Inverse curvature of space-time

[ScientificMind]

Assuming that my understanding is correct, I believe it was Einstein who proposed that gravity is the result of the warping or curving of space-time. My question is this: if gravity, which is solely attractive in nature, is the result of warped or curved space time, then is it possible for the inverse warping of space-time to produce a sort of anti-gravity effect. If not, why not, and if so, how?

 

[Matterwave]

What's an "inverse warping"? If I have a piece of paper I can bend it...how do I inverse bend it?

That being said, there is a way to produce a "sort of anti-gravity effect" in GR, this is the cosmological constant. It produces an acceleration in the expansion in the universe, akin to an "anti-gravity" effect.

 

[ScientificMind]

Ah, sorry about that. I was worried that I might not be explaining it well but I wasn't really sure how to do so better. Anyway, I suppose that I was trying to say something akin to warping -- or curving (still not to sure about which one, but whatever) -- in the opposite way. Sort of like a concave lens is to a convex lens, or how expansion is the opposite of the contraction

 

[PeterDonis]

gravity, which is solely attractive in nature

Spacetime curvature is not equivalent to "gravity" per se; it's equivalent to tidal gravity. So even for an ordinary mass, whose "gravity" in the Newtonian sense is purely attractive, the spacetime curvature produced is not all "convex" (or whatever word you were thinking of), because tidal gravity acts differently depending on the direction. Objects separated tangentially (at the same altitude but slightly apart horizontally) are moved towards each other by tidal gravity; objects separated radially are moved away from each other. The full description of tidal gravity/spacetime curvature requires the Riemann curvature tensor, which has 20 independent components in 4 dimensions.

 

[PeterDonis]

how do I inverse bend it?

Think of the curvature of a saddle as opposed to the curvature of a sphere. The latter is something like the spacetime curvature of an ordinary gravitating body in the tangential direction; the former is something like the spacetime curvature of that body in the radial direction.

 

[Matterwave]

Think of the curvature of a saddle as opposed to the curvature of a sphere. The latter is something like the spacetime curvature of an ordinary gravitating body in the tangential direction; the former is something like the spacetime curvature of that body in the radial direction.

I wasn't sure if the OP was talking about dark energy, or exotic matter, or simply in terms of geodesic deviation like you are talking about, so I asked for him to clarify. I am still not sure which direction he wants to take this discussion.

 

[jerromyjon]

I get what the OP is trying to explain, instead of a geodesic curving to the COM it would curve away from the center of mass. Similar to saying spacetime is warped everywhere except where there is "anti-mass" it is "less curved". The whole concept of GR is modeled around attractive mass because that is all that has ever been observed. I bet if there was "anti-mass" it could probably be modeled with GR, but we'd certainly have to find some to see how to model it...

 

[ScientificMind]

Spacetime curvature is not equivalent to "gravity" per se; it's equivalent to tidal gravity. So even for an ordinary mass, whose "gravity" in the Newtonian sense is purely attractive, the spacetime curvature produced is not all "convex" (or whatever word you were thinking of), because tidal gravity acts differently depending on the direction.

What is tidal gravity, and how does it act differently depending on the direction?

As for my original question, since I do not know the specific geometry how space-time is curved, I was more asking, whether it is curved in a convex way, a concave way, or something entirely different, if space-time can be made to curve . . . warp? . . . in the opposite way, or at least a different way. i.e. concave to convex, convex to concave, or unknown (to me) third option to unknown (to me) fourth option, and how/if that would affect gravity

 

[ScientificMind]

I get what the OP is trying to explain, instead of a geodesic curving to the COM it would curve away from the center of mass. Similar to saying spacetime is warped everywhere except where there is "anti-mass" it is "less curved". The whole concept of GR is modeled around attractive mass because that is all that has ever been observed. I bet if there was "anti-mass" it could probably be modeled with GR, but we'd certainly have to find some to see how to model it...

Yes, that sounds like essentially what I was trying to ask. Thank you, I believe that I was just unable to sufficiently explain that with my currently limited vocabulary and understanding.

 

[Dale]

In GR the kind of curvature that we are interested in is called intrinsic curvature. It is the curvature that can be measured entirely within the curved surface.

For instance, on a sphere if you draw a triangle you find that the sum of the angles is greater than 180°. A sphere is said to have positive curvature. So the opposite is also possible, a triangle on a saddle would have less than 180° and has negative intrinsic curvature.

 

[PeterDonis]

What is tidal gravity, and how does it act differently depending on the direction?

Imagine two small rocks that are momentarily at rest in space above the Earth, and are freely falling. Consider the following two scenarios:

(1) The two rocks are both at the same altitude, but separated by a small distance horizontally. As the rocks fall, their (horizontal) separation will decrease (because they are both falling towards the center of the Earth, and that is in a slightly different direction for each rock).

(2) The two rocks are at slightly different altitudes, but both lie along the same radial line from the center of the Earth. As the rocks fall, their (vertical) separation will increase (because the rock that starts out lower will fall slightly faster).

These two scenarios illustrate tidal gravity in the tangential and radial directions, respectively.

whether it is curved in a convex way, a concave way

It's either, depending on direction. But it can also change depending on what kind of matter-energy is present. In the case of the Earth, we have an isolated, roughly spherical body surrounded by vacuum, and the tidal gravity I described above is what is observed in the vacuum region. (Note that inside the Earth, if we could imagine "freely falling" objects--say we dig tunnels through the Earth for them to fall through--the tidal gravity in the radial direction would behave differently: the separation between two objects that start at slightly different altitudes would decrease, not increase, because objects "fall" slower as they get closer to the center of the Earth.) Other kinds of tidal gravity can be observed in other kinds of scenarios; for example, if we are talking about a uniform region of space filled with dark energy (or, equivalently, a positive cosmological constant), then two objects that start out at rest and close together will separate (like the radial case above the Earth) regardless of direction. (The fact that this is independent of direction is why Matterwave said it can be thought of as "anti-gravity"--there is no direction in which anything converges.)

 

[bahamagreen]

Everyone is getting so complicated... what about a simple case?

There is the Earth, and there is the Moon... you are between them and notice that at one position you are pulled to the Earth, but a bit further away and you are pulled to the Moon, and at a certain place the two pulls balance.

Would you think the curvature due to the Earth is "inverse" of that due to the Moon because of the change in direction of pull?
Would you assume there is not curvature at the balance point?

 

[PeterDonis]

Would you think the curvature due to the Earth is "inverse" of that due to the Moon because of the change in direction of pull?

There is a sense in which the curvature changes direction at the balance point, yes.

Would you assume there is not curvature at the balance point?

It's not an assumption, it's a fact; tidal gravity is zero exactly at the balance point.

 

[ScientificMind]

It's either, depending on direction. But it can also change depending on what kind of matter-energy is present. In the case of the Earth, we have an isolated, roughly spherical body surrounded by vacuum, and the tidal gravity I described above is what is observed in the vacuum region.

The concavity and convexity that I was talking about was just about the curvature and bending of space-time in a general sense in correlation to the point I had made previously in the post you had quoted that basically explained that I don't have the slightest clue as to the actual geometry of the bending of space-time as described by general relativity. Still, thank you very much for attempting to answer my questions so far, and I apologize if my limited vocabulary and understanding of the subject has resulted in my question becoming unclear. Luckily though, I believe that jerromyjon was actually able to portray my question even better than I was by saying, "instead of a geodesic curving to the COM it would curve away from the center of mass"

 

[A.T.]

"instead of a geodesic curving to the COM it would curve away from the center of mass"

This was discussed recently:
https://www.physicsforums.com/threads/negative-mass.777140/#post-4888410

 

[PeterDonis]

"instead of a geodesic curving to the COM it would curve away from the center of mass"

But this "curving" is not the same as spacetime curvature. In fact, in an invariant sense, geodesics are not "curved" at all; they are the equivalent of "straight lines" in spacetime. The "curvature" you are visualizing for the geodesics is in space, not spacetime, and the concept of "space" you are using only works in certain kinds of spacetimes. (For example, it doesn't work in a spacetime filled with dark energy.)

 

[ScientificMind]

But this "curving" is not the same as spacetime curvature. In fact, in an invariant sense, geodesics are not "curved" at all; they are the equivalent of "straight lines" in spacetime. The "curvature" you are visualizing for the geodesics is in space, not spacetime, and the concept of "space" you are using only works in certain kinds of spacetimes. (For example, it doesn't work in a spacetime filled with dark energy.)

As I have explained before, I do not know how space-time is curved to produce the phenomenon we call gravity, I only know that it is, and as such I am not "visualizing" any curvature of space-time. I was hoping that the quote I provided from an earlier post in this discussion would provide a better depiction of what I am trying to ask, however, it does not seem to have had that effect. Anyway, my question was essentially this: if space-time curves or bends in a certain way, and that produces gravity as a byproduct, then what would happen if space-time were to bend in the opposite way, and is that even possible in nature as we currently understand it. For example, if you were to graph the curvature of space-time caused by Earth with a geometric function (say f(x) or something like that), then what would happen if you were to happen if space-time were to curve in a way modeled by the inverse function of the one I have previously mentioned (say f^-(x) or something like that).

 

[PeterDonis]

my question was essentially this: if space-time curves or bends in a certain way, and that produces gravity as a byproduct, then what would happen if space-time were to bend in the opposite way

And the answer has already been given: spacetime that produces ordinary gravity as a by-product does not curve in one "way", so the concept of curving in "the opposite way" is not well-defined. So your question does not have an answer as you ask it.

if you were to graph the curvature of space-time caused by Earth with a geometric function (say f(x) or something like that), then what would happen if you were to happen if space-time were to curve in a way modeled by the inverse function of the one I have previously mentioned (say f-(x) or something like that).

You can't depict the curvature of spacetime caused by the Earth as just one function; you would need twenty different functions, one for each independent component of the Riemann curvature tensor. So the concept of the "inverse function" is not well-defined. So again, your question does not have an answer as you ask it.

(Also, the inverse of f(x) is not f(-x); the inverse of a function f(x) is a function g(y) such that g(f(x)) = x. But that's a whole separate topic that really belongs in the math forum.)

 

[PeterDonis]

I was hoping that the quote I provided from an earlier post in this discussion would provide a better depiction of what I am trying to ask

If you mean jerromyjon's post that you quoted in post #9, you have also been told several times that the curvature he describes is not spacetime curvature; it's something different. So if that's what you really want to ask about, you should not keep asking about spacetime curvature.

If jerromyjon's post captured what you wanted to ask, then the thread A. T. linked to in post #15 addresses it. But again, that's not the same thing as spacetime curvature.

 

[ScientificMind]

And the answer has already been given: spacetime that produces ordinary gravity as a by-product does not curve in one "way", so the concept of curving in "the opposite way" is not well-defined. So your question does not have an answer as you ask it.
You can't depict the curvature of spacetime caused by the Earth as just one function; you would need twenty different functions, one for each independent component of the Riemann curvature tensor. So the concept of the "inverse function" is not well-defined. So again, your question does not have an answer as you ask it.

Thank you, this has been very informative. I suppose that, for now at least, I may not have a sufficiently detailed understanding of this process to fully understand my own confusion, but this has still been very helpful, so thank you.

(Also, the inverse of f(x) is not f(-x); the inverse of a function f(x) is a function g(y) such that g(f(x)) = x. But that's a whole separate topic that really belongs in the math forum.)

As for the notation of an inverse function – at least according to how I was taught – I believe it can be written by adding a superscript of negative one to the "f" – or whatever letter is used in the original function, be it an f, a g, an h, or any other letter. As the "-" was not meant to signify a negative x, buts rather a negative superscript for the "f"

 

[PeterDonis]

the "-" was not meant to signify a negative x, buts rather a negative superscript for the "f"

Ah, ok. The usual notation isn't just a negative sign but a -1, as you noted earlier in your post; so the inverse of $f$ is $f^{-1}$.

 

[pervect]

Consider a sphere of test particles around a point. Then each test particle is following a geodesic through space-time. A geodesic is the closest thing to a straight line that exists entirely within space-time, which the test particles must remain in - they can't somehow go "outside of space and time", they are stuck in the framework of space-time like everyone and everything else.

Each particle is moving along a geodesic, but the geodesics can converge - or diverge. The reason that the geodesics converge or diverege is space-time curvature. Remember that the geodesics we are talking about are space-time geodesics, not just curves through space. We can look at the volume of the sphere at some time t, then, knowing that the worldlines progress through time, ask whether the volume of this sphere is increasing, or decreasing, as time progresses, and how fast it is changing (accelerating).

The answer is given by Einstein's field equations. If you look at "The Meaning of Einstein's Equations", http://math.ucr.edu/home/baez/einstein/einstein.html, you'll see a fuller description of the process. The second derivaitive, i.e. the rate of change of the rate of change of the volume, is given by a simple formula : the density at the center of the sphere (rho) plus three times the pressure at the center of the sphere (P).

To answer your other question, you can get a sort of anti-gravity effect where the geodesics diverge, this requires either a negative energy density at the center of the sphere, or a negative pressure, from the above formula. If they are both positive, then the geodesics converge, the sphere of test particles shrinks, and we have what you think of as attractive gravity.

The idea that "pressure causes gravity" isn't at all intuitive, but it does come out of Einstein's equations. Baez's paper, which I liked to above, is probably going to be the best explanation of this you'll find at a moderately low mathematical level.