How to take the integral of the gravitational time dilation formula

The formula for gravitational time dilation is $$t' = t\sqrt{1-\frac{2GM}{rc^2}}$$. That gives us a value of $$t'$$ at some distance $$r$$. But is it possible to find the integral value over two different radii? Let's say, for instance, between $$r = 10$$ and r$$= 5$$, in whatever units are appropriate.

Example: $$t'$$ at 1 AU from the sun $$t'= 0.9999999901t$$. $$t'$$ at 0.5 AU from the sun $$t'= 0.9999999803t$$. Is there an integral value between them?

• You would have to specify $r(t)$. Commented Nov 3, 2022 at 15:51
• That equation is derived from the Schwarzschild line element, so your t & t' are really differentials; see hepweb.ucsd.edu/ph110b/110b_notes/node75.html Using $r_s=2GM/c^2$, we can rewrite that time dilation equation as $$\frac{d\tau}{dt}=\sqrt{1-\frac{r_s}{r}}$$ where $t$ is the Schwarzschild time coordinate, i.e. the time of the observer "at infinity" outside the gravity well, and $\tau$ is the proper time of a static observer at Schwarzschild radial coordinate $r$. Commented Nov 3, 2022 at 16:16

The proper time of fall from point $$r_{0}$$ to point $$r$$ is: $$\tau-\tau_{0}=\frac{1}{c}\int_{r_{0}}^{r} \frac{1}{\sqrt{\frac{R_{s}}{{r}}-\frac{R_{s}}{r_{0}}}}\;dr\;\;\;\;\;\;(*)$$ where $$R_{s}=2GM/c^{2}$$

(*): Theoretical physics volume II: field theory , L.Landu, E.Lifchitz

Using the notation: instead of t' and ct instead of t, the problem with the equation $$c\tau=ct\sqrt{1-\frac{r_s}{r}}$$ is that it only applies to objects at rest.

A generalized version of the gravitational time dilation formula that takes motion into account comes from taking the integral of the magnitude of a worldline's tangent vectors parametrized by λ. $$c\tau=\int\bigg|\bigg|\frac{d}{d\lambda}\bigg|\bigg|d\lambda$$ Which upon expanding into our time and radius components gives: $$c\tau=\int\sqrt{\left(\frac{d(ct)}{d\lambda}\frac{\partial}{\partial(ct)}+\frac{dr}{d\lambda}\frac{\partial}{\partial{r}}\right)\cdot\left(\frac{d(ct)}{d\lambda}\frac{\partial}{\partial(ct)}+\frac{dr}{d\lambda}\frac{\partial}{\partial{r}}\right)}d\lambda$$ Setting λ = ct, we can rewrite the equation so that an observer's radius changes over time: r(t), thereby generalizing the formula to multiple dimensions in Schwarzschild spacetime.

r(t) is just a shorthand for r(ct)

(+,-,-,-) metric signature used

$$c\tau=\int\sqrt{\left(\frac{d(ct)}{d(ct)}\frac{\partial}{\partial(ct)}+\frac{dr(t)}{d(ct)}\frac{\partial}{\partial{r(t)}}\right)\cdot\left(\frac{d(ct)}{d(ct)}\frac{\partial}{\partial(ct)}+\frac{dr(t)}{d(ct)}\frac{\partial}{\partial{r(t)}}\right)}cdt$$ $$=\int\sqrt{\left(\frac{d(ct)}{d(ct)}\right)^2\frac{\partial}{\partial(ct)}\cdot\frac{\partial}{\partial(ct)}+\left(\frac{dr(t)}{d(ct)}\right)^2\frac{\partial}{\partial{r(t)}}\cdot\frac{\partial}{\partial{r(t)}}}cdt$$ $$=\int\sqrt{g_{tt}+\left(\frac{dr}{d(ct)}\right)^2(-g_{rr})}cdt$$ where r(t) is linear in the case of constant velocity and quadratic in the case of acceleration.

For example: $$r(t)=\frac{1}{2}t$$ where r(t) is measured in AU and t is measured in hours, making the velocity of the observer: $$\frac{dr}{dt}=\frac{1}{2}\frac{AU}{hr}$$ Plugging this into the equation, we get: $$c\tau=\int\sqrt{g_{tt}+\frac{1}{4}(-g_{rr})}cdt=\int\sqrt{\left(1-\frac{r_s}{r}\right)-\frac{1}{4}\left(1-\frac{r_s}{r}\right)^{-1}}cdt$$ $$=\int\sqrt{\frac{r-r_s}{r}-\frac{r}{4(r-r_s)}}cdt=\sqrt{\frac{r-r_s}{r}-\frac{r}{4(r-r_s)}}\int{cdt}$$ $$=ct\sqrt{\frac{r-r_s}{r}-\frac{r}{4(r-r_s)}}$$