Maybe this a dumb question, but, is the gravitational dilatation of time caused because a particle travelling through a geodesic in a curved space-time must cover a larger distance than the one travelling through empty space?


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The answer is positive if you replace "geodesic" with integral curve of the Killing vector field.

The phenomenon you are considering is usually discussed as follows. There is a timelibe Killing vector field $K$ in the region of spacetime one cosiders.

Physically speaking it means that the curves tangent to $K$ are the world lines of observers at rest in the gravitational field. In other words, according to them, when the Killing time flows (the Killing time is the parameter of the integral lines of $K$), the metric around them (= the gravitational field) does not change.

We can always arrange local coordinates $t,x^1,x^2,x^3$ in the studied region, where $K= \frac{\partial}{\partial t}$ and the remaining three coordinates are "spatial coordinates". In these coordinates $g_{ab}= g_{ab}(x^1,x^2,x^3)$.

You see that there is no dependence on $t$ as previously said.

The integral lines of $K$, in the considered coordinate patch are simply described by $$t \mapsto (t, x^1,x^2,x^3)\:,$$ for some constants $x^1,x^2,x^3$ depending on the curve.

Let us now focus on a pair of integral lines of $K$, denoted by $\gamma$ and $\gamma'$ and respectively referred to stationary positions $x_0^1,x_0^2,x_0^3$ and $x_0'^1,x_0'^2, x_0'^3$.

We want to compare the proper time measured along these two curves in a common interval of Killing time $[0,T] \ni t$.

$$\Delta \tau = \int_0^T \sqrt{-g_{ab} \frac{dx^a}{dt}\frac{dx^b}{dt}} dt = \int_0^T \sqrt{-g_{00}(x_0^1,x_0^2,x_0^3)} dt = T \sqrt{-g_{00}(x_0^1,x_0^2,x_0^3)}$$

$$\Delta \tau' = T \sqrt{-g_{00}(x_0'^1,x_0'^2,x_0'^3)}$$

As a consequence $$\frac{\Delta \tau' }{\Delta \tau} = \sqrt{\frac{g_{00}(x_0'^1,x_0'^2,x_0'^3)}{g_{00}(x_0^1,x_0^2,x_0^3)}}\:.$$

In practice, the phenomenon of gravitational dilatation of time is, in fact, due to a different distance measured in terms of proper time along two stationary curves with the same "stationary- time" length $T$.

This phenomenon can physically experienced. You can consider a pair of twins initially both staying at $(x_0^1, x_0^2,x_0^3)$. Next one of them moves to the place $(x_0'^1, x_0'^2,x_0'^3)$. Finally he/she comes back to $(x_0^1, x_0^2,x_0^3)$.

Assuming that the (proper) time spent along $\gamma$ and $\gamma'$ is very larger than the (proper) time spent during the two trips connecting the two stationary positions, the twins will turn out to have different ages on their second meeting, according to the formula above.

If $(x_0'^1, x_0'^2,x_0'^3)$ is close to a big stationary mass, a star, and $(x_0^1, x_0^2,x_0^3)$ is very far form every mass in the universe, so that the metric there is flat and $g_{00}=-1$, we have

$$\frac{\Delta \tau' }{\Delta \tau} = \sqrt{-g_{00}(x_0'^1,x_0'^2,x_0'^3)}\:.$$ In relatively small gravitationa fields, the form of $g_{00}$ is related to the classical gravitational potential $\varphi$: $$g_{00}(\vec{x})= -1 + \frac{2}{c^2}\varphi(\vec{x})\:.$$

Therefore, in that regime, the gravitational dilatation of time formula takes the semiclassical form $$\Delta \tau' = \Delta \tau \sqrt{\frac{1- \frac{2}{c^2}\varphi(x_0'^1,x_0'^2,x_0'^3)}{1- \frac{2}{c^2}\varphi(x_0^1,x_0^2,x_0^3)}}\:.$$


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