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

[PeterDonis]

The metric is

Grrr...I just realized that I used the wrong line element in my previous posts; it should be

$$
ds^2 = - \frac{\left( 1 - \frac{M}{2 a(t) \rho} \right)^2}{\left( 1 + \frac{M}{2 a(t) \rho} \right)^2} + a^2(t) \left( 1 + \frac{M}{2 a(t) \rho} \right)^4 \left( d\rho^2 + \rho^2 d\Omega^2 \right)
$$

Note the extra factor of $a^2(t)$ in the spatial part.

Constant area now means $A = 4 \pi a^2(t) \left( 1 + \frac{M}{2 a(t) \rho} \right)^4 \rho^2$ is constant, which means areal radius $R = \sqrt{A / 4 \pi} = a(t) \left( 1 + \frac{M}{2 a(t) \rho} \right)^2 \rho$ is constant, which gives

$$
\rho(t) = \frac{1}{2} \left[ \frac{R}{a(t)} \left( 1 + \sqrt{1 - \frac{2M}{R}} \right) - \frac{M}{a(t)} \right] \approx \frac{R - M}{a(t)}
$$

That gives corrected values for $U$ and $A$ as follows:

$$
U = \gamma \left( 1, - \frac{R - M}{a^2} \dot{a} \right)
$$

$$
A = \gamma \left( \dot{\gamma}, \frac{R - M}{a^2} \left[ - \dot{\gamma} \dot{a} + 2 \gamma \frac{\dot{a}^2}{a} + \gamma \ddot{a} \right] \right)
$$

So the only difference in $A$ is that we have $R - M$ instead of just $M$, which doesn't change any of the conclusions; but I wanted to correct the formulas that needed correcting.

 

[PAllen]

Physically, the predictions are saying that the fact that the universe outside, say, the Milky Way is expanding should affect the trajectories of objects inside the Milky Way. That appears to violate the shell theorem--in this idealized model, there is still spherical symmetry, so what is happening outside a given sphere should not affect what happens inside it.

I’m still reviewing all the calculations, thanks so much for the help on this. However, homogeneity means the FLRW curvature exists everywhere. That is, it is not matter outside a spherical shell influencing geometry inside. It is the FLRW geometric contribution that is everywhere homogeneous that is influencing local geodesic convergence or divergence. So, I am not convinced by this argument as far as the McVittie model goes. However, I might be convinced by the idea that the FLRW homogeneity is an averaged idealization. And that a model that literally makes a fluid penetrate everywhere (which I think McVittie does), is not realistic. So my view remains that McVittie is correct for a body embedded in idealized FLRW cosmology, but not necessarily descriptive of real solar system dynamics. For that we would need a model that includes inhomogeneity at small scales, at least in some simple form. I have no idea if such a model has been developed.

 

[PeterDonis]

homogeneity means the FLRW curvature exists everywhere

The McVittie metric is not homogeneous or isotropic. It is spherically symmetric, but only about $\rho = 0$, and its spatial curvature varies with $\rho$. However, you are correct that it has nonzero stress-energy everywhere, so stress-energy present in a given spherical region can affect trajectories there. So it's not violating the shell theorem as a matter of mathematics.

My concern with it on scales near the horizon scale (i.e., $\rho \rightarrow M / 2 a$) is that on those scales, the universe is not a fluid even approximately; it's lumps of matter surrounded by vacuum. So a more reasonable solution for, say, the solar system embedded in an FRW universe would be joining, say, a Schwarzschild metric centered on the Sun with a McVittie metric at a spherical shell, say, 1 light year from the Sun. Or similarly for, say, the Milky Way, or the Local Group of galaxies, or even a galaxy cluster.

The question in such a solution would then be, how should the joining surface be constrained? Should it have constant surface area? Should it be a surface on which observers whose worldlines stay within the surface have constant proper acceleration? Should it be a surface of constant comoving coordinate $\rho$? Or something else?

It's also worth noting that, at least according to an interesting Perimeter Institute talk I found [1], the causal structure near the horizon scale in the McVittie metric is not what one might expect. For example, the horizon at $\rho = M / 2a$ is a past horizon.

http://www.physics.ntua.gr/cosmo13/Paros2013/Talks/Guariento.pdf

 

[PAllen]

The McVittie metric is not homogeneous or isotropic. It is spherically symmetric, but only about $\rho = 0$, and its spatial curvature varies with $\rho$. However, you are correct that it has nonzero stress-energy everywhere, so stress-energy present in a given spherical region can affect trajectories there. So it's not violating the shell theorem as a matter of mathematics.

I know the metric as a whole is not homogeneous. I was referring to the idea of the cause of homogeneous curvature in FLRW being superposed, nonlinearly, on top of field of the body. However, this is an imprecise idea, and the crux of the matter, as we both agree, is that the metric is nowhere vacuum. And this does raise questions about how valid it is as a model of a quasi-local dynamics where the stress energy should be vacuum within most of the local region. Then, the question is whether average past universe density influences the local Weyl curvature in some way. Since the McVittie metric is clearly introducing Ricci curvature locally, it doesn't really seem to answer this in a valid way.

I found your link quite interesting.

 

[PAllen]

Reviewing your calculations, I still have a couple of issues:

1) Your calculation of proper acceleration uses an ordinary derivative by proper time, rather than an absolute (or covariant) derivative that would involve Christoffel symbols. While an ordinary derivative is correct for the tangent vector, it is not correct for the proper acceleration in general coordinates.

2) The constraint that proper acceleration is constant requires only that the norm per the metric is constant, not that the components on a coordinate basis are constant. The split into t and rho components can readily change over time, and have no physical meaning, especially for non-physical coordinates like these.

Unfortunately, these corrections would make the calculation substantially more complex. I would not be willing to trust anything for this case not checked with a gr tensor package of some type.

As an aside (nothing I see wrong with your implicit method), you could get the form of $\frac {dt} {d \tau}$ explicitly by normalizing the raw tangent vector per the 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.