According to the general theory of relativity, “gravitation is a consequence of the curvature of spacetime”. Does that mean that mass and/or gravity is equivalent to a curvature of spacetime?
Say I’ve got a device that can locally curve spacetime to make the inside of my apartment twice as big as the outside, would that result in an extra gravitational force (and would it pull in or push out)? Having to nail my furniture down would be sort of a drawback.
Basically, the presence of mass (or energy) makes spacetime curve, and this curvature causes masses to be attracted towards each other, and this attraction is gravitation. Does that answer your (first) question?
I’m not sure I follow you on the apartment question, but as far as push vs. pull is concerned, gravity is always attractive so far as I know.
My degree was in atomic physics, though, so maybe someone more well-versed in GR will have a better answer for you.
The only known way to cause space to curve is via mass (or, strictly speaking, things closely related to mass, such as momentum flux or stress). So your apartment-TARDIS would, so far as we know, need to have a very significant amount of mass built into it.
And in order to produce a TARDIS effect (inside bigger than the outside) in general relativity, you need to have a supply of exotic matter, which has a negative mass. So, interestingly enough, your apartment would actually cause a repulsion. And no, nobody knows of any way to obtain negative mass, though it’d have all sorts of really wild properties.
In order to make a local area “larger” (i.e. lengthen the time required for light to follow a geodesic path) would require negative pressure. This sort of “dark energy” is hypothesized to be the force which (under our current understanding of cosmology) causes the Universe to accelerate its expansion despite the stored gravitational energy that should be slowing the rate of expansion. In terms of General Relativity, this would be some kind of force (presumably gravitational or fundamentally indistinguishable from gravity) that causes the eigenvalues of the stress-energy tensor in the Einstein field equations to be negative. However, the mysterious “dark energy” aside, we don’t have any way to generate repulsive gravity.
If you could create this, it would essentially create a gradient at any point from which everything seemed to be continually expanding away from you, just as galaxies beyond our Local Supercluster are fleeing away from us as if we have cosmological halitosis. So in a sense it would pull away (or be pushed away).
I agree for the most part with what Chronos and Stranger have said: it’s correct that “exotic matter” is required to create an apartment which is “bigger on the inside.” However, it’s entirely possible to describe spacetimes in which there’s a spherical central region with no gravitational forces inside (i.e. the central region of the spacetime is flat), surrounded by an arbitrarily thin shell of exotic matter that warps space such that a sphere just outside of it has a smaller surface area radius than a sphere just inside of it. Since there’s no gravitational forces inside the sphere, you wouldn’t have to “nail your furniture down.” (The fact that there’s no gravity inside the sphere is closely analogous to the fact that there’s no electric field inside a uniformly charged sphere.) You can jigger things such that the sphere causes no exterior gravitational field, either, so your neighbours won’t complain. I think it’s not even required that the central apartment be spherical (good thing — have you ever tried arranging furniture in a room with a curved wall?)
By bigger on the inside, don’t we mean a circle drawn close around the apartment should have a radius that is bigger than its circumference over two pi? Wouldn’t that mean a large positive mass? This is from memory, and while I am a physicist, I’m not a very good one…
On the other hand, reading this thread and skimming some other articles it appears that while we don’t know of anything else that could cause the bending of spacetime, it’s not clear that that is “all there is” to mass. At least that’s what it looks like to me. Unless inertial mass is something fundamentally different than gravitational mass.
FWIW, that’s what I thought, too – if I drew a sphere around a massive body, surely I’d have a greater interior than exterior volume? Granted, that’d probably work as well with negative mass, but since that’s kinda hard to get one’s hands on…
OK, could you explain that to me like I’m a total idiot? Because, it seems to me, that if a yardstick is length contracted in a gravitational field, you’d need more of them to reach some boundary sphere, and thus measure a greater internal radius than you’d expect from measuring the sphere’s circumference externally…
I was trying to think of an intuitive reason for this. How about this:
A projectile will fly across your apartment more quickly if there’s a mass in the middle, because the mass attracts it and speeds it up. But from a GR perspective the projectile is still following its same “straight line” path, and it is spacetime itself that is distorted (changing the definition of “straight line”). The fact that the projectile gets their quicker must thus be due to the path length being shortened, meaning the interior of the room must have been shrunk by the presence of the massive body.
Again, GR isn’t my strong suit, so perhaps this explanation is way off the mark.
If you’re in the room you should measure the same shouldn’t you? I mean if you’re measuring a 10 foot wide floor with a 1 foot long ruler, and the ruler and floor are each contracted by 10%, then it would still take ten lengths of the ruler to span the width of the floor.
I would think that it’s only from the perspective of the person outside the room that the contents of the room are contracted. Just like the guy in the fast moving rocket ship doesn’t feel time slowing down, but the guy looking through the window as he whizzes by sees him moving in slow motion.
Ah, right, I see my mistake now – I thought that, if you had a stick as long as the room is deep, measured from the outside, for instance, and you took that stick into the room, it would not reach from back wall to front wall due to its being length contracted; that’s bullshit, of course, as the room’s interior will be equivalently ‘smaller’. What confused me was that a ray of light, sent across the room, actually will take longer than if it’s crossing the same distance along the exterior wall, due to the Shapiro effect. However, I’m not alone in this confusion, it seems; even at the Perimeter Institute, they claim:
Actually, having gone through the calculation (for a weakly curved spherically symmetric spacetime in isotropic coordinates), I’m pretty sure that whoever wrote the Perimeter Institute page is right. If you measure a sphere of radius R in such a spacetime, you find that its area is A = 4*pi R[sup]2[/sup] (1 - 2 Phi®), where Phi® is the Newtonian gravitational potential at radius r. Meanwhile, the volume inside the sphere is
V = 4*pi Integral[sub]0[/sub][sup]R[/sup] (1 - 3 Phi®) r[sup]2[/sup] dr
(apologies for the kludgy text-only equation.) If you calculate the “expected volume” V’ that you would obtain by assuming the sphere was in flat spacetime and extrapolating the volume from the measured area A (as described in the PI page), you find that the “expected volume” V’ is less than the actual volume V, as long as Phi® increases with radial distance. I’m pretty sure the same goes for imaginary hoops stretched about the Sun, too.
I think. If someone could check my math that’d be great.
So is that a contradiction to the claim above that it would be smaller inside? Or is there some nuance that distinguishes what the Perimeter Institute people are saying from what Chronos et al. are saying above.