- #1

Kashmir

- 468

- 74

For a clock moving along a worldline the above equation reduces to ##\begin{aligned} d s^2=&-d t^2\end{aligned}## , hence we can say that the time measured by the clock moving along the world line reads time dt such that ##\begin{aligned} d s^2=&-d t^2\end{aligned}## which is called proper time.

Then

*Hartle gravity*pg 126 while motivating curvature of space gives an example of geometry such that ##d s^2=-\left(1+\frac{2 \Phi\left(x^t\right)}{c^2}\right)(c d t)^2+\left(1-\frac{2 \Phi\left(x^i\right)}{c^2}\right)\left(d x^2+d y^2+d z^2\right)##

In this space what should the proper time be? I think that for a clock moving along a worldlines the spatial differential are zero thus the proper time should be ##\frac{d \tau^2}{c}=\frac{-d s^2}{c\left(1+\frac{2 \phi}{c^2}\right)}## but the author says that the proper time is

##d \tau^2=-d s^2 / c^2##.

Whats wrong with my thinking? Please help.