# Proper time in a curved space

In special relativity we've the invariant $$d s^2=-d t^2 +d x^2 + d y^2+d z^2$$

For a clock moving along the worldline in question 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

$$d s^2=-\left(1+\frac{2 \Phi\left(x\right)}{c^2}\right)(c d t)^2+\left(1-\frac{2 \Phi\left(x\right)}{c^2}\right)\left(d x^2+d y^2+d z^2\right)$$

In this space I want an expression for the proper time.

Suppose in one frame I measure two events $$A$$ and $$B$$. I calculate the interval $$ds$$ using $$d s^2=-\left(1+\frac{2 \Phi\left(x\right)}{c^2}\right)(c d t)^2+\left(1-\frac{2 \Phi\left(x\right)}{c^2}\right)\left(d x^2+d y^2+d z^2\right)$$

Any other frame will measure the same interval.

Now suppose we take a clock moving along the worldline of our events. In the clocks frame the spatial differentials $$dx, dy, dz$$ for the two events $$A$$ and $$B$$ are zero. Suppose in the moving clocks frame we measure time $$dt$$.

This time $$dt$$ is also the proper time $$d\tau$$

To find proper time we put $$dx, dy, dz=0$$ in $$d s^2=-\left(1+\frac{2 \Phi\left(x\right)}{c^2}\right)(c d t)^2+\left(1-\frac{2 \Phi\left(x\right)}{c^2}\right)\left(d x^2+d y^2+d z^2\right)$$ which gives me the proper time as $$d{ \tau^2}=\frac{-d s^2}{c^2\left(1+\frac{2 \phi}{c^2}\right)}$$

but the author says that the proper time is $$d \tau^2=-d s^2 / c^2$$

Where am I wrong?

• Isn't what you are defining as proper time just the coordinate time $dt$? Oct 29, 2022 at 7:36
• I've added more text to clarify about my doubt. Oct 30, 2022 at 2:26

the author says that the proper time is $$d \tau^2=-d s^2 / c^2$$

This is the correct general definition of proper time. This definition applies for all timelike worldlines (inertial or non-inertial) in all spacetimes (flat or curved) and with all coordinate systems (including ones without a time coordinate).

It automatically reduces to your other formula when you have an inertial coordinate system in flat spacetime: $$ds^2=-c^2 dt^2+dx^2+dy^2+dz^2$$ and the worldline is at rest $$dx=dy=dz=0$$ and you use units where $$c=1$$. Under those conditions it automatically and easily simplifies, but the simplified expression is not the definition in general.

• Please see the edit. Oct 30, 2022 at 2:26
• The edit doesn’t change the answer
– Dale
Oct 30, 2022 at 3:09
• Wikipedia : "proper time is defined as the time as measured by a clock following that line" In our case the clock is our frame of reference, if it measures time $d_t$ then it should be the proper time according to above definition. Oct 30, 2022 at 5:01
• Not in curved spacetime. The clocks run slower there as in flat spacetime by factor $\sqrt{g_{00}}$
– JanG
Oct 30, 2022 at 9:15
• @Kashmiri the frame of reference here is not inertial. It does not generally match proper time and is a coordinate not a clock. The $d\tau^2=-ds^2/c^2$ definition of proper time is correct and is the definition that matches Wikipedia’s definition
– Dale
Oct 30, 2022 at 11:58

Suppose in the moving clocks frame we measure time dt. This time dt is also the proper time dτ.

The last sentence is valid only in flat spacetime. In curved spacetime observer's clock runs slower as in flat one because of gravitation ($$g_{00}\ne 1$$).

In curved spacetime for $$dx=dy=dz=0$$ there is $$c^2 d\tau^2 \equiv ds^2=g_{00}~c^2 dt^2\approx (1+\frac{2\Phi}{c^2})~c^2 dt^2,\tag{1}$$ thus the proper time is $$d\tau \approx~(1+\frac{2\Phi}{c^2})~c~dt.\tag{2}$$

The difference in minus sign is due to Hartle's choice of other metric signature convention as in your reference. My answer is thought as a supplement to Photon's comment and Dale's answer to your question.