1
$\begingroup$

The Schwarzschild solution shows decreasing velocities with larger orbits - and needs help from dark matter or MOND to explain galaxies. Apparantly no Einstein vacuum solution with flat velocities is known but there seems to be also no proof that there cannot be one. Regardless of what other oddities or disadvantages such a solution may have, it is a question of principle: Does it exist?

In case a solution exists we will see two oddities:

  1. The flat solution which explains galactic rotation but fails in the solar system.
  2. The Schwarzschild solution which explains the solar system but fails for galactic rotation.
$\endgroup$

2 Answers 2

8
$\begingroup$

According to Birkhoff's theorem, Schwarzschild is the unique spherically symmetric vacuum solution. So there is no alternative solution that could give you different orbits.


The original question did not demand spherical symmetry. Without that requirement, you can obtain circular orbit velocities that are "flat" (independent of radius) in a vacuum in Newtonian gravity by considering orbits around an infinite line (or cylinder) of mass. Thus, the same is true in general relativity in the weak-field limit, since that reduces to Newtonian gravity.

As long as the linear mass density of the central mass is much smaller than $c^2/G$, Levi-Civita (1919) showed that the relativistic correction is indeed small.

$\endgroup$
6
  • $\begingroup$ I am sorry, I had solutions with spherical symmetry in mind. My post was not clear enough, I thus edited my question accordingly. And I added possible oddities. $\endgroup$
    – Tantal181
    Commented Mar 13 at 6:47
  • 1
    $\begingroup$ @Tantal181 According to Birkhoff's theorem Schwarzschild is the unique vacuum spherically symmetric solution $\endgroup$
    – Sten
    Commented Mar 13 at 6:53
  • $\begingroup$ Birkhoff requires asymptotic Minkowski and concludes it can only be Schwarzschild. Asymptotic Minkowski is questionable in case of flat rotation everywhere, out to infinity. So Birkhoff would not be applicable. $\endgroup$
    – Tantal181
    Commented Mar 13 at 7:46
  • $\begingroup$ No, it implies asymptotic flatness. $\endgroup$
    – Sten
    Commented Mar 13 at 8:30
  • $\begingroup$ Infinite cylinder of mass cannot have a weak field limit since it would not be asymptotically flat and could not be reduced to Newtonian gravity. $\endgroup$
    – A.V.S.
    Commented Mar 13 at 14:22
0
$\begingroup$

There is a spherically symmetric Einstein vacuum solution for flat circular velocities.

Its line element is the general spherically symmetric line element [1][2] $$\mathrm{d}s^2=\left(\frac{r_S}{C}-1 \right)c^2\mathrm{d}t^2 + \frac{{C'}^2 \mathrm{d}r^2}{1-r_S/C} + C^2\left(\mathrm{d}\theta^2+\sin ^2\theta \mathrm{d}\phi^2\right)$$ where $r_S$ is the Schwarzschild radius and $C(r)$ is the free function.

When $C(r)$ is defined by $$C(r)=\left(\frac{c^2 r^2r_S}{2{V_{\text{flat}}}^2}\right)^\frac{1}{3}$$ where $V_{\text{flat}}$ is a constant, then the resulting geodesic equations for circular orbits are solvable only when they all have $V_{\text{flat}}$ as velocity.

It shows another oddity: It is an example of a coordinate transformation which results in a different physical law. (Note that $C(r)=r$ is Schwarzschild)

Not satisfied with the mismatches? Then design your own $C(r)$: Gradually changing from solar system solution to galactic rotation solution to Minkowski at infinity. $C(r)$ is a free function.

How would the universe look like with it? Conjecture: Since $C(r)=r$ is a stretched version of $C(r)\sim r^{2/3}$, this may be a hint that our present view of the universe (and its expansion) is the view of a local observer.

References

[1] Kirk T. McDonald, Gravitational Acceleration of a Moving Object at the Earth’s Surface (Apr 4, 2016), http://kirkmcd.princeton.edu/examples/gravity_moving.pdf

[2] Leonard S. Abrams, Alternative Space-Time for the Point Mass, arXiv:gr-qc/0201044

$\endgroup$

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.