Curvature of space vs. curvature of spacetime

[arupel]

I have no problem understanding the curvature of spacetime. I read Misner. I have a big problem in understanding what the curvature of space is in modern physics. All I seem to know about it is that it is space, in a cosmological sense, that is expanding, not spacetime, but other than that I have no idea what is meant by the curvature of space. Can someone please elaborate. Is this still in some sense, a classical concept that got glued to GR?

Thanks,
Arhur Rupel

 

[facenian]

I have no problem understanding the curvature of spacetime. I read Misner. I have a big problem in understanding what the curvature of space is in modern physics. All I seem to know about it is that it is space, in a cosmological sense, that is expanding, not spacetime, but other than that I have no idea what is meant by the curvature of space. Can someone please elaborate. Is this still in some sense, a classical concept that got glued to GR?
Thanks,
Arhur Rupel

May be this example can help. Let's consider special relativiy(SR) only. In this case we have a flat space-time(zero curvature). I we use a non inertial reference frame to descrive this flat space-time manifold then we would find that ordinary three dimensional space is non-euclidean and the curvature of space is non zero. Am I correct?

 

[A.T.]

I have a big problem in understanding what the curvature of space is in modern physics.

For example:
https://en.wikipedia.org/wiki/Schwarzschild_metric#Flamm's_paraboloid

 

[PeterDonis]

I have no idea what is meant by the curvature of space

It means the curvature of a 3-dimensional spacelike hypersurface of constant coordinate time in a particular coordinate chart. In cosmology, the coordinate chart is the standard FRW chart, in which "comoving" observers have constant spatial coordinates. In the case given in A.T.'s post, the coordinate chart is the standard Schwarzschild chart on Schwarzschild spacetime.

we use a non inertial reference frame to descrive this flat space-time manifold then we would find that ordinary three dimensional space is non-euclidean and the curvature of space is non zero. Am I correct?

It depends on which non-inertial coordinates you pick. If you pick Rindler coordinates, space is Euclidean and the curvature of space is zero.

 

[johnkclark]

it takes 20 numbers to describe the curvature of spacetime.

I was under the impression that in 4D spacetime the Einstein Tensor had 16 components and only 10 of them are independent.

John K Clark

 

[PeterDonis]

I was under the impression that in 4D spacetime the Einstein Tensor had 16 components and only 10 of them are independent.

Yes, but the Einstein tensor does not describe all of the spacetime curvature. For that you need the Riemann tensor, which has 20 independent components. You can split those 20 up into the 10 independent components of the Einstein tensor, and the 10 independent components of the Weyl tensor; but you still need 20 total.

 

[johnkclark]

Yes, but the Einstein tensor does not describe all of the spacetime curvature. For that you need the Riemann tensor, which has 20 independent components. You can split those 20 up into the 10 independent components of the Einstein tensor, and the 10 independent components of the Weyl tensor; but you still need 20 total.

How did Einstein manage to get by with just 10? Did he cut corners or are some of those spacetime curvatures unphysical and so are of interest to a mathematician but not a physicist?

John K Clark

 

[PAllen]

I was under the impression that in 4D spacetime the Einstein Tensor had 16 components and only 10 of them are independent.

John K Clark

The Einstein tensor is only part of the curvature. For example, the Einstein tensor is identically zero for the Schwarzschild solution which describes gravity around spherically symmetric body. The referenced discussion is about the full curvature tensor which is not zero for the Schwarzschild solution.

 

[PAllen]

How did Einstein manage to get by with just 10? Did he cut corners or are some of those spacetime curvatures unphysical and so are of interest to a mathematician but not a physicist?

John K Clark

He didn’t get by on just ten. As noted in my prior post, Weyl curvature is a critical part of the theory - all of solar system motion is determined by Weyl curvature. Even gravitational radiation is all Weyl curvature I.e. vanishing Einstein tensor.

 

[PAllen]

Setting the Einstein tensor to zero is the equation describing gravity in vacuum. It may be vacuum, but it is hardly a vacuous equation - it is still a nonlinear system of partial differential equations for determining gravitation outside massive bodies.

 

[PeterDonis]

How did Einstein manage to get by with just 10?

He didn't. The fact that the Einstein tensor is named after Einstein and appears in the Einstein Field Equation does not mean it's the only tensor he used in GR. He used the metric tensor, the Riemann tensor, the Weyl tensor, the Ricci tensor, the Einstein tensor, and the stress-energy tensor.

 

[Nugatory]

How did Einstein manage to get by with just 10? Did he cut corners or are some of those spacetime curvatures unphysical and so are of interest to a mathematician but not a physicist?

They're all important and Einstein cut no corners. The Einstein field equations relate the stress-energy tensor to the Einstein tensor. However, the Einstein tensor is defined in terms of the metric tensor (directly from the $g_{\mu\nu}$ term and indirectly because the Ricci tensor $R_{\mu\nu}$ and constant $R$ are calculated from the Riemann tensor which in turn is calculated from the metric). Thus, solving the field equation means solving for the metric tensor (not the Einstein tensor).

Different metric tensors will lead to different Riemann tensors but different Riemann tensors can produce the same Ricci tensor and hence the same Einstein tensor, because Riemann has more degrees of freedom than Ricci. Thus, different spacetimes with different metrics and different Riemann tensors can satisfy the Einstein field equations for the same stress-energy tensor; or equivalently specifying the stress-energy tensor and hence the Einstein tensor does not completely specify the spacetime and the metric. This situation is no different than with any other problem involving differential equations: solving the differential equation gives you an entire family of possible solutions, and you have to use boundary conditions or other external constraints to select which of these corresponds to the physical situation you're working with.

One example: In vacuum the stress-energy tensor is zero, so the Einstein field equations reduce to $G=0$; among the many metric tensors that satisfy this equation are the Schwarzschild metric, the Kerr metric, a gravitatonal wave passing through empty space, and the ordinary Minkwowski metric of the flat spacetime of special relativity. The curvature and hence the Riemann tensors are different, but these differences appear in the ten degrees of freedom that aren't part of the Ricci and Einstein tensors. A corollary is that the Weyl tensors are different; the members of a family of solutions for a particular stress-energy tensor are distinguished by their Weyl tensors.

But despite this proliferation of tensors... The metric tensor is the foundational one. Know it, and you know everything there is to know about the spacetime. The Einstein field equations can be viewed as a constraint on the possible metric tensors, given a particular physical configuration described by a particular stress-energy tensor.

 

[pervect]

One approach I've recently seen (in a couple of different papers) is rather interesting. One of the papers is "Why the Riemann Curvature Tensor needs twenty independent components", David Melgin.

A short summary, which is not intended to be a complete exposition of the short paper, but simply to give an overview of the conclusions.

The metric tensor has 10 degrees of freedom, but the number of degrees of freedom in it's simplification via tensor coordinate transforms is 16.

If we assume the worst possible situation arises, ten of the independently
alterable $dx^a / dy^u$
are required to transform the arbitrary metric to the Minkowski
metric. After using these ten numbers we still have six remaining degrees of
freedom within the coordinate transformation. A little imagination leads one
to believe that there might be a six parameter family of local transformations
at every point which lead the form of the metric unchanged. Remember we are
still only working with a single point so don’t go looking for a six parameter
family of global transformations for any metric.

Next one considers the first derivatives of the metric. The conclusion here is:

Any arbitrary metric can locally be made into the Minkowski metric with
vanishing first derivatives, consistent with Riemann normal coordinates.

See the paper for the details of the argument, which involves a Taylor expansion of the metric to first order.

Now one considers the second derivatives of the metric. The author's conclusion is as follows:

An arbitrary metric will have a total of a hundred independent second deriva-
tives of the metric, but only eighty numbers to simplify them. Outside of special
cases we end up having twenty second derivatives of the metric be non-zero no
matter how cleverly we choose our coordinates. Information describing the es-
sential unsimplifiable nature of the second derivative of the metric is contained
in contained in these twenty functions. Any attempt to construct a tensor de-
scribing the second derivatives of the metric or any of their properties must
have at least twenty functions. Careful analysis shows the Riemann curvature
tensor has exactly twenty independent components. We can understand these
independent components as conveying the coordinate unsimplifiable nature of
the second derivatives of the metric.

So while the metric tensor has only 10 degrees of freedom, it's higher order derivatives can have more degrees of freedom. We can always find a coordinate system that makes the metric locally Minkowskian, and we can go beyond this to finding a coordinate system which also makes the first order derivatives of the metric tensor zero.. When we consider the second derivaitves of the metric, we cannot (in the general case) make them all zero, in general we are left with 20 numbers (degrees of freedom) that cannot be elimianted by coordinate changes. These numbers, which cannot be transformed away by coordinate changes, turn out to represent the degrees of freedom in the curvature.

 

[PAllen]

Well, I don't buy that argument. Not only Spivak, but also Riemann and Einstein and many others have argued that 6 arbitrary functions of the manifold are sufficient to specify the metric up to diffeomorphism. From that, the curvature tensor follows. The 20 independent components of the curvature tensor represent algebraic not functional degrees of freedom. The definition of the curvature tensor from the metric amounts to a huge number of differential constraints.

The argument I find simplest is that in some coordinate patch, you can specify e.g. harmonic gauge without loss of generality. This then leaves only 6 remaining functions to determine the metric in 4 d. Once you have the metric, all derivatives of any order are determined. I first encountered this argument in one of Einstein’s papers, but I assume he was’t the originator

 

[pervect]

Well, I don't buy that argument. Not only Spivak, but also Riemann and Einstein and many others have argued that 6 arbitrary functions of the manifold are sufficient to specify the metric up to diffeomorphism.

Do you have a reference? I'm not familiar with your argument about six arbitrary functions. I can, however, follow that there are 10 independent numbers in a metric. This excludes any considerations of diffeomorphisms, so it's not necessarily in conflict with what you said (though I'm not qutie following what you said). And the tensor transformation rules give us 16 possible linear transformations (a 4x4 matrix) , which gives us more than enough degrees of freedom to transform away all the components of the metric tensor at a single point. Which means that specifying the metric tensor alone at a single point can't tell us anything physical, as we can always find a coordinate system in which the metric tensor is diag(-1,1,1,1).

The argument "with a little imagination" made by Melgrin isn't quite rigorous enough for me to want to defend, though I thought it was interesting.

The conclusion that we can transform the metric to diag(-1,1,1,1) and also make all it's first-order derivatives vanish at a single point also seems reasonably obvious on physical grounds from the existence of Riemann normal coordinates.

From that, the curvature tensor follows. The 20 independent components of the curvature tensor represent algebraic not functional degrees of freedom. The definition of the curvature tensor from the metric amounts to a huge number of differential constraints.

The argument I find simplest is that in some coordinate patch, you can specify e.g. harmonic gauge without loss of generality. This then leaves only 6 remaining functions to determine the metric in 4 d. Once you have the metric, all derivatives of any order are determined. I first encountered this argument in one of Einstein’s papers, but I assume he was’t the originator

Sorry, I'm not following this at all. Do you have any references or can you explain further?

 

[PAllen]

Do you have a reference? I'm not familiar with your argument about six arbitrary functions. I can, however, follow that there are 10 independent numbers in a metric. This excludes any considerations of diffeomorphisms, so it's not necessarily in conflict with what you said (though I'm not qutie following what you said). And the tensor transformation rules give us 16 possible linear transformations (a 4x4 matrix) , which gives us more than enough degrees of freedom to transform away all the components of the metric tensor at a single point. Which means that specifying the metric tensor alone at a single point can't tell us anything physical, as we can always find a coordinate system in which the metric tensor is diag(-1,1,1,1).

The argument "with a little imagination" made by Melgrin isn't quite rigorous enough for me to want to defend, though I thought it was interesting.

The conclusion that we can transform the metric to diag(-1,1,1,1) and also make all it's first-order derivatives vanish at a single point also seems reasonably obvious on physical grounds from the existence of Riemann normal coordinates.
Sorry, I'm not following this at all. Do you have any references or can you explain further?

One reference, to Spivak, is given in post #31, for the specific application to sectional curvature. The most succinct online reference I can find is to math stack exchange:

https://math.stackexchange.com/ques...a-metric-have-on-a-psuedo-riemannian-manifold

 

[pervect]

One reference, to Spivak, is given in post #31, for the specific application to sectional curvature. The most succinct online reference I can find is to math stack exchange:

https://math.stackexchange.com/ques...a-metric-have-on-a-psuedo-riemannian-manifold

After some thought, I have to agree that the original article I quoted is probabl flawed. If we really had 10 degrees of freedom in the metric tensor, the metric tensor could be made all zero - there would be no constraints. But that isn't correct. I do think the idea of counting the degrees of freedom of the metric, it's first derivatives, and the second derivatives offers an explanation of how the curvature tensor winds up with more degrees of freedom than the metric does. But I have to agree that the detailed analysis appears to need a bit more work than the paper(s) I was looking at, and I'm not familiar enough with the topic to fix the argument.

 

[PAllen]

After some thought, I have to agree that the original article I quoted is probabl flawed. If we really had 10 degrees of freedom in the metric tensor, the metric tensor could be made all zero - there would be no constraints. But that isn't correct. I do think the idea of counting the degrees of freedom of the metric, it's first derivatives, and the second derivatives offers an explanation of how the curvature tensor winds up with more degrees of freedom than the metric does. But I have to agree that the detailed analysis appears to need a bit more work than the paper(s) I was looking at, and I'm not familiar enough with the topic to fix the argument.

You may find the following thread of interest. Focus only on the OP and page 4 when Ben Niehoff joins the discussion, and some final references Atty found ( a little of page 3 might be interesting, but much before the last page just goes in circles).

https://www.physicsforums.com/threads/independent-components-of-the-curvature-tenso.530857/