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I'm having an interpretation problem with the radial spacelike geodesics in the flat radiation dominated universe.

I'm using standard conformal coordinates $\eta \ge 0$ and $\chi \ge 0$ for the euclidian flat space metric (curvature parameter $k = 0$): $$\tag{1} ds^2 = a^2(\eta) \bigl( d\eta^2 - d\chi^2 - \chi^2 (d\vartheta^2 + \sin^2 \vartheta \, d\varphi^2) \bigr). $$ Take note that the radial coordinate $\chi$ is exactly the same as the standard comoving $r$, when $k = 0$. I'm considering the radiation universe; $a(t) \propto t^\frac{1}{2}$, or in conformal time; $a(\eta) = \alpha \, \eta$ ($\alpha$ is just an arbitrary constant). To get the radial geodesics equation, I use this lagrangian ($\dot{\vartheta} = \dot{\varphi} = 0$ for the radial curves): $$\tag{2} L = a^2(\eta) \, (\dot{\eta}^2 - \dot{\chi}^2). $$ The usual Euler-Lagrange equation give two equations, in the general case (valid for any $k = -1, 0, +1$): \begin{gather} \frac{d\, }{d\sigma} ( a \, \dot{\eta} ) = -\, \frac{d a}{d\eta} \, \dot{\chi}^2, \tag{3} \\[2ex] a^2 \, \dot{\chi} = \text{cste}. \tag{4} \end{gather} The lagrangian (2) itself gives another constant of motion, since it doesn't depend on $\sigma$. It's easy to show that $\chi = \text{cste}$ is a timelike geodesic solution to (3),(4), with $L = 1$, and its interpretation is obvious (local stationary - or comoving - observers are moving on their timelike geodesic). It's also easy to find the radial lightlike geodesics ($L = 0$): $\eta = \eta_0 \pm \chi$. Everything is nice and well, until I try to seek the spacelike radial geodesics that give $L < 0$. It is clear that $\eta = \text{cste}$ is NOT a geodesic of spacetime, since it doesn't solves the geodesics equation (3). Instead of using a parametrization $\sigma$, I may use $\chi$ as the parametrization. Inserting (4) in (3) gives the following differential equation: $$\tag{5} \frac{d\, }{d\chi} \Bigl( \frac{1}{a} \, \frac{d\eta}{d\chi} \Bigr) = -\, \frac{1}{a^2} \, \frac{d a}{d\eta}. $$ This equation is valid for $k = -1, 0, 1$ and any function $a(\eta)$. For the $k = 0$ radiation dominated universe; $a(\eta) = \alpha \, \eta$, I get this equation: $$\tag{6} \frac{d\, }{d\chi} \Bigl( \frac{1}{\eta} \, \frac{d \eta}{d\chi} \Bigr) = -\, \frac{1}{\eta^2}. $$ I'm looking for a solution that gives $L < 0$ (i.e a radial spacelike geodesic). The test function $$\tag{7} \eta(\chi) = A \cos (\omega \chi + \phi) $$ is a solution to (6) when $\omega = 1/A$, and $L < 0$ as it should. $A$ and $\phi$ are arbitrary constants. But then comes the interpretation problem:

This simple (but non-trivial) solution is vertically oscillating in the $\eta \, \chi$ plane, and cross the Big Bang singularity at $\eta = 0$, where it becomes lightlike. Since $\omega = 1/A$, the oscillation is stronger when the "amplitude" $A$ is small. The solution (7) appears to be valid only up to $\chi_{\text{max}} = \frac{\pi}{2} \, A$, when $\phi = 0$.

This spacelike solution is really weird! I have reasons to believe (from some very limited sources on the internet) that this odd spacelike geodesics behavior is generic for most FLRW models. What is happening here? How should we interpret the strange behavior of spacelike radial geodesics of the flat radiation FLRW universe? Are there other known analytic spacelike solutions?

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  • $\begingroup$ I think you might be missing a factor of $a$ in the RHS of eq. (3). $\endgroup$
    – Javier
    Jul 1, 2022 at 19:40
  • $\begingroup$ @Javier, I don't see it. I don't think there's a mistake in (3). $\endgroup$
    – Cham
    Jul 1, 2022 at 19:44
  • $\begingroup$ You're using unintuitive coordinates and variables, but the solution in comoving coordinates is at Output 11-14 in yukterez.net/f/einstein.equations/files/38.html where μ=0 for photons, μ=-1 for timelike and μ=+1 for spacelike geodesics $\endgroup$
    – Yukterez
    Jul 1, 2022 at 22:39
  • $\begingroup$ @Yukterez, this file isn't very informative, especially since it's in german! And the conformal coordinates are directly related to the comoving coordinates ($d\eta = \frac{1}{a(t)} \, dt$ and $\chi \equiv r$ for $k = 0$). $\endgroup$
    – Cham
    Jul 1, 2022 at 22:47
  • $\begingroup$ @Cham - the language doesn't matter, especially in the times of online translators and even more so if the german and english terms in question are almost the same. All you need to know to understand the link is that t'=dt/dτ, t''=d²t/dτ², tachyons have an imaginary τ and v is the local velocity relative to a comoving observer, so it's pretty much self explanatory $\endgroup$
    – Yukterez
    Jul 1, 2022 at 23:22

1 Answer 1

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There's nothing especially odd about solutions of this form given the shape of FLRW spacetimes and the nature of FLRW (or conformal) coordinates.

FLRW coordinates are essentially polar; you can think of $t$ as the distance from the pole and $χ,\vartheta,\varphi$ as the angle. Consider a sphere covered with (latitude,longitude) coordinates with arbitrarily chosen poles. The points at a constant latitude are extrinsically curved in the space (i.e. are not a great circle) except at the equator. A great circle plotted in latitude,longitude coordinates appears sinusoidal.

Other positive-definite manifolds with the appropriate symmetries to be covered by latitude, longitude coordinates include circular cones, paraboloids of revolution, the Euclidean plane, etc. The FLRW family of spacetimes is essentially the family of Lorentzian manifolds that can be covered by such coordinates. The locus of points at constant $t$ or $η$ is extrinsically curved in the spacetime (even if it's intrinsically flat), except at the "equator" where $\dot a=0$, if there is one. The direction of the extrinsic curvature is opposite in the Lorentzian and Euclidean cases, which is why your geodesic seems to curve back toward the origin when $\dot a>0$.

The geodesic doesn't actually become lightlike at the singularity, since the singularity isn't part of the manifold. Its peculiar speed goes to $c$ as $a\to 0$ for the same reason peculiar speeds go to zero as $a\to\infty$. This happens in the Euclidean case also. On a sphere, the angle between a general great circle and lines of longitude is smallest at the equator and largest (90°) where it's farthest from the equator. On a cone, the angle of any geodesic to the radial lines goes to zero as the distance from the apex goes to infinity.

I think the geodesic you found is the most general spacelike geodesic in the 1+1-dimensional version of this manifold, and with a few extra parameters the most general in 3+1 dimensions. As you noted, it attains every superlight speed, and it can be made to pass through any point with any speed by adjusting the parameters.

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  • $\begingroup$ This is interesting, and may be the answer to my query. Thanks! I'll mark it as the answer in a few some days, just in case someone else find another (simpler?) way of explaining this feature of the FLRW universes. $\endgroup$
    – Cham
    Jul 2, 2022 at 20:51
  • $\begingroup$ Also, on another - related - subject, I believe that it could be interesting to plot the spacelike geodesics on a Penrose diagram, to see the differences for various cosmological models. The Penrose diagrams are all having the same shape (a triangle), for all the models that satisfies the strong energy condition. The timelike and lightlike geodesics are all the same on these diagrams too. Yet, something in these diagrams should mark the differences across models. The spacelike geodesics could be that thing (the "inprint" of the energy distribution). $\endgroup$
    – Cham
    Jul 2, 2022 at 20:57

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