Does mass physically bend space or is time being bent?

1. Mar 20, 2013

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)?

2. Mar 20, 2013

Staff: Mentor

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

3. Mar 20, 2013

Seminole Boy

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

4. Mar 20, 2013

HomogenousCow

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

5. Mar 20, 2013

Bill_K

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".

6. Mar 20, 2013

GarageDweller

How bout you pick up a book

7. Mar 20, 2013

Naty1

How do we know time is one dimensional??

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.

8. Mar 20, 2013

1977ub

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?

9. Mar 20, 2013

Staff: Mentor

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

10. Mar 20, 2013

Staff: Mentor

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.

11. Mar 20, 2013

Staff: Mentor

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

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.)

12. Mar 20, 2013

1977ub

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?

13. Mar 20, 2013

Staff: Mentor

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.

14. Mar 20, 2013

1977ub

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?

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.

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15. Mar 20, 2013

Staff: Mentor

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.

16. Mar 20, 2013

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...

17. Mar 20, 2013

Staff: Mentor

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.

18. Mar 20, 2013

1977ub

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?

19. Mar 20, 2013

Passionflower

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.

Last edited: Mar 20, 2013
20. Mar 20, 2013

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.