I am reading Weinberg's Gravitation and Cosmology. In section 3.5, the author got a result$$\frac{dt}{\Delta t}=(-g_{00})^{-1/2}\tag{3.5.2}.$$Here $dt$ is the time interval of a stationary observer in the gravitational field and $\Delta t$ is the corresponding proper time in the freely falling frame. As $g_{00}=-1-2\phi$ in weak-field limit ($\phi$ is the Newtonian gravitational potential), it seems that, according to the book, $dt>\Delta t$, which means a clock runs faster in a gravitational field! But as far as I know, the correct result should be: a clock runs slower in a gravitational field. Where is the problem?
7
-
$\begingroup$ Now I'm confused. We need a proper definition of what these time intervals are. Is $dt$ (in contradiction with almost every GR text I've seen) a proper time interval for a stationary observer at some point in the gravitational field and $\Delta t$ the proper time interval for a free-falling body at the same position? $\endgroup$– ProfRobCommented Jul 8 at 16:33
-
$\begingroup$ Yes. $dt$ for stationary observer and $\Delta t$ for freely falling frame. $\endgroup$– rioiongCommented Jul 8 at 16:38
-
$\begingroup$ And Weinberg's derivation of (3.5.2) is simple and clear. $\endgroup$– rioiongCommented Jul 8 at 16:41
-
$\begingroup$ Well what do you mean by "a clock runs slower in a gravitational field" - it only runs slower compared to a stationary clock that is "higher up" - i.e. in a different place. Your clocks are in the same place. $\endgroup$– ProfRobCommented Jul 8 at 16:43
-
$\begingroup$ Imagine 2 stationary clocks, one higher up and one lower down. Compare both with the freely falling clock. Then you get $dt_1$, $dt_2$ and the ratio of them from (3.5.2). $\endgroup$– rioiongCommented Jul 8 at 16:48
|
Show 2 more comments
2 Answers
$\begingroup$
$\endgroup$
That the clock rate for a stationary clock deeper in the potential well is slower can easily be established from the Schwarzschild metric with $dr= d\theta = d\phi= 0$ for two clocks at different fixed radii.
What is being compared by Weinberg (I believe) is the clock rate for a stationary observer and that of a radially free-falling clock (from infinity) at the same position. (Or at least, that is the correct formula for such a scenario).
Indeed, this formula tells you that the ratio of the clock rates for two clocks at the same position but with one falling is simply given by $$\frac{dt}{\Delta t} = \left(1 - \frac{r_s}{r}\right)^{-1/2}$$ But the speed at which a stationary observer measures a passing free-falling object is given by $$ v^2 = c^2\left(\frac{r_s}{r}\right)\ . $$ Therefore we see that the ratio of clock rates is given by $$\frac{dt}{\Delta t} = \left(1 - \frac{v^2}{c^2}\right)^{-1/2}\ ,$$ and as expected, the ratio of clock rates is just the Lorentz factor of special relativity, since the clocks are at the same position in the graviational potential.
If you were to fix $\Delta t$ as 1 second of proper time on the falling clock then $dt$ gets larger as the falling clock passes a series of stationary clocks at decreasing radial coordinate. That is because the relative speeds between the falling and stationary clocks is increasing.
$\begingroup$
$\endgroup$
I think the confusion comes from my misunderstandings of coordinate time $dt$ and proper time $\Delta t$. Actually the reading on the stationary clock is the proper time. Therefore, to compare two stationary clocks on the Earth (clock 1 higher up, clock 2 lower down), we should use the proper time. We have $$\frac{\Delta t_1}{\Delta t_2}=(\frac{g_{00}(x_1)}{g_{00}(x_2)})^{1/2}.$$ Clock 2 runs slower than clock 1, which is correct.