What Is the Difference Between Spacetime Curvature and Spatial Curvature?

[Ranku]

Why do we call it spacetime curvature of gravitation and spatial curvature of the universe? Why don't we call it spacetime curvature of the universe?

 

[Ibix]

Why do we call it spacetime curvature of gravitation and spatial curvature of the universe? Why don't we call it spacetime curvature of the universe?

All FLRW models have non-zero spacetime curvature ($R^i{}_{jkl}\neq 0$). What distinguishes the models is the sign and magnitude of curvature of space in FLRW coordinates. So in the particular case of cosmology it is the spatial curvature that is the parameter of interest. You can, of course, measure the spacetime curvature if you wish.

 

[Orodruin]

Because when we talk about spatial curvature we are not referring to the spacetime curvature. We are referring to the curvature of the isotropic and homogenous spatial hypersurfaces of the Robertson-Walker metric.

 

[Orodruin]

All FLRW models have non-zero spacetime curvature (Rijkl≠0).

The Milne universe would like a word ...

 

[Ranku]

All FLRW models have non-zero spacetime curvature ($R^i{}_{jkl}\neq 0$). What distinguishes the models is the sign and magnitude of curvature of space in FLRW coordinates. So in the particular case of cosmology it is the spatial curvature that is the parameter of interest. You can, of course, measure the spacetime curvature if you wish.

So if the spatial curvature is flat, measurement of the spacetime curvature is also flat?

 

[Ibix]

The Milne universe would like a word ...

"All FLRW models with non-zero matter density", then.

 

[Ibix]

So if the spatial curvature is flat, measurement of the spacetime curvature is also flat?

No. As I said, all FLRW models have non-zero spacetime curvature (with Orodruin's exception of the everywhere empty Milne universe, which is just Minkowski spacetime in funny coordinates).

In fact, all spacetimes that aren't Minkowski spacetime (possibly in disguise) have non-zero spacetime curvature.

 

[Ranku]

No. As I said, all FLRW models have non-zero spacetime curvature (with Orodruin's exception of the everywhere empty Milne universe, which is just Minkowski spacetime in funny coordinates).

In fact, all spacetimes that aren't Minkowski spacetime (possibly in disguise) have non-zero spacetime curvature.

Is there a way to correlate non-zero spacetime curvature with spatial flatness?

 

[Ibix]

Is there a way to correlate non-zero spacetime curvature with spatial flatness?

I don't know what you mean.

 

[Ranku]

I don't know what you mean.

Is the time component the main differentiator between spatial curvature and spacetime curvature.

 

[Orodruin]

Spatial curvature is about properties of (in some sense arbitrary) hypersurfaces in spacetime. It does not a priori relate to spacetime curvature, which is a property of the spacetime itself. They are properties of two different objects.

 

[etotheipi]

The idea is that if the spacetime is homogeneous then that means that there exists a family of spacelike hypersurfaces $\Sigma_t$ parameterised by $t$ which foliate the whole spacetime, and further if the spacetime is isotropic then there is a congruence of timelike worldlines whose tangents $u^a$ are orthogonal to the $\Sigma_t$.

The rest is explained by Landau and Lifshitz in volume two. Let the metric on the $\Sigma_t$ be $\gamma_{ab}$ then the curvature tensor of $\Sigma_t$ is $P_{abcd} = \lambda(\gamma_{ac} \gamma_{bd} - \gamma_{ad} \gamma_{bc})$ for a constant $\lambda$. Contract on indices one and three $P_{bd} = {P^a}_{bad} = \lambda(\delta^a_a \gamma_{bd} - \delta^a_d \gamma_{ba}) = \lambda(3 \gamma_{bd} - \gamma_{bd}) = 2\lambda \gamma_{bd}$ then contract on the remaining indices ${P^b}_b = 2\lambda \delta^b_b = 6 \lambda$. Thus $\lambda$ fully characterises the curvature.

First consider the case $\lambda > 0$. To study the geometry L&L consider a hypersphere in $\mathbf{R}^4$ defined by $x_1^2 + x_2^2 + x_3^2 + x_4^2 = a^2$. Then eliminate $x_4$ in the usual line element of $\mathbf{R}^4$ in Cartesian coordinates, i.e. first write$$2x_1 dx_1 + 2x_2 dx_2 + 2x_3 dx_3 + 2x_4 dx_4 = 0 \implies dx_4 = \frac{-x_1 dx_1 - x_2 dx_2 - x_3 dx_3}{x_4}$$and then\begin{align*}

dl^2 &= dx_1^2 + dx_2^2 + dx_3^2 + \frac{ (-x_1 dx_1 - x_2 dx_2 - x_3 dx_3)^2}{a^2 - x_1^2 - x_2^2 - x_3^2} \\

dl^2 &= dx_1^2 + dx_2^2 + dx_3^2 + \frac{ x_1^2 dx_1^2 + x_2^2 dx_2^2 + x_3^2 dx_3^2 + 2x_1 x_2 dx_1 dx_2 + 2x_2 x_3 dx_2 dx_3 + 2x_1 x_3 dx_1 dx_3}{a^2 - x_1^2 - x_2^2 - x_3^2}

\end{align*}In some small neighbourhood $N$ of the origin of $\mathbf{R}^4$ we will write $a^2 - x_1^2 - x_2^2 - x_3^2 \rightarrow a^2$ so that the above reduces to
$$dl^2 = \left(\delta_{ab} + \frac{x_a x_b}{a^2} \right) dx^a dx^b \overset{!}{=} \gamma_{ab} dx^a dx^b \implies \gamma_{ab} = \delta_{ab} + \frac{x_a x_b}{a^2}$$Write$$\Gamma^i_{kl} = \frac{1}{2} \gamma^{im} \left\{ \frac{\partial \gamma_{mk}}{\partial x^l} + \frac{\partial \gamma_{ml}}{\partial x^k} - \frac{\partial \gamma_{kl}}{\partial x^m} \right\}$$these all vanish in the neighbourhood of the origin since the derivatives $\partial_c \gamma_{ab}$ vanish in the neighbourhood of the origin. The curvature tensor reduces to\begin{align*}

P_{ik} &= \frac{\partial \Gamma^l_{ik}}{\partial x^l} - \frac{\partial \Gamma^l_{il}}{\partial x^k} \\

&= \frac{\partial}{\partial x^l} \left( \frac{1}{2} \gamma^{lm} \left\{ \frac{\partial \gamma_{mi}}{\partial x^k} + \frac{\partial \gamma_{mk}}{\partial x^i} - \frac{\partial \gamma_{ik}}{\partial x^m} \right\}\right) - \frac{\partial}{\partial x^k} \left( \frac{1}{2} \gamma^{lm} \left\{ \frac{\partial \gamma_{mi}}{\partial x^l} + \frac{\partial \gamma_{ml}}{\partial x^i} - \frac{\partial \gamma_{il}}{\partial x^m} \right\}\right) \\ \\

&= \frac{1}{2} \gamma^{lm} \frac{\partial}{\partial x^l} \left\{ \frac{\partial \gamma_{mi}}{\partial x^k} + \frac{\partial \gamma_{mk}}{\partial x^i} - \frac{\partial \gamma_{ik}}{\partial x^m} \right\} - \frac{1}{2} \gamma^{lm} \frac{\partial}{\partial x^k} \left\{ \frac{\partial \gamma_{mi}}{\partial x^l} + \frac{\partial \gamma_{ml}}{\partial x^i} - \frac{\partial \gamma_{il}}{\partial x^m} \right\} \\\end{align*}but we may write$$\frac{\partial^2 \gamma_{ab}}{\partial x^i \partial x^j} = \frac{1}{a^2} \frac{\partial}{\partial x^i} \left( x_a \delta_{bj} + \delta_{aj} x_b \right) = \frac{1}{a^2} (\delta_{ai} \delta_{bj} + \delta_{aj} \delta_{bi} )$$so that (let's hope I didn't mess this up!)\begin{align*}

\frac{1}{2} \gamma^{lm} \frac{\partial}{\partial x^l} \left\{ \frac{\partial \gamma_{mi}}{\partial x^k} + \frac{\partial \gamma_{mk}}{\partial x^i} - \frac{\partial \gamma_{ik}}{\partial x^m} \right\} &= \frac{1}{2a^2} \gamma^{lm} \left( \delta_{ml} \delta_{ik} + \delta_{mk} \delta_{il} + \delta_{ml} \delta_{ki} + \delta_{mi} \delta_{kl} - \delta_{il} \delta_{km} - \delta_{im} \delta_{kl}\right) \\

&= \frac{1}{2a^2} \gamma^{lm} \left( \delta_{ml} \delta_{ik} + \delta_{ml} \delta_{ki} \right) \\

&= \frac{1}{a^2} \gamma^{lm} \left( \delta_{ml} \delta_{ik} \right) \\

&= \frac{1}{a^2} \delta^m_m \delta_{ik} = \frac{3}{a^2} \delta_{ik}

\end{align*}and similarly\begin{align*}

\frac{1}{2} \gamma^{lm} \frac{\partial}{\partial x^k} \left\{ \frac{\partial \gamma_{mi}}{\partial x^l} + \frac{\partial \gamma_{ml}}{\partial x^i} - \frac{\partial \gamma_{il}}{\partial x^m} \right\} &= \frac{1}{2a^2} \gamma^{lm} \left( \delta_{mk} \delta_{il} + \delta_{ml} \delta_{ik} + \delta_{mk} \delta_{li} + \delta_{mi} \delta_{lk} - \delta_{ik} \delta_{lm} - \delta_{im} \delta_{lk}\right) \\

&= \frac{1}{2a^2} \gamma^{lm} \left( \delta_{mk} \delta_{il} + \delta_{mk} \delta_{li} \right) \\

&= \frac{1}{a^2} \gamma^{lm} \delta_{mk} \delta_{il} \\

&= \frac{1}{a^2} \delta^l_k \delta_{il} = \frac{1}{a^2} \delta_{ik}

\end{align*}Thus doing the subtraction,\begin{align*}
P_{ik} &= \frac{\partial \Gamma^l_{ik}}{\partial x^l} - \frac{\partial \Gamma^l_{il}}{\partial x^k} \\

&= \frac{2}{a^2} \delta_{ik}

\end{align*}we know that $P_{ik} = 2 \lambda \gamma_{ik}$ which in the neighbourhood of the origin should reduce to $2 \lambda \delta_{ik}$ in which case we can identify$$\lambda = \frac{1}{a^2}$$as the curvature. For the case of $\lambda < 0$ you can obtain similar results by using the formal substitution $a \mapsto ia$, i.e. you will have $\lambda = - \frac{1}{a^2}$.

 

[Ibix]

Is the time component the main differentiator between spatial curvature and spacetime curvature.

No. You can have flat space in curved spacetime (e.g. spatially flat FLRW) and curved space in flat spacetime (e.g. the Milne cosmology).

As Orodruin says, the choice of what you call "space" is more or less arbitrary. Sure, you might well pick a definition of space that has some relation to some directly measurable quantity and this will probably reflect something of the spacetime. But it won't necessarily reflect the flatness or otherwise of the spacetime, and anyway you're not obligated to use a definition of "space" that makes sense globally. There may not even be such a thing.

 

[Ranku]

The idea is that if the spacetime is homogeneous then that means that there exists a family of spacelike hypersurfaces $\Sigma_t$ parameterised by $t$ which foliate the whole spacetime, and further if the spacetime is isotropic then there is a congruence of timelike worldlines whose tangents $u^a$ are orthogonal to the $\Sigma_t$.

The rest is explained by Landau and Lifshitz in volume two. Let the metric on the $\Sigma_t$ be $\gamma_{ab}$ then the curvature tensor of $\Sigma_t$ is $P_{abcd} = \lambda(\gamma_{ac} \gamma_{bd} - \gamma_{ad} \gamma_{bc})$ for a constant $\lambda$. Contract on indices one and three $P_{bd} = {P^a}_{bad} = \lambda(\delta^a_a \gamma_{bd} - \delta^a_d \gamma_{ba}) = \lambda(3 \gamma_{bd} - \gamma_{bd}) = 2\lambda \gamma_{bd}$ then contract on the remaining indices ${P^b}_b = 2\lambda \delta^b_b = 6 \lambda$. Thus $\lambda$ fully characterises the curvature.

First consider the case $\lambda > 0$. To study the geometry L&L consider a hypersphere in $\mathbf{R}^4$ defined by $x_1^2 + x_2^2 + x_3^2 + x_4^2 = a^2$. Then eliminate $x_4$ in the usual line element of $\mathbf{R}^4$ in Cartesian coordinates, i.e. first write$$2x_1 dx_1 + 2x_2 dx_2 + 2x_3 dx_3 + 2x_4 dx_4 = 0 \implies dx_4 = \frac{-x_1 dx_1 - x_2 dx_2 - x_3 dx_3}{x_4}$$and then\begin{align*}

dl^2 &= dx_1^2 + dx_2^2 + dx_3^2 + \frac{ (-x_1 dx_1 - x_2 dx_2 - x_3 dx_3)^2}{a^2 - x_1^2 - x_2^2 - x_3^2} \\

dl^2 &= dx_1^2 + dx_2^2 + dx_3^2 + \frac{ x_1^2 dx_1^2 + x_2^2 dx_2^2 + x_3^2 dx_3^2 + 2x_1 x_2 dx_1 dx_2 + 2x_2 x_3 dx_2 dx_3 + 2x_1 x_3 dx_1 dx_3}{a^2 - x_1^2 - x_2^2 - x_3^2}

\end{align*}In some small neighbourhood $N$ of the origin of $\mathbf{R}^4$ we will write $a^2 - x_1^2 - x_2^2 - x_3^2 \rightarrow a^2$ so that the above reduces to
$$dl^2 = \left(\delta_{ab} + \frac{x_a x_b}{a^2} \right) dx^a dx^b \overset{!}{=} \gamma_{ab} dx^a dx^b \implies \gamma_{ab} = \delta_{ab} + \frac{x_a x_b}{a^2}$$Write$$\Gamma^i_{kl} = \frac{1}{2} \gamma^{im} \left\{ \frac{\partial \gamma_{mk}}{\partial x^l} + \frac{\partial \gamma_{ml}}{\partial x^k} - \frac{\partial \gamma_{kl}}{\partial x^m} \right\}$$these all vanish in the neighbourhood of the origin since the derivatives $\partial_c \gamma_{ab}$ vanish in the neighbourhood of the origin. The curvature tensor reduces to\begin{align*}

P_{ik} &= \frac{\partial \Gamma^l_{ik}}{\partial x^l} - \frac{\partial \Gamma^l_{il}}{\partial x^k} \\

&= \frac{\partial}{\partial x^l} \left( \frac{1}{2} \gamma^{lm} \left\{ \frac{\partial \gamma_{mi}}{\partial x^k} + \frac{\partial \gamma_{mk}}{\partial x^i} - \frac{\partial \gamma_{ik}}{\partial x^m} \right\}\right) - \frac{\partial}{\partial x^k} \left( \frac{1}{2} \gamma^{lm} \left\{ \frac{\partial \gamma_{mi}}{\partial x^l} + \frac{\partial \gamma_{ml}}{\partial x^i} - \frac{\partial \gamma_{il}}{\partial x^m} \right\}\right) \\ \\

&= \frac{1}{2} \gamma^{lm} \frac{\partial}{\partial x^l} \left\{ \frac{\partial \gamma_{mi}}{\partial x^k} + \frac{\partial \gamma_{mk}}{\partial x^i} - \frac{\partial \gamma_{ik}}{\partial x^m} \right\} - \frac{1}{2} \gamma^{lm} \frac{\partial}{\partial x^k} \left\{ \frac{\partial \gamma_{mi}}{\partial x^l} + \frac{\partial \gamma_{ml}}{\partial x^i} - \frac{\partial \gamma_{il}}{\partial x^m} \right\} \\\end{align*}but we may write$$\frac{\partial^2 \gamma_{ab}}{\partial x^i \partial x^j} = \frac{1}{a^2} \frac{\partial}{\partial x^i} \left( x_a \delta_{bj} + \delta_{aj} x_b \right) = \frac{1}{a^2} (\delta_{ai} \delta_{bj} + \delta_{aj} \delta_{bi} )$$so that (let's hope I didn't mess this up!)\begin{align*}

\frac{1}{2} \gamma^{lm} \frac{\partial}{\partial x^l} \left\{ \frac{\partial \gamma_{mi}}{\partial x^k} + \frac{\partial \gamma_{mk}}{\partial x^i} - \frac{\partial \gamma_{ik}}{\partial x^m} \right\} &= \frac{1}{2a^2} \gamma^{lm} \left( \delta_{ml} \delta_{ik} + \delta_{mk} \delta_{il} + \delta_{ml} \delta_{ki} + \delta_{mi} \delta_{kl} - \delta_{il} \delta_{km} - \delta_{im} \delta_{kl}\right) \\

&= \frac{1}{2a^2} \gamma^{lm} \left( \delta_{ml} \delta_{ik} + \delta_{ml} \delta_{ki} \right) \\

&= \frac{1}{a^2} \gamma^{lm} \left( \delta_{ml} \delta_{ik} \right) \\

&= \frac{1}{a^2} \delta^m_m \delta_{ik} = \frac{3}{a^2} \delta_{ik}

\end{align*}and similarly\begin{align*}

\frac{1}{2} \gamma^{lm} \frac{\partial}{\partial x^k} \left\{ \frac{\partial \gamma_{mi}}{\partial x^l} + \frac{\partial \gamma_{ml}}{\partial x^i} - \frac{\partial \gamma_{il}}{\partial x^m} \right\} &= \frac{1}{2a^2} \gamma^{lm} \left( \delta_{mk} \delta_{il} + \delta_{ml} \delta_{ik} + \delta_{mk} \delta_{li} + \delta_{mi} \delta_{lk} - \delta_{ik} \delta_{lm} - \delta_{im} \delta_{lk}\right) \\

&= \frac{1}{2a^2} \gamma^{lm} \left( \delta_{mk} \delta_{il} + \delta_{mk} \delta_{li} \right) \\

&= \frac{1}{a^2} \gamma^{lm} \delta_{mk} \delta_{il} \\

&= \frac{1}{a^2} \delta^l_k \delta_{il} = \frac{1}{a^2} \delta_{ik}

\end{align*}Thus doing the subtraction,\begin{align*}
P_{ik} &= \frac{\partial \Gamma^l_{ik}}{\partial x^l} - \frac{\partial \Gamma^l_{il}}{\partial x^k} \\

&= \frac{2}{a^2} \delta_{ik}

\end{align*}we know that $P_{ik} = 2 \lambda \gamma_{ik}$ which in the neighbourhood of the origin should reduce to $2 \lambda \delta_{ik}$ in which case we can identify$$\lambda = \frac{1}{a^2}$$as the curvature. For the case of $\lambda < 0$ you can obtain similar results by using the formal substitution $a \mapsto ia$, i.e. you will have $\lambda = - \frac{1}{a^2}$.

Are there explicit equations that describe spacetime curvature and spatial curvature? Comparing them visually would help me see the difference.

 

[bhobba]

Why do we call it spacetime curvature of gravitation and spatial curvature of the universe? Why don't we call it spacetime curvature of the universe?

Well, gravitation is space-time curvature or at least modeled by it anyway. You can talk about the spacetime curvature of the universe if you wish. I think at the start of your studies in GR, thinking of it as spacetime curvature in both cases is likely best, even though for the whole universe, space is often talked about. After all, Hubble did show that space is expanding. The following from Scientific American may help:

https://blogs.scientificamerican.co...what-do-you-mean-the-universe-is-flat-part-i/

Is the time component the main differentiator between spatial curvature and spacetime curvature.

Interestingly in the small gravitational limit of GR (such as here on earth), it turns out that the curvature of time is the most important thing, the curvature of space is tiny:
https://en.wikipedia.org/wiki/Spacetime#Curvature_of_time

Hopefully, as an I level thread, those reading this can understand the above.

Are there explicit equations that describe spacetime curvature and spatial curvature? Comparing them visually would help me see the difference.

That post is more at the A than I level. So if you find it difficult to understand, don't worry. I have studied GR for many years now, with Wald (an advanced text) as my main reference. I have also studied Landau's Classical Theory of Fields referenced in that post. I found it hard going - which is reflected s my 'wow' rating. So if you find it hard going - do not worry. I hope my explanation and the link to Wikipedia and Scientific American helped you understand what is going on.

Added Later:

It occurred to me knowing about the FLRW metric often used in cosmology may also help:
https://en.wikipedia.org/wiki/Friedmann–Lemaître–Robertson–Walker_metric

Thanks
Bill

 

[Orodruin]

Interestingly in the small gravitational limit of GR (such as here on earth), it turns out that the curvature of time is the most important thing, the curvature of space is tiny:

That’s a bit if a misnomer in my mind. A single dimension cannot have curvature. In the end, curvature is described by the Riemann curvature tensor where all components with the same index everywhere are identically zero because of the antisymmetries of the tensor.

What is appearing in the Wiki article is more about the metric components than the actual curvature. It is true that the tidal effects, described by the geodesic deviation equation, mainly depend on the cuvature component $R_{0i0j}$ so the time index is in as many places as it can be without causing a zero by definition.

 

[bhobba]

That’s a bit if a misnomer in my mind. A single dimension cannot have curvature.

Yes - the article could have been worded better. Spacetime is curved - not time - which makes no sense, as you correctly say. Of course, what is meant here is the curvature of space is so tiny it can be considered flat. It is necessary to use space-time to account for gravity here on earth. It is the time component that is important.

Thanks
Bill

 

[etotheipi]

Are there explicit equations that describe spacetime curvature and spatial curvature? Comparing them visually would help me see the difference.

Using Landau's notation, the former is described by the tensor $R$ whilst the latter is described by the tensor $P$. They are just the curvature tensors ascribed to different manifolds, and actually in this case any $\Sigma_t$ is a submanifold of $M$.

 

[kimbyd]

Are there explicit equations that describe spacetime curvature and spatial curvature? Comparing them visually would help me see the difference.

One way to visualize the space-time curvature is to use the definition of the Ricci Scalar, which can be written as $\sum_\mu R_\mu^\mu$, where the index $\mu$ goes over all four coordinates. This value is independent of choice of coordinates, and is a fundamental property of the manifold.

The spatial curvature can be extracted by doing the above sum over only the spatial coordinates. This comes from doing a time-slicing of the universe, and looking at the curvature of each slice. Such a construction is inherently coordinate-specific, and therefore isn't fundamental in the same way that space-time curvature is fundamental: spatial curvature is critically dependent upon how we write down the equations for the universe. In particular, it depends upon which slices of the universe we define to be equal-time slices.

The question, then, is whether or not there is a way to create equal-time slices to extract useful information? The answer to that is very much yes, due to the symmetries our observable universe observes: homogeneity and isotropy. That homoegeneity and isotropy is only present with a particular choice of time slicing, which turns out to be an equal-time slice where the temperature of the CMB is the same across the entire slice. With that choice, we can calculate a useful spatial curvature.