Does mass physically bend space or is time being bent?

[Seminole Boy]

I know I've separated space and time, which is the opposite of what Einstein was trying to do. But do planets and even human bodies (and my golden retriever) actually warp the physical properties of space, or is it that time (whatever that is, though it seems to exist in space, thus spacetime) is being bent (curved, warped)?

Thanks in advance.

 

[PeterDonis]

Yes. Both space and time are curved by the presence of matter.

 

[Seminole Boy]

Oh, goodness, Peter! This stuff is getting complicated!

 

[HomogenousCow]

Well, spacetime can be curved in the absence of matter as well, there just has to some mass or a field somewhere.

 

[Bill_K]

Both space and time are curved by the presence of matter.

I hear what you're saying, Peter, but I wonder if there isn't a better way to express it. Since time is one-dimensional, it can't really be curved. How about, "In the vicinity of a mass the passage of time is altered".

 

[GarageDweller]

How bout you pick up a book

 

[Naty1]

Since time is one-dimensional, it can't really be curved.

How do we know time is one dimensional??

"In the vicinity of a mass the passage of time is altered".

better, but I thought what altered the passage of time was gravitational potential.

If the Lorentz group 'unifies' space and time so they cannot be disentangled, seems like
time must have some similar attributes as space.

 

[1977ub]

Yes. Both space and time are curved by the presence of matter.

Can you say more about how space is curved? I understand that spacetime being curved is experienced as for instance gravitational force or acceleration. But distances in the neighborhood of a gravitational body, such as are measured by straight rods, will not deviate from their straight arrangements?

 

[PeterDonis]

Well, spacetime can be curved in the absence of matter as well, there just has to some mass or a field somewhere.

Yes, I probably should have said "stress-energy" instead of "matter" since that covers all possibilities.

 

[PeterDonis]

Since time is one-dimensional, it can't really be curved. How about, "In the vicinity of a mass the passage of time is altered".

If we're focusing on just the effects in the time dimension, so to speak, then yes, I agree that "time curvature" is probably not the best way to express it. I was trying to focus on the more general point that it is *spacetime* that is curved; you can't say that either just space is curved, or that just time is curved.

 

[PeterDonis]

Can you say more about how space is curved?

Space is not Euclidean near a gravitating mass. See further comments below.

But distances in the neighborhood of a gravitational body, such as are measured by straight rods, will not deviate from their straight arrangements?

Locally, you can always find a local inertial frame in which straight lines look straight, so to speak. But globally, you can't do that, and that applies to space as well as to time. For example, space around a gravitating mass is not Euclidean: if you take two 2-spheres enclosing the body with slightly different areas, the radial distance between them will be larger than you would expect from the formulas of Euclidean geometry. (Note that the "radial distance" I'm talking about here is the distance that would be measured by static observers; this has to be specified since distance is frame-dependent.)

 

[1977ub]

Locally, you can always find a local inertial frame in which straight lines look straight, so to speak. But globally, you can't do that, and that applies to space as well as to time. For example, space around a gravitating mass is not Euclidean: if you take two 2-spheres enclosing the body with slightly different areas, the radial distance between them will be larger than you would expect from the formulas of Euclidean geometry. (Note that the "radial distance" I'm talking about here is the distance that would be measured by static observers; this has to be specified since distance is frame-dependent.)

So, from the surface of the Earth, if I extended a pole for a mile, attached a 90 degree joint, extended another mile, another joint, etc, I would expect to complete a perfect square back at my location by classical reckoning. By GR does this not happen?

 

[PeterDonis]

So, from the surface of the Earth, if I extended a pole for a mile, attached a 90 degree joint, extended another mile, another joint, etc, I would expect to complete a perfect square back at my location by classical reckoning. By GR does this not happen?

This is a bad example because the surface of the Earth is curved regardless of whose theory of gravity you adopt. The construction you describe would *not* work (at least not if you could make accurate enough measurements) because the Earth is a sphere, but the Earth is a sphere even if the space around it is perfectly Euclidean. (In fact, the non-Euclideanness caused by the Earth's gravity is much too small to matter for this scenario.)

Here's a better scenario: construct two spherical shells enclosing the Earth, one with area A and another with area A + dA, where dA << A. Make sure both shells are exactly centered on the Earth and are exactly at rest relative to the Earth. (We're assuming that the Earth is a sphere and neglecting its rotation; the corrections due to the Earth's non-sphericity and rotation are much smaller than the effect I'm about to describe.) Then measure, very carefully, the radial distance between the two shells.

By Euclidean geometry, we would expect this measurement to yield a distance

dr = \frac{dA}{8 \pi r}

where r = \sqrt{A / 4 \pi}. But according to GR, the radial distance we will actually measure is

\frac{dr}{\sqrt{1 - 2 G M / c^2 r}}

Supposing that the inner sphere, at radius r, is just above the Earth's surface, this gives a radial distance of dr ( 1 + 6.9 * 10^{-10}), i.e., a larger distance than the Euclidean prediction by about 7 parts in 10 billion.

 

[1977ub]

This is a bad example because the surface of the Earth is curved regardless of whose theory of gravity you adopt. The construction you describe would *not* work (at least not if you could make accurate enough measurements) because the Earth is a sphere, but the Earth is a sphere even if the space around it is perfectly Euclidean. (In fact, the non-Euclideanness caused by the Earth's gravity is much too small to matter for this scenario.)

Actually no. Perhaps you aren't understanding. This is pretty straightforward. The first leg extends upward by a mile. We affix a perfect 90 degree joint (or reflect an upward laser beam by 90 degrees). The next leg begins perpendicular to the plumb line down to me but ends up at a higher altitude (obviously because the Earth is spherical). Then another joint / mirror and the third leg extends generally downward (not toward the center of the Earth obviously because the Earth is spherical) and then after another locally perfect 90 degree turn, the last leg comes back to me. So does it get exactly back to me or doesn't it?

(In fact, the non-Euclideanness caused by the Earth's gravity is much too small to matter for this scenario.)

Big enough to "matter" or not, I'm asking if under ideal circumstances, it could be detected.

All of my google searches for "curved space" under GR yield descriptions of curved spacetime, not space itself.

 

[PeterDonis]

The first leg extends upward by a mile.

Ah, I see, you were talking about a vertical construction. Sorry for the confusion on my part. Now the issue is that the Earth's curvature will make the top and bottom legs of the square not sit flat on the ground, if you make accurate enough measurements. (The Earth's surface curves about 1/8 of an inch per mile, IIRC, which is well within our current capabilities to detect.) Also, the vertical legs of the square will not be exactly radial--that is, they won't be pointing exactly at the center of the Earth. But let's ignore all that since it's a different issue than the one you are raising.

The non-Euclideanness of space due to the Earth's mass will not prevent you from constructing this square, no. But here's what it *will* do. Suppose the Earth were a perfect sphere; then its area is equal to the area of a 2-sphere that is exactly tangent to the bottom side of the square. The area of a 2-sphere exactly centered on the Earth and exactly tangent to the *top* side of the square will then be slightly *less* than what you would expect based on Euclidean geometry.

In other words, we have the area of Earth's surface, A, and the area of the 2-sphere centered on the Earth and tangent to the top side of the square, A + dA. Based on the vertical sides of the square being of length s each (s = 1 mile, but I'll just use s in the formulas below), we expect to find

dA = 4 \pi \left[ \left( r + s \right)^2 - r ^2 \right] = 8 \pi R s

where R is the radius of the Earth, and we have used the fact that s << r. But what we will actually find is

dA = 8 \pi R s \sqrt{1 - 2 G M / c^2 R}

which, again, will be smaller than the Euclidean result by about 7 parts in 10 billion.

 

[1977ub]

This is a very interesting reasoning which shows a difference between the radius of what is expected for the 2 spheres.

I would expect from your reasoning that the standard formula for the surface area of a sphere given the radius doesn't exactly apply to the Earth in GR? If that is true, I would expect there to be a slight adjustment in my example...

 

[PeterDonis]

I would expect from your reasoning that the standard formula for the surface area of a sphere given the radius doesn't exactly apply to the Earth in GR?

Yes, that's correct; the "radius" of the Earth, meaning the physical distance you would measure if you carefully laid rulers end to end from the surface to the center, is longer than you would expect from the Euclidean formula for its surface area. But that's because the Earth's mass is in the interior of the 2-sphere described by its surface. A 2-sphere that has vacuum inside would still obey the Euclidean formula.

 

[1977ub]

Yes, that's correct; the "radius" of the Earth, meaning the physical distance you would measure if you carefully laid rulers end to end from the surface to the center, is longer than you would expect from the Euclidean formula for its surface area. But that's because the Earth's mass is in the interior of the 2-sphere described by its surface. A 2-sphere that has vacuum inside would still obey the Euclidean formula.

Interesting thanks - can you direct me to some resources which discuss this? Also I still am not sure if this effect applies to my example or only kicks in when surfaces perpendicular to the gravitational gradient come up?

 

[Passionflower]

Interesting thanks - can you direct me to some resources which discuss this?

Lookup "the interior Schwarzschild solution" and with it integrate the distance from the center to the edge and compare it with the area. The result will be non-Euclidean.

 

[1977ub]

Lookup "the interior Schwarzschild solution" and with it integrate the distance from the center to the edge and compare it with the area. The result will be non-Euclidean.

Thanks will do. BTW is any such effect found with the ratio of the radius to the circumference of a great circle such as the equator? Or only once surface becomes involved.

 

[Passionflower]

BTW is any such effect found with the ratio of the radius to the circumference of a great circle such as the equator? Or only once surface becomes involved.

Yes.

Also the volume inside a given sphere is larger than expected if we assume the space is Euclidean.

 

[PeterDonis]

Interesting thanks - can you direct me to some resources which discuss this?

I don't know of an online reference offhand that discusses this specifically, unfortunately.

I still am not sure if this effect applies to my example or only kicks in when surfaces perpendicular to the gravitational gradient come up?

Sure, the effect applies to your example. It's just that in your example, you specified that the vertical sides of the square were 1 mile in length, so the difference in areas of the 2-spheres, top and bottom, was *less* than expected; whereas in my version, I specified the difference in areas of the 2-spheres, so the radial distance between them was *more* than expected.

 

[1977ub]

Sure, the effect applies to your example. It's just that in your example, you specified that the vertical sides of the square were 1 mile in length, so the difference in areas of the 2-spheres, top and bottom, was *less* than expected; whereas in my version, I specified the difference in areas of the 2-spheres, so the radial distance between them was *more* than expected.

How long has this particular effect been known?

It's not clear to me that "areas" are involved in my example... I chose this example as something very simple that would demonstrate a deviation or distortion from classical expectation of Euclidian flat space.

 

[PeterDonis]

How long has this particular effect been known?

It's an obvious consequence of the structure of Schwarzschild spacetime (at least, it's "obvious" today), but I don't know who first pointed it out.

It's not clear to me that "areas" are involved in my example...

You didn't mention them directly, true, but without them, or something like them, there is no way to point out any consequence of "space curvature" or space being non-Euclidean that would be observable in your example. There is no distortion of the square itself due to the non-Euclideanness of space; the non-Euclideanness is not a local phenomenon, it's a global phenomenon.

 

[1977ub]

Thanks will do. BTW is any such effect found with the ratio of the radius to the circumference of a great circle such as the equator? Or only once surface becomes involved.

Yes.

There is no distortion of the square itself due to the non-Euclideanness of space; the non-Euclideanness is not a local phenomenon, it's a global phenomenon.

If I can draw a circle around the Earth and the diameter is of a different length than what euclidean geometry would suggest, then I would imagine that instead of a circle I drew a square which touched the Earth at four points we would come upon a similar phenomenon, and that furthermore some similar effect would come into play with any suitably large square which I drew in the neighborhood of the Earth.

I don't see how a "global" deviation from euclidean flatness can fail to show itself in a local area, albeit of extraordinarily tiny magnitude.

 

[PeterDonis]

I would imagine that instead of a circle I drew a square which touched the Earth at four points we would come upon a similar phenomenon

Yes; for example, the diagonals of the square would be slighty longer than expected.

and that furthermore some similar effect would come into play with any suitably large square which I drew in the neighborhood of the Earth.

If the square doesn't enclose the Earth, then no, the effect won't show up the same way. The key thing that makes the diagonals of the square enclosing the Earth longer than expected is that the diagonals go through a region of different curvature than the sides (because the Earth is there in the center). That's not the case for a square in the neighborhood of the Earth but not enclosing it.

The non-Euclideanness of space might still show up in some other way with the square not enclosing the Earth; I would have to think about it some more.

I don't see how a "global" deviation from euclidean flatness can fail to show itself in a local area, albeit of extraordinarily tiny magnitude.

If you have a specific observable you're looking at, then yes, for any finite region that observable will deviate by *some* amount, however small, from the expected Euclidean value if space is not Euclidean. For example, if you constructed your square deep inside the Earth, with sides 1 mile in length and enclosing the Earth's center, then the diagonals would be longer than expected by a very small amount.

 

[A.T.]

Lookup "the interior Schwarzschild solution" and with it integrate the distance from the center to the edge and compare it with the area. The result will be non-Euclidean.

Here is an embedding diagram of the spatial geometry (2D spatial slice) that combines interior and exterior Schwarzschild solution:

From http://commons.wikimedia.org/wiki/File:Schwarzschild_interior.jpg

How long has this particular effect been known?

Einstein was initially only thinking about a distortion of the time dimension, depending on spatial position (gravitational time dilation). But he soon realized that this also implies spatial distortion.

The non-Euclideanness of space might still show up in some other way with the square not enclosing the Earth; I would have to think about it some more.

I think near a spherical mass, over an area non-including the mass, you will find negative spatial curvature. Inside of the mass or over an area including the mass you have positive spatial curvature.

 

[1977ub]

I'm surprised that this is a bit buried... I didn't find it in google searches or wikipedia. It seems like a type of effect which would come to play in "near a black hole" scenarios.

So if I measured the diameter of the Earth from afar by watching it to eclipse other bodies, and then I brought an iron bar of that length, and drove it through the Earth, there would be no surprises, correct? However if I had built a circle around the bar out in space so that the bar was a diameter of the circle, and then brought this construction to the Earth with the intention of the diameter bar going through the Earth and the circle going around the idealized equator, it sounds like there would be stretching stress on the circle piece?

 

[PeterDonis]

It seems like a type of effect which would come to play in "near a black hole" scenarios.

It would be a much larger effect near a black hole, yes.

So if I measured the diameter of the Earth from afar by watching it to eclipse other bodies, and then I brought an iron bar of that length, and drove it through the Earth, there would be no surprises, correct?

No, not correct. Measuring the angle subtended by the Earth from far away doesn't measure the diameter of the Earth directly; it's really a measure of the circumference of the Earth's surface--more precisely, it's a measure the area on the sky occupied by the outline of the Earth's surface, which is only a function of the "Euclidean radius" of the Earth, i.e., of the circumference (or area) of its surface.

if I had built a circle around the bar out in space so that the bar was a diameter of the circle, and then brought this construction to the Earth with the intention of the diameter bar going through the Earth and the circle going around the idealized equator, it sounds like there would be stretching stress on the circle piece?

No. If you specifically built the diameter bar to be the correct length for the Earth's actual diameter (i.e., a bit longer than the Euclidean expectation based on the circumference of the equator), then there would be *compression* of the circle piece when you fit it to the equator (since the ratio of circumference to diameter is a bit smaller than the Euclidean expectation). If you specifically built the circle piece to exactly fit the equator, then there would be stretching of the *diameter bar*, since the diameter is a bit longer than the Euclidean expectation.

 

[1977ub]

No, not correct. Measuring the angle subtended by the Earth from far away doesn't measure the diameter of the Earth directly; it's really a measure of the circumference of the Earth's surface--more precisely, it's a measure the area on the sky occupied by the outline of the Earth's surface, which is only a function of the "Euclidean radius" of the Earth, i.e., of the circumference (or area) of its surface.

I'm thinking of the linear arc from side to side - from one end to the other as a star dips behind it and re-emerges. I don't see what that has to do with the circumference, since I'm measuring in a line. How about giant calipers? What is the most reliable way for someone out in space to determine the diameter of the Earth?

 

[Passionflower]

I think A.T. brings up a great point.

The interior Schwarzschild solution describes a space with positive curvature while the exterior solution describes a space with negative curvature.

Thus in the Schwarzschild solution the radius between two shells in empty space is larger than the Euclidean equivalent while the radius between two shells inside an object like a star is shorter than the Euclidean equivalent.

E.g for a negative curvature we have:
\rho > \sqrt{A / 4 \pi}
and for a positive curvature we have:
\rho < \sqrt{A / 4 \pi}
Where \rho is the physical distance not the r-coordinate difference.

Agreed or am I mistaken?

[Edited (a million times to get the darn Latex to work) ]

 

[A.T.]

I think A.T. brings up a great point.

The interior Schwarzschild solution describes a space with positive curvature while the exterior solution describes a space with negative curvature.

Thus in the Schwarzschild solution the radius between two shells in empty space is larger than it would be if the space where Euclidean while the radius between two shells inside an object like a star is shorter[/i[ than it would be if the space where Euclidean.

Agreed?

I find is simpler to think in terms of inner triangle angle sum or circumference/radius ratio.

Local curvature in a small area:

If the triangle or circle is completely inside the mass, you have positive spatial curvature:
angle_sum > π, circumference < 2πr

If the triangle or circle is completely outside the mass (doesn't enclose or intersect it), you have negative spatial curvature:
angle_sum < π, circumference > 2πr


Average curvature over a larger area:

If the triangle or circle completely encloses the mass, you have positive spatial curvature:
angle_sum > π, circumference < 2πr

If the triangle or circle just partially encloses the mass, it depends. You might even have cases with
angle_sum = π, circumference = 2πr
when on average the curvatures cancel.

 

[PeterDonis]

I'm thinking of the linear arc from side to side - from one end to the other as a star dips behind it and re-emerges. I don't see what that has to do with the circumference, since I'm measuring in a line.

But the "line" is defined by light coming from one side and then the other, not by light coming from in between--i.e., it is defined by the outline of the surface of the Earth, not by the actual diameter going through the interior.

Furthermore, the "line" is defined by the angles at which the light rays come into your eye, which is not determined just by the Earth, but by the spacetime in between. See further comments below.

What is the most reliable way for someone out in space to determine the diameter of the Earth?

Send someone down to measure it with rulers, and have them report the answer back to you.

Seriously, in a curved spacetime there is, in general, *no* way to "measure things at a distance" reliably. There is no way to see the actual diameter of the Earth "from the outside"; you have to go down and measure it locally, "from the inside".

 

[PeterDonis]

Agreed or am I mistaken?

You're mistaken. The radial distance between shells, compared to its Euclidean value, is given by the metric coefficient g_{rr} in a chart where the radial coordinate r is defined as A / 4 \pi for a 2-sphere of surface area A. Since g_{rr} is greater than 1 inside the massive body (except at the exact center, r = 0, where it becomes 1), the radial distance between shells will be larger than the Euclidean value there.

What you and A.T. are calling the "sign" of the curvature determines, roughly speaking, the *gradient* of g_{rr}, not g_{rr} itself. In the vacuum exterior region, g_{rr} gets larger as you move inward from infinity (where it is 1) to the surface of the body. In the non-vacuum interior region, g_{rr} gets *smaller* as you move inward, until it is 1 again at the center.

(I say "roughly speaking" because the curvature is the second derivative of the metric coefficients, not the first. But in this particular case the sign of the curvature happens to match up with the gradient of g_{rr} in the way I said.)

 

[1977ub]

If I measure the circumference on the equator, and then extend rods down through the Earth to the other side will show me a diameter > than circumference / pi, correct? utilizing more iron than classically expected.

Then I move this contraption into space, there will be stress from the center rod being longer than would snugly fit in the circle.

And if I built my contraption in space - using measurement I had made in person on the equator [and then rod = circumference / pi], then moved it to the earth, the rod would not be long enough.

Have I got all that right?

 

[PeterDonis]

Have I got all that right?

Yes, all this looks right.

 

[1977ub]

The "warped graph paper" diagrams of GR - they seem almost entirely intended to convey that that space-TIME is warped, i.e. to suggest gravitation as a result of curvature of space-time. But the space graph paper is warped as well it seems. Structures brought to the presence of high gravitation are subjected to forces other than what is expected from gravity itself, i.e. acceleration toward the body and tidal forces. I think it's odd that this isn't mentioned more often.

 

[PeterDonis]

Structures brought to the presence of high gravitation are subjected to forces other than what is expected from gravity itself, i.e. acceleration toward the body and tidal forces.

Yes, this is true (provided "tidal forces" means forces due to objects *resisting* tidal deformation, not the tidal deformation itself--tidal deformations, not resisted by internal forces of the body, *are* due to gravity). We've assumed that any effects from these forces would be negligible in our discussion here, but of course they often won't be.

 

[Passionflower]

You're mistaken.

No problem, but could you demonstrate it? It would be helpful.
During integration how do we handle the pressure component?
Also for the interior Schwarzschild solution there is actually more mass for a given area compared to the Ricci flat Schwarzschild solution. Thus the Schwarzschild radius r_s < 2M/c² for the interior solution, how does that tie in if at all? After all in a region where the Ricci tensor does not vanish, and that is the case for the interior solution, do we not have a volume reducing effect?

 

[Naty1]

Structures brought to the presence of high gravitation are subjected to forces other than what is expected from gravity itself,

I don't understand why Peter says 'yes', then describes effects of gravity.

Gravity can be so strong as to cause electron and neutron degeneracy...and even
crush mass to the point of a black hole...all part of GR...

 

[PeterDonis]

No problem, but could you demonstrate it?

I thought I had done so in my post #34.

For the internal Schwarzschild solution there is actually more mass for a given area.

I don't understand what you mean by this.

Thus the Schwarzschild radius r_s < 2M/c², how does that tie in if at all?

It doesn't. We're talking about a stable massive object with a radius greater than 2M (actually it has to be greater than 2.25M to be stable), so there is no horizon anywhere.

After all in a region where the Ricci tensor does not vanish do we not have a volume reducing effect?

The volume of a small sphere of freely falling test particles decreases with time, yes. But that's not the effect being described by the non-Euclideanness of space we've been talking about.

 

[A.T.]

What you and A.T. are calling the "sign" of the curvature determines, roughly speaking, the *gradient* of g_{rr},

Actually I was talking in terms of triangle angles and circumference/radius ratio, not distance between two shells. Du you agree with the cases I describe in post #32?

 

[PeterDonis]

I don't understand why Peter says 'yes', then describes effects of gravity.

I was trying to describe both. If I lower an object into a gravity well and then hold it static, the force it feels is not an effect of gravity; it's an effect of whatever is holding it static. It's true that it only takes a force to hold it static because it's in a gravity well, but that's not the same as saying the force itself is an "effect of gravity"; objects can be in a gravity well and not be static.

Similarly, if I lower an object into a gravity well and it deforms, the deformation may not be due to gravity; it may be due to the forces I am applying to it to place it where I want it.

Gravity can be so strong as to cause electron and neutron degeneracy...and even crush mass to the point of a black hole...all part of GR...

Yes, I wasn't intending to deny any of this.

 

[A.T.]

The "warped graph paper" diagrams of GR - they seem almost entirely intended to convey that that space-TIME is warped,

Actually most diagarms show the spatial curvature, which misleading because it has only minor effect, compared to the time dimension. Here a super simple overview:
http://www.physics.ucla.edu/demoweb..._and_general_relativity/curved_spacetime.html

to suggest gravitation as a result of curvature of space-time.

The major effects come from the the space-time curvature. Space curvature doesn't even affect objects at rest.

Structures brought to the presence of high gravitation are subjected to forces other than what is expected from gravity itself, i.e. acceleration toward the body and tidal forces..

Acceleration towards the mass and tidal forces are also mainly due to space-time curvature. Only objects moving fast relative to the gravity source are significantly affected by the spatial curvature. For photons, which move very fast, the effects are comparable. Huge rigid structures build in flat space would be crushed by curved space, but this is not what is typically meant by tidal forces.

 

[Passionflower]

I don't understand what you mean by this.
...
It doesn't. We're talking about a stable massive object with a radius greater than 2M (actually it has to be greater than 2.25M to be stable), so there is no horizon anywhere.

In the Schwarzschild solution mass is actually a length unit of measure, do you agree with that? This length is related to mass in the following formula: r_s = 2M/c² because the Ricci tensor vanishes in empty space.

However inside an object that is no longer the case because the Ricci tensor does not vanish. Unless there is exotic negative pressure the proper mass is always larger than the mass as described by the above relationship.

To get the proper mass for the interior Schwarzschild solution we need to integrate.

Of course a star does not have an event horizon but it does have a Schwarzschild radius.

See for instance: http://en.wikipedia.org/wiki/Tolman–Oppenheimer–Volkoff_equation#Total_Mass

Am I mistaken?

 

[PeterDonis]

Do you agree with the cases I describe in post #32?

I don't think so. I think you are confusing space curvature with spacetime curvature. To briefly recap the general facts about space curvature vs. spacetime curvature:

Outside the gravitating body, spacetime curvature is negative in the radial direction (freely falling objects increase their radial separation with time) and positive in the tangential direction (freely falling objects decrease their tangential separation with time). (Note that this is Weyl curvature we're talking about; there is zero Ricci curvature, since we're in vacuum. That means a small sphere of freely falling particles will distort radially and tangentially, as above, but will not change its volume.)

Inside the gravitating body, spacetime curvature is positive everywhere. Ricci curvature is positive since there is positive stress-energy present; that means a small sphere of freely falling particles will decrease in volume. Weyl curvature is positive tangentially for the same reason it is outside the body: freely falling radial geodesics converge. But it is also positive radially because the gradient of g_{rr} is reversed; freely falling particles that are radially separated will converge, not diverge, because the "acceleration due to gravity" gets smaller as they fall, not larger.

However, neither of the above tells us about *space* curvature, which is what your triangle scenarios refer to. To talk about space curvature, we have to first decide how we will slice up the spacetime into space and time. The most obvious way to do that is to use the slicing that matches up with the static nature of the spacetime; i.e., we use slices of constant time according to static observers, who stay at the same radius forever. This is the slicing that produces the Flamm paraboloid that you gave a picture of.

If we use that slicing, then space curvature is *positive* everywhere; a triangle's angles sum to greater than \pi, and the circumference of a circle is less than 2 \pi times its proper radius. This may seem confusing since the paraboloid does appear to "change the direction it curves" at the body's surface. But that only changes the *gradient* of g_{rr}; it doesn't change the curvature, which is, as I noted before, the *second* derivative of the metric, not the first.

Draw some triangles on the paraboloid; you will see that regardless of whether you are inside or outside the body's surface, the triangles will work like triangles drawn on a spherical surface. A negatively curved surface would be shaped like a saddle, bending one way in one direction and the other way in a perpendicular direction at a single point. The paraboloid doesn't do that.

(Actually, a triangle that straddles the body's surface *may* behave differently, since it would straddle the change in gradient of g_{rr}. I have been unable to find an explicit computation of the spatial curvature of a constant-time slice in Schwarzschild coordinates, and I haven't done the full computation myself.)

 

[A.T.]

This is the slicing that produces the Flamm paraboloid that you gave a picture of. If we use that slicing, then space curvature is *positive* everywhere;

Are you sure about this? Here they say Gaussian curvature of Flamm's paraboloid is negative:
http://books.google.de/books?id=75r...flamm's paraboloid gaussian curvature&f=false

A negatively curved surface would be shaped like a saddle, bending one way in one direction and the other way in a perpendicular direction at a single point. The paraboloid doesn't do that.

To me it looks like it does exactly that.

 

[WannabeNewton]

Geometrically, the Gaussian curvature at a point of an embedded 2 - surface is the product of the principal curvatures at that point, which are the eingenvalues of the shape operator (classically called the Weingarten map). You can picture the principal curvatures at a point $p$ on the surface $S$ as follows: choose a normal $N$ to $S$ at $p$ and consider a plane going through $p$ that contains $N$. This plane (called a normal plane) will "slice" out a curve $\gamma$ on $S$. At $p$, $\gamma$ has some curvature and we assign it a sign based on if it is turning away or towards $N$; this is called a signed curvature. Do this for all such normal planes through $p$ containing $N$; the principal curvatures are the maximum and minimum signed curvatures so obtained.

You can picture this for Flamm's paraboloid and see that there is one direction where the curve would locally turn away from the normal and another direction where it would locally turn towards it (the principal directions); it isn't too different locally from a hyperbolic paraboloid. I'm being hand wavy here but take a look at chapter 3 in "Differential Geometry of Curves and Surfaces" - Do Carmo if you are interested.

 

[1977ub]

The major effects come from the the space-time curvature. Space curvature doesn't even affect objects at rest.

I'm understanding that a structure can be brought into the presence of larger gravitation, held in place, and experience 2 types of deformation: one from tidal forces, and a 2nd due to these "space bending" effects.

Oh, do the tidal forces only come into play as the structure is moving toward the gravitational body?

 

[A.T.]

I'm understanding that a structure can be brought into the presence of larger gravitation, held in place, and experience 2 types of deformation: one from tidal forces, and a 2nd due to these "space bending" effects.

Yes. In curved 4D-space-time this gets mangled together a bit. But if you have a long free falling stick aligned with the radial direction it will be stretched by tidal forces, while spatial curvature imposes no strains on it. You need a 2d of 3d structure to experience deformation from spatial curvature.