In my opinion, “stored in” is a bit of a misnomer. There are already good leads on current theories accounting for the so-called “vacuum energy”, but I think seeing a hint of the evolution of the Einstein equation may help understand what’s going on before you throw quantum theories into the mix.
Starting with the idea that matter and energy “generate” the gravitational field, we must come up with some way of talking about them that doesn’t matter what coordinate system we use. It turns out that this was already known long ago: the stress-energy tensor T. Basically, it’s an object attatched to each point in spacetime that measures “how much stuff” is there. When you pick a coordinate system on a patch of spacetime, the tensor looks like a symmetric 4x4 matrix. Since it is a function on spacetime, you can take its divergence (just like you did with vector fields in calc 3), which turns out to vanish. This fact encodes the conservation of mass-energy, of momentum, and of angular momentum among a couple other things. What a useful little gadget. We’ll put it on the right hand side of the equation.
On the left hand side, we need the field, which will be completely determined by the “metric”, which is a gadget at each point of spacetime that calculates the length of a vector at that point. “Curvature” is a fancy word for how this metric changes from point to point. The metric also looks (when you pick local coordinates) like a 4x4 symmetric matrix field whose divergence vanishes. Unfortunately, it can’t ever vanish completely (like stress-energy can), so it’s no good on its own for the left hand side. We need something else that’s built up out of the metric and its first and second derivatives, but defined in such a way that it doesn’t depend at all which coordinates you choose to look at it in. The natural candidate that sums up the interesting second derivatives is the “Riemannian curvature tensor”, which (unfortunately) looks like a 4x4x4x4 matrix in a local coordinate system. Relativists usually write it “R[sub]abcd[/sub]”, where a, b, c, and d all run from 0 to 3. This can’t be what we’re looking for, but we’re getting closer. If we define
R[sub]ab[/sub] = -R[sub]0a0b[/sub] + R[sub]1a1b[/sub] + R[sub]2a2b[/sub] + R[sub]3a3b[/sub]
the “Ricci curvature tensor”, we have a 4x4 matrix again, and it’s symmetric to boot. The problem is that its divergence doesn’t vanish like we need it to. Not to worry, though. The Riemannian curvature satisfies a certain differential equation that says that if we define
G[sub]ab[/sub] = R[sub]ab[/sub] - 1/2 g[sub]ab[/sub] (-R[sub]00[/sub] + R[sub]11[/sub] + R[sub]22[/sub] + R[sub]33[/sub])
where g[sub]ab[/sub] is the matrix expressing the metric in our local coordinates, then G[sub]ab[/sub] (called the “Einstein tensor”) will have vanishing divergence and stay symmetric. Of course, any scalar multiple will also have the same properties, so we set the left hand side to be kG. Comparing the predictions on a very simple system with Newtonian gravitation (which the new theory should agree with in simple cases), we see that k should equal 1/8pi, which is usually put on the other side, giving the Einstein equation G = 8pi T. G is essentially the only symmetric, rank 2 tensor with vanishing divergence (what we need to match properties of T) that can be made of only the metric and its first and second derivatives and which vanishes for a flat geometry (since no matter means no curvature).
The problem is that there are no solutions of this equation that don’t have any solutions that don’t describe spacetime shrinking to a point as we run time backwards. Einstein was horrified by this and slapped another term on the left – the only one he could that wouldn’t ruin the symmetry, rank, or divergence properties – a scalar (the “cosmological constant”, usually written with a capital lambda but I’ll use an L) times the metric, giving
G + Lg = 8pi T
as the Einstein equation. Later Hubble discovered that the universe seems to be expanding. Einstein realized that his theory really had predicted something like this in its original form and erased the cosmological constant term, calling it “the worst blunder of my life”.
But why is the term such a bad thing? All it says is that in empty space (T vanishes), the Einstein tensor G doesn’t vanish (since the metric g doesn’t). Why must empty space be flat? Since then, the community has generally still liked to think of flatness as the norm, so they slightly rewrote the equation
G = 8pi T - Lg
to move the cosmological constant term to the right hand side. The interpretation is that the right hand side measures “stuff” (mass-energy) with T being the contribution of matter and photons and such and g being the “energy of the vacuum”, while the left hand side is a measure of the curvature of spacetime. The upshot is that even in “empty space” (T vanishing), the metric never goes away and so spacetime will still be curved as if there were a certain amount of stress-energy there.
Various current theories try to explain this extra term in various ways, but IMHO the most important thing is to know the semi-classical result they’re trying to explain.
