What Would Happen in a Universe with Negative Curvature of Spacetime?

[mieral]

I know negative curvature of spacetime is close to impossible.. but reading about dark energy and how it is repulsive... I'm trying to find illustrations of what would happen when spacetime curvature is negative (locally as whole cosmos having negative curvature is different concept than local negative curvature, right?)..

or beside negative curvature.. what form must the stress energy tensor be so instead of objects attracting via gravity.. they become repulsive...

I'm not trying to spread disinformation but just want to imagine how the geodesics and curvature would look like if there were antimass and negative curvature locally.

I know antimass doesn't exist. But I just want to visualize how the spacetime diagram would look like when it is negative. Or does the question make any sense at all? Just give me a youtube link for some illustration.. thank you.

 

[A.T.]

I know negative curvature of spacetime is close to impossible.. but reading about dark energy and how it is repulsive

Tidal gravity pulls free falling objects apart radially. No need for dark energy to have some negative spacetime curvature.

 

[mieral]

Tidal gravity pulls free falling objects apart radially. No need for dark energy to have some negative spacetime curvature.

I'm not argung about tigal graviy pulls free falling objects radiclly or other stuf. What I'd like to know are what variables and changes in equation for that to happen.. producing negative curvature and objects are repelled. Does it make to have really negative curvature and the stress energy tensor adjusted certain way. In the end, the question is whether it is feasible or not

 

[PAllen]

I'm not argung about tigal graviy pulls free falling objects radiclly or other stuf. What I'd like to know are what variables and changes in equation for that to happen.. producing negative curvature and objects are repelled. Does it make to have really negative curvature and the stress energy tensor adjusted certain way. In the end, the question is whether it is feasible or not

It is trivially feasible. In 'empty space' connect two large masses by a sufficiently rigid very low mass rod. Then, in between them will be an area of negative curvature where test bodies will separate from each other.

 

[pervect]

I suspect that the OP may be thinking of negative spatial curvature as is talked about in cosmology, rather than negative space-time curvature.

So I suspect that AT and PAllen may be talking about a different notion of curvature than the OP is. Especially since the OP doesn't recognize the fact that particles being pulled apart by tidal forces is associated with a negative component of the Riemann curvature tensor, which is what is motivating PAllen's and AT's response.

Spatial curvature is often confused with space-time curvature, and it's frequently talked about in the context of cosmology as a single number (though I'm hazy on why it's only a single number, I'm not recalling the details at the moment.) But I think there may be some confusion in that AT & PAllen may be answering what the OP asked using the technical definitions of the terms he used, but the intent of the OP might well be different.

 

[mieral]

I suspect that the OP may be thinking of negative spatial curvature as is talked about in cosmology, rather than negative space-time curvature.

No. I'm talking about local negative space-time curvature. My thinking (wrong or not) is that local positive space-time curvature can make moon be closer to the Earth in geodesics with positive curve spacetime causing "gravity"... so could local negative space-time curvature make the moon be farther away from the Earth in geodesics with negative cuved spacetime causing repel of gravity. Or is this not enough and the stress energy tensor must be in certain way? How.

Was Pallen describing this repelling of gravity. If, as he described, you have two larges objects connected with a sufficiently rigid very low mass rod, would the area between them have negative curvature.. so if you send a ball to the rod.. would it repel the rod.

Or if you still can't understand the above. What must you do to the spacetime curvature, stress energy tensor to make the moon be repel from the Earth where although the geodesics is straight but the curve is negative and objects got separated (creating the effect of gravity repulsion).

Again, I know gravity repulsion is not possible. But just want to understand what conditions or adjustments in General Relativity to make this occur.

So I suspect that AT and PAllen may be talking about a different notion of curvature than the OP is. Especially since the OP doesn't recognize the fact that particles being pulled apart by tidal forces is associated with a negative component of the Riemann curvature tensor, which is what is motivating PAllen's and AT's response.

Spatial curvature is often confused with space-time curvature, and it's frequently talked about in the context of cosmology as a single number (though I'm hazy on why it's only a single number, I'm not recalling the details at the moment.) But I think there may be some confusion in that AT & PAllen may be answering what the OP asked using the technical definitions of the terms he used, but the intent of the OP might well be different.

 

[Nugatory]

So could local negative space-time curvature make the moon be farther away from the Earth in geodesics with negative cuved spacetime causing repel of gravity.

Sure - the point on the moon that is closest to us (right in the middle of the disk during a full moon) and the point on the opposite side are experiencing a repulsive force between them. The moon is made of solid rock which doesn't stretch much, but we've been able to measure the distortion. If we could saw the moon in half along the line between the side that faces us and the hidden side, the more distant hemisphere would move away from the earth.

You may object that this is not a "repulsive" force; both sides of the moon are attracted towards the earth, and all that's going on is that the near side is attracted more strongly than the far side. That is indeed the standard Newtonian explanation... but note that you have (probably without noticing) selected coordinates in which the the Earth is at rest to come up with that explanation. Work this problem in coordinates in which the center of the moon is at rest and you'll come up with a different explanation, one that recognizes that the negative curvature between the two halves of the moon.

 

[mieral]

Is there a website with java where you can adjust the curvature of space from positive to negative and see parallel geodesics converge (gravity) or diverge (repel of gravity). I know the latter doesn't occur.. but I just want to know if spacetime diagram can theoreticaly cause diverging of geodesics (how does this look like.. there must be one website amongst hundreds out there with this).

Remember we always have opposites.. like positive vs negative, day vs night.. man vs woman.. so i just wonder how positive vs negative curvature looks like locally. Thank you.

 

[Nugatory]

Remember we always have opposites.. like positive vs negative, day vs night.. man vs woman.. so i just wonder how positive vs negative curvature looks like locally. Thank you.

Imagine yourself in a small windowless room that is in freefall towards the surface of a planet. It's freefall (at least until the room hits the surface) so if you hold an object out in front of you and let go, it will float in front in you, just as you've seen in so many videos from astronauts in space. Now suppose you position one small object near the ceiling, another near the floor, and four more near each of the four walls. Look closely though, and you will see the objects near the walls moving towards one another and the objects at the floor and the ceiling moving away from one another... The spacetime around the planet is has negative curvature in the "vertical" direction and positive curvature in the "horizontal" direction, and this is about as local of a situation as we can imagine.

(And if you're wondering why I put the scare-quotes around the words "horizontal" and "vertical"? Someone watching from outside would call those directions "radial" and "tangential", a better way of thinking about the global spacetime around the planet).

 

[A.T.]

.. but I just want to know if spacetime diagram can theoreticaly cause diverging of geodesics (how does this look like.. there must be one website amongst hundreds out there with this).

See case C in post below. That is negatively curved space-time along a radial line outside of a massive body.

This is my own non-animated way of looking at it:

• A. Two inertial particles, at rest relative to each other, in flat spacetime (i.e. no gravity), shown with inertial coordinates. Drawn as a red distance-time graph on a flat piece of paper with blue gridlines.
  
• B₁. The same particles in the same flat spacetime, but shown with non-inertial coordinates. Drawn as the same distance-time graph on an identical flat piece of paper except it has different gridlines.
  
  B₂. Take the flat piece of paper depicted in B₁, cut out the grid with some scissors, and wrap it round a cone. Nothing within the intrinsic geometry of the paper has changed by doing this, so B₂ shows exactly the same thing as B₁, just presented in a different way, showing how the red lines could be perceived as looking "curved" against a "straight" grid.
  
• C. Two free-falling particles, initially at rest relative to each other, in curved spacetime (i.e. with gravity), shown with non-inertial coordinates. This cannot be drawn to scale on a flat piece of paper; you have to draw it on a curved surface instead. Note how C looks rather similar to B₂. This is the equivalence principle in action: if you zoomed in very close to B₂ and C, you wouldn't notice any difference between them.

Note the diagrams above aren't entirely accurate because they are drawn with a locally-Euclidean geometry, when really they ought to be drawn with a locally-Lorentzian geometry. I've drawn it this way as an analogy to help visualise the concepts.

 

[mieral]

See case C in post below. That is negatively curved space-time along a radial line outside of a massive body.

How small can negative curvature be induced in middle of positive curvature? Supposed, for example, just for sake of illustration.. you have a tiny stone made of antimass. If it floats up or repel gravity of earth.. can we say it can induce tiny negative curvature in middle of the positive curvature created by the earth. Again this is just an example.. I know you will say there is no antimass and I'm not arguing anything. .

If localized negative curvature has minimum size. Does it mean a tiny stone of antimass that can be held inside the hands can induce mile size negative curvature so everything a mile around it would float up or repel earth? Let me emphasize this is just for sake of illustration to understand how negative curvature can occur in middle of positive curvature and not to spread disinformation which this website so hate. Thanks.

 

[A.T.]

it can induce tiny negative curvature in middle of the positive curvature created by the earth

The curvature created by the Earth (outside of the Earth) is already negative. See above.

 

[martinbn]

There are already several answers and in my opinion they are all good and illuminating, but it seems that none satisfies you. I think it is because you questions a very vague and probably you don't know what you are asking. Can you clarify what you mean by curvature and by it being negative? Do you mean the Riemann tensor and some components being negative, do you mean scalar curvature, do you mean something else?

 

[mieral]

There are already several answers and in my opinion they are all good and illuminating, but it seems that none satisfies you. I think it is because you questions a very vague and probably you don't know what you are asking. Can you clarify what you mean by curvature and by it being negative? Do you mean the Riemann tensor and some components being negative, do you mean scalar curvature, do you mean something else?

Well.. I'm avoiding a word from the start because it is a taboo word and physicists hate it. But for now let's use the word and just try to hold off any automatic hate for it.

Gravity = mass = positive curvature = normal stress energy tensor
Antigravity = antimass = negative curvature = altered stress energy tensor

I know physicists hate the word "antigravity" because it is not the mainstream and it's a word for crackpots. But I need to understand how General Relativity can accommodate matter to repel each other (and why not if it can't). You may say there is no such thing as antigravity. But I want to understand why General Relativity can forever only accommodate gravity that attracts. I've searched for this all over the net but couldn't find the explanations. A Scientific American article or others that explain why it can't would be illuminating.. this would enable more understanding of General Relativity. So can anyone give a good summary why it can't? You may say there is no antimass. But just supposed you directly make the curvature be certain shape (like negative curvature).. can it cause opposite of gravity attraction or repulsion?

 

[A.T.]

But just supposed you directly make the curvature be certain shape (like negative curvature).. can it cause opposite of gravity attraction or repulsion?

Yes, that's what you see locally as tidal effects, along the radial direction.

See also:
http://burro.astr.cwru.edu/Academics/Astr328/Notes/Metrics/metrics.html

 

[martinbn]

You still haven't explained what you mean by curvature.

You have the spacetime, which is a four dimensional manifold. You talk about its curvature. What do you mean by curvature (mathematically)?

 

[mieral]

You still haven't explained what you mean by curvature.

You have the spacetime, which is a four dimensional manifold. You talk about its curvature. What do you mean by curvature (mathematically)?

I'm still trying to visualize or imagine how a four dimensional manifold has positive or negative curvature. I'll search in the webs more before asking so my questions would make more sense.

But remember Einstein Field Equations are kinda like solutions. You can enter certain input and you have black hole solutions. So I'm kinda asking what if you input matter that is opposite to matter.. would it have repulsive gravity. If there is a arxiv paper about this.. do share it.

I just read now so repulsive gravity is not a fringe thing at all.

https://phys.org/news/2011-04-antimatter-gravity-universe-expansion.html

If antimatter or stuff really antigravitate.. what actually happens to the geodesics.. illustrations appreciated.

 

[A.T.]

I'm still trying to visualize or imagine how a four dimensional manifold has positive or negative curvature.

I can't even visualize a four dimensional manifold without curvature. But for simple cases like radial fall you don't need all four dimensions.

 

[mieral]

I can't even visualize a four dimensional manifold without curvature. But for simple cases like radial fall you don't need all four dimensions.

http://sns.ias.edu/~malda/sciam-maldacena-3a.pdf

In the above.. Maldacena mentioned anti-di Sitter space has negative curvature. So negative curvature automatically has repulsive gravity right? But how do you imbed small localized negative curvature within a big positive curvature like the earth.. say the negative curvature is caused by a basketball size antimatter (protective by say some kind of bubble to avoid detonation when it is in contact with matter) that just shoots up from the ground on earth? I just need an illustration how the imbedding occurs. I'll draw it tomorrow to correct where I could have imagined it wrong.

 

[timmdeeg]

But I need to understand how General Relativity can accommodate matter to repel each other (and why not if it can't). You may say there is no such thing as antigravity. But I want to understand why General Relativity can forever only accommodate gravity that attracts. I've searched for this all over the net but couldn't find the explanations. A Scientific American article or others that explain why it can't would be illuminating.. this would enable more understanding of General Relativity. So can anyone give a good summary why it can't? You may say there is no antimass. But just supposed you directly make the curvature be certain shape (like negative curvature).. can it cause opposite of gravity attraction or repulsion?

It seems you are making your life harder than this issue requires.
Imagine two particles which are falling radially towards a central mass, being separated by some radial distance. The lower particle being closer to the mass falls faster than the upper one at a given time. By this their distance increases accelerated or in other words their geodesics are accelerating away from each other, just as shown in #15 "Negative Curvature".
Now imagine two particles with the same radial distance, which are falling towards the center of a mass, being separated in the tangential direction. By this their geodesics are accelerating towards each other, see #15 "Positive Curvature".

Note that most presumably antimass behaves like mass regarding the curvature of spacetime. Two anti-hydrogen atoms would attract each other (a few have been created so far). According to present knowledge repelling gravity is caused by vacuum energy, the cosmological constant or dark energy resp.

 

[PeterDonis]

I'm still trying to visualize or imagine how a four dimensional manifold has positive or negative curvature

It has both. You are thinking of "curvature" as a single number, but it's not. In a general four-dimensional spacetime, twenty independent numbers (the independent components of the Riemann curvature tensor) are needed to describe curvature. These numbers can be split into two sets of ten numbers, the Weyl curvature and the Ricci curvature; only the latter is tied to stress-energy by the Einstein Field Equation (see below).

Even in a vacuum, spherically symmetric spacetime, such as the spacetime around a massive body like the Earth (i.e., not in the interior but the vacuum region around it), there are at least two curvature numbers (the radial and tangential tidal gravity), and they have different signs (the radial number is negative and the tangential number is positive). This curvature is entirely Weyl curvature, because Ricci curvature is zero in any vacuum region (in the absence of a cosmological constant--see below).

Gravity = mass = positive curvature = normal stress energy tensor
Antigravity = antimass = negative curvature = altered stress energy tensor

These statements apply specifically to Ricci curvature; they do not apply to Weyl curvature. Also, "stress energy tensor" here has to include the cosmological constant: a positive cosmological constant (in the usual sign convention) corresponds to negative curvature, i.e., "antigravity".

 

[mieral]

It seems you are making your life harder than this issue requires.
Imagine two particles which are falling radially towards a central mass, being separated by some radial distance. The lower particle being closer to the mass falls faster than the upper one at a given time. By this their distance increases accelerated or in other words their geodesics are accelerating away from each other, just as shown in #15 "Negative Curvature".
Now imagine two particles with the same radial distance, which are falling towards the center of a mass, being separated in the tangential direction. By this their geodesics are accelerating towards each other, see #15 "Positive Curvature".

Note that most presumably antimass behaves like mass regarding the curvature of spacetime. Two anti-hydrogen atoms would attract each other (a few have been created so far). According to present knowledge repelling gravity is caused by vacuum energy, the cosmological constant or dark energy resp.

Does dark energy only occur in between galaxies because it's the only place where spacetime can be entirely negative curved (so gravity entirely repulsive)?

For sake of discussion. What would happen if concentrated dark energy were acquired and send to Earth and stored in a lab. Would it anti-gravitate upward to the sky or fall to the ground? and how to you draw the geodesics of these. This is asked just to understand more about the versatility of spacetime and the different form it may take.. not because of science fiction or crackpotty.

 

[PeterDonis]

Does dark energy only occur in between galaxies

No. It's everywhere (at least in the simplest model, which right now we have no reason to think is not the correct one). But its effects are only measurable over large distances.

What would happen if concentrated dark energy were acquired

You can't concentrate dark energy (again, in the simplest model). It's the same density everywhere.

Would it anti-gravitate upward to the sky or fall to the ground?

Dark energy itself doesn't move at all. Its density is the same everywhere. It causes objects that are in free fall (i.e., acted on by no forces) to accelerate away from each other. More precisely, it causes a form of spacetime curvature (tidal gravity) that does that (negative curvature).

 

[Comeback City]

It causes objects that are in free fall (i.e., acted on by no forces) to accelerate away from each other. More precisely, it causes a form of spacetime curvature (tidal gravity) that does that (negative curvature).

Would this only be observable on VERY large scales, or is it noticeable if, say, 2 objects are free-falling to Earth side-by-side?

 

[PeterDonis]

is it noticeable if, say, 2 objects are free-falling to Earth side-by-side?

No. By "large scales" I mean "hundreds of millions of light-years or more".

 

[Comeback City]

No. By "large scales" I mean "hundreds of millions of light-years or more".

Ahhhhhhh okay it's starting to come together a little bit more for me now. This would explain why specifically galaxies (being separated by those long of distances) on the "large scales" are affected by universal expansion and dark energy (rather than at the local level), or do I have this wrong?

 

[mieral]

I have actually tried to google the following question but can't get answer either here or elsewhere.. it is a very general question that needs to be answered.

Supposed 2 billion years from now.. we have mastered spacetime and can create technology that can create artificial synthetic dark energy-like stuff (meaning not exactly dark energy but artificially made yet can cause tidal gravity or repel test objects or cause negative curvature weyl wise). If it would be made portable and deployed in aircraft. Can this synthetic dark energy be used as propulsion by changing directly spacetime curvature *locally*? Most stuff about antigravity speaks of shielding mass and we know this is nonsense in the language of General Relativity. But how about engineering spacetime itself. Note this is my original question and main question of this thread. Would a localized engineered dark energy-like spacetime entity be used to act locally and make any object float or would it only cause a mile wide effect.. can you cause tiny localize tidal gravity within a large positive curvature?

Do you get my question. We on surface of Earth is attracted to Earth because of positive curvature. To make artificial negative curvature.. can this occur locally in tiny part of it (like in aircraft on big earth) or can this only occur with minimum area.. meaning you need to make mile wide distance into negative curvature as minimum? Hope someone gets my question.
For those who understood my question. Please rephrase my question so it would make sense to the other experts. Thank you.

 

[PeterDonis]

Supposed 2 billion years from now.. we have mastered spacetime and can create technology that can create artificial synthetic dark energy-like stuff

We don't know if this is even allowed by the laws of physics, so this is a very shaky speculation.

Also, making artificial dark energy-like stuff would be making artificial stress-energy, not artificial spacetime.

Can this synthetic dark energy be used as propulsion by changing directly spacetime curvature *locally*?

Any stress-energy will affect spacetime curvature locally. But the effect depends on the density of the stress-energy. Ordinary dark energy has no measurable effect on scales smaller than hundreds of millions of light years because its density is so tiny (something like 30 orders of magnitude smaller than the density of water), so its effect on ordinary distance scales is similarly tiny (meaning much tinier than we can possibly measure, and also much tinier than the effects of other stress-energy in our vicinity). If we could make synthetic dark energy that had a density more like ordinary densities, it would have effects on spacetime curvature that were measurable locally just like the spacetime curvature due to an ordinary mass like the Earth. But, as I said, we don't know if this is even allowed by the laws of physics.

 

[mieral]

Please see graphics above. Are you saying it is possible to have sharp turns in spacetime rather than just smooth transition? All we are exposed to are smooth transition. The red sharp turn or graph depicts "for_sake_of discussion" dark energy with the density of ordinary matter..

 

[PeterDonis]

Are you saying it is possible to have sharp turns in spacetime rather than just smooth transition?

I'm not sure how you got that from what I said. To answer the question as you ask it, GR assumes that spacetime is always and everywhere a smooth manifold, so no.

 

[mieral]

I'm not sure how you got that from what I said. To answer the question as you ask it, GR assumes that spacetime is always and everywhere a smooth manifold, so no.

If it's a smooth manifold.. then it's outlawed by General Relativity to have dark energy-like stuff with density of matter because it won't be a smooth manifold if you have suddenly have stress energy tensor that is repulsive to matter in middle of normal density matter.. because if someday we could engineer artificial dark energy with the density of matter and you put this underneath your chair.. your chair would shoot up straight to the sky. But if you or another person can draw graphics of what it looks like to have this negative curvature occurring inside a large positive curvature without sharp turn.. please do.. as they say.. a picture is worth a thousand words... and I'd like to have idea how the spacetime diagram of this chair (propulsioned by artificial dark energy) shooting up the sky look like.

 

[PeterDonis]

If it's a smooth manifold.. then it's outlawed by General Relativity to have dark energy-like stuff with density of matter

No, it is not. Dark energy, in its simplest form, appears as a cosmological constant, and any finite value of the cosmological constant is compatible with a smooth manifold.

what it looks like to have this negative curvature occurring inside a large positive curvature

That's not what would happen if you had dark energy with a density similar to ordinary matter near the Earth. The negative curvature due to the dark energy would not be localized any more than the curvature due to the Earth (which is not just positive--you need to go back and read my post #21) is localized.

You seem to be trying to use vague intuition to imagine what your scenario would be like, instead of using the actual math. I suggest using the actual math.

a picture is worth a thousand words

You can't draw a correct picture until you've done the underlying math. I don't have time to do it here (I don't get paid for this), but consider: the Einstein Field Equation, including dark energy in the form of a cosmological constant, can be written as follows:

$$
R_{\mu \nu} - \frac{1}{2} g_{\mu \nu} R = 8 \pi T_{\mu \nu} - \Lambda g_{\mu \nu}
$$

The simplest model for dense dark energy near (but not inside) the Earth would be to have a spherical region with the radius of the Earth where $T_{\mu \nu}$ is nonzero (and its components will depend on the internal structure of the Earth, but we don't need to go into detail about that) and $\Lambda$ is zero, then a region of vacuum (or near vacuum--the Earth has an atmosphere, but we can use the Moon if we want to remove that complication, it won't change anything important) in which both $T_{\mu \nu}$ and $\Lambda$ are zero, then a region where $\Lambda$ is nonzero (and positive, since it's just a scalar) and $T_{\mu \nu}$ is zero. It is easy to model a smooth transition between each of these regions, with no discontinuity. Therefore, the spacetime curvature will also have a smooth transition.

Also, since the Einstein Field Equation is nonlinear (if you expand out the LHS above you will see that it involves products of the metric and its derivatives), you can't just assume that the solution for the model I described above will be a simple sum of solutions for the Earth by itself and the dark energy by itself. You have to come up with a single solution that includes both, as described above. So, again, there will be no "sharp turn" where one solution stops and the other takes over. There is just a single smooth solution.

 

[mieral]

No, it is not. Dark energy, in its simplest form, appears as a cosmological constant, and any finite value of the cosmological constant is compatible with a smooth manifold.
That's not what would happen if you had dark energy with a density similar to ordinary matter near the Earth. The negative curvature due to the dark energy would not be localized any more than the curvature due to the Earth (which is not just positive--you need to go back and read my post #21) is localized.

Thanks a lot for the explanation. But you can't seem to get my specific question. I was not asking about what would happen if you had dark energy with a density similar to ordinary matter *near* the Earth. I was asking what would happen if you had dark energy with a density similar to ordinary matter *underneath an airplane parked* right on earth. Would the airplane just float up.

You seem to be trying to use vague intuition to imagine what your scenario would be like, instead of using the actual math. I suggest using the actual math.
You can't draw a correct picture until you've done the underlying math. I don't have time to do it here (I don't get paid for this), but consider: the Einstein Field Equation, including dark energy in the form of a cosmological constant, can be written as follows:

$$
R_{\mu \nu} - \frac{1}{2} g_{\mu \nu} R = 8 \pi T_{\mu \nu} - \Lambda g_{\mu \nu}
$$

The simplest model for dense dark energy near (but not inside) the Earth would be to have a spherical region with the radius of the Earth where $T_{\mu \nu}$ is nonzero (and its components will depend on the internal structure of the Earth, but we don't need to go into detail about that) and $\Lambda$ is zero, then a region of vacuum (or near vacuum--the Earth has an atmosphere, but we can use the Moon if we want to remove that complication, it won't change anything important) in which both $T_{\mu \nu}$ and $\Lambda$ are zero, then a region where $\Lambda$ is nonzero (and positive, since it's just a scalar) and $T_{\mu \nu}$ is zero. It is easy to model a smooth transition between each of these regions, with no discontinuity. Therefore, the spacetime curvature will also have a smooth transition.

Your above explanation assumes the hypothetical dark matter with density that of matter is big and near the earth. I was specifically asking since yesterday what if it were just small and say put beneath an automobile parked on ground. Would there still be smooth transition of the curvatures (which is opposite) and make the automobile float up. I know you are tired typing all the long replies. I appreciate them so much. So I hope others can answer this specific question in case you are tired already. I need to someone to answer this because I'll share the concept with laymen and don't want to be wrong. Thanks.

Also, since the Einstein Field Equation is nonlinear (if you expand out the LHS above you will see that it involves products of the metric and its derivatives), you can't just assume that the solution for the model I described above will be a simple sum of solutions for the Earth by itself and the dark energy by itself. You have to come up with a single solution that includes both, as described above. So, again, there will be no "sharp turn" where one solution stops and the other takes over. There is just a single smooth solution.

Again you are assuming I was describing the Earth beside dark energy with density that of Earth and if there was a smooth transition. But I was asking of dark matter with density that of matter but very SMALL and say just baseball size.. and this is stored on a lab or put underneath automobile. Would it float up and would the transition be smooth. Anyone else?

 

[PeterDonis]

I was asking what would happen if you had dark energy with a density similar to ordinary matter *underneath an airplane parked* right on earth.

That is near the Earth; you would model it basically the same as the way I described. The region with $\Lambda > 0$ would be small, but that doesn't change the basic model, just the details.

Your above explanation assumes the hypothetical dark matter with density that of matter is big and near the earth. I was specifically asking since yesterday what if it were just small and say put beneath an automobile parked on ground.

Same answer.

Would there still be smooth transition of the curvatures (which is opposite)

Yes. That's always going to be true in the general model I described--just as with any other model that's consistent with GR.

and make the automobile float up

That will depend on how the spacetime curvature in the actual model comes out. You can't just assume it will be negative because the dark energy is there. You have to actually solve the Einstein Field Equation for the model I described. My intuitive guess is that, if the region of dark energy is much smaller than the Earth, the automobile won't float up; the dark energy's effect will just be to decrease its weight some, by canceling out some of the effect of Earth's gravity. But I haven't done the math.

 

[mieral]

That is near the Earth; you would model it basically the same as the way I described. The region with $\Lambda > 0$ would be small, but that doesn't change the basic model, just the details.
Same answer.
Yes. That's always going to be true in the general model I described--just as with any other model that's consistent with GR.
That will depend on how the spacetime curvature in the actual model comes out. You can't just assume it will be negative because the dark energy is there. You have to actually solve the Einstein Field Equation for the model I described. My intuitive guess is that, if the region of dark energy is much smaller than the Earth, the automobile won't float up; the dark energy's effect will just be to decrease its weight some, by canceling out some of the effect of Earth's gravity. But I haven't done the math.

Yes, this is what I was asking since yesterday, whether it would float up or not. Your intuitive guess would be that "if the region of dark energy is much smaller than the Earth, the automobile won't float up; the dark energy's effect will just be to decrease its weight some, by canceling out some of the effect of Earth's gravity". That's brilliant comment. Now is there a General Relativity student here who can actually solve the Einstein Field Equation for the model.. for exercise and practice? Please solve whether the automobile with dark energy-like fuel would float or just decrease the weight by some percentage. Also compute how large should be the dark matter with density of matter so the automobile would float up.
Thank you.
In case there is a paper already written of similar calculations or sorta.. please share it. Ty.