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In curved spacetime, the spacetime interval $ds^2=g_{ab}dx^a dx^b$ between two infinitesimally events $(t,x^i)$ and $(t+dt,x^i)$ happening at the same space point w.r.t an observer is given by $ds=\sqrt{g_{00}}cdt$ since $dx^1=dx^2=dx^3=0$. How is the proper time interval $d\tau$ equal to $d\tau=\sqrt{g_{00}}dt$? How does $d\tau$ become the proper time instead of $dt$? Why not $dt$ itself? $dt$ is the time measured by the clock at rest w.r.t the observer.

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There are two different $dt$s here. One, $dt_{you}$ is the time interval measured on your clock while the other $dt_{me}$ is the time interval measured on my clock.

In your coordinates, i.e. your rest frame, you are at rest at the origin and your spacetime looks locally like flat spacetime. So in your coordinates $g_{00}=1$ and the time you measure on your clock is equal to the proper time:

$$d\tau = dt_{you} \tag{1}$$

In my coordinates you are stationary (we both agree you are stationary) but in my coordinates the spacetime around you is curved so $g_{00}$ is no longer one. Then I have to write:

$$ d\tau^2 = g_{00} dt_{me}^2 \tag{2} $$

But the proper time is an invariant i.e. both you and I must agree on its value, so the $d\tau$ in equations (1) and (2) must have the same value. That means we can equate the two equations to get:

$$ dt_{you} = dt_{me} \sqrt{g_{00}} $$

And this is how we get the relative time dilation between us i.e. it relates time intervals measured by you to the same time interval as measured by me. I suspect the confusion has arisen because in your rest frame the time intervals measured by you are equal to the proper time.

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  • $\begingroup$ Yes. When you say 'you' and 'me', are you assuming that we are at rest w.r.t each other? $\endgroup$ Sep 28 '20 at 16:44
  • $\begingroup$ @mithusengupta123 yes, otherwise in your frame events that happened at the same point in space would happen at different points in space in my frame. $\endgroup$ Sep 28 '20 at 17:13
  • $\begingroup$ Thanks. "And this is how we get the relative time dilation between us" Is this gravitational time dilation effect? $\endgroup$ Sep 28 '20 at 17:36
  • $\begingroup$ @mithusengupta123 yes. If you are interested in how we use this approach to calculate time dilation due to motion have a look at my answer to Is gravitational time dilation different from other forms of time dilation? $\endgroup$ Sep 28 '20 at 17:43
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In GR, curves are described by a parameter $\tau$: $x^{\mu}(\tau)$ which traces out a path in the real world more generally called the worldline.
But, any parameter suffices to describe this curve, one can namely use another parameter $\lambda$ which can be related to the first one by $\lambda(\tau)$.
What we call proper time $\tau$ is the parameter for the wordline such that $u^2 = u^\mu u_\mu = c^2$ for a timelike geodesic (massive particle). Where the 4-velocity is defined as $u^\mu = \frac{dx^\mu}{d\tau}$.
So in your example we would have $ds^2 = g_{00} c^2 dt^2$ which means that
\begin{equation} u^2 = g_{00} u^0 u^0 = g_{00} \left( \frac{cdt}{d\tau} \right)^2 = c^2 \end{equation} Only if
\begin{align} g_{00} \left( \frac{dt}{d\tau} \right)^2 &= 1 \\ \Rightarrow \frac{d\tau}{dt} &= \sqrt{g_{00}} \\ \Rightarrow d\tau &= \sqrt{g_{00}}dt \end{align}

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