# Does the curvature of spacetime theory assume gravity?

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
I really loath that "explanation" for exactly this reason... –  dmckee Mar 30 '11 at 4:13
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
Possible duplicate: physics.stackexchange.com/a/13839/2451 –  Qmechanic Jan 14 '13 at 19:05

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
@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.physics.ucla.edu/demoweb/demomanual/modern_physics/principal_of_equivalence_and_general_relativity/curved_spacetime.html

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

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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