gravity - Existence of gravitons?

The metric describes the curvature of the space. For space around a massive object this is the Schwarzchild metric

$$
ds^2 = -left(1-frac{r_s}{r}right)dt^2 + left(1-frac{r_s}{r}right)^{-1}dr^2 + r^2(dtheta^2 + sin^2theta dphi^2)
$$

Clearly, if $r>>r_s$ this looks like

$$
ds^2=-dt^2+dr^2+r^2(dtheta^2 + sin^2theta dphi^2)
$$
which is the metric for flat space. So effectively the space gets flatter and flatter at a rate of $1/r^2$, which is the inverse square that you're looking for.

But where does the Schwarzchild metric come from? Without getting into the gritty maths, it can be proven that it is the unique metric that possesses spherical symmetry, without which nothing would make much sense. This is called Birkhoff's theorem.

I want to talk about where gravitons come from, but first lets talk about curvature.

If you want to measure the curvature of a space one way of doing it is to move in some closed loop, ending up back where you started. If the space is curved though, you won't be facing the same direction (this idea is called parallel transport)

Let's say we're parallel-transporting a tangent vector, as in the picture. We get a tangent vector from the derivative at a point (a slighly special derivative called a covariant derivative, because the space is curved). Let's take the tangent vector and move forwards then left. And we try again this time moving left then forwards. We end up the same point both ways, but like in the picture, the derivatives will be different in some way. We summarise this up with a commutator (where $D$ is the covariant derivative) like so

$$
[D_mu,D_nu] = D_mu D_nu - D_nu D_muneq 0
$$
It basically means "doing it one way is not the same as doing it the other".

Now let's back up a tiny bit and talk about how electromagnetism and other forces are typically discussed, using quantum field theory.

We describe the theory in terms of a Lagrangian, for a fermion (like an electron) it looks like this

$$
mathcal{L}=barpsi(igamma^mu D_mu-m)psi
$$

If I take the field $psi$ and give it a transformation
$$
psitopsi'=e^{ixi(x)}psi
$$
then the Lagrangian will remain unchanged. This type of transformation belongs to a group called $U(1)$. We say that the Lagrangian possesses $U(1)$ symmetry. Notice that this $D_mu$ is in there again? Its the same thing, a covariant derivative, here in QED as well. We can try taking a commutator again

$$
[D_mu,D_nu] = -iF_{munu}psi
$$
where
$$
F_{munu} = partial_mu A_nu - partial_nu A_mu
$$

From this we form the complete QED (the quantum theory of electrodynamics) Lagrangian
$$
mathcal{L}=barpsi(igamma^mu D_mu-m)psi-frac{1}{4}F_{munu}F^{munu}
$$

Don't get bogged down in the maths. The point is very simple. See the $A_mu$? Its a new field, we had to introduce it to make things work. In QED this field corresponds to a photon (particles are quanta of a field, like a small bump in the field). We had to introduce it becuase we had curvature. How do I know we have curvature? Becuase the covariant derivatives do not commute, just like in GR, above. This time though, the curvature is not of phyical space, its of an abstract object called the $U(1)$ gauge bundle.

So you're totally on the right track when you say that other forces may curve space. Its nice that gravity curves space-time, its very physical and easy to imagine, for the other forces its not so simple to picture, even though its fundamentally the same.

If you want the full picture of Einstein's gravity you do some maths and arrive at something called the Einstein-Hilbert action (an action is just an integral over a Lagrangian), one tidy object that sums up the whole theory

$$
S=int R sqrt{g} d^4x
$$
where $R$ comes (more or less) from the commutator of covariant derivatives we saw at the top. When talking about QED I brushed over the fact that its a quantum theory (it is). This EH action, however, does not describe a quantum theory. So, you might say, lets make it one! Hold on a second though, because it doesn't actually work. The problem is something called renormalisability - QED is renormalisable, GR is not. This is the root of the incompatibility between GR and quantum field theory. If we could carry out the resultant qunatum particle would be a graviton. You're right to doubt their existance as they haven't yet been observed, however...

We saw QED, which desribes particles of light, photons. They are quantised. Then we saw how in many ways GR and QED are very similar. We can't properly quantise GR but if we could we would have gravitons, exactly like photons popped out in QED. The duality between QED (and other gauge theories, QCD, etc) is clear, which leads a lot of people to believe that probably should have gravitons, even if they have not yet been observed, nor consistently formulated.

A note on other theories

There are many theories where gravitons are present from first principles without the problems of renormalisability, string theory or supergravity for instance.

A note on errors in the above

Sorry, I'm tired and rambling. Please point them out if you find them!