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?