1
$\begingroup$

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.

$\endgroup$
0
0
$\begingroup$

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

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.

$\endgroup$
1
  • $\begingroup$ 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 '20 at 17:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.