How exactly does curved space-time describe the force of gravity?

I understand that people explain (in layman's terms at least) that the presence of mass "warps" space-time geometry, and this causes gravity. I have also of course heard the analogy of a blanket or trampoline bending under an object, which causes other objects to come together, but I always thought this was a hopelessly circular explanation because the blanket only bends because of "real" gravity pulling the object down and then pulling the other objects down the sloped blanket. In other words, to me, it seems that curved space wouldn't have any actual effect on objects unless there's already another force present.

So how is curved space-time itself actually capable of exerting a force (without some source of a fourth-dimensional force)?

I apologize for my ignorance in advance, and a purely mathematical explanation will probably go over my head, but if it's required I'll do my best to understand.

-
In many "video" explanations of general relativity curvature of Time is omited, time is certainly not easy to graph with the blanket example, but sometimes it's not even mentioned, perhaps lack of self-questioning of the explainer, then it's a good question +1 –  HDE May 16 '11 at 17:28

Luboš's answer is of course perfectly correct. I'll try to give you some examples why the straightest line is physically motivated (besides being mathematically exceptional as an extremal curve).

Image a 2-sphere (a surface of a ball). If an ant lives there and he just walks straight, it should be obvious that he'll come back where he came from with his trajectory being a circle. Imagine a second ant and suppose he'll start to walk from the same point as the first ant and at the same speed but into a different direction. He'll also produce circle and the two circles will cross at two points (you can imagine those circles as meridians and the crossing points as a north resp. south poles).

Now, from the ants' perspective who aren't aware that they are living in a curved space, this will seem that there is a force between them because their distance will be changing in time non-linearly (think about those meridians again). This is one of the effects of the curved space-time on movement on the particles (these are actually tidal forces). You might imagine that if the surface wasn't a sphere but instead was curved differently, the straight lines would also look different. E.g. for a trampoline you'll get ellipses (well, almost, they do not close completely, leading e.g. to the precession of the perihelion of the Mercury).

So much for the explanation of how curved space-time (discussion above was just about space; if you introduce special relativity into the picture, you'll get also new effects of mixing of space and time as usual). But how does the space-time know it should be curved in the first place? Well, it's because it obeys Einstein's equations (why does it obey these equations is a separate question though). These equations describe precisely how matter affects space-time. They are of course compatible with Newtonian gravity in low-velocity, small-mass regime, so e.g. for a Sun you'll obtain that trampoline curvature and the planets (which will also produce little dents, catching moons, for example; but forget about those for a moment because they are not that important for the movement of the planet around the Sun) will follow straight lines, moving in ellipses (again, almost ellipses).

-
Thanks heaps guys, it's starting to make some sense. So that makes sense to me with moving objects, but I still don't quite understand how it causes objects to accelerate. For example, with your analogy, what if the ants were stationary on the ball? When we lift something off the ground and let go, it accelerates toward the earth. Is this just because that's the "straighest" line through the curved spacetime around the earth? Why must it always be "moving" through a straight line, and what does it mean in terms of curved spacetime for something to be stationary? –  Zac Jan 16 '11 at 12:08
@Zac: good questions. First, geodesics contain more information than just shape of the path (which we call trajectory). They also contain information about how fast the particle will move along it. This is what will give you acceleration towards the Earth for free falling object but it has to be computed from the equations (actually, in this case conservation of energy is enough to derive that acceleration). –  Marek Jan 16 '11 at 14:33
Also @Zac, for something to be stationary in spaceTIME means that it only exists for a single instant in time! Even something that stays stationary in space for all time moves on a curve in spacetime. (think about what an x vs. t plot looks like for a stationary object) –  wsc Jan 16 '11 at 18:33
@wsc: it's really called perihelium in my language (Slovak) so I never imagined it might be something different in English. Anyway, thank you :) –  Marek Jan 16 '11 at 19:34
@Marek: nor did I realize it was different in Slovak; always nice to learn! Anyway, I was using that meaning of stationary since that was what @Zac was using: his question seemed to me to be: 'Sure you have geodesics on curved manifolds, but why do the ants have to move?' Which is a very good question, you just have to remember that time is a coordinate too. –  wsc Jan 16 '11 at 20:07

There are actually two different parts of general relativity. They're often stated as

1. Spacetime tells matter how to move
2. Matter tells spacetime how to curve

Point #1 is actually straightforward to explain: objects simply travel on the straightest possible paths through spacetime, called geodesics. The paths only seem curved because of the warping of spacetime. If you're a physicist, then you would want to know that that fact can be derived from the principle of extremal action (with all the requisite mathematical details), but if you don't want to wade through the math, hopefully it at least makes sense that objects move on "straight" lines. There is no actual force involved when a massive (or even a massless) object's trajectory curves in response to gravity, because it doesn't take any force to keep something moving on a straight line. (I can definitely expand on this point if you want)

Now, I mentioned that spacetime needs to be warped in order for objects' trajectories to appear curved to us despite them actually being "straight." So the essence of point #2 is, why is spacetime warped in the first place? Physics doesn't have a good answer to that. Technically, we don't have an answer to point #1 either, but the "straight line" argument at least makes it seem plausible; unfortunately, there's no equivalent plausibility argument for why spacetime warps itself around matter. (Perhaps someday we will come up with one) All we can do right now is produce equations that describe how spacetime behaves around matter, namely the Einstein equations which can be written $G_{\mu\nu} = 8\pi T_{\mu\nu}$ among other ways.

-

the trampoline analogy needs an extra source of gravity - because this is what the laymen, the recipients of the explanation, intuitively understand - but the real general relativity doesn't need any extra "external" gravity.

Instead, general relativity says that the space is getting curved by Einstein's equations, $$G=T$$ where the left-hand side are numbers describing the curvature at a given point and the right hand side is the density of matter and momentum. I omit indices and constants haha. So general relativity says how the spacetime is curved under the influence of matter.

The second part of the story is that general relativity also says how matter moves in external geometry. It moves along "geodesics", lines that are as straight as you can get. $$\delta S_{action\,ie\,proper\,length} = 0$$ This actually means that the objects move along the predicted, seemingly curved trajectories. These trajectories are actually as straight in the curved spacetime as you can get.

Imagine that there is a hemisphere replacing a disk in the trampoline. So there exists a (nearly) straight line on the hemisphere - namely the equator near the junction with the rest of the trampoline. Note that the equator on the Earth is a maximum circle - so it is one of the straightest lines you can draw on the surface of Earth. The same is true for all actual trajectories that objects choose in spacetime of general relativity.

So in the hemisphere-above-trampoline example, particles can orbit around the equator of the attached hemisphere, just like planets, because it is the straightest and most natural line they can choose. I don't use any external gravity to explain the real gravity; instead, I use the principle that particles choose the most natural - the straightest - line they can find in the curved spacetime.

Best wishes Lubos

-

The other answers are more or less correct, but pehaps I can say something more to the point of the question, «¿how is curved space-time itself actually capable of exerting a force? ».

No force whatsoever is needed.

Gravity is not a force. What is a force, anyway? Newton clarified for almost the first time in Science what a force is: First I will say it, then explain it: A force is something which makes the motion of a body deviate from uniform straightline motion.

Newton pointed out that bodies have a tendency, inertia, to continue in whatever direction they are already going, with whatever velocity they have at the moment. That means uniform, rectilineal motion: steady velocity, same direction. Newton actually knew this was what would be later called a geodesic, since « a straight line is the shortest distance between two points ».

Newton then went on to say that to overcome inertia, to overcome this tendency, requires a force: force is what makes a body depart from the geodesic it is (even momentarily) headed on (its direction and speed).

It was then Einstein (and partly Mach before him) who said this does not get to the essence of the question. For Einstein, any coordinate system had to be equally allowable, and in fact, space-time is curved (as already explained by other posters). A body or particle under the influence of gravity actually does travel in a geodesic....i.e., it does what a free particle does. I.e., it does what a particle not under the influence of any force does. So gravity is not a force.

Newton did not realise that space-time could be curved and that then the geodesics would not appear to our sight to be straight lines when projected into space alone. That ellipse you see in pictures of planetary orbits? It is not really there of course since the planet only reaches different points of the ellipse at different times...that ellipse is not what the planet really traverses in space-time, it is the projection of the path of the planet onto a slice of space, it is really only the shadow of the true path of the planet, and seems much more curved than the true path really is.

( ¡ The curvature of space-time in the neighbourhood of the earth is really very small ! The path of the earth in space-time would even appear to be nearly straight to an imaginary Euclidean observer who, in a flat five-dimensional space larger than ours, was looking down on us in our slightly curved four dimensional space-time embedded in their world. It's $ct$, remember, so the curving around the ellipse gets distributed over an entire light-year, and appears to be nearly straight...and is straight when one takes into account the slight curvature of space-time.)

Since every particle under the influence of gravity alone moves in a geodesic, it does not experience any force that would make it depart from its inertia and make it depart from this geodesic. So gravity is not a force, but electric forces still do exist. They could overcome the inertia of a charged body and make it deviate from the geodesic it is headed on: change its speed and direction (when speed and direction are measured in curved space-time).

Einstein (and me too) did not want to change the definition of force in this new situation, since after all electric forces are known to exist and are still forces in GR. So the old notion of force still retains its usefulness for things other than gravity. To repeat: if a body is not moving in a geodesic in space-time, you go looking for a force that is overcoming its inertia....but since gravity and curvature of space-time do not make a body depart from a geodesic, neither of them is a force.

See also http://www.einstein-online.info/elementary/generalRT/GeomGravity which avoids the trampoline fallacy and has a great image of the great circle.

-

As others mentioned, the main problem with the common visualization is, that it omits the time dimension. In the animation linked below the time-dimension is included to explain how General Relativity differs form Newton's model.

-

The Postulate is that the scalar curvature (Ricci scalar) is proportiortional the Lagrangian density of gravity. From that, we can deduce the Einstein Field Equations. \begin{gathered} {{\mathcal{L}}_{G + M}} = \lambda R + {{\mathcal{L}}_M} \\ {S_{G + M}} = \int_{}^{} {\left( {\lambda R + {{\mathcal{L}}_M}} \right)\sqrt { - \det {g_{\mu \nu }}} {\text{d}}{x^4}} {\text{ }} \\ \delta S = 0 \\ \delta \left( {{S_G} + {S_M}} \right) = 0 \\ \int_{}^{} {\delta \left( {\left( {{{\mathcal{L}}_M} + \lambda R} \right)\sqrt { - \det {g_{\mu \nu }}} } \right){\text{d}}{x^4}} = 0{\text{ }} \\ \int_{}^{} {\left( {\frac{{\delta \left( {\left( {{{\mathcal{L}}_M} + \lambda R} \right)\sqrt { - \det {g_{\mu \nu }}} } \right)}}{{\delta {g_{\mu \nu }}}}} \right)\delta {g_{\mu \nu }}{\text{d}}{x^4}} = 0\\ \int_{}^{} {\left( {\sqrt { - \det {g_{\mu \nu }}} \frac{{\delta {{\mathcal{L}}_M}}}{{\delta {g_{\mu \nu }}}} + \lambda \sqrt { - \det {g_{\mu \nu }}} \frac{{\delta R}}{{\delta {g_{\mu \nu }}}} + \left( {{{\mathcal{L}}_M} + \lambda R} \right)\frac{{\delta \sqrt { - \det {g_{\mu \nu }}} }}{{\delta {g_{\mu \nu }}}}} \right)\delta {g_{\mu \nu }}{\text{d}}{x^4}} = 0 \\ \sqrt { - \det {g_{\mu \nu }}} \frac{{\delta {{\mathcal{L}}_M}}}{{\delta {g_{\mu \nu }}}} + \lambda \sqrt { - \det {g_{\mu \nu }}} \frac{{\delta R}}{{\delta {g_{\mu \nu }}}} + \left( {{{\mathcal{L}}_M} + \lambda R} \right)\frac{{\delta \sqrt { - \det {g_{\mu \nu }}} }}{{\delta {g_{\mu \nu }}}} = 0 \ \frac{{\delta R}}{{\delta {g_{\mu \nu }}}} + \frac{R}{{\sqrt { - g} }}\frac{{\delta \sqrt { - g} }}{{\delta {g_{\mu \nu }}}} = - \frac{1}{\lambda }\left( {\frac{1}{{\sqrt { - g} }}{{\mathcal{L}}_M}\frac{{\delta \sqrt { - g} }}{{\delta {g_{\mu \nu }}}} + \frac{{\delta {{\mathcal{L}}_M}}}{{\delta {g_{\mu \nu }}}}} \right) \\ {R_{\mu \nu }} - \frac{1}{2}R{g_{\mu \nu }} = \frac{1}{{2\lambda }}{T_{\mu \nu }}\\ {G_{\mu \nu }} = \kappa {T_{\mu \nu }}\\ \end{gathered}

To fix the value of kappa, we try to impose the fact that Newtonian gravity is the non-relativistic limit of the General Theory of Relativity!.

$$\begin{gathered} {G_{00}} = \kappa c_0^4\rho \\ {R_{00}} = {G_{00}} - \frac{1}{2}G_\mu ^\mu {g_{00}} \\ {R_{00}} = \kappa \left( {{T_{00}} - \frac{1}{2}T{g_{00}}} \right) \\ {R_{00}} \approx \kappa \left( {c_0^4\rho - \frac{1}{2}{g^{00}}{T_{00}}c_0^2} \right) \\ {R_{00}} \approx \kappa \left( {c_0^4\rho - \frac{1}{2}\frac{1}{{c_0^2}}c_0^4\rho c_0^2} \right) \\ {R_{00}} \approx \frac{\kappa }{2}c_0^4\rho \\ {\nabla ^2}\Phi \approx \Gamma _{00,\alpha }^\alpha \approx {R_{00}} \approx \frac{\kappa }{2}c_0^4\rho \\ 4\pi G\rho = \frac{\kappa }{2}c_0^4\rho \\ 4\pi G = \frac{\kappa }{2}c_0^4 \\ \kappa = \frac{{8\pi G}}{{c_0^4}} \\ {G_{\mu \nu }} = \frac{{8\pi G}}{{c_0^4}}{T_{\mu \nu }} \\ \end{gathered}$$

So, finally, this means that postualting that the Ricci Scalar is the "cause" of gravity reduces to Newtonian gravity at the non-relativistic limit. Experiments have shown that it is more accurate!.

-
This is not an explanation in "layman's terms"... –  Comp_Warrior Jun 22 '13 at 1:20
@Comp_Warrior: The layman's terms are the last line. And the explanation is quite layman. meta.physics.stackexchange.com/questions/429/… –  Dimensio1n0 Jun 22 '13 at 4:19
@Comp_Warrior: Not really. I was very interested in this elegant relationship when I first learnt GR. So people who just learnt GR would be interested in this. –  Dimensio1n0 Jun 22 '13 at 13:21
@Comp_Warrior: After that, he also wrote, " , but if it's required I'll do my best to understand." –  Dimensio1n0 Jun 22 '13 at 17:09
@Comp_Warrier what dimension10 says, and the SE system is exactly desined that the op can accept the answer he likes best, maybe a popular one, whereas there can legitimately be other more technical answers too, that are liked by other members of the community. Answers to a question are not only meant to serve the op, but the whole community. So there is absolutely nothing wrong with a question getting answers of different level. It would be nice if you stop discouraging good technical posts which are perfectly legitimate. –  Dilaton Jun 22 '13 at 22:20

I think the problem for the layman is understanding why there is motion in spacetime and I think a sort of answer is that we already accept motion through time when we think of time and space as separate. Well we are in motion through spacetime where time and space are not separable and when we move through a region of spacetime that contains matter the shortest spacetime path between two events is the one that includes motion through the space bit as well as the time bit (ie not orthogonal to the space axes). That is experienced as falling under gravity.

-

What Einstein's equation tell us, at a basic level, is that the curvature of space-time and stress-energy are the same thing.

In order for this law to be respected it is clear that the stress-energy of a test particle cannot be constant in a space-time with changing curvature.

So, if you can choose a coordinate set in which the stress energy tensor is represented by the mass-energy of the particle, then the practical effect you can observe is changing energy and momenta of the test particle.

When you therefore observe the test particle, you will see it as having changing energy and momenta, and therefore derive a force driving these changes. This is what we call gravity.

However, general relativity gives a much deeper picture of gravity as a description of the curvature of space-time, so, in a way, gravity is an observed effect of the curvature of space-time, or, if you like, an observed effect of the distribution of mass and energy.

-

protected by Qmechanic♦Aug 17 '13 at 11:27

Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.