# How does the curvature of spacetime induce gravitational attraction?

I don't know how to ask this more clearly than in the title.

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Consider posting where you heard about this, what your current understanding is, and what, specifically, you find confusing about it. Right now we have no idea if we need to write answers assuming you understand basic space time concepts, or whether this is some phrase you read somewhere and you're starting from elementary physics. –  Adam Davis Mar 30 '11 at 5:24
I have a minor in physics and have taken 4 semesters of Physics at the University level (all As and Bs). –  JoeHobbit Mar 30 '11 at 13:39

I'm a bit worried about getting a reputation for citing myself too much, but I'll go for it anyway. (In my defense, I always admit it when I'm doing it!)

John Baez's and my pedagogical paper The Meaning of Einstein's Equation aims to address exactly this question. We describe the meaning of spacetime curvature and the way that Einstein's equation connects spacetime curvature to the matter content of a region of space. As one example, we use this description to heuristically "derive" Newtonian gravity.

I think the most important point is that, in "ordinary" situations involving particles moving at speeds much less than $c$, the "time" part of spacetime curvature is by far the most important part. Intuition about curved space (as opposed to spacetime) only gets you so far.

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Hey man, if it is possible to cite yourself, then by all means do so. Not many people can claim such a privilege. –  user346 Mar 30 '11 at 15:05
This is actually a very nice paper. –  MBN Mar 30 '11 at 15:27
Thanks! (I only mentioned the self-citing issue because I noticed there was a discussion about it on Meta. I'm not really worried about it, at least in this case, because the paper really is relevant.) –  Ted Bunn Mar 30 '11 at 15:45
I love that paper - never put two and two together that you were an author on it. Thanks. –  Mark Eichenlaub Mar 30 '11 at 18:51
As David's answer there suggests, it is absolutely acceptable to self-cite if that is disclaimed (you did so) and is really helpful to answer the question (it does). I wouldn't see the point of you having to repeat what is stated there (especially since you mention the main point here) or searching another citation that you are probably not as acquainted with as with your own (and John Baez's of course) work –  Tobias Kienzler Mar 31 '11 at 7:38

The simple answer is that the curvature of spacetime does not induce gravitational attraction. It describes it.

In all of classical physics (not only general relativity), the motion of bodies can be described as motion along the geodesics of a suitable differential geometry. Body A acts on body B by influencing the curvature of the geodesic along which body B moves. See the illustrations on this page of my personal website for both classical electrodynamics and general relativity.

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why is the link about "citeing yourself" here? –  JoeHobbit Mar 30 '11 at 13:44
@user2843 Because Koantum's answer links to his own web page without identifying it as such. –  Mark Eichenlaub Mar 30 '11 at 18:52
Apologies, @Mark. I now made it clear that the link is to my personal website. –  Koantum Mar 31 '11 at 2:52
Great, thanks. –  Mark Eichenlaub Mar 31 '11 at 4:18

The best way to see this, I think, is using Einstein's 1907 approximation. Einstein noted that gravity slows down clocks, so that where the potential is more negative, clocks run slow, and where the potential is small, clocks run fast, so that the rate at which a clock ticks at position $x$ is $\sqrt{(1-2\phi(x))}$.

Assume that space is completely flat, and only the clock rate changes from point to point. This is a curved space-time, but the space-part is flat. Then you ask: "what is the curve in space-time which is extremal for the relativistic distance from point A to point B?". This is the analog of the question "what is the shortest path from A to B" in geometry.

The integral for this is:

$\int \sqrt{(1-2\phi(x)) - |v|^2)} dt = \int -\phi(x(t)) + {v^2\over 2}$

Which, if you multiply by the mass m, gives you the action for a Newtonian particle in a gravitational potential. So Einstein's law reduces to Newton's. The remaining work is to see that the equation for the time-time component of the metric reduces to Laplace's equation for static masses, and this is not too hard to do, but annoying to do twice or to write up. That this reduction works was one of Einstein's criteria for a good field equation.

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