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In General Relativity Theory, mass can warp spacetime. However, in my view interaction only occurs between pieces of matter. Spacetime is not matter; how can it be affected by matter?

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How do you understand the concept of spacetime? – elcojon Jan 20 '13 at 16:23
@elcojon In my view, spacetime is a frame which has capability to contain realities, but itself is not a reality. – Popopo Jan 20 '13 at 16:52
Your 'view' is an arbitrarily arrived at proposition. Spacetime can be affected by matter through gravity, why? --because that's what observations and theory suggest to be the case. – DilithiumMatrix Jan 20 '13 at 23:04
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This is dangerously close to philosophy rather than physics, but your question has no answer because you appear to have a different concept of spacetime to the rest of us.

The concept of spacetime arose from special relativity and was proposed by Minkowski as a way to understand special relativity. There have been various questions on this site that expand on this. See for example What's the difference between space and time? for a discussion of why spacetime is a useful concept in SR.

The point of this is that we didn't start with an idea of spacetime and then try and work back to special relativity. Spacetime is a concept that emerged from SR. Exactly the same is true in general relativity, except that the metric is now variable rather than fixed and as you say in your question, the metric is determined by the presence of matter or more precisely the stress-energy tensor.

So in general relativity spacetime is defined as that which obeys the Einstein equation. That's why it doesn't make sense to ask how matter can warp spacetime, because that is the way that spacetime is defined.

You wouldn't be the first to find this slightly unsatisfactory and indeed Einstein himself was uneasy with an equation that had geometry on one side and matter on the other. Hence his famous comment about marble and wood. Nevertheless, as far as we know GR works perfectly. You are certainly at liberty to start with a different concept of spacetime and see where it leads, but you will find it hard to improve on GR!

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In GR, is there no space-time independent from matters&fields? – Popopo Jan 21 '13 at 3:56
This simple answer is that there is no spacetime independant from the stress-energy tensor. GR assumes that a manifold exists, and that it is four dimensional, however without a metric this hardly corresponds to a spacetime. It's the metric that depends on the stress-energy tensor. – John Rennie Jan 21 '13 at 8:26

The OP seems to be aligned with the non-absolute space(-time) championed by Leibniz. A little bit of history: Newton and Leibniz debated more than just calculus. The former had an idea of space and time as real substances, while the latter thought of them more as conveniences for mathematically modeling the things we actually care about. Leibniz (and after a fashion Mach and Einstein) felt all we had was relations between objects, anything else (e.g. spacetime) was an invention that got the right answer. See for instance the Stanford philosophy encyclopedia for more details.

If you want, you can think of GR in this way too. Start with a distribution of real mass, energy, momentum, etc., bundled in the stress-energy tensor $\mathbf{T}$. This tells you the Einstein tensor $\mathbf{G}$: $$ \mathbf{G} = 8\pi \mathbf{T}. $$ With $\mathbf{G}$ you can solve a horrendous mess of coupled, nonlinear partial differential equations to find the metric $\mathbf{g}$. Whether you take this to be a physical thing or not is the crux of your question, but it doesn't really matter in solving the problem.

In the reference frame of your choosing, you can define the Christoffel symbols $\Gamma^\rho_{\mu\nu}$ from $\mathbf{g}$. With these you get the set of partial differential equations of motion together known as the geodesic equation: $$ \frac{\mathrm{d}^2x^\rho}{\mathrm{d}\lambda^2} + \Gamma^\rho_{\mu\nu} \frac{\mathrm{d}x^\mu}{\mathrm{d}\lambda} \frac{\mathrm{d}x^\nu}{\mathrm{d}\lambda}. $$ Plug in initial conditions and you have $\vec{x}$ as a function of some affine parameter $\lambda$. That is, you have predicted the motion (relative to some frame) of an object - it goes along the spacetime path $\vec{x}(\lambda)$ - starting with the "real stuff" in the universe.

We used the metric as a convenient waypoint, and it's form was certainly affected by what we included in $\mathbf{T}$. You are free to say $\mathbf{g}$ is just a mathematical symbol that encapsulates how all the matter in $\mathbf{T}$ affected the thing described by $\vec{x}$. However, you must admit that $\mathbf{g}$ is determined by the matter and energy in the universe; if it were constant we wouldn't have GR.

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