Tuesday, 10 April 2018

general relativity - Relationship between Coordinate Time and Proper Time


While I was reading Ta-Pei Cheng's book on relativity, I was unable to derive the correct relationship between coordinate time $dt$ (the book defined it as the time measured by a clock located at $r=\infty$ from the source of gravity) and proper time $d\tau$ from the definition of metric.


The book states that for a weak and static gravitational field, $g_{00}(r)=-\left(1+\frac{2\Phi(r)}{c^2}\right)$ (with the metric signature $(-1,1,1,1)$ and $\Phi(r)$ is the gravitational potential) and the proper time $d\tau=\sqrt{-g_{00}}\,dt$.


From the gravitational redshift result I know that the above result is correct (in a more unambiguous form $d\tau=\sqrt{-g_{00}(r_\tau)}\,dt$).


However, if I simply use the formula for spacetime interval $ds^2=g_{\mu\nu}dx^\mu dx^\nu$ (assuming two clocks that measure proper time and coordinate time are at rest relative to each other), I have


$$ ds^2=g_{00}(r_\tau)c^2d\tau^2=g_{00}(r_t)c^2dt^2=-c^2dt^2\\ \implies \sqrt{-g_{00}(r_\tau)}\,d\tau=dt$$ This suggests that time flows faster with a lower gravitational potential which is incorrect.


I'm not sure why the above method lead to a wrong conclusion, did I misunderstood the the definition of proper time, coordinate time or spacetime interval?




Update:




  1. One mistake I've made is letting $ds^2=g_{00}(r_\tau)c^2d\tau^2$, which should be $ds^2=-c^2d\tau^2$ by definition. However, I'm confused about two definitions of $ds^2$ now. $ds^2=-c^2d\tau^2=g_{\mu\nu}dx^\mu dx^\nu$, this suggests that $g_{00}$ is always $-1$ for the frame that measures proper time, but in my problem $g_{00}$ is a function of $r$ which is only equals to $-1$ if $r=\infty$, how could two both be true at the same time?

  2. Assuming $ds^2=-c^2d\tau^2$ is true, as all the answers pointed out that $d\tau=\sqrt{-g_{00}}\,dt$. But by the definition of $g_{00}$ and $ds^2$ the $g_{00}$ used here must be $-(1+2\Phi(r_t)/c^2)=-1$, but I want $g_{00}$ here to be $-(1+2\Phi(r_\tau)/c^2)$ so that $$d\tau=\sqrt{-g_{00}}\,dt=\sqrt{1+2\Phi(r_\tau)/c^2}\,dt\approx (1+\Phi(r_\tau)/c^2)\,dt\\ \implies \frac{d\tau-dt}{dt}=\frac{\Phi(r_\tau)}{c^2}=\frac{\Phi(r_\tau)-\Phi(r_t)}{c^2}$$


Please correct me if I've made any mistakes!




No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...