\Gamma^\nu_{\mu t}u^\mu)=u^t(\partial_t u^\nu+\Gamma^\nu_{t t}u^t)=u^t\partial_t u^\nu$$
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.
• 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. – martin Apr 24 '20 at 17:00