0
$\begingroup$

I'm solving the Schwarzschild geodesics numerically, and I'm wondering if the result I'm getting makes physical sense. The parameter of the path $x^{\mu}=(t,r,\theta,\phi)$ is the proper time $\tau$, and when I graph the coordinate time $t$ against the proper time, I get a really straight, linear relation, which, considering the wild oscillation in the radial coordinate (ie: 6+ oscillations between 6.5 and 15 $GM$), I have a hard time believing.

Granted, the coordinate time is always greater than the proper time, so time-dialation IS taking place.

Below I've attached the 2 corresponding plots.

Asymptotic coordinate time vs proper time

Radial coordinate vs proper time

For reference, I'm integrating the equations from the wiki page https://en.wikipedia.org/wiki/Schwarzschild_geodesics#Geodesic_equation with maple's dsolve(numeric), rk45 .

Improve this question
$\endgroup$
5
  • $\begingroup$ I'm using the equations from the wiki page en.wikipedia.org/wiki/Schwarzschild_geodesics#Geodesic_equation , and solving them via maple's dsolve(numeric), rk45. $\endgroup$
    – Johnny
    Commented Mar 28, 2021 at 19:00
  • $\begingroup$ Calculate the maximum and minimum orbital speed as a fraction of $c$, and the maximum and minimum gravitational potential as a fraction of $c^2$, to see whether the time dilation should vary much. It looks like there is a slight wobble in the straightness of the plot. $\endgroup$
    – G. Smith
    Commented Mar 28, 2021 at 19:05
  • $\begingroup$ Without the initial conditions, it’s hard to know. $\endgroup$
    – G. Smith
    Commented Mar 28, 2021 at 19:09
  • $\begingroup$ For what it's worth, my initial conditions were $x^{\mu}(0)=(0,6.5,\pi /2, 0)$ and $ v^{\mu}(0) = (1.38,0,0, 0.088) $, with $c=G=M=1$ . $\endgroup$
    – Johnny
    Commented Mar 28, 2021 at 19:15
  • $\begingroup$ The initial conditions and units belong in the question. $\endgroup$
    – G. Smith
    Commented Mar 28, 2021 at 19:20

1 Answer 1

Reset to default
2
$\begingroup$

Let’s ignore the kinematic time dilation and estimate the gravitational time dilation. The dilation factor, approximately $1+\frac{GM}{rc^2}$ at large $r$, varies between $1+\frac{1}{6.5}\approx 1.15$ to $1+\frac{1}{15}\approx 1.07$. That’s about a 4% variation around the average, and it seems consistent with the slight visible wobble around the plot’s straightness.

If you want to see a larger effect, use an eccentric orbit that gets closer to the event horizon.

Wikipedia has an exact formula for the time dilation factor, taking both kinematic time dilation and gravitational time dilation into account, and not using the large-$r$ approximation.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.