B Curvature of space vs. curvature of spacetime

[Buckethead]

Regarding curvature of spacetime/space: At some given point in a gravitational field, spacetime is curved at that point and this is a constant. (I'm assuming this is true).

Although we can talk about the curvature of spacetime, I never hear anyone talking about the curvature of space. Can curvature of space (not spacetime) be talked about if one takes into account the velocity of an inertial object. My line of thinking is this: An inertial object passing by a large mass will trace a curve as seen by an observer at some prime vantage point, but will be tracing a different curve than another object passing at the same distance but with a different velocity. Both are inertial and therefore are both following the same path through spacetime, but tracing different paths through space. It seems to me that the curvature of spacetime is simply a more compact way of expressing the combination of space and velocity. Is this true? In other words, can the curvature of spacetime be calculated from the trajectory of an inertial object and its velocity?

 

[PeterDonis]

At some given point in a gravitational field, spacetime is curved at that point and this is a constant. (I'm assuming this is true).

By "point" do you mean "point in spacetime" or "point in space"?

If you mean "point in spacetime", then the spacetime curvature at a point is what it is--"constant" isn't really the right word because it implies that it could be changing but just doesn't happen to be.

If you mean "point in space", then it depends on how you split spacetime up into space and time. See below.

Although we can talk about the curvature of spacetime, I never hear anyone talking about the curvature of space.

Actually, there is plenty of talk about curvature of space, but pretty much all of it glosses over the key reason why it's better to talk about curvature of spacetime: because curvature of spacetime is a tensor--the Riemann tensor--and doesn't depend on how you choose coordinates, i.e., how you split up spacetime into "space" and "time". But curvature of space does depend on that; i.e., it's not a coordinate-independent thing, and the physics can only depend on coordinate-independent things.

Both are inertial and therefore are both following the same path through spacetime,

No, they aren't. "Different velocity" means "different paths through spacetime". The paths just happen to intersect at a particular point (the point where both are at the same distance from the large mass).

can the curvature of spacetime be calculated from the trajectory of an inertial object and its velocity?

Not from just one object, no. You need enough different ones to determine all of the components of the Riemann tensor, and there are twenty independent components. IIRC you need at least four inertial objects in a local region of spacetime, and they have to have their motions arranged just right to probe the Riemann tensor components properly. And that's just to determine the Riemann tensor in a small local region of spacetime.

 

[Dale]

Can curvature of space (not spacetime) be talked about

Yes, but you have to be careful to identify what you mean by “space”. Any smooth spacelike slice of spacetime is itself a manifold, so you can compute its curvature.

Both are inertial and therefore are both following the same path through spacetime,

This is not correct. I am not sure where you are getting it from.

It seems to me that the curvature of spacetime is simply a more compact way of expressing the combination of space and velocity. Is this true?

The curvature of spacetime is mostly what you think of as tidal effects.

 

[Buckethead]

If you mean "point in spacetime", then the spacetime curvature at a point is what it is--"constant" isn't really the right word because it implies that it could be changing but just doesn't happen to be.

Right, I should have used "value". So a point in spacetime will have a specific value. When we talk about curvature of spacetime, are we only talking about a curved line in a spacetime diagram, or does it have a real physical meaning as in a curve through space? You mentioned a deep mathematical relationship between velocity of objects through space and curvature of spacetime (which is very eye opening) so it does seem there is a translation to the physical, but its certainly much more involved than I thought.

No, they aren't. "Different velocity" means "different paths through spacetime". The paths just happen to intersect at a particular point (the point where both are at the same distance from the large mass).

I see. I'll explain further down why I came to this conclusion.

This is not correct. I am not sure where you are getting it from.

Bad logic on my part. It comes from me trying to think more into the definition of a "straight line". I think of a straight line as the path an object takes when its moving inertially, but that can't quite be true because objects in curved spacetime will move in different "straight lines" depending on their velocity so concluded instead that they DO follow the same path through spacetime. You gave a good definition of a straight line in another thread, but I can't find it. In any case, my conclusion here is also not correct. I would like a good definition of a straight line.

The curvature of spacetime is mostly what you think of as tidal effects.

Ah yes, I remember you mentioning this before and I must remember this!

 

[PeterDonis]

a point in spacetime will have a specific value.

The curvature of spacetime at a point is not a single number, it's 20 numbers: the 20 independent components of the Riemann tensor at that point.

When we talk about curvature of spacetime, are we only talking about a curved line in a spacetime diagram, or does it have a real physical meaning as in a curve through space?

Neither. Spacetime curvature is tidal gravity, as @Dale pointed out in an earlier post.

You mentioned a deep mathematical relationship between velocity of objects through space and curvature of spacetime

Did I? Can you quote the specific thing I said that gave you that impression?

I think of a straight line as the path an object takes when its moving inertially

That's correct.

that can't quite be true because objects in curved spacetime will move in different "straight lines" depending on their velocity

Yes, because "different velocities" means "straight lines pointed in different directions in spacetime". At any given point in spacetime, there are an infinite number of straight lines (more precisely, timelike geodesics) passing through that point. Each distinct straight line corresponds to a different possible velocity that an object can have at that point.

I would like a good definition of a straight line.

A geodesic in spacetime. Since we are only talking about timelike geodesics, you can think of geodesics as curves that maximize the proper time between two nearby events. The infinite number of possible timelike geodesics at a given point in spacetime corresponds to the infinite number of possible nearby points, each in a distinct timelike direction.

 

[Dale]

I would like a good definition of a straight line.

There is an “I” level mathematical definition that can be used rigorously. But at a “B” level the best definition is probably “the shortest path between two points is a straight line”. For spacetime a straight line (geodesic) doesn’t necessarily minimize the interval, it extremizes it, but the core concept remains.

 

[PeterDonis]

at a “B” level the best definition is probably “the shortest path between two points is a straight line”.

For timelike lines, it's longest, though, not shortest. At least, that's the best "B" level definition.

 

[A.T.]

I think of a straight line as the path an object takes when its moving inertially,

When it moves inertially, then it's path in space-time is a locally straight line (geodesic).

but that can't quite be true because objects in curved spacetime will move in different "straight lines" depending on their velocity

And what's problem with different straight paths starting a the same initial point?

so concluded instead that they DO follow the same path through spacetime.

Why should they, if their initial directions through space-time are different?

 

[Buckethead]

The curvature of spacetime at a point is not a single number, it's 20 numbers: the 20 independent components of the Riemann tensor at that point.

Interesting. A spacetime line cannot be drawn through space simply based on an objects velocity or trajectory but involves much more.

Neither. Spacetime curvature is tidal gravity, as @Dale pointed out in an earlier post.

Right, a curved line in a spacetime diagram is simply an object changing it's speed.

That's correct.

Wow! It's not often I make a correct comment in PF. I should quit right now and go celebrate!

es, because "different velocities" means "straight lines pointed in different directions in spacetime". At any given point in spacetime, there are an infinite number of straight lines (more precisely, timelike geodesics) passing through that point. Each distinct straight line corresponds to a different possible velocity that an object can have at that point.

Ah, wonderful. Yes, of course. But I have to be careful that I clearly understand this. When you say "straight lines ... in spacetime" you are referring to lines on a spacetime (worldline) diagram correct? And this is why one can say an inertial object even near a massive body is moving in a straight line. One other thing. How is a line in a spacetime diagram drawn so it represents gravity (curved spacetime)?

Did I? Can you quote the specific thing I said that gave you that impression?

Sure, you said:

Not from just one object, no. You need enough different ones to determine all of the components of the Riemann tensor, and there are twenty independent components. IIRC you need at least four inertial objects in a local region of spacetime, and they have to have their motions arranged just right to probe the Riemann tensor components properly. And that's just to determine the Riemann tensor in a small local region of spacetime.

That to me is a deep mathematical relationship although I think I jumped the gun when I said "between velocity of objects through space and curvature of spacetime" since I guess you were talking about something entirely different not related to that at all.

But at a “B” level the best definition is probably “the shortest path between two points is a straight line”.

That's the shortest path on a spacetime diagram right? For a straight line in space though I now understand it means the path traced by an inertial object which can vary depending on the speed of the object. Correct?.

 

[PeterDonis]

A spacetime line cannot be drawn through space simply based on an objects velocity or trajectory but involves much more.

A "spacetime line" is not the same thing as spacetime curvature. At any given point in spacetime, if you know an object's velocity, and you know it's in free fall, then that is indeed sufficient to determine its entire worldline, i.e., the "spacetime line" that describes its entire history.

a curved line in a spacetime diagram is simply an object changing it's speed.

Only if "curved" means "non-geodesic". But if you are imagining that a spacetime diagram of a curved spacetime works just like the spacetime diagrams in SR, where straight lines--geodesics, worldlines of freely falling objects--are actually straight lines on the diagram, you are imagining it wrong. In a curved spacetime there is no way to draw such a spacetime diagram that covers the entire spacetime.

I think I jumped the gun when I said "between velocity of objects through space and curvature of spacetime" since I guess you were talking about something entirely different not related to that at all.

Yep.

That's the shortest path on a spacetime diagram right?

No. Go back and read my previous posts where I explained how timelike lines, the kinds of lines we are talking about, work differently.

a straight line in space

Forget straight lines in space. You'll only confuse yourself. Go back and read my post #2, carefully.

 

[Buckethead]

A "spacetime line" is not the same thing as spacetime curvature. At any given point in spacetime, if you know an object's velocity, and you know it's in free fall, then that is indeed sufficient to determine its entire worldline, i.e., the "spacetime line" that describes its entire history.

When I said spacetime line I actually meant the shape of spacetime in space and had earlier thought that for objects of different velocities this "shape" would simply change to conform to the objects speed, but apparently its not that simple.

Actually, there is plenty of talk about curvature of space, but pretty much all of it glosses over the key reason why it's better to talk about curvature of spacetime: because curvature of spacetime is a tensor--the Riemann tensor--and doesn't depend on how you choose coordinates, i.e., how you split up spacetime into "space" and "time". But curvature of space does depend on that; i.e., it's not a coordinate-independent thing, and the physics can only depend on coordinate-independent things.

I don't quite understand this part. By spitting up into space and time are you talking about just different velocities, how the curvature of space (not spacetime) depends on the velocity of the object or do you mean something else?

Yes, because "different velocities" means "straight lines pointed in different directions in spacetime". At any given point in spacetime, there are an infinite number of straight lines (more precisely, timelike geodesics) passing through that point. Each distinct straight line corresponds to a different possible velocity that an object can have at that point.

I'm only familiar with what I guess is the SR spacetime (light cone) diagram, and when I think of a timelike geodesic I understand that to mean a line within the light cone and different straight lines going through a single point will represent different velocities.

I'm misunderstanding too much and I don't want to take up any more of your time as I can tell I'm getting in way over my head. But thank you for your help.

 

[A.T.]

I'm misunderstanding too much...

For an intuitive explanation see:
http://demoweb.physics.ucla.edu/content/10-curved-spacetime

Also check out the animations in this thread and the links in their descriptions on youtube:
https://www.physicsforums.com/threads/question-about-gravity-according-to-gr.937514/

 

[PeroK]

Bad logic on my part. It comes from me trying to think more into the definition of a "straight line". I think of a straight line as the path an object takes when its moving inertially, but that can't quite be true because objects in curved spacetime will move in different "straight lines" depending on their velocity so concluded instead that they DO follow the same path through spacetime. You gave a good definition of a straight line in another thread, but I can't find it. In any case, my conclusion here is also not correct. I would like a good definition of a straight line.

It seems to me that outside of Euclidean geometry, the concept of a "straight line" is particularly unhelpful. I would, therefore, confine the definition to Euclidean geometry, hence flat spacetime, outside of which the concept simply does not exist.

 

[A.T.]

It seems to me that outside of Euclidean geometry, the concept of a "straight line" is particularly unhelpful. I would, therefore, confine the definition to Euclidean geometry...

It generalizes in non-Euclidean geometry to "geodesic", but people are sloppy with terminology.

 

[Dale]

How is a line in a spacetime diagram drawn so it represents gravity (curved spacetime)?

You would have to use a curved piece of paper.

That's the shortest path on a spacetime diagram right?

No, the distance in spacetime is given by the metric. In flat spacetime it can be written $ds^2=-c^2 dt^2 +dx^2 + dy^2+dz^2$. A straight line extremizes (minimizes or maximizes) that distance, which is also called the spacetime interval.

 

[PeterDonis]

had earlier thought that for objects of different velocities this "shape" would simply change to conform to the objects speed, but apparently its not that simple.

It's not that it's not that simple, it's that you have things backwards. The shape (geometry) of spacetime is what it is. Objects have worldlines--curves in this geometry--that describe their histories. The geometry of spacetime is the same for all objects; it doesn't change based on an object's speed.

By spitting up into space and time are you talking about just different velocities, how the curvature of space (not spacetime) depends on the velocity of the object or do you mean something else?

Something else. It's a choice of coordinates. This is a "B" level thread so that's about all that can be said; but I strongly suggest taking some time to work through a good textbook presentation of coordinate charts. Carroll's online lecture notes would be one good choice.

I'm only familiar with what I guess is the SR spacetime (light cone) diagram

Which only works the way it does in SR if spacetime is flat.

when I think of a timelike geodesic I understand that to mean a line within the light cone and different straight lines going through a single point will represent different velocities.

This part is fine, but it's important to remember that in a curved spacetime, this viewpoint only works in a small local patch of the spacetime.

 

[pervect]

For timelike lines, it's longest, though, not shortest. At least, that's the best "B" level definition.

At the B level, I also prefer "longes", it's accurate enough and gets the idea across nicely. With a B level approach applied to the simplest case (the flat space-time of special relativity), we are left with saying that a straight line in space is the path through space along which the length of the path is minimized, while a "straight line" (i.e. a geodesic) path through space-time is the path along which time measured by a clock traveling the path is maximized. Taylor calls this "the principle of maximal aging", the principle says that if two observers take different paths through space-time and meet up again later, it is the inertial observer that will age the most.

This in this B level approach, we treat the twin paradox not as a paradox, but as a necessary feature of relativistic physics that allows us to define what we mean by geodesics ("straight lines") through space-time.

There is of course much merit in treating space and time in a uniform manner. This leads some authors to prefer "extremizing" or even "stationary" to describe the necessary condition on the space-time interval. Taylor, for instance, replaces the principle of maximal aging with the principle of extremal aging in some papers. I don't recall the titles exactly at this point.

At the B level, it may be a mystery why we need to maximize time intervals and minimize distances to get straight lines or their space-time equivalent, geodesics. At higher levels, we note that there are occasional sign differences in our treatment of space and time as a unified entity which we call space-time, but we see the advantages of the unified treatment.

One thing to beware of. The principle of maximal aging needs a few tweaks to apply in more general cases where we have gravity (i.e. curved space-time). There is a local region in which the principle holds, but it won't hold over arbitrarily large regions. The same can be said for the principle of straight lines being the shortest path between two points - on a curved surface, this holds in a local region, but not necessarily over a longer region. For instance, if we have two towns separated by a very tall hill, there is a straight line path that connects the towns that goes over the top of the hill, but there are shorter paths (that may or may not be straight) that go around the hill instead of over it.

 

[SiennaTheGr8]

There is of course much merit in treating space and time in a uniform manner. This leads some authors to prefer "extremizing" or even "stationary" to describe the necessary condition on the space-time interval. Taylor, for instance, replaces the principle of maximal aging with the principle of extremal aging in some papers. I don't recall the titles exactly at this point.

https://books.google.com/books?id=Z...lor&focus=searchwithinvolume&q=extremal+aging

 

[Buckethead]

Hey, thank you everyone for the last several posts which certainly has minimized a lot of my confusion. It's very helpful. I sometimes feel that I must really be pushing patience but as always, all of you are very generous with your explanations and your time and I sincerely appreciate it. This is truly the very finest forum when you just got to have an answer about something in physics and the usual resources don't quite cut it.

For an intuitive explanation see:
http://demoweb.physics.ucla.edu/content/10-curved-spacetime

Also check out the animations in this thread and the links in their descriptions on youtube:
https://www.physicsforums.com/threads/question-about-gravity-according-to-gr.937514/

Thank you for the links. I look forward to studying them.

It seems to me that outside of Euclidean geometry, the concept of a "straight line" is particularly unhelpful. I would, therefore, confine the definition to Euclidean geometry, hence flat spacetime, outside of which the concept simply does not exist.

Interesting, but try and tell that to an architect that has to construct a building with straight walls! But seriously, I'm actually OK with calling a straight line the path that an object takes when traveling inertially or as A.T. said, "geodesic", although I'm also comfortable with limiting it to only something that travels at c. Either way, I'm good!

It's not that it's not that simple, it's that you have things backwards. The shape (geometry) of spacetime is what it is. Objects have worldlines--curves in this geometry--that describe their histories. The geometry of spacetime is the same for all objects; it doesn't change based on an object's speed.

Ah...perfect. This is the answer I needed.

This part is fine, but it's important to remember that in a curved spacetime, this viewpoint only works in a small local patch of the spacetime.

Right. I remember watching a great video recently by Eugene Khutoryansky called "Einsteins Field Equations of General Relativity Explained" which talks about GR being just SR in small chunks of space and if feels like this might be what you are referring to in a way.

we are left with saying that a straight line in space is the path through space along which the length of the path is minimized, while a "straight line" (i.e. a geodesic) path through space-time is the path along which time measured by a clock traveling the path is maximized.

I finally understand this now. Had me confused earlier.

 

[deRoy]

The curvature of spacetime at a point is not a single number, it's 20 numbers: the 20 independent components of the Riemann tensor at that point.

Why so? I was thinking up to now, that for any 4-dimensional manifold in x,y,z,t coordinates say, the Riemann tensor was just an elegant formula prescription way to calculate the Gaussian curvature at a point in 6 distinct planar sections of space-time, namely at the xy, xz, xt, yz, yt, zt planes passing through a point. Am I wrong?

 

[PeterDonis]

Why so?

Counting the independent components of the Riemann tensor is rather complicated. The tensor has four indexes, so in $d$ dimensions the total number of components is $d^4$. The number of independent components is reduced by the symmetries of the tensor, so it ends up having only one independent component in 2 dimensions, 6 in 3 dimensions, and 20 in 4 dimensions. But I don't know of any simple way of describing the symmetries in terms of things like Gaussian curvature.

I was thinking up to now, that for any 4-dimensional manifold in x,y,z,t coordinates say, the Riemann tensor was just an elegant formula prescription way to calculate the Gaussian curvature at a point in 6 distinct planar sections of space-time, namely at the xy, xz, xt, yz, yt, zt planes passing through a point. Am I wrong?

Yes. See above.

 

[deRoy]

Counting the independent components of the Riemann tensor is rather complicated. The tensor has four indexes, so in $d$ dimensions the total number of components is $d^4$. The number of independent components is reduced by the symmetries of the tensor, so it ends up having only one independent component in 2 dimensions, 6 in 3 dimensions, and 20 in 4 dimensions. But I don't know of any simple way of describing the symmetries in terms of things like Gaussian curvature.
Yes. See above.

I think you mean that 20 numbers define the Riemann curvature tensor of a 4-hypersurface.

My point was that for an n-hypersurface there are n(n-1)/2 numbers that define its intrinsic Curvature completely, namely the Gaussian Curvatures of n(n-1)/2 sectional 2-planes through a point.

And of course there is a generalization of Theorem Egregium to any d-hypersurface yielding just one number to define its intrinsic curvature. I am not sure, I think I am correct...

 

[pervect]

Using the Bel decomposition of the Riemann tensor, https://en.wikipedia.org/w/index.php?title=Bel_decomposition&oldid=814235481, we can gain some insight. I don't have a really good textbook reference for this, alas, though MTW touches on some of the same points without calling it the Bel decomposition. There's also a section in MTW that uses curvature 2-forms to do the job of counting the degrees of freedom (they also arrive at the number 20), but I never quite got an intuitive grasp on said curvature two forms. Wiki's terse entry on these is https://en.wikipedia.org/w/index.php?title=Curvature_form&oldid=807183414

In any event we wind up with 20 independent components, which I tend to think of as 21 numbers and one constraint equation that reduces the number of degrees of freedom from 21 to 20. Using the Bel decomposition formalism, we decompose the 4x4x4x4 Riemann tensor into three 3x3 tensors, the electrogravitic tensor (3x3), the magnetogeravitic tensor (3x3), and the topogravitic tensor (3x3). We can think of this as resulting from a specific way of splitting space-time into space and time at some particular point.

An intermediate step in the decomposition is writing the nonzero components of the 4x4x4x4 Riemann tensor as a 6x6 matrix by throwing out components that must be zero by symmetry. The number six comes from the number of combinations of four indices (t,x,y,z) taken two at a time (tx, ty, tz, xy, xz, yz). Symmetry says that if the first two or last two indices of the Riemann (assumed to be in an orthonormal basis) are the same, the component is zero. Thus R(tt**), R(xx**), R(yy**), R(zz**) are all zero because of the repeated index and the symmetries.

The next step divides this 6x6 matrix into 4 3x3 matrices

EH
HT

E, H, and T are the electrogravitic, magnetogravitic, and topogravitc 3x3 tensors, respectively.

The two H's in the above diagram are either equal or transposes (I forget which). This is how we get the three 3x3 matrices in the decomposition. The details of this process are documented in MTW without invoking the name "Bel decomposition".

The electrogravitic and topogravitic tensors are symmetric , meaning they only have six degrees of freedom, 3 diagonal elements, and 3 off-diagonal elements. The magnetogravitic tensor has no special symmetry, so it has 9 degrees of freedom. 6+6+9 = 21, so if the numbers were all independent we would have 21 degrees of freedom. However, the numbers are not all independent, there is one additional constraint, so we only have 20 degrees of freedom.

 

[PeterDonis]

I think you mean that 20 numbers define the Riemann curvature tensor of a 4-hypersurface.

Yes, which is what spacetime is. That means it takes 20 numbers to describe the curvature of spacetime.

My point was that for an n-hypersurface there are n(n-1)/2 numbers that define its intrinsic Curvature completely, namely the Gaussian Curvatures of n(n-1)/2 sectional 2-planes through a point.

Where are you getting this from? The Riemann tensor is the intrinsic curvature of the manifold.

 

[PeterDonis]

An intermediate step in the decomposition is writing the nonzero components of the 4x4x4x4 Riemann tensor as a 6x6 matrix by throwing out components that must be zero by symmetry. The number six comes from the number of combinations of four indices (t,x,y,z) taken two at a time (tx, ty, tz, xy, xz, yz). Symmetry says that if the first two or last two indices of the Riemann (assumed to be in an orthonormal basis) are the same, the component is zero. Thus R(tt**), R(xx**), R(yy**), R(zz**) are all zero because of the repeated index and the symmetries.

It's worth noting that this logic can be generalized to any number of dimensions. For example, here is how it works in two and three dimensions:

Two dimensions: the tensor is a 2x2x2x2 tensor, but the independent components can be written as a 1x1 matrix (i.e., a single number), because that's how many ways you can take two indices two at a time. Or, to put it another way, since R(11**) and R(22**) are zero by symmetry, the only possible nonzero component is R(1212) and the components related to it by symmetry.

Three dimensions: the tensor is a 3x3x3x3 tensor, and the independent components can be written as a single 3x3 matrix which must be symmetric, so there are 6 independent components. The number 3 is the number of ways you can take three indices two at a time (with no two being the same). The matrix must be symmetric for the same reason that the electrogravitic and topogravitic tensors in 4 dimensions must be symmetric (in fact, AFAIK this 3x3 matrix can basically be thought of as the "electrogravitic" tensor, which is the only nontrivial component of the Bel decomposition in 3 dimensions).

The magnetogravitic tensor has no special symmetry

AFAIK it must be traceless, which reduces the independent components from 9 to 8 (and reduces the total independent components from 21 to 20).

 

[deRoy]

Yes, which is what spacetime is. That means it takes 20 numbers to describe the curvature of spacetime.
Where are you getting this from? The Riemann tensor is the intrinsic curvature of the manifold.

Riemann's lecture: " On the hypotheses which lie...".

Haven't done it by hand yet, but it shouldn't pose any problem to find the Gaussian Curvature of any 2-plane of a hypersurface. And without tensors involved...

 

[PAllen]

The relation between sectional curvature of a plane through a point on a riemannian manifold and the curvature tensor is covered in the sectional curvature part of the following article. The two formulations are shown to be equivalent, but what you cannot conclude is that only 6 sectional curvatures characterize the curvature at a point in a riemannian 4 manifold.

https://en.m.wikipedia.org/wiki/Curvature_of_Riemannian_manifolds

 

[deRoy]

The relation between sectional curvature of a plane through a point on a riemannian manifold and the curvature tensor is covered in the sectional curvature part of the following article. The two formulations are shown to be equivalent, but what you cannot conclude is that only 6 sectional curvatures characterize the curvature at a point in a riemannian 4 manifold.

https://en.m.wikipedia.org/wiki/Curvature_of_Riemannian_manifolds

What I am concluding is that sectional curvature defines the curvature tensor completely. And so does Riemann when he says that there are n(n-1)/2 functions that define the metric ( hence the curvature ) completely. Each differential equation is just the sectional curvature given by Gauss in his Theorema Egregium yielding a single number and there are n(n-1)/2 of them equal to the number of 2-sectional planes as formed by the n number of coordinates of the hypersurface at a point.

If I am wrong I would like you to point out where I am mistaken.

 

[pervect]

What I am concluding is that sectional curvature defines the curvature tensor completely. And so does Riemann when he says that there are n(n-1)/2 functions that define the metric ( hence the curvature ) completely. Each differential equation is just the sectional curvature given by Gauss in his Theorema Egregium yielding a single number and there are n(n-1)/2 of them equal to the number of 2-sectional planes as formed by the n number of coordinates of the hypersurface at a point.

If I am wrong I would like you to point out where I am mistaken.

One plus the number of degrees of freedom in the Riemann, D+1, is the number of degrees of freedom in a n(n-1)/2 by n(n-1)/2 symmetric matrix. When n=4, D+1 is is the number of degrees of freedom in a 6x6 symmetric matrix. This is a number greater than 6. In fact it is D+1 = 1+2+3+4+5+6. As previously mentioned this means that D+1=21, or D=20.

The number of degrees of freedom of the sectional curvature approach must be the same as the number of degrees of freedom in the Riemann because, as you say, the two approaches are equivalent.

The Bel decomposition approach I mentioned gives us some good insight into the physical significance of these 21 numbers. We have 21 numbers in this approach, but only 20 degrees of freedom, because there is one constraint relation on the 21 numbers that comes from the Binachi identites.

However, it is not necessary to perform the second step of the Bed decomposition where we split the 6x6 symmetric matrix into different parts. We can count the number of degrees of freedom in the 6x6 symmetric matrix just fine without performing this last step.

For n=4, the number six is important in calculating D, but it is not true that D=6. Rather it is true that D+1 is the number of degrees of freedom in a 6x6 symmetric matrix.

 

[PeterDonis]

One plus the number of degrees of freedom in the Riemann, D+1, is the number of degrees of freedom in a n(n-1)/2 by n(n-1)/2 symmetric matrix.

I'm not sure the "+1" is correct for 2 or 3 dimensions; the Riemann tensor in 2 dimensions has 1 degree of freedom, and in 3 dimensions it has 6, but the D+1 rule you give would result in 0 and 5, respectively, for those cases. I think this is because the extra constraint (which in 4 dimensions can be thought of as tracelessness of the magnetogravitic tensor, as I noted in a previous post) only becomes non-vacuous in 4 or more dimensions; in 2 or 3 it turns out to just be a restatement of other identities.

 

[deRoy]

Here, I have found a reference in Spivak's book:

The metric, is determined if Q(W) ( Gaussian Curvature ) is known for n(n-1)/2 independent 2-dimensional subspaces at each point q.
Michael Spivak: "A comprehensive Introduction to Differential Geometry" Vol. 2 page 179

I wish I had the time to go through his calculations in how he proves Riemann's assertion.

As I understand it, for practical reasons in order to calculate the metric we must work with 2-sections. But the Gaussian Curvature of a hypersurface is only one number. There is a generalization of the Theorema Egregium in dimensions > 2 and I have already done work how to find it for d=3.

 

[PeterDonis]

The metric, is determined if Q(W) ( Gaussian Curvature ) is known for n(n-1)/2 independent 2-dimensional subspaces at each point q.

The metric is not the same as the Riemann tensor; the number of degrees of freedom in the two is not the same.

 

[pervect]

I'm not sure the "+1" is correct for 2 or 3 dimensions; the Riemann tensor in 2 dimensions has 1 degree of freedom, and in 3 dimensions it has 6, but the D+1 rule you give would result in 0 and 5, respectively, for those cases. I think this is because the extra constraint (which in 4 dimensions can be thought of as tracelessness of the magnetogravitic tensor, as I noted in a previous post) only becomes non-vacuous in 4 or more dimensions; in 2 or 3 it turns out to just be a restatement of other identities.

I know it's right for 4 dimensions, but I haven't worked with other dimensions or checked it against textbooks or online sourcesfor that case, so let the reader beware.

For the 4 dimensional case, Wolfram gives the same argument I did in my post (see http://mathworld.wolfram.com/RiemannTensor.html) quoting Misner.

 

[PeterDonis]

I know it's right for 4 dimensions, but I haven't worked with other dimensions or checked it against textbook

Checking, it appears that the correct textbook formula is

$$
\frac{1}{12} n^2 \left( n^2 - 1 \right)
$$

This happens to match your formula for n = 4, and happens to match my heuristic "leave out the +1 in pervect's formula" for n = 2 and n = 3, but neither of our formulas work for n > 4. For n = 5, the correct formula gives 50, while your formula gives 54; the difference gets larger for larger n.

 

[PAllen]

Here, I have found a reference in Spivak's book:

The metric, is determined if Q(W) ( Gaussian Curvature ) is known for n(n-1)/2 independent 2-dimensional subspaces at each point q.
Michael Spivak: "A comprehensive Introduction to Differential Geometry" Vol. 2 page 179

I wish I had the time to go through his calculations in how he proves Riemann's assertion.

As I understand it, for practical reasons in order to calculate the metric we must work with 2-sections. But the Gaussian Curvature of a hypersurface is only one number. There is a generalization of the Theorema Egregium in dimensions > 2 and I have already done work how to find it for d=3.

Thanks, I had seen some reference to a result like this, but could not track anything down (in a math very lite book on Perelman's proof of geometrization conjecture).

I should also note that it pretty well known that in 4 d, you only need 6 functions to determine the metric, and generally, that you need d less than the number of algebraically independent components of rank 2 symmetric tensor. The reason is the ability to impose up to d coordinate conditions on a chart without loss of generality of the geometry, due to diffeomorphism invariance. So, in a certain sense, this result should not be totally surprising to those (like myself) who have only seen differential geometry as used in GR, which rarely (if ever) uses notions of gaussian curvature.

Note that this indeed implies to an indirect, complex, way to get the curvature tensor from 6 numbers supplied at each point of the manifold. You first get the metric, by whatever implicit procedure is described in Spivak. Then you derive the curvature tensor from the metric.

We've had some other threads on this type of issue. A key point is that algebraic independence is not the same as (to coin a term) analytic independence. Specfically, there are integrability conditions on the metric compatible connection resulting from its dependence on a metric with 6 functional degrees of freedom. These imply similar conditions for the curvature tensor.

But what you can't do is get the complete curvature tensor in any algebraic way from 6 numbers at a point. Interestingly, the reverse is possible, which shows the utility of the curvature tensor. That is, the Gaussian curvature for any plane, at a point, is algebraically computable from curvature tensor at that point. This formula is given in the wikipedia article I linked earlier. Going the reverse, algebraically, all you can do is:

Given a scalar function of pairs of vectors in the tangent space at a point giving the gaussian curvature for the plane they determine, and given any 4 specific linearly independent vectors at a point, then using 6 evaluations of gaussian curvature for different combinations of those 4 vectors, you get the the contraction of curvature tensor to a scalar corresponding to those specific 4 vectors.

 

[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?