- #1
latentcorpse
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Consider the spacetime metric
[itex]ds^2=-(1+r)dt^2+\frac{dr^2}{(1+r)} + r^2 ( d \theta^2 + \sin^2{\theta} d \phi^2)[/itex]
where [itex]\theta, \phi[/itex] are polar coordinates on the sphere and [itex]r \geq 0[/itex].
Consider an observer whose worldline is [itex]r=0[/itex]. He has two identical clocks, A and B. He keeps clock A with himself and throws clock B away which returns to him after an interval of 4 minutes according to clock B. What time interval has elapsed on clock A?
So by setting [itex]r=0[/itex] the mteric simplifies to
[itex]ds^2=-dt^2+dr^2[/itex]
Now I said that we can assume that the clock will travel on a timelike geodesic (since it is essentially a massive particle). And so using [itex]g_{ab}u^au^b=-1[/itex] for timelike geodesics we get
[itex]-1= \left( \frac{dt}{d \tau} \right)^2 + \left( \frac{dr}{d \tau} \right)^2[/itex].
Now I'm stuck. We know A is measuring proper time I think and so I imagine we want to solve this equation for [itex]\frac{dt}{d \tau}[/itex] and then use that to get an equation for t in terms of tau and then solve for tau when t is equal to 4. Am I right?
Also, is [itex]g_{ab}u^au^b=-1[/itex] true for any timelike curve or just for timelike geodesics, and if so, why?
Thanks a lot.
[itex]ds^2=-(1+r)dt^2+\frac{dr^2}{(1+r)} + r^2 ( d \theta^2 + \sin^2{\theta} d \phi^2)[/itex]
where [itex]\theta, \phi[/itex] are polar coordinates on the sphere and [itex]r \geq 0[/itex].
Consider an observer whose worldline is [itex]r=0[/itex]. He has two identical clocks, A and B. He keeps clock A with himself and throws clock B away which returns to him after an interval of 4 minutes according to clock B. What time interval has elapsed on clock A?
So by setting [itex]r=0[/itex] the mteric simplifies to
[itex]ds^2=-dt^2+dr^2[/itex]
Now I said that we can assume that the clock will travel on a timelike geodesic (since it is essentially a massive particle). And so using [itex]g_{ab}u^au^b=-1[/itex] for timelike geodesics we get
[itex]-1= \left( \frac{dt}{d \tau} \right)^2 + \left( \frac{dr}{d \tau} \right)^2[/itex].
Now I'm stuck. We know A is measuring proper time I think and so I imagine we want to solve this equation for [itex]\frac{dt}{d \tau}[/itex] and then use that to get an equation for t in terms of tau and then solve for tau when t is equal to 4. Am I right?
Also, is [itex]g_{ab}u^au^b=-1[/itex] true for any timelike curve or just for timelike geodesics, and if so, why?
Thanks a lot.