<p>Gravity is a <a href="https://en.wikipedia.org/wiki/Fictitious_force" rel="nofollow noreferrer">fictitious force</a>, actually, much like centrifugal force.  In a free falling frame of reference it disappears.  In general relativity (GR) gravity is just a result of (differential) geometry: space-time curvature.  The inverse square law is just the low energy approximation, but the actual equation for gravity derived from GR is more complex than that.  The massive success of Newtonian gravity tells us that any model of gravity must be approximated by the classical inverse square law at low energies.</p>

<p>Whether GR does that by (Einstein's) design or something else is a matter of personal opinion.  Einstein definitely knew he had to get approximately Newtonian gravity at low energies, so he would have discarded or modified any ideas that failed this criterion.  However, there are <a href="https://en.wikipedia.org/wiki/Inverse-square_law#Justification" rel="nofollow noreferrer">standard arguments for why gravity must obey an inverse square law</a>, at least in low energy situations.</p>

<p>Now in GR <a href="https://en.wikipedia.org/wiki/Stress%E2%80%93energy_tensor" rel="nofollow noreferrer">a lot more than just mass contributes to the gravitational field</a>.  Spin and electric charge, for example, and quite importantly: energy (the famous $E=mc^2$ tells us how to express a mass as energy, so we can treat these things on a common footing).  So, yes, all forces and particles contribute to gravity.  <a href="https://physics.stackexchange.com/a/22878/55483">Even photons</a>.</p>

<p>GR itself makes no predictions (or requirements) for the existence of any new particles outside of the standard model, such as gravitons.  GR and quantum mechanics (QM) are famously incompatible: in extreme situations where both GR and QM are relevant (neutron stars and black hole formation, for example), they stop making sense rather quickly.  <a href="https://en.wikipedia.org/wiki/Graviton#Gravitons_and_renormalization" rel="nofollow noreferrer">Especially GR</a>.  "Gravitons" and assorted variations are hypothetical particles that are proposed to resolve this issue by creating a quantum theory of gravity.  The only "evidence" we have for them at this stage is that our two most massively successful theories about the workings of the universe, GR and QM, are so painfully incompatible.  So we know these theories are flawed (aka wrong) and that some other theory is necessary that can handle these situations, while also incorporating all of the successes of QM and GR&mdash;they're amazingly accurate when only one of them is particularly relevant, after all.</p>

<p>Exactly what that theory <em>is</em> is an ongoing and substantial research area.</p>

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2016-02-09T11:15:30.347Z
Does that really mean quantum gravity is the correct solution to the problem, though? Is there some reason to believe that GR is the part that needs to be fixed? For example, GR is background independent while QM isn't - absent other evidence/problems, you'd assume QM is the incomplete theory, rather than GR. Do you know about something that shows GR (or both GR and QM, of course) is the "broken" theory?
Luaan
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2016-02-09T15:42:21.933Z
@Luaan GR is horribly non-renormalizable.  QM has lots of infinity "issues", too, but the theory is renormalizable and this basically resolves the problem.  The divergences in GR are just unmanageable.  In a vague sense, quantum theories are intrinsically immune to such unmanageable divergences--everything is constructed to either mitigate or disallow them.  So it's natural to be inclined to try to quantize GR. Both theories are known to have issues, so really both need to be fixed in some sense or another.  How and in what way is a major and unresolved question.
zibadawa timmy
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2016-02-17T15:09:47.407Z
@zibadawatimmy .. silly question: Has the result that gravity doesn't behave as an inverse square law in high energy situations been verified by experiment? I'm sure that the equations that contain this were used in computer simulations that gave us a pretty good idea of the physical process that created the gravitational waves that LIGO saw.
Jack R. Woods
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2016-10-22T16:20:01.080Z
I've asked a slightly related [naive question](http://physics.stackexchange.com/q/288185/83380).
uhoh
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2021-11-29T08:48:30.270Z
@Jack R. Woods  gravitational force markedly deviates from the inverse square law in collisions of BH+BH, NS+NS that were observed through the gravitational waves. The observed signal is in good agreement with the prediction based on GR. 
Simulations of EM pulses from pulsars are in better agreement with observations if GR is used; geometry of the bright spots on the NS surface varies weirdly as it rotates, from the point of view of GR-skeptic observer at infinity.
Precession of the perihelia of Mercury and Venus orbits are explained by GR; r<sup>‒2<sup> law is violated.
matteroffact
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2016-02-08T16:33:52.787Z
2016-02-08T16:33:52.787Z
2017-04-13T12:40:15.377Z
zibadawa timmy
<h1>First, the fact that gravity falls off $1/r^2$ is visible in the metric.</h1>

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

<p>$$
\mathrm{d}s^2 = -\left(1-\frac{r_s}{r}\right)\mathrm{d}t^2 + \left(1-\frac{r_s}{r}\right)^{-1}\mathrm{d}r^2 + r^2(\mathrm{d}\theta^2 + \sin^2\theta  \ \mathrm{d}\phi^2)
$$</p>

<p>Clearly, if $r \gg r_s$ this looks like</p>

<p>$$
\mathrm{d}s^2=-\mathrm{d}t^2+\mathrm{d}r^2+r^2(\mathrm{d}\theta^2 + \sin^2\theta \ \mathrm{d}\phi^2)
$$
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.</p>

<p>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.</p>

<h1>The little afterthought on your question takes some more thought</h1>

<p>I want to talk about where gravitons come from, but first lets talk about curvature.</p>

<p>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)</p>

<p><img src="https://upload.wikimedia.org/wikipedia/commons/3/3e/Connection-on-sphere.png" alt="Parallel Transport"></p>

<p>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</p>

<p>$$
[D_\mu,D_\nu] = D_\mu D_\nu - D_\nu D_\mu\neq 0
$$
It basically means "doing it one way is not the same as doing it the other".</p>

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

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

<p>$$
\mathcal{L}=\bar\psi(i\gamma^\mu D_\mu-m)\psi
$$</p>

<p>If I take the field $\psi$ and give it a transformation
$$
\psi\to\psi'=e^{i\xi(x)}\psi
$$
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</p>

<p>$$
[D_\mu,D_\nu] = -iF_{\mu\nu}\psi
$$
where
$$
F_{\mu\nu} = \partial_\mu A_\nu -  \partial_\nu A_\mu
$$</p>

<p>From this we form the complete QED (the quantum theory of electrodynamics) Lagrangian
$$
\mathcal{L}=\bar\psi(i\gamma^\mu D_\mu-m)\psi-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}
$$</p>

<p>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 <strong>becuase we had curvature</strong>. How do I know we have curvature? <strong>Becuase the covariant derivatives do not commute</strong>, 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.</p>

<p>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.</p>

<h1>Anyway, back to GR</h1>

<p>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</p>

<p>$$
S=\int R \sqrt{g} \ \mathrm{d}^4x
$$
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...</p>

<h1>Two versions of the same thing</h1>

<p>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.</p>

<h2>A note on other theories</h2>

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

<h2>A note on errors in the above</h2>

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

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2016-02-09T00:54:33.367Z
2016-10-20T18:18:19.433Z
2016-10-20T18:18:19.433Z
Sam
<blockquote>
  <p>For a lot of my uninformed life, I have doubted the existence of gravitons or even that gravity is an actual "force" (like electromagnetism).</p>
</blockquote>

<p>Gravity is a force like electromagnetism, but it does have a special property in that all test particles fall the same way in a gravitational field, no matter their composition. This means that inertial masses and gravitational masses are the same (or at least universally proportional, so we can use units in which they are equal), and we are free to interpret gravitational freefall as inertial motion.</p>

<p>In terms of quantum field theory, it is actually a theorem that at low energies, massless spin-2 particles must couple to all energy-momentum equally, regardless of particle species. In other words, the equivalence principle of general relativity is a provable theorem for gravitons.</p>

<p>Conversely, we can also interpret general relativity as a massless spin-2 field on a flat background spacetime, but because of this universality, the background will be unobservable by any experiment. That's why relativists don't tend to do this, as it makes the geometric interpretation is more convenient.</p>

<p>Unfortunately, quantized general relativity is very badly behaved if one tries to take them to arbitrarily energy scales. Physically, this means that some new physics must come in before then to fix it. However, this kind of situation is hardly unique to gravity, quantizing which still makes sense as an effective field theory at lower energies; cf. <a href="http://relativity.livingreviews.org/Articles/lrr-2004-5/">living review by Cliff P. Burgess</a>. The tension between general relativity and quantum mechanics is often overstated in popular descriptions.</p>

<blockquote>
  <p>My question would be: Does the fact that gravity follows an inverse square law naturally fall out of general relativity equations or is it an assumption used when developing the equations?</p>
</blockquote>

<p>The inverse-square part falls out by itself, but the specific constant of proportionality needs an additional assumption.</p>

<p>If one considers a general field equation $G_{\mu\nu} = \kappa T_{\mu\nu}$, where $T_{\mu\nu}$ is the stress-energy tensor that is assumed to be symmetric and covariantly conserved, then the Einstein tensor $G_{\mu\nu} \equiv R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R$ is the unique scale-invariant solution that can built from the metric. This requirement means that only terms that are second-order in derivatives of the metric are allowed, and it is broken by e.g. cosmological constant term $\Lambda g_{\mu\nu}$, as this introduces a length $\Lambda^{-1/2}\sim 10^{10}\,\mathrm{ly}$ into the theory.</p>

<p>There are other ways of developing the Einstein field equation, e.g. via the Einstein-Hilbert action, which don't need specific assumptions about the stress-energy tensor. Regardless, the role of the Newtonian limit is in fixing the value of the otherwise undetermined constant $\kappa = 8\pi G/c^4$. If you're only interested in a Newton-like inverse-square relationship, then <em>that alone</em> doesn't need any additional assumptions about trying to match Newtonian gravity.</p>

<p>Given a timelike vector field $u$, which can be interpreted as the four-velocities of some family of observers, we can write the time-time projection of an equivalent form of the Einstein field equation, $R_{\mu\nu} = \kappa(T_{\mu\nu}-\frac{1}{2}g_{\mu\nu}T)$, as
$$R_{00} \equiv R_{\mu\nu}u^\mu u^\nu = \frac{1}{2}\kappa(\rho+3p)\text{,}$$
where $\rho$ is the energy density and $p$ is the average of the principal stresses as measured by an observer with four-velocity $u$. For non-relativistic matter, the stress terms are negligible compared to the energy density.</p>

<p>The way the Newtonian limit is typically discusses is to use the weak-field approximation, $g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu}$ with $|h_{\mu\nu}|\ll 1$, to show that
$$\frac{1}{2}\kappa\rho\approx R_{00} = R^\alpha{}_{0\alpha 0}\approx\partial_\alpha\Gamma^{\alpha}_{00}\approx-\frac{1}{2}\nabla^2h_{00}\text{,}$$
which then has the form of Poisson's equation for Newtonian gravitational potential in terms of matter density $\rho_\text{m}$, i.e. $\nabla^2\Phi = 4\pi G\rho_\text{m}$. For slow-moving test particles, the geodesic equation reduces to the Newtoni an equation of motion:
$$\frac{\mathrm{d}^2\mathbf{x}}{\mathrm{d}t^2} = \frac{1}{2}\nabla h_{00} = -\nabla\Phi\text{.}$$
Another way to think of this is to write down the proper time of freefalling particle and show that extremizing it is equivalent to extremizing  $\int\left(\frac{1}{2}v^2+\frac{1}{2}h_{00}\right)\mathrm{d}t$, which is the action action (per mass) of a particle subject to Newtonian gravity whenever $h_{00} \approx -2\Phi/c^2$.</p>

<p>You may be interested in <a href="http://math.ucr.edu/home/baez/einstein/node6a.html">this simpler derivation</a> of Newton's law of gravitational around a spherically symmetric body, based on the geometric interpretation of the Ricci curvature as the acceleration of the volume of a small ball of initially comoving test particles.</p>

<blockquote>
  <p>And, just now, I had the thought that other forces may curve space also (just in higher dimensions).</p>
</blockquote>

<p>This was done for electromagnetism by Kaluza and Klein shortly after GTR, but it turns out that it's not a directly useful way to think about other forces.</p>

<p>Instead, we can think of Riemann curvature in general relativity as the <a href="https://en.wikipedia.org/wiki/Curvature_form">curvature form</a> of the Levi-Civita connection on the tangent bundle of a given manifold, with structure group $\mathrm{O}(1,n)$. But in this language, the electromagnetic field strength is the curvature of the connection $ieA_\mu$ over a line bundle with structure group $\mathrm{U}(1)$. The other non-gravitational forces are similarly described by <a href="https://en.wikipedia.org/wiki/Yang%E2%80%93Mills_theory">Yang-Mills theory</a>.</p>

<p>In other words, the other forces already have a description in which they're caused by a curvature, just not of spacetime. So while gravity is different from them, it's not different enough to consider it in some sense 'less real' than the others.</p>

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2016-10-22T16:07:40.480Z
The direction that antimatter falls in a gravitational field [has not been directly measured yet](http://physics.stackexchange.com/a/241181/83380), although I think most people expect it to fall the same way as the regular stuff.
uhoh
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2016-02-09T02:19:09.790Z
2016-02-09T02:19:09.790Z
Stan Liou
<p>For a lot of my uninformed life, I have doubted the existence of gravitons or even that gravity is an actual "force" (like electromagnetism). This is because my vision of general relativity was that mass curves space such that objects are still traveling in a "straight line" when being acted upon by "gravity", so that no "force" is needed. I know now that this is a naive view, but I'm not 100% sure why. I was thinking the other day that just the fact that gravity follows an inverse square law implies that it is a force carried by particles (falling off in flux intensity due to the geometry of 3D space). </p>

<p>My question would be: Does the fact that gravity follows an inverse square law naturally fall out of general relativity equations or is it an assumption used when developing the equations?</p>

<p>And, just now, I had the thought that other forces may curve space also (just in higher dimensions).</p>

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2016-02-09T11:27:05.050Z
Note that GR doesn't describe gravity as inverse square force - that's just the low-energy approximation. All the "solutions" to the field equations discovered by Einstein we have are approximations for some specific scenario, e.g. the Schwarzschild solution which describes gravity around a spherically-symetric, uncharged and non-rotating objects, or the Kerr solution which handles rotating objects. To get the complete solution, you'd have to account for every bit of energy in the universe - not quite possible or practical. Since gravity is so weak, the approximation works very well, though :)
Luaan
<gravity><cosmology><graviton>
Existence of gravitons?
Jack R. Woods
# Existence of gravitons?

(Asked by: username_0 on 2016-02-08)

For a lot of my uninformed life, I have doubted the existence of gravitons or even that gravity is an actual "force" (like electromagnetism). This is because my vision of general relativity was that mass curves space such that objects are still traveling in a "straight line" when being acted upon by "gravity", so that no "force" is needed. I know now that this is a naive view, but I'm not 100% sure why. I was thinking the other day that just the fact that gravity follows an inverse square law implies that it is a force carried by particles (falling off in flux intensity due to the geometry of 3D space).

My question would be: Does the fact that gravity follows an inverse square law naturally fall out of general relativity equations or is it an assumption used when developing the equations?

And, just now, I had the thought that other forces may curve space also (just in higher dimensions).

**Tags:** gravity, cosmology, graviton
> **Comments:**
> **username_1**: Note that GR doesn't describe gravity as inverse square force - that's just the low-energy approximation. All the "solutions" to the field equations discovered by Einstein we have are approximations for some specific scenario, e.g. the Schwarzschild solution which describes gravity around a spherically-symetric, uncharged and non-rotating objects, or the Kerr solution which handles rotating objects. To get the complete solution, you'd have to account for every bit of energy in the universe - not quite possible or practical. Since gravity is so weak, the approximation works very well, though :)  

---

## 3 answers

### Answer by username_6 on 2016-02-09. Score: 10 | Accepted answer

> For a lot of my uninformed life, I have doubted the existence of gravitons or even that gravity is an actual "force" (like electromagnetism).

Gravity is a force like electromagnetism, but it does have a special property in that all test particles fall the same way in a gravitational field, no matter their composition. This means that inertial masses and gravitational masses are the same (or at least universally proportional, so we can use units in which they are equal), and we are free to interpret gravitational freefall as inertial motion.

In terms of quantum field theory, it is actually a theorem that at low energies, massless spin-2 particles must couple to all energy-momentum equally, regardless of particle species. In other words, the equivalence principle of general relativity is a provable theorem for gravitons.

Conversely, we can also interpret general relativity as a massless spin-2 field on a flat background spacetime, but because of this universality, the background will be unobservable by any experiment. That's why relativists don't tend to do this, as it makes the geometric interpretation is more convenient.

Unfortunately, quantized general relativity is very badly behaved if one tries to take them to arbitrarily energy scales. Physically, this means that some new physics must come in before then to fix it. However, this kind of situation is hardly unique to gravity, quantizing which still makes sense as an effective field theory at lower energies; cf. [living review by Cliff P. Burgess](http://relativity.livingreviews.org/Articles/lrr-2004-5/). The tension between general relativity and quantum mechanics is often overstated in popular descriptions.

> My question would be: Does the fact that gravity follows an inverse square law naturally fall out of general relativity equations or is it an assumption used when developing the equations?

The inverse-square part falls out by itself, but the specific constant of proportionality needs an additional assumption.

If one considers a general field equation $G\_{\mu\nu} = \kappa T\_{\mu\nu}$, where $T\_{\mu\nu}$ is the stress-energy tensor that is assumed to be symmetric and covariantly conserved, then the Einstein tensor $G\_{\mu\nu} \equiv R\_{\mu\nu} - \frac{1}{2}g\_{\mu\nu}R$ is the unique scale-invariant solution that can built from the metric. This requirement means that only terms that are second-order in derivatives of the metric are allowed, and it is broken by e.g. cosmological constant term $\Lambda g\_{\mu\nu}$, as this introduces a length $\Lambda^{-1/2}\sim 10^{10}\,\mathrm{ly}$ into the theory.

There are other ways of developing the Einstein field equation, e.g. via the Einstein-Hilbert action, which don't need specific assumptions about the stress-energy tensor. Regardless, the role of the Newtonian limit is in fixing the value of the otherwise undetermined constant $\kappa = 8\pi G/c^4$. If you're only interested in a Newton-like inverse-square relationship, then *that alone* doesn't need any additional assumptions about trying to match Newtonian gravity.

Given a timelike vector field $u$, which can be interpreted as the four-velocities of some family of observers, we can write the time-time projection of an equivalent form of the Einstein field equation, $R\_{\mu\nu} = \kappa(T\_{\mu\nu}-\frac{1}{2}g\_{\mu\nu}T)$, as
$$R\_{00} \equiv R\_{\mu\nu}u^\mu u^\nu = \frac{1}{2}\kappa(\rho+3p)\text{,}$$
where $\rho$ is the energy density and $p$ is the average of the principal stresses as measured by an observer with four-velocity $u$. For non-relativistic matter, the stress terms are negligible compared to the energy density.

The way the Newtonian limit is typically discusses is to use the weak-field approximation, $g\_{\mu\nu} = \eta\_{\mu\nu} + h\_{\mu\nu}$ with $|h\_{\mu\nu}|\ll 1$, to show that
$$\frac{1}{2}\kappa\rho\approx R\_{00} = R^\alpha{}\_{0\alpha 0}\approx\partial\_\alpha\Gamma^{\alpha}\_{00}\approx-\frac{1}{2}\nabla^2h\_{00}\text{,}$$
which then has the form of Poisson's equation for Newtonian gravitational potential in terms of matter density $\rho\_\text{m}$, i.e. $\nabla^2\Phi = 4\pi G\rho\_\text{m}$. For slow-moving test particles, the geodesic equation reduces to the Newtoni an equation of motion:
$$\frac{\mathrm{d}^2\mathbf{x}}{\mathrm{d}t^2} = \frac{1}{2}\nabla h\_{00} = -\nabla\Phi\text{.}$$
Another way to think of this is to write down the proper time of freefalling particle and show that extremizing it is equivalent to extremizing $\int\left(\frac{1}{2}v^2+\frac{1}{2}h\_{00}\right)\mathrm{d}t$, which is the action action (per mass) of a particle subject to Newtonian gravity whenever $h\_{00} \approx -2\Phi/c^2$.

You may be interested in [this simpler derivation](http://math.ucr.edu/home/baez/einstein/node6a.html) of Newton's law of gravitational around a spherically symmetric body, based on the geometric interpretation of the Ricci curvature as the acceleration of the volume of a small ball of initially comoving test particles.

> And, just now, I had the thought that other forces may curve space also (just in higher dimensions).

This was done for electromagnetism by Kaluza and Klein shortly after GTR, but it turns out that it's not a directly useful way to think about other forces.

Instead, we can think of Riemann curvature in general relativity as the [curvature form](https://en.wikipedia.org/wiki/Curvature_form) of the Levi-Civita connection on the tangent bundle of a given manifold, with structure group $\mathrm{O}(1,n)$. But in this language, the electromagnetic field strength is the curvature of the connection $ieA\_\mu$ over a line bundle with structure group $\mathrm{U}(1)$. The other non-gravitational forces are similarly described by [Yang-Mills theory](https://en.wikipedia.org/wiki/Yang%E2%80%93Mills_theory).

In other words, the other forces already have a description in which they're caused by a curvature, just not of spacetime. So while gravity is different from them, it's not different enough to consider it in some sense 'less real' than the others.

> **Comments:**
> **username_3**: The direction that antimatter falls in a gravitational field [has not been directly measured yet](http://physics.stackexchange.com/a/241181/83380), although I think most people expect it to fall the same way as the regular stuff.  

---

### Answer by username_2 on 2016-02-08. Score: 13

Gravity is a [fictitious force](https://en.wikipedia.org/wiki/Fictitious_force), actually, much like centrifugal force. In a free falling frame of reference it disappears. In general relativity (GR) gravity is just a result of (differential) geometry: space-time curvature. The inverse square law is just the low energy approximation, but the actual equation for gravity derived from GR is more complex than that. The massive success of Newtonian gravity tells us that any model of gravity must be approximated by the classical inverse square law at low energies.

Whether GR does that by (Einstein's) design or something else is a matter of personal opinion. Einstein definitely knew he had to get approximately Newtonian gravity at low energies, so he would have discarded or modified any ideas that failed this criterion. However, there are [standard arguments for why gravity must obey an inverse square law](https://en.wikipedia.org/wiki/Inverse-square_law#Justification), at least in low energy situations.

Now in GR [a lot more than just mass contributes to the gravitational field](https://en.wikipedia.org/wiki/Stress%E2%80%93energy_tensor). Spin and electric charge, for example, and quite importantly: energy (the famous $E=mc^2$ tells us how to express a mass as energy, so we can treat these things on a common footing). So, yes, all forces and particles contribute to gravity. [Even photons](https://physics.stackexchange.com/a/22878/55483).

GR itself makes no predictions (or requirements) for the existence of any new particles outside of the standard model, such as gravitons. GR and quantum mechanics (QM) are famously incompatible: in extreme situations where both GR and QM are relevant (neutron stars and black hole formation, for example), they stop making sense rather quickly. [Especially GR](https://en.wikipedia.org/wiki/Graviton#Gravitons_and_renormalization). "Gravitons" and assorted variations are hypothetical particles that are proposed to resolve this issue by creating a quantum theory of gravity. The only "evidence" we have for them at this stage is that our two most massively successful theories about the workings of the universe, GR and QM, are so painfully incompatible. So we know these theories are flawed (aka wrong) and that some other theory is necessary that can handle these situations, while also incorporating all of the successes of QM and GR—they're amazingly accurate when only one of them is particularly relevant, after all.

Exactly what that theory *is* is an ongoing and substantial research area.

> **Comments:**
> **username_1**: Does that really mean quantum gravity is the correct solution to the problem, though? Is there some reason to believe that GR is the part that needs to be fixed? For example, GR is background independent while QM isn't - absent other evidence/problems, you'd assume QM is the incomplete theory, rather than GR. Do you know about something that shows GR (or both GR and QM, of course) is the "broken" theory?  
> **username_2**: @Luaan GR is horribly non-renormalizable.  QM has lots of infinity "issues", too, but the theory is renormalizable and this basically resolves the problem.  The divergences in GR are just unmanageable.  In a vague sense, quantum theories are intrinsically immune to such unmanageable divergences--everything is constructed to either mitigate or disallow them.  So it's natural to be inclined to try to quantize GR. Both theories are known to have issues, so really both need to be fixed in some sense or another.  How and in what way is a major and unresolved question.  
> **username_0**: @zibadawatimmy .. silly question: Has the result that gravity doesn't behave as an inverse square law in high energy situations been verified by experiment? I'm sure that the equations that contain this were used in computer simulations that gave us a pretty good idea of the physical process that created the gravitational waves that LIGO saw.  
> **username_3**: I've asked a slightly related [naive question](http://physics.stackexchange.com/q/288185/83380).  
> **username_4**: @Jack R. Woods  gravitational force markedly deviates from the inverse square law in collisions of BH+BH, NS+NS that were observed through the gravitational waves. The observed signal is in good agreement with the prediction based on GR.  Simulations of EM pulses from pulsars are in better agreement with observations if GR is used; geometry of the bright spots on the NS surface varies weirdly as it rotates, from the point of view of GR-skeptic observer at infinity. Precession of the perihelia of Mercury and Venus orbits are explained by GR; r<sup>‒2<sup> law is violated.  

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### Answer by username_5 on 2016-02-09. Score: 6

# First, the fact that gravity falls off $1/r^2$ is visible in the metric.

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

$$
\mathrm{d}s^2 = -\left(1-\frac{r\_s}{r}\right)\mathrm{d}t^2 + \left(1-\frac{r\_s}{r}\right)^{-1}\mathrm{d}r^2 + r^2(\mathrm{d}\theta^2 + \sin^2\theta \ \mathrm{d}\phi^2)
$$

Clearly, if $r \gg r\_s$ this looks like

$$
\mathrm{d}s^2=-\mathrm{d}t^2+\mathrm{d}r^2+r^2(\mathrm{d}\theta^2 + \sin^2\theta \ \mathrm{d}\phi^2)
$$
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.

# The little afterthought on your question takes some more thought

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)

![Parallel Transport](https://upload.wikimedia.org/wikipedia/commons/3/3e/Connection-on-sphere.png)

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

$$
[D\_\mu,D\_\nu] = D\_\mu D\_\nu - D\_\nu D\_\mu\neq 0
$$
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

$$
\mathcal{L}=\bar\psi(i\gamma^\mu D\_\mu-m)\psi
$$

If I take the field $\psi$ and give it a transformation
$$
\psi\to\psi'=e^{i\xi(x)}\psi
$$
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

$$
[D\_\mu,D\_\nu] = -iF\_{\mu\nu}\psi
$$
where
$$
F\_{\mu\nu} = \partial\_\mu A\_\nu - \partial\_\nu A\_\mu
$$

From this we form the complete QED (the quantum theory of electrodynamics) Lagrangian
$$
\mathcal{L}=\bar\psi(i\gamma^\mu D\_\mu-m)\psi-\frac{1}{4}F\_{\mu\nu}F^{\mu\nu}
$$

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.

# Anyway, back to GR

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

$$
S=\int R \sqrt{g} \ \mathrm{d}^4x
$$
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...

# Two versions of the same thing

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!


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