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Whenever I read about the curvature of spacetime as an explanation for gravity, I see pictures of a sheet (spacetime) with various masses indenting the sheet to form "gravity wells." Objects which are gravitationally attracted are said to roll down the curved sheet of spacetime into the gravity well. This is troubling to me, because, in-order for objects on the locally slanted spacetime sheet to accelerate, gravity must be assumed. Therefore I ask; does the explanation of gravity as the curvature of spacetime assume gravity? If yes, what is the point of the theory? If No, what am I missing?

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It is just an attempt to illustrate the concept. It is not great and can be confusing. The sheet is also embedded in three space, for space-time that is not needed. So I guess my answer is no, you don't have space-time embedded in another space which has 'gravity' pull downwards that dents space-time. –  MBN Mar 30 '11 at 3:55
    
I have a follow up question: How does the curvature of spacetime induce attraction? I am new here. Should I post it here or make another thread? –  JoeHobbit Mar 30 '11 at 4:00
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I really loath that "explanation" for exactly this reason... –  dmckee Mar 30 '11 at 4:13
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Another reason that explanation is very misleading: under these circumstances, the gravitational attraction is essentially entirely due to the "time" part of spacetime curvature: ordinary "spatial" curvature has essentially nothing to do with it. Intiuition about curvature of space as opposed to spacetime only gets you so far and can be confusing. (For cognoscenti: I know that that statement has no invariant meaning, but under the circumstances here of nonrelativistic particles in a static background, there's a natural notion of which direction is "time".) –  Ted Bunn Mar 30 '11 at 14:18
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Possible duplicate: physics.stackexchange.com/a/13839/2451 –  Qmechanic Jan 14 '13 at 19:05

6 Answers 6

up vote 14 down vote accepted

I greatly sympathize with your question. It is indeed a very misleading analogy given in popular accounts. I assure you that curvature or in general, general relativity (GR) describe gravity, they don't assume it. As you appear to be uninitiated I shall try to give you some basic hints about how gravity is described by GR.

In the absence of matter/energy the spacetime (space and time according to the relativity theories are so intimately related with each other it makes more sense to combine them in a 4 dimensional object called space-time) is flat like a table top. This resembles closely with (not completely) Euclidean geometry of plane surfaces. We call this spacetime, Minkowski space. In this space the shortest distance between any two points are straight lines.

However as soon as there is some matter/energy the geometry of the surrounding spacetime is affected. It no longer remains Minkowski space, it becomes a (pseudo) Riemannian manifold. By this I mean the geometry is no longer like geometries of a plane surface but rather like geometries of a curved surface. In this curved spacetime the shortest distance between any two points are not straight lines in general, rather they are curved lines. It is not very hard to understand. Our Earth is a curved surface and the shortest distance between any two points are great circles rather than straight lines. Similarly the shortest distance between any two points in the 4 dimensional spacetime are curved lines. An object like sun makes the geometry of spacetime curved in such a way that the shortest distance between any two points are curved. This is called a geodesic. A particle follows this curved geometry by moving along this geodesic. Einstein's equations are mathematical descriptions of the relation of the geometry to the matter/energy.

This is how gravity is described in general relativity.

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Can a curved line in the (pseudo) Reimannian manifold correspond to a straight line in real life? –  JoeHobbit Mar 30 '11 at 13:50
    
Are you asking if the shortest distance between two points in our everyday three dimensional space is a straight line? Yes; in the dimensions/sizes we live in, the special relativity 4 dimensional Minkowski space is the local limiting case of the 4 dimensional GR space. In the 3 space dimensions the shortest distance is a straight line. GR needs large masses and energies to manifest its curvature. –  anna v Mar 30 '11 at 14:41
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@sb1 an editorial observation - if you broke your answers up into paragraphs they would be a lot easier to read. –  user346 Mar 30 '11 at 15:06
    
@Deepak: Thanks Deepak for the suggestion :) –  user1355 Mar 30 '11 at 15:42
    
I was asking about how the geometries of the Space-time correspond to the geometries of everyday life. I think that probably deserves its own thread though. –  JoeHobbit Mar 30 '11 at 21:10

Those sheets with dips having stars at their centers illustrate gravity by illustrating that freely falling objects move along geodesics, and that geodesics are curved by the dips in a way that looks as if freely falling objects were attracted by the stars. (Needless to say, the 2-dimensional sheets remain poor substitutes for the real McCoy, which is a (3+1)-dimensional pseudo-Riemannian manifold.)

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Therefore I ask; does the explanation of gravity as the curvature of spacetime assume gravity?

No, it just a bad analogy. Checkout the links below for a much better visualization of General Relativity.

http://www.youtube.com/watch?v=DdC0QN6f3G4

http://www.physics.ucla.edu/demoweb/demomanual/modern_physics/principal_of_equivalence_and_general_relativity/curved_spacetime.html

http://www.relativitet.se/spacetime1.html

http://www.adamtoons.de/physics/gravitation.swf

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No. Firstly, see my answer here. That is the reason why the Stress-Energy-Momentum (SEM) tensor curves spacetime (induces a ricci scalar in the "background" spacetime). For fun, let me copy it here too: \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!.

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

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!.

Note that all the $\approx$ signs don't mean any innacuraccy in Genara;l Rrelativity. They mean an inaccuracy in Newtonian Mechanics, which is only an approximation to the real General Relativity of Gravity.

Now, how does this curvature of spacetime result in gravitational attraction? Start with: $$\nabla_{v^\nu}v^\nu=0$$ To make things easy, write it in a simpler to simplify way: $$v^\mu\nabla_\mu v^\nu=0$$ Recall the definition of the Christoffel symbols and express the scary-looking covari;ant derivative as the sum of the partial derivative and a Christoffel symbols: $$v^\mu\partial_\mu v^\nu+\Gamma^\nu_{\rho\mu}v^\mu v^\rho=0$$ $$\frac{1}{c_0^2}\frac{\mbox{d}^2x^\nu}{\mbox{d}\tau^2}=-\frac{1}{c_0^2}\Gamma_{\rho\mu}^\nu \frac{\mbox{d}x^\mu}{\mbox{d}\tau} \frac{\mbox{d}x^\rho}{\mbox{d}\tau}$$ $$\frac{\mbox{d}^2x^\nu}{\mbox{d}\tau^2}=-\Gamma_{\rho\mu}^\nu \frac{\mbox{d}x^\mu}{\mbox{d}\tau} \frac{\mbox{d}x^\rho}{\mbox{d}\tau}$$ This is the amazing geodesic equation. This predicts motion due to a gravitational field. However, how do we know it is due to a gravitational field? By taking it to the Newtonian Limit.! But first, rewrite it in a nicer form: $$\frac{\mbox{d}^2x^\rho}{\mbox{d}\tau^2}=-\Gamma_{\mu\nu}^\rho \frac{\mbox{d}x^\mu}{\mbox{d}\tau} \frac{\mbox{d}x^\nu}{\mbox{d}\tau}$$ At the Newtonian Limit, we assume that the particle is moving very slowly in space. $$\frac{\mbox{d}x^\mu}{\mbox{d}\tau}\approx\left[ \begin{array} \mbox{ }\\ \gamma\\ 0\\ 0\\ 0 \end{array} \right]$$

Here, $\gamma=\frac{\operatorname{d}t}{\operatorname{d}\tau}$ is the "general relativistic lorentz factor". Since in Newtonian Mechanics, this is approximately 1, $$\frac{\mbox{d}x^\mu}{\mbox{d}\tau}\approx\left[ \begin{array} \mbox{ }\\ 1\\ 0\\ 0\\ 0 \end{array} \right]$$ And thus, $$\frac{\mbox{d}^2x^\nu}{\mbox{d}\tau^2}\approx-\Gamma_{00}^\rho\approx-\partial_\mu \Phi=-\nabla\Phi$$ In the last equation, we adopted the old Calculus notation for the gradient. Thus, the Newtonian approximation (Adopting even more old notation) is: $$\vec g\approx-\nabla\Phi$$ This is precisely the Newtonian equation for the gravitational field is! Thus, this force predicted by General Relativity IS gravity!

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Dear DImension10 Abhimanyu PS l: it is usually frown upon to directly copy-paste an answer into another answer. Linking to the other answer (possibly summarizing the main point line of the other answer) should be enough. (The problem is if everybody start to copy-paste identical answers en mass.) –  Qmechanic Oct 25 '13 at 20:11

A better analogy for curvature is to imagine ants walking on bowling ball in space. They start out at the same point exchange information in some ant fashion, then take off in different directions and even if they try to walk as straight as possible on the surface, they end up converging towards that point opposite where they started (And if you were at the north pole and started flying as straight as possible at constant altitude, you head to the south pole, no matter which direction you headed out).

Those kinds of things are what the curvature is supposed to do, to create paths that are the natural ones that converge, and that converge in a way that didn't depend on say if you were small and light or if you were more massive, just the path you take.

That said there is an important further point, which is that curvature exists even far from sources because curvature begets curvature. For instance, mass, energy, momentum, stress, and pressure are sources of curvature, but they are not the only things that create curvature, curvature itself can create further and additional curvature. A gravitational wave can propagate or even spread in a vacuum of empty space devoid of all mass, energy, momentum, stress, and pressure.

The region outside a symmetric nonrotating static star is curved, even the parts far from any mass or energy or momentum or stress or pressure. The space remains curved because the existing curvature is exactly shaped so as to persist (or otherwise cause future curvature exactly like itself).

So curvature allows and sometimes requires more and/or future curvature, just as a travelling electromagnetic wave allows and/or even requires there be more electromagnetic waves elsewhere and/or later. The vacuum allows curvature far from gravitational sources just as it allows electromagnetic waves far from electromagnetic sources. What electromagnetic sources allow is for electromagnetic fields to behave differently (namely to gain or lose energy as well as move in different ways and gain and lose momentum and stress). Similarly what gravitational sources do is allow curvature to react differently to itself than it otherwise would.

Imagine a flat region of space shaped like a ball, then imagine a funnel type curved space where two regions of surface area are farther apart than they would be if flat (like a higher dimensional version of a funnel and on a funnel surface two circles of a particular circumference are farther away as measured along the funnel then if two similarly sized circles were in a flat sheet). On its own, spacetime doesn't allow itself to connect those two kinds of regions together, but that mismatch is exactly the kind or not-lining-up that putting some mass or energy right there on the boundary fixes. So without mass those two regions can't line up, with mass they can. Just like an electromagnetic field can have a kink if there is a charge there.

So your curvature likes to propagate a certain way, and if you want it to deviate from that, you need mass, energy, momentum, stress, and/or pressure. And you'd need the right kind to get it to match up, the kind you want might be available, and might not even exist, so not all kinds of curvature will be allowed. But the point of a source is that it changes the balance between nearby curvature and not that affects future curvature. So there is a kind of balance, and there are things that can warp that a balance. Those things that warp that natural vacuum balance are called gravitational sources.

So that means you want to depict two things, firstly nearby curvature affects nearby curvature, that current curvature determines future curvature, and secondly that gravitational sources allow the curvature to be different than what it would be on its own. The curvature itself is supposed to remind you about paths converging regardless of whether a light thing or a heavy thing sets out on the path. But when you look at the analogy for gravity look for those three features, but don't take any other aspect of the analogy too seriously, the curvature of time is important and not pictured, and the curvature of space is different than as pictured, and things don't move within the space because of an external space or because of an external force and finally the sources themselves make different curvature by allowing pieces that otherwise wouldn't fit to fit together.

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I will admit that my understanding of General Relativity at the moment is limited, but this is my understanding:

When dealing with curved coordinate systems, like space-time curved by a massive object, you can calculate what is known as a "geodesic" which is a straight line in curved coordinates. If you are dealing with a spherical space, all geodesics will be "great circles" which wrap around the circumference of the sphere. While this will be the shortest path between 2 points, it can seem oddly curved to anyone on the sphere (if you're confused about this than here is a program someone made that should help explain.)

Where this comes into general relativity is that objects will follow geodesic paths through space-time, and as you may have learned: everything must move through space-time at the speed of light which is why when you move at the speed of light then you experience no time because all your motion is through space and none of your motion can be through time. A massive object will create what is basically a shortcut through time, so objects will move through space as it is a shorter path. With this in mind, feel free to refer to any form of descending whether it be (walking down stairs, jumping down or taking an elevator) as "taking a shortcut to the future".

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