I’m in the middle of reading Cosmic Catastrophes by J. Craig Wheeler, and I just read, for the nth time in my life, a description of gravity as curved spacetime. The author states that an orbit, for example, is in Einstein’s view a straight line in curved spacetime. If you are following the shortest path between any two points along the curvature, that is defined as the “straight” line for that curvature. Fine. I get that. The author goes on to state that gravity, again in Einstein’s view, is not a force acting on a body as Newton claimed, but simply the curvature of local spacetime around mass concentrations, which deflect the paths of moving bodies. Again, I get it.
It seems to me, though, that there is something missing from this description. Suppose you have a large mass concentration creating a nice hefty spacetime curvature. Suppose a much smaller mass comes ambling along and passes in the immediate vicinity of the large mass. The “straight line” path of the smaller body is deflected. Okay so far. But now suppose that the smaller body is instead at rest with respect to the larger mass, but very close to it. There is no straight-line motion to be affected, so the description as I understand it states that the smaller mass should stay where it is.
But that’s not what happens. The smaller mass will accelerate towards the larger mass (I know, I know, the larger mass also accelerates toward the smaller, but let’s keep it simple). How does this acceleration figure into the “straight line in curved spacetime” description? It clearly represents an input of energy into the smaller mass, which doesn’t sound like it can be encompassed by a simple geometric argument.
While I’m sure some physicist will be along to explain how it all works, I want to put in that the best explanation of curved space-time I’ve ever heard was the analogy of the bowling ball on a mattress. Makes a big, hyperbolic dent. Try to roll a marble past it and the marble will A.) make it through after curving along the inside lip of the depression or B.) become trapped in the depression and begin to spiral inward. Of course, if we ignore the force of real gravity acting on the marble, it would go into orbit around the bowling ball rather than spiral inward if it had sufficient velocity.
Try putting the same marble ‘at rest’ in the depression. Falls right in.
As for the math behind it all, a better scientist than I will have to explain.
Exactly my point. In the bowling ball analogy, if we ignore the “real” force of gravity, and the initial velocity of the marble was zero with respect to the bowling ball, then it should remain zero. No acceleration would take place.
I think I’d have to ask, “How did it come that the marble is at rest near the bowlingball”. It could be at rest, as long as it’s out of the immediate gravitational force of the larger mass. I’m thinking of this problem like a marble sitting on the apex of a hill, at rest and not moving. But if you could move the hill under it, so that it’s now resting on the slope of the hill, it will roll down it. One of the two objects has to come near the other for them to be that close. They can’t be created at rest that close to each other is what I’m saying I guess.
The smaller object could be at rest close to it, if it had an equal and opposing force, but in either the absence of this, or an increase in the force of the larger object (i.e. it moves toward the smaller object) they will begin to accelerate towards each other. Right?
I think we’re talking past each other. If gravity is only a curvature of spacetime, and objects under the influence of gravity are only following the shortest possible path through that spacetime, then where does the acceleration come from? It doesn’t seem to me as though geometric curvature alone could impart kinetic energy to a body, but that’s the effect of acceleration under the influence of gravity.
I think you are forgetting about the time portion of spacetime. You must include the temporal component of the spacetime path. You will then find that the path (including all four components of spacetime) taken by a falling object starting from rest will be shorter than the path of a object maintaining its position.
Maybe an object which encounters a “Depression in the fabric of space/time” created by gravity, suddenly has an easier path through normal space/time, thus creating a tangenital acceleration in the direction of the gravitational body.
I know enough physics to be confused by it all. So be warned.
I’m wondering: in Einsteinian mechanics, can you havea body at absolute rest? Cuz your scenario assumes the smaller body is at absolute rest. Otherwise, it would have a path through space-time; as that path enters the curved space-time around the larger mass, the path bends.
Or maybe you’re thinking that the smaller mass is at relative rest to the larger one, but close by. All this means is that you’re starting with a hypothetical initial state in which both masses have the same velocity vector. OK, now let’s run the projector! Immediately the smaller mass (still moving remember, but apparently not moving with respect to the larger mass) has a path that is affected by the space-time around the larger mass. The smaller mass’ path starts curving.
I wonder if I’ve argued myself into postulating that gravity doesn’t exist independent of time. If you stop the universe cold, so that nothing is moving, gravity doesn’t exist.
I’m wondering: in Einsteinian mechanics, can you havea body at absolute rest? QUOTE]
Good question. Even if it’s spacial position isn’t changing, its temporal position is always changing. But I’ve never seen a temporal component associated with the force of gravity, it may be time independent. Force can change over time, but taken at an infinitly small period, it should still have a value, shouldn’t it?
Yes, I was thinking “at rest relative to the larger mass.” You make an intersting point, but your statement “still moving remember, but apparently not moving with respect to the larger mass” would seem to imply some kind of state of absolute motion on the part of the two masses. Are we allowed to assume that? I thought one reference frame was as good as another.
Hmmm. The situation is a helluva lot more complex than I first thought, but I think that Pleonast answered the question correctly.
Short answer: Pleonast did indeed answer the question correctly. Long answer: Hold on to your hats, folks, ‘cause I’m about to open a can o’ relativistic whoop-ass on this.
I need to explain the notion of a “translation symmetry” first. Suppose you had an infinitely long cylinder stretched out in a straight line. This cylinder has a “translation symmetry”: if you go from point A within the cylinder to point B, such that point A and point B can be connected by a straight line parallel to the axis of the cylinder, then when you’re at point B everything will look the same as it did at point A.
Now to the spacetime case. For the sake of simplicity, let’s assume that we’re talking about a spacetime containing a single massive body, and we’ll consider the paths “test particles” take in this spacetime. This spacetime has a “time-translation symmetry”: If you start at point A in the spacetime and go along a particular curve in the spacetime to point B, the spacetime will look exactly the same as it did at point A. The only difference this time is that the curve goes in the “time” direction.
Since a particle following one of these curves always looks like it’s at the same point relative to the rest of the spacetime, it’s legitimate to say that such a particle is “at rest.” However, these curves are not geodesics. So if I’m floating in space in a spacesuit and I try to remain “at rest” by following one of these curves, I have to turn on my rockets and apply some force to maintain my position. If I’m holding on to a screwdriver and I let go of it, it will accelerate away from me – or, more accurately, I’ll accelerate away from it, since the screwdriver will be following a geodesic in the spacetime and I won’t be. In other words, my path in the spacetime is curved, even though I’m staying in the same place; if I let go of something, it’ll follow a geodesic path instead, which will appear to move relative to someone like me who’s “at rest”.
At first, I thought you meant “relative ot the rest of space” but now I’m not sure.
One of the best illustrations of this is in the early pages of the MTW bible of Gravitation. If you shoot a bullet towards the horizon, its path is fairly flat compared to a tossed tennis ball. How can the “curvature” be the same?
Their answer is to point out that it is spacetime that is curved, not space, the same point that Pleonast made. But they calculate the curvature for you, and point out that the tennis ball flight takes much longer than the flight of the bullet, and so it is stretched out much much more in spacetime, as the conversion factor is to multiply the extra time by c, the speed of light.
One thing to remember is, you’re never at rest. You’re always moving at some rate in space and/or time. If you move more in space over a “space-time interval”, you move less over time. But whatever direction you’re going in, you can’t stop. Because masses warp space and time, the only way you can travel the shortest space-time interval in the vicinity of a massive object is to have your world line converge on it. If you stayed the same distance in space from the object, it would increase the length of your world line in total because you’d have to try to move away in the time direction. In other words, you would have to “speed up” your passage through time (by sitting still in space). This would violate the principle of least action. Things always travel the shortest path through space and time, if left on their own. Since time is “slowing down” the closer you get to the object, the path that takes the least amount of time takes you through space towards the object. Given that ct^2=x^2+y^2+z^2, it’s a much shorter space-time path to move through space towards the object than to move through time away from it.
I should say, of course, the above holds for classical (i.e. big) objects. Quantum theory lets particles take all kinds of whacky paths, but those mostly cancel out to yield the classical space-time trajectory we observe, especially the bigger the object is. For macroscopic objects, weird quantum behavior can be ignored
Yeah, the phrasing I used is a little awkward. What I was trying to get across is that just as the entire space looks the same under the translation of the cylinder (my first example), the enitre spacetime looks the same under the time-translation. Since this spacetime is static, it’s also true that there exists a way to take “snapshots” of the spacetime such that every “snapshot” looks the same. (More formally, the existence of a timelike Killing vector field with time-reversal symmetry means that you can choose some time coordinate t such that the hypersurfaces of constant t are diffeomorphic.)
You’re thinking in three dimensions. There’s really no such thing as “at rest”, as every massive particle is moving with a constant 4-velocity (in its own reference frame). As such, there is always an initial 4-velocity to be continued “straight”.
