Question re spacial curvature K(r) w/r/t the Shwarzchild metric

[PeterDonis]

Unfortunately, these corrections would make the calculation substantially more complex.

Yes, indeed. However, they are simpler in the "canonical" coordinates given in this paper (referenced in one of the slides in the talk I linked to previously):

https://arxiv.org/pdf/1003.4777.pdf

Equation (7) is the metric in "canonical" coordinates, which are "Schwarzschild-type" coordinates in that the radial coordinate is the areal radius. This makes "constant area" much easier since it's just "constant $r$".

Another interesting paper is this one:

https://arxiv.org/pdf/1710.07373.pdf

 

[PeterDonis]

they are simpler in the "canonical" coordinates

I was lucky enough to find still another paper that actually gives the proper acceleration of an observer at constant $r$ in these coordinates:

https://arxiv.org/pdf/1104.4447.pdf

Equation (86) is the one; the paper calls it "force", but it's clear from the units and the context that it's actually the force per unit mass in an orthonormal tetrad frame, i.e., proper acceleration. The formula is (note that this isn't components, it's the norm of the 4-acceleration vector):

$$
A = \frac{\frac{M}{r^2} - H^2 r}{\sqrt{1 - \frac{2M}{r} - H^2 r^2}} - \frac{r \frac{dH}{dt} \sqrt{1 - \frac{2M}{r}}}{\left( 1 - \frac{2M}{r} - H^2 r^2 \right)^{\frac{3}{2}}}
$$

For $H = 0$ this reduces to the familiar formula for the proper acceleration of a hovering observer in Schwarzschild spacetime. For $M = 0$ this does not reduce to $0$, but that's because we are working in Schwarzschild-type coordinates, so a curve of constant $r$ is not comoving and will not be a geodesic.

The key point for this discussion, though, is that $A$ is indeed not independent of $t$ (except for the special case $dH / dt = 0$, which is just Schwarzschild-de Sitter spacetime), so once again, a curve of constant areal radius cannot also be a curve of constant proper acceleration (except for that one known special case).

This paper also notes that the McVittie metric we have been working with is for a mass embedded in a spatially flat universe, and gives generalizations to the cases of spatially closed and spatially open universes. I haven't had a chance to look at those in any detail.

 

[PAllen]

...For $H = 0$ this reduces to the familiar formula for the proper acceleration of a hovering observer in Schwarzschild spacetime. For $M = 0$ this does not reduce to $0$, but that's because we are working in Schwarzschild-type coordinates, so a curve of constant $r$ is not comoving and will not be a geodesic.

...

The result for M=0 bothers me. The problem is that the Milne metric does not have a zero Hubble parameter, but must have constant radial curves be geodesics. I don't know how to resolve this discrepancy. (Of course, you know that there are many geodesics in any FLRW metric that are not comoving world lines, and didn't mean to imply this). Note also, that the time derivative of the Hubble parameter for the Milne case is also not zero or constant.

[edit: the resolution is obvious on reading the referenced paper. Formula (86) is for the case of the asymptotic FLRW metric having flat spatial slices, which rules out the Milne case. Later, they give a formula for hyperbolic cosmologies, and it, indeed, has the right properties in the Milne limit.]

 

[PeterDonis]

The result for M=0 bothers me.

It's just the FRW metric in "outgoing Painleve" type coordinates:

$$
ds^2 = - \left( 1 - H^2 r^2 \right) dt^2 - 2 H r dt dr + dr^2 + r^2 d\Omega^2
$$

Just define $\beta = H r$ and you can see that a worldline with outward speed $\beta$ is a geodesic, and coordinate time in these coordinates is equal to proper time for an observer following such a worldline. In other words, such an observer is a "comoving" observer, and surfaces of constant $t$, which are spatially flat, are easily shown to be orthogonal to the worldlines of such observers. So we have a congruence of timelike geodesics all orthogonal to a family of spatially flat 3-surfaces, and all moving outward from a chosen point at the Hubble speed. That's spatially flat FRW spacetime.

I haven't derived an explicit coordinate transformation to take the above metric to the known spatially flat FRW form, but it should be straightforward to do so.

the Milne metric

$M = 0$ in this case is not the same as the Milne metric. The Milne metric is a vacuum solution (since it's just Minkowski spacetime in unusual coordinates); the above metric with $M = 0$ is not.

 

[PAllen]

See my edit above, I figured out the issue with M=0 for the Milne case.

 

[PeterDonis]

Later, they give a formula for hyperbolic cosmologies, and it, indeed, has the right properties in the Milne limit

The metric for the open case is given earlier in the paper; for $M > 0$ it is quite messy, but the $M = 0$ case is (using the spacelike signature convention I've been using instead of the timelike one used in the paper):

$$
ds^2 = - \frac{1 - H^2 r^2 + \frac{r^2}{R^2}}{1 + \frac{r^2}{R^2}} dt^2 - \frac{2 H r}{1 + \frac{r^2}{R^2}} dt dr + \frac{dr^2}{1 + \frac{r^2}{R^2}} + r^2 d\Omega^2
$$

where $R(t)$, as far as I can tell, is comoving distance. Plugging this metric into Maxima gives a nonzero Einstein tensor, so I don't think this metric is the Milne metric either.

 

[PAllen]

No, R(t) is the scale factor, for which most authors use a(t). See equations (1) and (2) and the surrounding discussion. So the Milne case has H(t) = 1/t, and R(t) =t, for t as cosmological time.

 

[PeterDonis]

R(t) is the scale factor

It's the "physical" scale factor--it has units of distance, at least as this paper is defining it. That's why I used the term "comoving distance", although I realize now that that term is also used to denote a "distance" with the scale factor taken out, so it was a bad choice of words on my part.

the Milne case has H(t) = 1/t, and R(t) =t, for t as cosmological time

Ah, yes, that's right, this is the edge case where all the terms in the Einstein tensor cancel.