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

[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.

 

[PeterDonis]

is there a General Relativity student here who can actually solve the Einstein Field Equation for the model.. for exercise and practice?

This goes beyond the scope of a "B" level thread; the math required is at least "I" level. However, a starting point for a heuristic solution is the Schwarzschild-de Sitter solution, which is described here:

https://en.wikipedia.org/wiki/De_Sitter–Schwarzschild_metric

The metric for this solution is:

$$
ds^2 = - \left( 1 - \frac{2M}{r} - 3 \Lambda r^2 \right) dt^2 + \frac{1}{1 - 2M / r - 3 \Lambda r^2} dr^2 + r^2 \left( d\theta^2 + \sin^2 \theta d\phi^2 \right)
$$

This metric describes a vacuum region around a central mass $M$ which is spherically symmetric, in which there is a cosmological constant (dark energy) with density $\Lambda$. The proper acceleration of an object at rest in these coordinates (i.e., at constant $r$) is given by

$$
a = \frac{1}{\sqrt{1 - 2M / r - 3 \Lambda r^2}} \left( \frac{M}{r^2} - 3 \Lambda r \right)
$$

where positive $a$ points radially outward (i.e., normal weight) and negative $a$ points radially inward (i.e., negative "weight"). As you can see, the terms in $M$ and $\Lambda$ have opposite signs--this is why dark energy is often said to cause "antigravity".

The above metric isn't quite what we want, though, because it assumes that the cosmological constant $\Lambda$ is present throughout the region out to radius $r$. So if we take $r$ to be just above the surface of the Earth, this would require the entire Earth to be permeated with dark energy of density $\Lambda$, where that density is some normal density rather than the actual dark energy density in our universe (which is so tiny that it has no measurable effect, by many orders of magnitude, on length scales like the size of the Earth--the product $\Lambda r^2$ for $r$ the radius of the Earth is many orders of magnitude smaller than $2M / r$ for the Earth). What we really want is a metric where, heuristically, the $2M / r$ term is as above, but the $3 \Lambda r^2$ term uses an $r$ that is much, much smaller--the size of the small bundle of dark energy that gets placed under the automobile. Such a metric, as I have stated it, is not, on its face, a solution of the Einstein Field Equation (since we can't just arbitrarily change $r$ in one term). But it shows the general kind of thing that I based my intuitive guess on in my previous post: basically, the relative magnitudes of $M / r^2$ for the Earth and $3 \Lambda R$, where $R$ is the radius of the dark energy bundle and $\Lambda$ is its density, will, heuristically, determine whether the automobile "falls upward" or just has less weight.

If you want to develop your understanding of GR, I suggest that you try working out a solution yourself.

 

[jedishrfu]

I think it's time to close this thread now. Thanks to all who contributed answers to the OPs question.

It's always difficult to answer these kinds of questions regarding something like dark energy of which we have no solid understanding beyond the cosmological context. Perhaps someday soon we will have a theory that can be applied to this question but until then we can only wonder.

Jedi