Newtonian gravitational potential for one mass gives oscillating motions to test particles.

Can a somewhat similar movement be produced by ($2$-dimensional $+-$) curvature? I mean an explicit metric and its geodesics.

  • $\begingroup$ Wouldn't any of the Schwarzschild metrics that are singularity free in 1+1 dimension do the job here? $\endgroup$ – Slereah Jun 23 '18 at 9:22

Take the Schwarzschild equivalent in $1+1$ dimensions, that would be

$$ds^2 = -f(r) dt^2 + \frac{1}{f(r)} dr^2$$

We'll also require that $f(r)$ be asymptotically flat (just for being a "realistic" potential) and that it is non-singular (so that the particle can't crash in the singularity). The gravity should also be attractive here, so we should require the geometric equivalent of the strong energy condition, $R_{ab} X^a X^b \geq 0$ (I say the geometric equivalent because in 2D as usual gravity isn't dynamic with the matter content).

The inverse metric, as it is diagonal, is fairly simply $g^{rr} = f$, $g^{tt} = f^{-1}$

The connection is

\begin{eqnarray} {\Gamma^t_{tt}} &=& {\Gamma^t}_{rr} = {\Gamma^r}_{tr} = {\Gamma^r}_{rt} = {\Gamma^t}_{rt} = {\Gamma^t}_{tr} = 0\\ {\Gamma^r_{rr}} &=& - \frac{1}{2} \frac{f'}{f} \\ {\Gamma^r_{tt}} &=& -\frac{1}{2} f f'\\ \end{eqnarray}

Fortunately, in $1+1$ dimensions, the Riemann tensor only has one independant component. That component would be, up to interchange

\begin{eqnarray} {R^r}_{trt} &=& {\Gamma^r}_{tt,r} - {\Gamma^r}_{tr,t} + {\Gamma^a}_{tt} {\Gamma^r}_{ar} - {\Gamma^a}_{tr} {\Gamma^r}_{at} \\ &=& -(\frac{1}{2} ff'' + \frac{1}{4} (f')^2) \end{eqnarray}

The Ricci tensor is $R_{ab} = {R^c}_{acb}$, so that

\begin{eqnarray} R_{tt} &=& {R^r}_{trt} = -(\frac{1}{2} ff'' + \frac{1}{4} (f')^2) \\ R_{rr} &=& {R^t}_{rtr} = -(\frac{1}{2} ff'' + \frac{1}{4} (f')^2) \\ R_{tr} &=& R_{rt} = 0 \end{eqnarray}

A good function to simulate some gravity would simply be a function around $r = 0$ decreasing with distance. Let's pick for instance

$$f(r) = 1 + \frac{1}{r^2 + 1}$$

As it is $$f(r) \underset{r \to \infty}{\to} 1$$

and it is bounded. We have

\begin{eqnarray} f'(r) &=& -\frac{2r}{(r^2 + 1)^2} \\ f''(r) &=& -2 \frac{r^4 + 1}{(r^2 + 1)^3} \end{eqnarray}

Then the geodesic equation :

\begin{eqnarray} \ddot{r} -\frac{1}{2} (\frac{f'}{f}\dot{r}^2 + ff' \dot{t}\dot{t}) &=& 0\\ \ddot{t} &=& 0 \end{eqnarray}

We have fairly easily $t = K \tau + \tau_0$, which we'll fix as simply $t = \tau$. The equation to solve is then

\begin{eqnarray} \ddot{r}[(r^4 + 3r^2 + 2) (1 + r^2)^2 + 2] + \dot{r}^2 r (r^2 + 1)^2 + r(r^4 + 3r^2 + 2) + r^3 + 2r &=& 0 \end{eqnarray}

or something of that form. I'll admit, I don't know what the analytical solution may be. Here's an attempt from Wolfram Alpha to solve it numerically :

enter image description here

which seems periodic to me. It might be interesting to try some Fourier methods here, though as it is nonlinear I'm not sure it will be quite nice.

you can find other similar periodic numerical solutions to the geodesic equation by using other localized functions such as $e^{-r^2}$ or $\operatorname{sech}(r)$, although none of them seem particularly solvable due to their very nonlinear nature.


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