Solve General Geodesics in FLRW Metric w/ Conformal Coordinates

[Onyx]

TL;DR Summary

Solving for General Geodesics in FLRW Metric

Once having converted the FLRW metric from comoving coordinates $ds^2=-dt^2+a^2(t)(dr^2+r^2d\phi^2)$ to "conformal" coordinates $ds^2=a^2(n)(-dn^2+dr^2+r^2d\phi^2)$, is there a way to facilitate solving for general geodesics that would otherwise be difficult, such as cases with motion in both $r$ and $\phi$? I'm curious because there seems to be an $n$ killing vector evident in the new form, but I don't know if that makes it any easier.

 

[PeterDonis]

there seems to be an killing vector evident in the new form

No, there isn't, because none of the metric coefficients are independent of $n$ (which is actually the Greek letter eta, $\eta$, in most treatments in the literature).

 

[Onyx]

No, there isn't, because none of the metric coefficients are independent of $n$ (which is actually the Greek letter eta, $\eta$, in most treatments in the literature).

It appeared to me that there was, because of $\frac{\partial K_\eta}{\partial \eta}-\Gamma^{\eta}_{\eta\eta}K_\eta=0$. That's strange.

 

[Onyx]

Anyways, I feel like there must be some straightforward way to handle calculating any geodesic in FLRW, but I'm not sure.

 

[PeterDonis]

It appeared to me that there was, because of $\frac{\partial K_\eta}{\partial \eta}-\Gamma^{\eta}_{\eta\eta}K_\eta=0$. That's strange.

It's not strange at all. You test for a Killing vector field using Killing's equation, which the equation you wrote down is not. The equation you wrote down is the geodesic equation (well, a somewhat garbled version of it, anyway), and (when properly written) shows that a worldline of constant $r$, $\theta$, $\phi$ is a geodesic. Which is not at all strange since that is the worldline of a comoving observer.

 

[PeterDonis]

I feel like there must be some straightforward way to handle calculating any geodesic in FLRW

There are two general approaches for computing geodesics. One is the brute force way of writing down all of the components of the general geodesic equation and then eliminating terms which are known to be zero until you have something manageable. The other way, which is considerably faster for metrics like this one where the metric coefficients are only functions of one or two of the coordinates, is the geodesic Lagrangian method, which is described briefly here:

https://en.wikipedia.org/wiki/Solving_the_geodesic_equations#Solution_techniques

 

[PeterDonis]

Once having converted the FLRW metric from comoving coordinates $ds^2=-dt^2+a^2(t)(dr^2+r^2d\phi^2)$ to "conformal" coordinates $ds^2=a^2(n)(-dn^2+dr^2+r^2d\phi^2)$, is there a way to facilitate solving for general geodesics that would otherwise be difficult

One obvious way to make solving for the geodesics easier is to not switch to conformal coordinates. All that does is add one more metric coefficient that depends on a coordinate ($g_{nn}$ depends on $n$, whereas in the original form $g_{tt}$ does not depend on $t$), and that is going to make more work for you in solving for geodesics no matter what method you use.

 

[Onyx]

One obvious way to make solving for the geodesics easier is to not switch to conformal coordinates. All that does is add one more metric coefficient that depends on a coordinate ($g_{nn}$ depends on $n$, whereas in the original form $g_{tt}$ does not depend on $t$), and that is going to make more work for you in solving for geodesics no matter what method you use.

Perhaps if I consider the form that the metric comes in when switching to $R=a(t)r$, where there is an $\frac{a'}{a}$ term, setting the scale factor to $e^t$ would help. But it adds a cross-term, so maybe not.

 

[Onyx]

Never mind, I actually found a source that provides a way to do it. Apparently, the shape of the orbits is independent of the choice of scale factor, which seems bizarre.

 

[PeterDonis]

I actually found a source that provides a way to do it.

Can you give a reference?

Apparently, the shape of the orbits is independent of the choice of scale factor

I'm not sure what you mean by "the choice of scale factor".

 

[Onyx]

Can you give a reference?I'm not sure what you mean by "the choice of scale factor".

"The shape of the orbit in FLRW spacetimes," by D Garfinkle.

 

[Ibix]

"The shape of the orbit in FLRW spacetimes," by D Garfinkle.

A link would have been helpful: https://arxiv.org/abs/1808.06683.

 

[Ibix]

The cited paper notes that the spatial projection of a geodesic in spacetime on to a constant curvature spatial slice (the usual FLRW coordinates' spatial slices) is also a geodesic of that space. Thus the paths depend only on the sign of the curvature and not the details of $a(t)$.

That doesn't seem to me to be completely true, in that the amount of time for which one can follow a geodesic does depend on $a(t)$ in a closed universe, but that might be overly pedantic.

I note that the paper does not appear to use conformal coordinates in its analysis.

 

[Onyx]

The cited paper notes that the spatial projection of a geodesic in spacetime on to a constant curvature spatial slice (the usual FLRW coordinates' spatial slices) is also a geodesic of that space. Thus the paths depend only on the sign of the curvature and not the details of $a(t)$.

That doesn't seem to me to be completely true, in that the amount of time for which one can follow a geodesic does depend on $a(t)$ in a closed universe, but that might be overly pedantic.

I note that the paper does not appear to use conformal coordinates in its analysis.

I was thinking only of a $k=0$ case.

 

[Onyx]

I was thinking only of a $k=0$ case.

For future reference, is orbit the most frequently used word to describe a geodesic that has angular momentum?