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

[Buzz Bloom]

TL;DR Summary

I understand that K(infinity) = zero, and K(Schwarzchild radius) = infinity, but what is K(r) between these limits?

I understand that

K(∞) = 0,​

and

K(r_s) = ∞​

where

r_s = 2GM/c².​

​

What is an equation for K(r) when

r_s < r < ∞?​

​

I have tried the best I can to search the Internet to find the answer, but I came up empty. I would very much appreciate the answer, or a reference that discusses the desired answer. I have seen some tensor equations, but my math skills cannot deal with those.

ADDED
In

https://en.wikipedia.org/wiki/Schwarzschild_metric​

in the section

Flamm's Paraboloid​

the following equation is derived:

w(r) = 2r_s√((r/r_s)-1).​

The coordinate system is described as

The Euclidean metric in the cylindrical coordinates (r, φ, w) is written:​

ds² = dw² + dr² + r²dφ².​

I am guessing (and may certainly be mistaken) that w(r) is related to the spatial curvature. In particular, I am guessing that w(r) is the radius of curvature, and the curvature is

k(r) = 1/w(r).​

I am hoping that some reader can confirm that my guesses are correct or incorrect, and if incorrect, what the curvature might be.

I am also wondering if this thread might be better placed in the Special and General Relativity forum. I also apologize for mispelling "spatial" in the title.

 

[PeterDonis]

I am also wondering if this thread might be better placed in the Special and General Relativity forum.

Yes. Moved.

 

[PeterDonis]

Summary:: I understand that K(infinity) = zero, and K(Schwarzschild radius) = infinity, but what is K(r) between these limits?

I understand that

K(∞) = 0,

and

K(r_s) = ∞

The first is correct, the second is not. See below.

I am guessing (and may certainly be mistaken) that w(r) is related to the spatial curvature.

$w(r)$ is related to the extrinsic curvature of the Flamm paraboloid, considered as a 2-surface embedded in Euclidean 3-space.

This is not the same as the intrinsic curvature of a spacelike hypersurface of constant Schwarzschild coordinate time. First, the 2-surface is not the same as the 3-surface. Second, extrinsic curvature is not the same as intrinsic curvature.

The intrinsic curvature of a spacelike hypersurface of constant Schwarzschild coordinate time can't be described by a single number. For a general 3-surface the Riemann curvature tensor has 6 independent components. Spherical symmetry in the Schwarzschild case cuts that down to two, which end up having the same $r$ dependence, namely $M / r^3$, one just has a factor of $2$ in front of it.

Note that, while this $r$ dependence means the curvature goes to zero as $r \rightarrow \infty$, the curvature does not increase without bound as $r \rightarrow 2M$. Again, one should not confuse the extrinsic curvature (which does increase without bound as $r \rightarrow 2M$) with the intrinsic curvature.

 

[Buzz Bloom]

Again, one should not confuse the extrinsic curvature (which does increase without bound as r→2M with the intrinsic curvature ...)

Hi Peter:

Thank you for your post.

I think I understand your post, but I want to make sure.

Your "2M" is intended to refer to r_s where you assume G and c to be unit constants. Is this correct?

If this is correct, then the two intrinsic curvature components are r_s/r³ and (1/2)r_s/r³. Wikipedia

https://en.wikipedia.org/wiki/Schwarzschild_metric#Curvatures​

says the Schwartzschild spatial curvature has these two components.

I vaguely remember reading somewhere that for some purposes a single curvature can be calculated as some simple mathematical operation on the product of these two components, but the details fail me, and I am unable to find this again. Do you know of such a relationship?

Regards,
Buzz

 

[Ibix]

Note that that is six non-zero components, not two, and is not using regular Schwarzschild coordinates, in which there are twelve non-zero components of the Riemann tensor.

The scalar invariant you are talking about, $K$, is the Kretschmann scalar, and is defined as $K=R_{abcd}R^{abcd}$. The Wikipedia page on it has that written out as eight nested sums starting from the all-lower index Riemann tensor (note that the version you quote above is not the all-lower version).

Note also that Peter seems to be calculating the curvature of space (by the Schwarzschild coordinate definition of space) whereas the process above will give you a curvature of spacetime. I'm not clear which one you actually want.

 

[Buzz Bloom]

Note also that Peter seems to be calculating the curvature of space (by the Schwarzschild coordinate definition of space) whereas the process above will give you a curvature of spacetime. I'm not clear which one you actually want.

Hi Ibix:

Thank you for your post.

I am interested in the spatial curvature of the Schwartzschild metric (perhaps also the McVitte metric) at a coordinate radius that does not change with time. I am not interested (for this thread) in the spacetime curvature. The reason for my interest in the spatial curvature is because it has an effect on the proper distance between concentric spherical shells, and I want to choose values of r for various cases to study at which the curvature influence is negligible.

Regards,
Buzz

 

[PeterDonis]

Your "2M" is intended to refer to r_s where you assume G and c to be unit constants. Is this correct?

Yes.

the two intrinsic curvature components are r_s/r³ and (1/2)r_s/r³.

Yes.

 

[PeterDonis]

Peter seems to be calculating the curvature of space

I am giving the space-space Riemann tensor components (i.e., those components that do not have a $t$ index). These are components of spacetime curvature, as you say, but they can also be viewed as the spatial curvature of a spacelike surface of constant $t$.

 

[PeterDonis]

The reason for my interest in the spatial curvature is because it has an effect on the proper distance between concentric spherical shells

You don't need to know the Riemann tensor to calculate this, just the metric. The proper distance between two shells at $r_1$ and $r_2$ is

$$
s = \int_{r_1}^{r_2} \sqrt{g_{rr}} dr
$$

 

[Buzz Bloom]

Hi Peter:
Here is my try using Latex to elaborate on the integral in your post #9.

$$ s=\int_a^b \sqrt{ \frac {dr}{\sqrt{ 1-r_s/r}} $$

I tried to have the limits r₁ and r₂, but I could not make that work. Here is my second try.

$$ s=\int_{r_1}^{r_2} \sqrt{ \frac {dr}{\sqrt{ 1-r_s/r}} $$

Thanks for your post to my first try.

My second try seemns to have made both tries unsuccessful. I will make a new post with the second try to see if i can make it work without using Preview, which caused me difficulties the last time I tried to use Latex.

Regards,
Buzz

 

[PeterDonis]

Here is my try using Latex to elaborate on the integral in your post #9.

Yes, that's correct, where your $a$ and $b$ are my $r_1$ and $r_2$.

 

[Buzz Bloom]

Here is my third try, this time without using Preview.

$$ s=\int_a^b \sqrt{ \frac {dr}{\sqrt{ 1-r_s/r}} $$

Now my first try (re-tried here) with "a" and "b" no longer works.

I think this exercise makes it clear why I do not like Latex.

I will now make one last try with "a" and "b".

 

[Buzz Bloom]

$$ s=\int_a^b \frac {dr}{\sqrt{ 1-r_s/r}} $$

This was successful because I found another error.

$$ s=\int_{r_1}^{r_2} \frac {dr}{\sqrt{ 1-r_s/r}} $$

I'm now on a roll! I think my best chance at a successful use of Latex is to include a Statement with a post saying that the post is a "Work in Progress and not ready for review". I can then delete failures and continue making new versions until it is what I want to be displayed. Then I will remove the "Work in Progress" statement.

Any comments?

 

[PAllen]

Hi Ibix:

Thank you for your post.

I am interested in the spatial curvature of the Schwartzschild metric (perhaps also the McVitte metric) at a coordinate radius that does not change with time. I am not interested (for this thread) in the spacetime curvature. The reason for my interest in the spatial curvature is because it has an effect on the proper distance between concentric spherical shells, and I want to choose values of r for various cases to study at which the curvature influence is negligible.

Regards,
Buzz

There is an important difference in this regard for Schwarzschild versus McVittie metrics. The former is a stationary metric - there is a family (congruence) of (timelike) world lines along which the metric doesn't change. More precisely, there is a timelike killing vector field, whose integral curves are stationary radial world lines. The McVittie metric is not stationary, so the whole notion of a coordinate radius that doesn't change with time is (physically) suspect, as is the notion of a stationary shell.

What I think might be meaningful is to consider cases of the two metrics with the same mass parameter. Then, in each find a 2-sphere congruence (where each world line in the congruence has constant angular coordinates) in time such that two things hold:

- proper acceleration is fixed at e.g. 1 g
- area of the 2-sphere computed using the respective metrics remains constant.

This is trivial to do in the Schwarzschild case, but not so in the McVittie case.

One may then use this as a reasonable physical definition of an equivalent planetary surface in each. Then, investigate radial geodesics of increasing relative velocity to the surface define above. Find the escape velocity that leads to escape to infinity.

As in the last thread, I don't see how proper distance between stationary shells has anything to do with escape velocity. This is besides the point, that in the mathematical sense, there are no stationary shells in the McVittie metric - stationary referring to being integral curves of a timelike killing vector field.

 

[PeterDonis]

I think my best chance at a successful use of Latex is to include a Statement with a post saying that the post is a "Work in Progress and not ready for review". I can then delete failures and continue making new versions until it is what I want to be displayed. Then I will remove the "Work in Progress" statement.

Any comments?

Just use the "Preview" button to check how your post will display. That's what it's for. It's very helpful for debugging LaTeX errors before you actually post.

 

[PeterDonis]

without using Preview, which caused me difficulties the last time I tried to use Latex.

What difficulties? Preview shows you the exact same thing you would see if you hit "Post", but without posting, so you get a chance to fix errors without having to go through a cumbersome process of "oops, edit post and try again" or "oops, delete post and try again".

 

[PeterDonis]

What I think might be meaningful is to consider cases of the two metrics with the same mass parameter.

While the $M$ in the metric is a constant, it doesn't mean the same thing in both metrics. In Schwarzschild, the horizon is at $M / 2$ (because we're in isotropic coordinates), which is constant. In McVittie, the horizon is at $M / 2 a$, which is a function of $t$ (since $a$ is). This is another thing that makes the interpretation of McVittie as "a mass embedded in an FRW spacetime" problematic; an actual massive object would not have a horizon radius that varied with cosmological time.

(Note that this difference also affects things like the worldlines of observers trying to "hover" above the "mass" with constant proper acceleration; I believe it makes it impossible to find a congruence of observers in the McVittie case that has both of the properties you describe.)

 

[PAllen]

While the $M$ in the metric is a constant, it doesn't mean the same thing in both metrics. In Schwarzschild, the horizon is at $M / 2$ (because we're in isotropic coordinates), which is constant. In McVittie, the horizon is at $M / 2 a$, which is a function of $t$ (since $a$ is). This is another thing that makes the interpretation of McVittie as "a mass embedded in an FRW spacetime" problematic; an actual massive object would not have a horizon radius that varied with cosmological time.

(Note that this difference also affects things like the worldlines of observers trying to "hover" above the "mass" with constant proper acceleration; I believe it makes it impossible to find a congruence of observers in the McVittie case that has both of the properties you describe.)

I think it is possible to find such a congruence. The issues you point out just mean that constant radial coordinate in McVittie metric has different physical meaning than in a Schwarzshild metric. The congruence I describe would have shrinking (for normal choices of a(t), I think) radial coordinate in McVittie, but constant proper acceleration and area, per the metric. This is not fundamentally different from stationary world lines having non-constant radial coordinate in e.g. Lemaitre coordinates for the Shcwarzschild spacetime (and the horizon is not coordinate stationary in either Lemaitre or Kruskal coordinates).

I admit I have not calculated such a congruence, only made some estimations, so take my claim with a grain of salt. But I also suspect you have not proven the nonexistence of such a congruence.

[edit: it is clearly possible to generate a family of constant area spherical congruences for the McVittie metric. If none have constant proper acceleration, then I would take this as a prediction of the metric that for the given choice of scale factor evolution, either the "surface gravity" must change very slowly with time, or that maintaining hydrodynamic equilibrium requires change in radius with time.]

 

[PeterDonis]

I also suspect you have not proven the nonexistence of such a congruence.

I had not done explicit calculations either, but they're straightforward to do, so let's do them.

The metric is:

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

Constant area means $A = 4 \pi \left( 1 + \frac{M}{2 a(t) \rho} \right)^4 \rho^2$ is constant, or, if we want to use "areal radius" instead, it means $R = \sqrt{A / 4 \pi} = \left( 1 + \frac{M}{2 a(t) \rho} \right)^2 \rho$ is constant. Solving this for $\rho$ gives

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

For $M << R a(t)$, which is the approximation I will use henceforward, this reduces to

$$
\rho(t) \approx R - \frac{M}{a(t)}
$$

A purely radial trajectory of constant area therefore has 4-velocity (giving only the $t$ and $\rho$ components)

$$
U = \frac{dt}{d\tau} \left( 1, \frac{d\rho}{dt} \right) = \frac{dt}{d\tau} \left( 1, \frac{M}{a^2(t)} \right)
$$

The function $dt / d\tau$ is left unspecified, but we can treat it as a function of $t$ and call it $\gamma(t)$. We then have $d^2 t / d\tau^2 = (d \gamma / dt) (dt / d\tau) = \gamma (d\gamma / dt)$.

The 4-acceleration of this trajectory is then

$$
A = \frac{dU}{d\tau} = \gamma \frac{d \gamma}{d t} \left( 1, \frac{M}{a^2(t)} \right) + \gamma^2 \left( 0, - \frac{2M}{a^3(t)} \right) = \gamma \left( \frac{d\gamma}{dt}, \frac{M}{a^2} \left[ \frac{d \gamma}{d t} - \frac{2 \gamma}{a} \right] \right)
$$

For this to be constant, independent of $t$, we must have $\gamma$ constant, so $d \gamma / dt = 0$, and we must have $d \gamma / dt - 2 \gamma /a$ constant. But these conditions are clearly not consistent unless $\gamma = 0$, which cannot be the case for a timelike curve. Hence, no congruence that has both constant area and constant proper acceleration is possible for the approximation given. I don't think that changes if we use the exact formula for $\rho$, since that just adds another term varying with $t$ to $d\rho / dt$, but I don't have time to do that calculation explicitly right now.

 

[PeterDonis]

If none have constant proper acceleration, then I would take this as a prediction of the metric that for the given choice of scale factor evolution, either the "surface gravity" must change very slowly with time, or that maintaining hydrodynamic equilibrium requires change in radius with time.

Neither of these predictions makes sense to me, so to me this is another indication that interpreting the McVittie metric as describing "a mass embedded in an FRW spacetime" is problematic in general. It might still be warranted for the specific case of analyzing gravitational lensing due to faraway objects.

 

[Buzz Bloom]

What difficulties? Preview shows you the exact same thing you would see if you hit "Post", but without posting, so you get a chance to fix errors without having to go through a cumbersome process of "oops, edit post and try again" or "oops, delete post and try again".

Hi Peter:

My difficulties I am pretty sure are related to my lack of good habits. Also, at my advanced age, I find myself able to learn new things only very slowly. One problem I have with the Preview displayed below the in-work text is that sometimes I accidentally edit the Preview rather than the in-work version. If I take a break and then edit some more, the in-work version does not get all the edits. This creates a lot of confusion for me. One approach I have been using to form better habits it do all my editing in another independent document which after editing I copy to the to-be-posted post-in-progress area of a thread. Sometimes this creates a formatting problem which requires additional editing. As you can probably tell from this exposition, I am not very competent these days.

Regards,
Buzz

 

[PeterDonis]

sometimes I accidentally edit the Preview rather than the in-work version

You can't edit the preview. It's not editable. So I don't understand what you're describing here.

 

[PAllen]

Neither of these predictions makes sense to me, so to me this is another indication that interpreting the McVittie metric as describing "a mass embedded in an FRW spacetime" is problematic in general. It might still be warranted for the specific case of analyzing gravitational lensing due to faraway objects.

1) in your calculation post (thanks!), it seems to me you miss several places where your terms should be multiplied by time derivatives of a(t). This suggests that the dual constraint can be exactly met for certain choices of a(t).

2) I don’t find these predictions implausible, as long as they are tiny. And multiplying scale factor derivatives will make them really tiny on the scale of a solar system.

 

[PeterDonis]

in your calculation post (thanks!), it seems to me you miss several places where your terms should be multiplied by time derivatives of a(t).

Hm, yes, let me go back and look at that.

 

[Buzz Bloom]

I don't see how proper distance between stationary shells has anything to do with escape velocity.

Hi Paul:

The integral I want to calculate is more complicated than the distance D between two shells. I wanted to make sure I had the correct way of making this calculation as a first step before I tackle a more complicated problem. I think I should start a new thread for this more complicated problem since it is not part of the topic of this thread. I certainly hope you will post your insights regarding this more complicated problem when I start the new thread.

Regards,
Buzz

 

[PeterDonis]

let me go back and look at that.

Ok, here's what the corrected equations look like:

$$
\frac{d\rho}{dt} = \frac{M}{a^2} \frac{da}{dt} = \frac{M}{a^2} \dot{a}
$$

(I'll use overdots to denote derivatives with respect to $t$ from now on.)

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

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

We now see that $A = \text{constant}$ requires $\dot{\gamma} = 0$, i.e., $\gamma = \text{constant}$, as before, but that the other condition is now:

$$
\frac{M}{a^2} \left[ \dot{\gamma} \dot{a} - \gamma \frac{2 \dot{a}^2}{a} + \gamma \ddot{a} \right] = \text{constant} = K
$$

Eliminating the $\dot{\gamma}$ term and rearranging gives

$$
\ddot{a} - 2 \frac{\dot{a}^2}{a} - a^2 \frac{K}{M \gamma} = 0
$$

This is a differential equation in $a$ that, by itself, could be solved; but we also have another condition we have to meet, that $A \cdot U = 0$. Plugging in $\dot{\gamma} = 0$ to $A$ gives for this condition:

$$
\frac{M^2}{a^4} \dot{a} \gamma^2 \left[ \ddot{a} - 2 \frac{\dot{a}^2}{a} \right] = 0
$$

This means the first two terms in the differential equation above have to vanish, leaving only the third term, which can only be satisfied if $a = 0$. So with the corrected equations it's still true that a congruence with both constant area and constant proper acceleration is impossible.

 

[PAllen]

Ok, here's what the corrected equations look like:

$$
\frac{d\rho}{dt} = \frac{M}{a^2} \frac{da}{dt} = \frac{M}{a^2} \dot{a}
$$

(I'll use overdots to denote derivatives with respect to $t$ from now on.)

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

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

We now see that $A = 0$ requires $\dot{\gamma} = 0$, i.e., $\gamma = \text{constant}$, as before, but that the other condition is now:

$$
\frac{M}{a^2} \left[ \dot{\gamma} \dot{a} - \gamma \frac{2 \dot{a}^2}{a} + \gamma \ddot{a} \right] = \text{constant} = K
$$

Eliminating the $\dot{\gamma}$ term and rearranging gives

$$
\ddot{a} - 2 \frac{\dot{a}^2}{a} - a^2 \frac{K}{M \gamma} = 0
$$

This is a differential equation in $a$ that, by itself, could be solved; but we also have another condition we have to meet, that $A \cdot U = 0$. Plugging in $\dot{\gamma} = 0$ to $A$ gives for this condition:

$$
\frac{M^2}{a^4} \dot{a} \gamma^2 \left[ \ddot{a} - 2 \frac{\dot{a}^2}{a} \right] = 0
$$

This means the first two terms in the differential equation above have to vanish, leaving only the third term, which can only be satisfied if $a = 0$. So with the corrected equations it's still true that a congruence with both constant area and constant proper acceleration is impossible.

But we don’t want proper acceleration, A, to be zero, just constant.

 

[Buzz Bloom]

You can't edit the preview. It's not editable. So I don't understand what you're describing here.

Hi Peter:

I will (after a while) make an attempt to reconstruct my issues regarding the Preview, and I will keep notes about what happens. It is certainly possible that I am just misremembering what the problems were I was experiencing. I have not tried (until today) to use Latex for quite a long time.

Is there a way to display a Latex string in a post with it not being translated into equations?

Regards,
Buzz

 

[PeterDonis]

we don’t want proper acceleration, A, to be zero, just constant.

That was a simple typo. Fixed now. Please read the post again.

 

[PeterDonis]

I don’t find these predictions implausible, as long as they are tiny.

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.

 

[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.