Does mass physically bend space or is time being bent?

[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.

 

[PeterDonis]

Are you sure about this? Here they say Gaussian curvature of Flamm's paraboloid is negative

...

To me it looks like it does exactly that.

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.

Yes, I see what you mean. I'm working on computing the principal curvatures from the metric to verify this.

 

[1977ub]

Yes, all this looks right.

I don't suppose this bending of space has experimental verification? I'm still stunned that it is not more widely popularized.

 

[PeterDonis]

I don't suppose this bending of space has experimental verification? I'm still stunned that it is not more widely popularized.

The effect, as I said, is about 7 parts in 10 billion at the Earth's surface, and you would have to make measurements covering a substantial fraction of that surface to see it at that magnitude. So I don't think we'll get a direct measurement any time soon.

 

[Nugatory]

I don't suppose this bending of space has experimental verification? I'm still stunned that it is not more widely popularized.

Considering that the most common popularization is the infamous rubber-sheet picture... I wish it weren't as widely popularized

 

[1977ub]

Considering that the most common popularization is the infamous rubber-sheet picture... I wish it weren't as widely popularized

In every case I have seen, this is a popularization of the curvature of space-TIME, not of space. Can you find an example where it is curvature of space which is being described?

 

[A.T.]

Considering that the most common popularization is the infamous rubber-sheet picture...

Nugatory probably refers to pictures like this:
http://2.bp.blogspot.com/_vqRjMx6Pk...C3M/iVVaGjIJ17U/s1600/Spacetime_curvature.png

In every case I have seen, this is a popularization of the curvature of space-TIME, not of space.

And which dimension of the above surface is supposed to be time? They might talk about "space-time curvature" as they show you that picture, but both dimensions of the sheet are spatial. So the picture obviously just shows the spatial curvature with dents like the one in the picture I posted in post #27. That's why it is so misleading.

Space-time embedding diagrams look like this:
http://www.relativitet.se/spacetime1.html

 

[A.T.]

I don't suppose this bending of space has experimental verification?

I don't know about an isolated test, just for space curvature. But the spatial distortion around the Sun affects the bending of light and the orbit precession of Mercury. To get the best predictions compared to observed values you have to take space curvature into account.

 

[Nugatory]

Nugatory probably refers to...

Yep - said it better than I would have - thanks.

 

[Passionflower]

Here is a paper on calculating the interior size:

http://www.johnstonsarchive.net/relativity/stcurve.pdf

Question for me remains that since the proper mass is larger due to pressure should we not adjust the Schwarzschild radius accordingly in the formulas?

 

[1977ub]

I don't know about an isolated test, just for space curvature. But the spatial distortion around the Sun affects the bending of light and the orbit precession of Mercury. To get the best predictions compared to observed values you have to take space curvature into account.

Have you ever seen it done this way - i.e. spatial curvature added with the space-TIME curvature in discussing perihelion movement or light deflection?