Is Minkowski spacetime a solution of the Friedmann Equations?

[timmdeeg]

The empty FRW-universe with curvature parameter $k = -1$ and expanding linearly is well known. Also that it is mathematically equivalent (after a coordinate transformation) with the Milne universe which also expands linearly.

I wonder if the Friedmann Equations have another solution (I remember something like that, but perhaps wrongly): The static Minkowski spacetime. If true this seems to require $H = 0$ and $k = 0$ and thus $\rho/\rho_c = 1$. However this doesn't seem to make sense because in the empty universe the densities are zero and thus $k$ has no defined value.

Any help appreciated.

 

[Orodruin]

Minkowski space is not another solution, it is the same solution, just expressed in different coordinates with a different foliation into homogeneous and isotropic space-like slices.

The situation is very reminiscent of expressing regular Euclidean $\mathbb R^2$ in polar coordinates versus Cartesian coordinates. In both coordinate systems, you have isotropic and homogeneous coordinate surfaces, planes in the case of Cartesian coordinates and circles in the case of polar coordinates. However, both coordinate systems describe the same space.

 

[timmdeeg]

Thanks. Hm, after some search, I found here a classification of Friedmann vacuum models noting the solutions (?) $k = 0$; $a = 1$ and $k = -1$; $a = t$. I should have done that before. Do you say these "solutions" are the same, just "expressed in different coordinates"?

 

[Orodruin]

Do you say these "solutions" are the same, just "expressed in different coordinates"?

Yes, both are Minkowski space. It is just that the Milne coordinates are curvilinear.

 

[timmdeeg]

Yes, both are Minkowski space. It is just that the Milne coordinates are curvilinear.

The problem I still have is that a static empty FRW-Universe requires $H = 0$ which then yields $k = 0$ although $k$ seems to have no definite value in this case as I mentioned in the OP. And why is this solution the same one than the $k = -1$ solution? Probably the whole reasoning isn't correct, but why? And does the $k = 0$ solution yield - after coordinate transformation - a static Minkowski spacetime while the $k = -1$ solution yields an expanding Minkowski spacetime after coordinate transformation?

 

[PeterDonis]

a static empty FRW-Universe requires $H = 0$ which then yields $k = 0$

Yes, and for that case--Minkowski spacetime in standard Minkowski coordinates--$k = 0$ is correct since the spacelike slices of constant time are indeed flat.

why is this solution the same one than the $k = -1$ solution?

It's the same spacetime geometry, but different coordinates; in the coordinates in which $k = -1$, the spacelike surfaces of constant coordinate time, if you look at them in standard Minkowski coordinates, are hyperboloids of constant proper time from the origin. Because they are hyperboloids, they have constant negative curvature, i.e., $k = -1$. But these are, of course, different spacelike surfaces than the $k = 0$ spacelike surfaces of constant time in Minkowski coordinates.

 

[kimbyd]

The problem I still have is that a static empty FRW-Universe requires $H = 0$ which then yields $k = 0$ although $k$ seems to have no definite value in this case as I mentioned in the OP. And why is this solution the same one than the $k = -1$ solution? Probably the whole reasoning isn't correct, but why? And does the $k = 0$ solution yield - after coordinate transformation - a static Minkowski spacetime while the $k = -1$ solution yields an expanding Minkowski spacetime after coordinate transformation?

I think the confusion here is that you seem to be thinking of the parameter $k$ as if it were a physical property of the universe. It is not. The parameter $k$ is a coordinate-dependent parameter.

The somewhat complicated reason for this is that the parameter $k$ is a function only of the spatial coordinates of the metric, but if you change coordinates the spatial coordinates themselves change, so you won't necessarily get the same answer. It is only parameters drawn from all four coordinates that can ever be independent of coordinates.

One way of understanding this is that this space-time, no matter the coordinates, has zero curvature. In the Milne universe, coordinates are chosen which create spatial curvature, which is balanced by an equal (and opposite) amount of space-time curvature, such that the total remains zero. For Minkowski you just get zeroes all around.

 

[PeterDonis]

coordinates are chosen which create spatial curvature, which is balanced by an equal (and opposite) amount of space-time curvature

This is a misleading way to phrase it. A better way would be that the coordinates are chosen so that spacelike surfaces of constant time have intrinsic curvature, which is balanced by an equal and opposite amount of extrinsic curvature, so that the overall spacetime curvature is zero, as it must be for Minkowski spacetime.

 

[Orodruin]

The parameter k is a coordinate-dependent parameter.

It is not a priori clear that there are several different coordinate systems where foliations are isotropic and homogeneous and I am not sure whether it is true in general or not. However, what is true is that there are two different ways of foliating Minkowski space with such spacelike surfaces.

To make it clearer to the OP that the spacetimes are indeed equivalent, start from the linearly expanding Universe where the metric is given by
$$
ds^2 = dt^2 - k^2 t^2 [dr^2 + k^{-2}\sinh^2(kr) d\Omega^2].
$$
Introducing new coordinates by choosing $\tau = t \cosh(kr)$ and $\xi = t \sinh(kr)$, we would find that
$$
d\tau = \cosh(kr) dt + k t \sinh(kr) dr, \qquad d\xi = \sinh(kr) dt + k t \cosh(kr) dr
$$
and therefore
$$
d\tau^2 - d\xi^2 = dt^2 - k^2 t^2 dr^2.
$$
Furthermore, $t^2 \sinh^2(kr) = \xi^2$ and thus
$$
ds^2 = d\tau^2 - d\xi^2 - \xi^2 d\Omega^2,
$$
which is just the Minkowski line element in polar coordinates for the spatial part.

It is also relatively easy to see that the linearly expanding universe is not geodesically complete as spatial geodesics hit $t = 0$ for a finite value of the affine parameter. This follows from the fact that the Milne coordinates just describe a patch of Minkowski space (the future light-cone of the origin).

Edit: It should also be clear from the above that the comoving observers in Milne coordinates are actually moving at velocity $v = \tanh(kr)$ relative to the comoving observer at the origin (a statement that is fine to make as Minkowski space is flat).

 

[timmdeeg]

Got it. Many thanks to everybody for clarifying this question.

 

[Orodruin]

Related to this topic, I wrote a PF Insight about two years back that discusses coordinate dependent statements in a RW universe. You might find it to be of interest.

 

[timmdeeg]

Related to this topic, I wrote a PF Insight about two years back that discusses coordinate dependent statements in a RW universe. You might find it to be of interest.

Thanks, very interesting whereby I am a bit overtaxed to understand some technical details though.

By the way I found among others the paragraph "Dopplershift Versus Cosmological Redshift" in your Insight article very helpful. The coordinate dependence of the usual interpretation isn't mentioned so often. I have just checked Peacock's "Cosmological Physics" and am quite astonished that he writes at page 72: "Photon wavelenghts therefore stretch with the universe, as is intuitively reasonable; see fig. 3.3. This is the only correct interpretation of the redshift at large distances." I don't hear any alarm bells ringing.

 

[PeterDonis]

This is the only correct interpretation of the redshift at large distances.

For "at large distances" read "over regions of spacetime that are too large to be covered by a single local inertial frame". The Doppler Shift interpretation that @Orodruin gives for the redshift only works within a single local inertial frame, i.e., over a region of spacetime small enough that spacetime curvature is negligible. "At large distances" that is not the case.

That said, Peacock does leave out a crucial point: the "stretching with the universe" interpretation of the redshift is only valid for comoving observers. In our actual universe, real objects like galaxies and galaxy clusters are not exactly comoving; their spatial coordinates in the standard FRW coordinate chart do not stay the same for all time. Cosmologists often say that this means the actual observed redshift of light signals emitted by one such object and observed by another has two pieces: a "cosmological" piece due to the comoving part of their motion, and a "Doppler" piece due to the non-comoving part of their motion. But in general we don't know how to break up the motion of distant galaxies and galaxy clusters into a comoving and non-comoving part; the best we can do is to try to estimate it from other data, in order to subtract out the non-comoving part to get just the comoving part, since that's what cosmologists are interested in (because it is what is needed to estimate the expansion history of the universe). This is one of the main sources of error in cosmological observations and models.

 

[Orodruin]

This is the only correct interpretation of the redshift at large distances.

This is wrong and coordinate dependent. It is exactly the kind of misstatement that I wrote the insight to avoid. The only ”correct” interpretation in the sense of being coordinate invariant is viewing the frequencies as products of 4-velocities and 4-frequencies, which is a geometrical concept. To claim that a coordinate dependent statement is the ”only correct interpretation” is misleading, wrong, and should make you seriously doubt whoever makes the statement.

 

[timmdeeg]

For "at large distances" read "over regions of spacetime that are too large to be covered by a single local inertial frame". The Doppler Shift interpretation that @Orodruin gives for the redshift only works within a single local inertial frame, i.e., over a region of spacetime small enough that spacetime curvature is negligible. "At large distances" that is not the case.

But I refer to @Orodruin's statement regarding the cosmological redshift, not regarding the redshift "over a region of spacetime small enough that spacetime curvature is negligible":

It is sometimes ascribed to the expansion “stretching” the wavelength of the light signals, alternatively the distance between consecutive signals. After reading up to this point, your warning bells should all already be sounding loudly and recognise this as a coordinate dependent statement. Indeed, in RW coordinates, the expansion of the universe does increase the distance between consecutive signals, but this does not a priori tell us that the same interpretation will explain the redshift in other coordinate systems.
Source https://www.physicsforums.com/insights/coordinate-dependent-statements-expanding-universe/

That said, Peacock does leave out a crucial point: the "stretching with the universe" interpretation of the redshift is only valid for comoving observers. In our actual universe, real objects like galaxies and galaxy clusters are not exactly comoving;

What I intended to say is that Peacock doesn't mention the coordinate dependence of the ""stretching with the universe" interpretation of the redshift" (@Orodruin's warning bells).

This is the only correct interpretation of the redshift at large distances.

Please note I didn't say this, the source is Peacock.

 

[timmdeeg]

This is wrong and coordinate dependent. It is exactly the kind of misstatement that I wrote the insight to avoid. The only ”correct” interpretation in the sense of being coordinate invariant is viewing the frequencies as products of 4-velocities and 4-frequencies, which is a geometrical concept. To claim that a coordinate dependent statement is the ”only correct interpretation” is misleading, wrong, and should make you seriously doubt whoever makes the statement.

It does, see #15.

 

[PAllen]

This is a misleading way to phrase it. A better way would be that the coordinates are chosen so that spacelike surfaces of constant time have intrinsic curvature, which is balanced by an equal and opposite amount of extrinsic curvature, so that the overall spacetime curvature is zero, as it must be for Minkowski spacetime.

Can you explain what you mean here? I don't see this balancing extrinsic curvature. Consider the trivial example of foliating Euclidean 3-space (except for 1 point) by a family of concentric 2-spheres. The intrinsic and extrinsic curvature are both constant and positive, in this case. No balancing is required.

In the Milne case, both the intrinsic and extrinsic curvature of the hyperbolic slices are both negative, so far as I understand.

[edit: I wonder if you are thinking about the Gauss-Codazzi equations which relate extrinsic and intrinsic curvature forms. If so, then we just have a case of the impossibility of summarizing nontrivial mathematics in words. 'Balancing extrinsic curvature and intrinsic curvature to be equal and oppposite' is certainly not how I would summarize these equations. However, I can't offer any alternative verbal formulation either. ]

 

[PeterDonis]

I don't see this balancing extrinsic curvature.

I'm actually no longer sure that's true in general, even if it turns out to be true in this particular case (see further comments below). So I would not want to hang my hat on the way I put it before.

In the Milne case, both the intrinsic and extrinsic curvature of the hyperbolic slices are both negative, so far as I understand.

The intrinsic curvature is negative, but the extrinsic curvature is not. More precisely, the two obvious curvature invariants you can compute from the extrinsic curvature are not. (The extrinsic curvature itself is a second-rank tensor.)

The easiest way to compute the extrinsic curvature is to use Minkowski coordinates, since in those the connection coefficients vanish so covariant derivatives become simple partial derivatives. I will do the computation for the 1+1 case since that makes things much simpler, but I think everything I'm doing should generalize to the 1+3 case by symmetry. I will be using the general treatment given on slide 17 and 18 here:

http://www.maths.qmul.ac.uk/~jav/LTCCmaterial/LTCCSlidesLecture2.pdf

First we need an expression for the normal $n^a$ to the spacelike slices. This will be the same as the 4-velocity of a comoving observer, and in Minkowski coordinates such an observer is moving with ordinary velocity $v = x / t$ for any coordinates $(t, x)$ in the upper "wedge" (i.e., the future light cone of the origin). So the 4-velocity of such an observer will be

$$
n^a = \gamma \partial_t + \gamma v \partial_x
$$

which expands out to

$$
n^a = \frac{t}{\sqrt{t^2 - x^2}} \partial_t + \frac{x}{\sqrt{t^2 - x^2}} \partial_x
$$

The covector corresponding to this just has the sign flipped in the first component (since we are using the $-+++$ signature convention):

$$
n_a = - \frac{t}{\sqrt{t^2 - x^2}} \text{d}t + \frac{x}{\sqrt{t^2 - x^2}} \text{d} x
$$

We will need the partial derivatives of the covector components, so we compute them now:

$$
\partial_0 n_0 = - \frac{1}{\sqrt{t^2 - x^2}} + \frac{t^2}{\left( t^2 - x^2 \right)^{3/2}} = \frac{x^2}{\left( t^2 - x^2 \right)^{3/2}}
$$

$$
\partial_0 n_1 = - \frac{t x}{\left( t^2 - x^2 \right)^{3/2}} = \partial_1 n_0
$$

$$
\partial_1 n_1 = \frac{1}{\sqrt{t^2 - x^2}} + \frac{x^2}{\left( t^2 - x^2 \right)^{3/2}} = \frac{t^2}{\left( t^2 - x^2 \right)^{3/2}}
$$

The key observation that now simplifies the computation is that the "acceleration" of this 4-velocity is zero:

$$
a_a = n^b \nabla_b n_a = n^b \partial_b n_a
$$

The two components of this are:

$$
a_0 = n^0 \partial_0 n_0 + n^1 \partial_1 n_0
$$

$$
a_1 = n^0 \partial_0 n_1 + n^1 \partial_1 n_1
$$

If you work these out from the above computations of the partial derivatives you will see that they both vanish. (Physically, this just means that the worldlines of comoving observers are geodesics, so it makes sense, but it was still worth checking explicitly to see that it is true.) This means that we can use a simple formula for the extrinsic curvature:

$$
K_{ab} = \partial_a n_b
$$

where we have used the fact that $\nabla \rightarrow \partial$. So we have already computed all of the components!

As we can see from the above, the diagonal components of $K_{ab}$ are positive and the off-diagonal components are negative. But if we compute the two obvious curvature invariants, we find that they are both positive; these are the trace:

$$
K = K_a{}^a = \eta^{ab} K_{ab} = - K_{00} + K_{11} = \frac{t^2 - x^2}{\left( t^2 - x^2 \right)^{3/2}} = \frac{1}{\sqrt{t^2 - x^2}}
$$

and one that I don't know the name for, but it's the next order scalar invariant:

$$
K_{ab} K^{ab} = \left( K_{00} \right)^2 - 2 \left( K_{01} \right)^2 + \left( K_{11} \right)^2 \\
\ \ \ \ \ \ \ \ \ = \frac{x^4}{\left( t^2 - x^2 \right)^3} - \frac{2 t^2 x^2}{\left( t^2 - x^2 \right)^3} + \frac{x^4}{\left( t^2 - x^2 \right)^3} \\
\ \ \ \ \ \ \ \ \ = \frac{\left( t^2 - x^2 \right)^2}{\left( t^2 - x^2 \right)^3} \\
\ \ \ \ \ \ \ \ \ = \frac{1}{t^2 - x^2}
$$

These are both positive. For the 1+3 case I would expect them also to be positive, since for that case we can express the 4-velocities entirely in terms of a radial coordinate $r$ that ends up looking like $x$ in the above formulas (the angular coordinates remain constant along each comoving worldline).

However, I'm not sure whether this will lead to the extrinsic curvature canceling out the intrinsic curvature of the surfaces of constant time; I have not done that computation.

 

[PAllen]

Ok, but to me what characterizes extrinsic negative curvature is that the curvature in different directions has different sign. Consider some examples:

2 sphere embedded in Euclidean 3 space: intrinsic curvature positive constant, extrinsic curvature positive constant In all directions at all points.

Cylinder embedded in Euclidean 3 space: intrinsic curvature zero, extrinsic principle curvatures zero in one direction, positive in another. Mean curvature positive.

Hyperboloid embedded inEuclidean 3 space: extrinsic curvature in different directions changes sign. I take this as the definition of negative extrinsic curvature. It is true that an invariant like mean curvature will be zero, positive or negative at different points of the embedded hyperboloid. However, the extrinsic curvature will change sign in different directions, at all points. This is what I characterize as negative curvature - it is characteristic of saddle shapes.

 

[PeterDonis]

It is true that an invariant like mean curvature will be zero, positive or negative at different points of the embedded hyperboloid.

The extrinsic curvature invariants I calculated are positive and have the same value everywhere on a given hyperboloid. That's evident from the formulas I gave, and it's also evident when you consider that these spatial 3-surfaces are surfaces of constant curvature, as they must be to satisfy homogeneity and isotropy. That means all intrinsic and extrinsic curvature invariants must have the same values at every point on a given hyperboloid.

 

[PeterDonis]

Hyperboloid embedded inEuclidean 3 space

The hyperboloids we are talking about are spacelike 3-surfaces embedded in 4-D Minkowski spacetime. That makes a difference.

 

[PAllen]

The extrinsic curvature invariants I calculated are positive and have the same value everywhere on a given hyperboloid. That's evident from the formulas I gave, and it's also evident when you consider that these spatial 3-surfaces are surfaces of constant curvature, as they must be to satisfy homogeneity and isotropy. That means all intrinsic and extrinsic curvature invariants must have the same values at every point on a given hyperboloid.

The hyperboloid embedded in Euclidean 3 space is a surface of constant intrinsic curvature. That does not prevent the extrinsic mean curvature invariant changing from point to point.

 

[PeterDonis]

The hyperboloid embedded in Euclidean 3 space

...is not what we're talking about. See post #21.

 

[Orodruin]

The hyperboloid embedded in Euclidean 3 space is a surface of constant intrinsic curvature.

I do not think that this is true. Far away from the origin, the embedding very well approximates a cone, which has zero intrinsic curvature. At the point closest to the origin, intrinsic curvature is non-zero and can be pretty large depending on the hyperboloid you choose.

Edit: To at least somewhat back that up, I ran the induced metric for the hyperboloid $1 = z^2 - x^2 - y^2$ through my Mathematica script for computing everything from Christoffel symbols to the Ricci scalar based on the metric components. The Ricci scalar for that surface is $2/\cosh^2(2\theta)$, where $z = \cosh(\theta)$. This clearly is 2 when $\theta = 0$ and goes to zero as $\theta \to \infty$, as expected.

On the other hand, the hyperboloid embedded in Minkowski space is a surface of constant intrinsic curvature, much like a sphere embedded in Euclidean space.

 

[Tendex]

I do not think that this is true. Far away from the origin, the embedding very well approximates a cone, which has zero intrinsic curvature. At the point closest to the origin, intrinsic curvature is non-zero and can be pretty large depending on the hyperboloid you choose.

Edit: To at least somewhat back that up, I ran the induced metric for the hyperboloid $1 = z^2 - x^2 - y^2$ through my Mathematica script for computing everything from Christoffel symbols to the Ricci scalar based on the metric components. The Ricci scalar for that surface is $2/\cosh^2(2\theta)$, where $z = \cosh(\theta)$. This clearly is 2 when $\theta = 0$ and goes to zero as $\theta \to \infty$, as expected.

On the other hand, the hyperboloid embedded in Minkowski space is a surface of constant intrinsic curvature, much like a sphere embedded in Euclidean space.

Actually, the complete hyperboloid not only is not a surface of constant curvature in Euclidean 3-space, it cannot be embedded at all in this space by Hilbert's theorem(it can be locally immersed but there it won't ever approximate a cone as it can't go far from the origin).

 

[PAllen]

Actually, the complete hyperboloid not only is not a surface of constant curvature in Euclidean 3-space, it cannot be embedded at all in this space by Hilbert's theorem(it can be locally immersed but there it won't ever approximate a cone as it can't go far from the origin).

Actually, I was referring to exactly the type of surface @Orodruin guessed, which is certainly embedded in Euclidean 3 space. I was mistaken in thinking this surface is the same as a constant curvature hyperboloid, forgetting the result you mentioned (which I did know previously, but forgot, for the moment). This surface (quadratic hyperboloid, for lack of a better term) has exactly the properties @Orodruin claimed. It is, then, precisely this asymptotically decreasing negative curvature which makes it possible to embed the quadratic hyperboloid.

 

[Tendex]

Actually, I was referring to exactly the type of surface @Orodruin guessed, which is certainly embedded in Euclidean 3 space. I was mistaken in thinking this surface is the same as a constant curvature hyperboloid, forgetting the result you mentioned (which I did know previously, but forgot, for the moment). This surface (quadratic hyperboloid, for lack of a better term) has exactly the properties @Orodruin claimed. It is, then, precisely this asymptotically decreasing negative curvature which makes it possible to embed the quadratic hyperboloid.

I can't figure out what surface you might have been referring to then. There is only a hyperboloid of revolution with negative curvature, the one sheet hyperboloid, and has got constant curvature.
The Ricci curvature of the usual one-sheet hyperboloid Orodruin computed is in fact -2 constantly and it is independent of any embedding.

 

[PAllen]

I can't figure out what surface you might have been referring to then. There is only a hyperboloid of revolution with negative curvature, the one sheet hyperboloid, and has got constant curvature.
The Ricci curvature of the usual one-sheet hyperboloid Orodruin computed is in fact -2 constantly and it is independent of any embedding.

I am referring to the surface @Orodruin mentioned in post #24, except he accidentally quoted the equation of the 2 sheet hyperboloid with positive curvature; the one sheet is 1 = x² + y² - z² . However, the rest of the calculation described should come out the same except for sign, with the same conclusion. This is consistent with the result you mention that the different 2d manifold of constant negative curvature cannot be fully embedded in Euclidean 3 space (smoothly, isometrically). The one sheet hyperboloid is embeddable precisely because its negative curvature is asymptotically zero.

If you think there is a mistake in @orodruin’s calculation, could you please point it out?

 

[Orodruin]

I can't figure out what surface you might have been referring to then. There is only a hyperboloid of revolution with negative curvature, the one sheet hyperboloid, and has got constant curvature.
The Ricci curvature of the usual one-sheet hyperboloid Orodruin computed is in fact -2 constantly and it is independent of any embedding.

This is only true in Minkowski space, not in Euclidean space. When we talk about a hyperboloid, we are implicitly talking about submanifolds of Minkowski/Euclidean space and the metric induced from the embedding. This metric will obviously be different depending on whether you talk about the hyperboloid in Euclidean or Minkowski space, simply because the metric that is being pulled back to the hyperboloid by the embedding is different. What I did was the 2-sheet hyperboloid (because these correspond to the space-like slices in the original question) but it really does not matter for the fact that the curvature is not constant. For the 1-sheet hyperboloid $1 = x^2 + y^2 - z^2$, we can just introduce coordinates $x = \cosh(\theta)\cos(\phi)$, $y = \cosh(\theta)\sin(\phi)$, $z = \sinh(\theta)$ instead. The line element induced from Euclidean space is then
\begin{align*}
ds^2 &= dx^2 + dy^2 + dz^2 \\&= [\sinh(\theta) \cos(\phi) d\theta - \cosh(\theta) \sin(\phi) d\phi]^2
+ [\sinh(\theta) \sin(\phi) d\theta + \cosh(\theta) \cos(\phi) d\phi]^2 + \cosh^2(\theta) d\theta^2
\\&=
[\sinh^2(\theta) + \cosh^2(\theta)] d\theta^2 + \cosh^2(\theta) d\phi^2
\\&=
\cosh(2\theta) d\theta^2 + \cosh^2(\theta) d\phi^2.
\end{align*}
This leads (as @PAllen expected) to the Ricci scalar being given by
$$
R = - \frac{2}{\cosh^2(2\theta)},
$$
which is clearly not constant.

The underlying reason for the hyperboloids having constant curvature when using the metric induced by the embedding in Minkowski space is the same as why the sphere has constant curvature when embedded in Euclidean space; they are invariant under a symmetry transformation of Minkowski space (i.e., Lorentz boosts) and the Lorentz boosts in the embedding Minkowski space therefore induce a Killing field on the hyperboloids. This is not the case when viewing the hyperboloids as embedded in Euclidean space because Lorentz boosts are not a symmetry of Euclidean space.

 

[Orodruin]

If you think there is a mistake in @orodruin’s calculation, could you please point it out?

There would have to be a mistake in the computation of the line element (which I do not think there is, but it should be quite easy to check by hand). The computation of the Ricci scalar itself I left to my Mathematica code (that takes the metric components and spits out Christoffel symbols, Riemann curvature, the Ricci tensor, and the Ricci scalar), which I have tested extensively to reproduce the results for a large number of space-times and other manifolds.

 

[Orodruin]

This is consistent with the result you mention that the different 2d manifold of constant negative curvature cannot be fully embedded in Euclidean 3 space (smoothly, isometrically).

I think it should be stressed that that manifold is a hyperbolic space, not a hyperboloid. These are not the same thing.