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As far as I’m aware, gravity in general relativity arises from the curvature of spacetime and is equivalent to an accelerated reference frame. Objects accelerating in a gravitational field are in fact inertial and are moving through geodesics in spacetime.

So it could be said then that it is not really a force, but a pseudoforce much like the Coriolis effect. If so, why is it necessary to quantise gravity with a gauge boson, the graviton? And why is it necessary to unify it with the other forces?

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    $\begingroup$ I like this question. I have seen many sparse answers, but haven't seen it addressed head-on. $\endgroup$ – InertialObserver Dec 23 '18 at 7:30
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    $\begingroup$ Possible duplicate of Is quantizing acceleration equivalent to quantizing gravity? $\endgroup$ – knzhou Dec 23 '18 at 11:40
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    $\begingroup$ There is a purely classical fact that you seem not to have recognized, which is that curvature is not fictitious. What is fictitious is the gravitational acceleration. $\endgroup$ – Ben Crowell Dec 23 '18 at 19:23
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    $\begingroup$ I presume because a graviton is necessary in Quantum Mechanics, not in General Relativity. But I'm not a physicist. $\endgroup$ – immibis Dec 24 '18 at 3:24
  • $\begingroup$ I would avoid both the term “pseudoforce” and “fictious force” in context of General Realtivity. General relativity is general, because it considers all reference frames equal and in some of them (in most of them, in fact), inertial forces are real. $\endgroup$ – Jan Hudec Dec 24 '18 at 22:32
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While it's common to describe gravity as a fictitious force we should be cautious about the use of the adjective fictitious as this is a technical term meaning the gravitational force is not fundamental but is the result of an underlying property. The force itself most certainly exists as anyone who has been sat on by an elephant can attest.

There is a sense in which all forces are fictitious since they are all the emergent long range behaviour of quantum fields, so gravity is not unique in this respect. For more on this see Can all fundamental forces be fictitious forces?

Anyhow, the object responsible for the gravitational force is a tensor field called the metric, and when we quantise gravity we are quantising the metric not the force. The graviton then emerges as the excitation of the quantum field that describes the metric. As with other quantum fields we can have real gravitons that are the building blocks of gravitational waves and virtual gravitons used in scattering calculations.

Finally, you ask why it's necessary to quantise gravity, and this turns out to be a complicated question and one that ignites much debate about what it means to quantise gravity. However the question has already been thoroughly discussed in Is the quantization of gravity necessary for a quantum theory of gravity? While it's not directly related I can also recommend A list of inconveniences between quantum mechanics and (general) relativity? as interesting reading.

The principle reason that we want to quantise gravity is because Einstein's equation relates the curvature to the matter/energy distribution and the matter/energy is quantised. Einstein's equation tells us:

$$ \mathbf G = 8 \pi \mathbf T $$

where $\mathbf G$ is the Einstein tensor that describes the spacetime curvature while $\mathbf T$ is the stress-energy tensor that describes the matter/energy distribution. The problem is that $\mathbf T$ could describe matter that is in a superposition of states or an entangled state, and that implies that the curvature must also be in a superposition of states or entangled. And this is only possible if the spacetime curvature is described by a quantum theory, or some theory whose low energy limit is quantum mechanics.

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    $\begingroup$ But isn't OP asserting the possibility that gravity is different in this sense? That the description of spacetime is more fundamental than the Yang-Mills gauge theories that we define on that spacetime, and so perhaps shouldn't be subject to the same intuition that it should be "quantized"? $\endgroup$ – InertialObserver Dec 23 '18 at 7:54
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    $\begingroup$ The term "fictitious" force, I believe, here, is actually best understood as being not in the sense of "not real" but rather in the sense of "centrifugal force", and what it's saying is that gravity is exactly the same kind of "force" as centrifugal force. That is, it is something which may be better called an "inertial force", which arises from operating in a non-inertial reference frame. In particular, the key is that a free-fall frame is an inertial frame, in that you cannot do any experiments at least "locally" to tell that you are falling versus simply moving through empty space. $\endgroup$ – The_Sympathizer Dec 23 '18 at 8:25
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    $\begingroup$ And in gravitation, the inertial force appears when you consider an observer fixed to, say, the surface of a gravitating planet, who sees things falling as being "pulled" by a force. This is because the surface of a gravitating planet is not an inertial frame. $\endgroup$ – The_Sympathizer Dec 23 '18 at 8:28
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    $\begingroup$ "The force itself it most certainly not fictitious as anyone who has been sat on by an elephant can attest." An elephant in a centrifuge would produce a "centrifugal force", that doesn't mean centrifugal force isn't "fictitious". $\endgroup$ – Acccumulation Dec 24 '18 at 17:42
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    $\begingroup$ I'm not convinced that anyone who has been sat on by an elephant can attest to anything. $\endgroup$ – JBentley Dec 25 '18 at 3:26
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Gravity is not equivalent to an accelerated frame. It's locally equivalent to an accelerated frame. That means that a point-like observer will never be able to tell whether he/she is in a gravitational field or in an accelerating spaceship. But an observer that has some characteristic size will experience tidal forces. Tidal forces are a result of a non zero curvature of space. On the other hand going to an accelerated frame does not curve spacetime because it's just a change of variables.

The mathematical statement is that you can always find a change of coordinates which, at a given point, sets to zero the first derivatives of the metric. Namely you can make the Christoffel symbols $\Gamma^\mu_{\nu\rho}$ vanish at a given point. On the other hand the second derivatives of the metric, which encode the curvature, cannot be set to zero.

There is a nice exercise that demonstrates the difference between accelerating frames and curved spacetime. Suppose you have two particles following two parallel geodesics. In curved spacetime the geodesics will not stay parallel. This is discussed in the book Schutz - A first course in general relativity at the end of Section 6.5. I'll summarize the punch line but I encourage you to check the derivation there.

Call $\vec{\xi}$ the vector that connects two geodesics $\vec{V}$ and $\vec{V}'$ initially parallel. In the book the following equation is proven $$ \nabla_V\nabla_V \xi^\alpha = R^\alpha_{\phantom{a}\mu\nu\beta}V^\mu V^\nu \xi^\beta\,. $$ where $\nabla_V = V^\mu \nabla_\mu$, $\nabla$ being the covariant derivative and $R^\alpha_{\phantom{a}\mu\nu\beta}$ is the Riemann tensor, which measures the curvature. The equation above states that the variation of $\vec{\xi}$ along the geodesic is not zero but proportional to the curvature, thus showing that geodesics deviate from being parallel in curved space. This is a frame-independent effect and cannot be realized by going to an accelerated frame.

An observer with a characteristic size of the order of $1/\sqrt{R^\mu_{\phantom{a}\nu\rho\beta}}$ would be able to notice this effect.

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    $\begingroup$ Thank you very much for your comment, although I was already aware of the equivalence being local; i really should have phrased my question better. $\endgroup$ – Thatpotatoisaspy Dec 23 '18 at 11:48
  • $\begingroup$ Wouldn't a sufficiently large object also experience something similar to tidal forces from normal acceleration, due to the fact that whatever force is acting on the object isn't doing so uniformly across the entire object and the resulting force can't propagate through the object faster than light? $\endgroup$ – Michael Dec 24 '18 at 23:16
  • $\begingroup$ Sure but he was referring to an acceleration that can be canceled by a suitable change of coordinates. Which would make any phenomenon stemming from it fictitious. Such a kind of acceleration can't create tidal forces. $\endgroup$ – MannyC Dec 25 '18 at 11:17
  • $\begingroup$ Does this really answer the question? Why does the fact that gravity entails tidal forces mean that it has to be quantized? $\endgroup$ – Peter Shor Dec 25 '18 at 18:15
  • $\begingroup$ The question wasn't about why does gravity have to be quantized. He was asking why can't we just regard gravity as a frame dependent phenomenon. This is at least how I understood it. $\endgroup$ – MannyC Dec 25 '18 at 18:21
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None of the answers current explain an aspect of the question that I am interested in: what goes wrong if you try to construct a theory of physics where gravity is not quantized?

There are various arguments that strongly suggest that everything coupled to a quantum system should, fundamentally, also be quantum.

We know that the stress-energy tensor sources curvature for the gravitational field, $$G_{\mu\nu} \sim T_{\mu\nu}$$ but in a quantum theory the stress-energy tensor does not have a definite value, but rather may be in superposition. So then how do we describe the curvature? If you say the curvature may be in superposition too, so that $G_{\mu\nu} = T_{\mu\nu}$ holds for each branch of the superposition, then you've just quantized gravity -- quantization is exactly the process where we treat the set of classical physical states of a system as separate quantum states which may be superposed.

The only other option which reduces to the classical result when the matter is nearly classical is $$G_{\mu\nu} \sim \langle T_{\mu\nu} \rangle.$$ However, this is extremely strange for many reasons. For example, consider a particle of mass $m$ which is in an equal superposition of being here or in Andromeda. Then the classical gravitational field would be that of two masses $m/2$, each in one galaxy. If the particle is measured, the wavefunction collapses, and the gravitational field instantaneously changes, so the observed mass in Andromeda becomes either $m$ or zero. This nonlocal change in the field allows superluminal signalling by somebody in the Milky Way. (There's nothing special about gravity here; it would also hold if we insisted on a classical electromagnetic field. In either case, when the field is quantized, this problem is avoided by the usual way in quantum field theory.)

One could argue that collapse is really unphysical; all branches of the wavefunction exist and we should sum over all of them. If we take this interpretation, then non-quantized gravity is already ruled out experimentally. See Page and Geilker (1981), where the result of a radioactive decay is used to determine the positioning of a mass in a Cavendish experiment. If all branches of the wavefunction count, then the pendulum should point at the midpoint of the two possible positions for the mass, but it doesn't.

As a separate issue, energy conservation may be violated. This is easier to see with the electromagnetic field. If one starts with an excited atom in an empty cavity, in state $|e \rangle$, after some time it will be in the superposition $(|e \rangle + |g \rangle) / \sqrt{2}$. If you insist the electromagnetic field have a definite classical configuration, then the branches of the wavefunction do not have equal energy. When you measure the energy, you'll generally find a different result than the initial energy; it can only match on average.

This is essentially the erroneous BKS theory which was rendered obsolete with the quantization of the electromagnetic field. In this case the wavefunction is $(|e\rangle \otimes |0 \rangle + |g \rangle \otimes |1 \rangle) / \sqrt{2}$ where the second factor indicates the number of photons, and the two branches of the wavefunction have exactly the same energy as they must. Similarly, if one couples to classical gravity, one must allow violations of energy conservation that only cancel out on the average, but there's no problem for quantized gravity.

I'm sure the mathematicians can come up with more sophisticated, complicated reasons that classical and quantum theories don't mesh, but these immediate issues are already bad enough.

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    $\begingroup$ Wave function collapse does not usually allow for superluminal signaling. And wave function collapse does not usually violate energy. Why would it when gravity is involved? $\endgroup$ – Peter Shor Dec 26 '18 at 1:37
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    $\begingroup$ @PeterShor It isn't specifically about gravity, but rather about a coupling of a quantum system to a classical field. You could use the same arguments for electromagnetism, and in fact they were used, as that was the justification for BKS theory in the first place. $\endgroup$ – knzhou Dec 26 '18 at 1:52
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You could also ask why the photon is necessary, if electromagnetism is a classical force based on Yang-Mills fields with gauge group U(1). Or also, why the gluons, the W, Z AND the Higgs boson are necessary, since non-abelian Yang-Mills fields are also meaningful as classical fields. In my opinion, the answer to this question, and why fields are to be quantized, must include two subtle issues:

  1. Quanta are not fundamental, but, as previous questions remark, are excitations from vacuum of certain FIELDS on space-time. What is relevant is the quantization of the action, that generally implies the quantization of energy and other magnitudes like angular momentum.
  2. Gravity has a different status with respect to other forces due to its universality, not due to it being a "pseudo-force". Gravity couples to everything, while other fields couple to certain properties of space-time like electric (magnetic) charge, flavor or color.

Moreover, the question of the need of the quantization of the gravitational field is evident when seeing the Einstein field equations for gravity: one side is the matter-energy having mass, energy, and quantum numbers, the other side is the geometry or metric of space-time. If identical, well, one should wonder if the metric itself has these features. String theory or loop quantum gravity show differently how the space-time itself could handle with quantum numbers. The problem with quantum gravity is not that we don't need gravitons. Indeed, Newton's gravity itself imply certain field theory in the form of Poisson equation that Einstein himself used as model to reproduce an analogy for building up his equations for gravity. The problem with quantum gravity and gravitons is in the heart of your question: if we model space-time like a metric and geometry, why do we need gravitons? We need gravitons because they must be there. Quantum theory is correct, even if some day is proved to be uncomplete or it must be modified to include gravity. Maxwell's equations are superseded by QED and the electroweakt theory at high energies, there new particles appear: the W, Z bosons and the Higgs (for consistency). Conceptually, maybe, the issue is understando how a set of gravitons could determine the geometry of the metric? No, the issue with gravitons is that General Relativity in a canonical quantum theory behaves badly. Calculations diverge. By the other hand, the space-time metric, the one in General Relativity, can not be the whole story...Just we know the Standard Model is not the whole story...The spacetime metrics in some concrete circumstances also diverge CLASSICALLY! Every theoretical physicist know that space-time singularities are a problem in most of the classical theories of gravity. You get singularities in black holes (hidden under event horizon, due to the cosmic censorship hypothesis), and you get singularities at the beginning of the time...In both cases, you have a very dense object in a very tiny space. Such extreme density conditions make us think that General Relativity and the description of space-time with a metric is only an approximation or a very good model excepting extreme cases (black holes, the Big Bang,...or similar). There, enter quantum gravity and gravitons. Graviton scattering must domine in such regime or produce some kind of extreme "matter"/object whose description with a metric is bad. Of course, some people work on the idea that black holes and space-time is some kind of "condensate" of gravitons or superfluid made of some preonic substance yet to discover (the nature of the microstates of black holes is only approached in some extreme cases with superstring theory). In summary:

1) A graviton is necessary due to universality description of all the forces as interchanging force carriers.

2) A graviton is necessary since we believe graviton excitations, maybe Wheeler's space-time foam in some form or alike, must dominate the description of very dense objects (microscopic black holes, the beginning of the time, and other similar examples as space-time singularities).

However, graviton scattering behaves badly in general relativity. Taking a conservative canonical quantum gravity approach provides divergent results. Only string theory and loop quantum gravity, and some minor third ways to quantum gravity, shed light on how to calculate these divergences. String theory provides a unifying framework to handle with all the "fundamental forces" and matter field. However, after two revolutions, and no hints of extra dimensions in experiments and detectors (and a critical 4D value from gravitational wave observations to date), we have no evidence from strings or p-branes yet. Loop quantum gravity (a modification of the canonical quantum gravity approach) provides an example of the quantization of geometry using a different technique than that in string theory. Area and volume are quantized in LQG. What are gravitons then? Gravitons in string theory are certain kind of excitations of the fundamental string (or brane). This fact is also remarked in the emergence of a symmetrical tensor when calculating the excitations of the string from the "vacuum". Gravitons in LQG are more subtle, I imagine them like polymer-like excitations from the area and volume operators, derived from spin networks and other discrete structures of the theory (I am not expert on that field, so I am being imprecise quite likely...).

3) Gravitons, photons, Higgs bosons, gluons, are likely not fundamental...Why do we need them? Because quantum fields can be represented as entities whose excitations produce particles. It happens with fermions as well. There is only a single electron field in all the Universe. However, the excitations in that field are the electrons we observe, reverberation of the beginning of the time...Just like gold atoms are produced in supernovae, electrons (or quarks) in the Universe were produced in the farthest past, and what remains is a rest from the annihilation with vacuum billions of years ago.

Gravitons, like photons and other particles, were produced in the beginning of the time. We don't understand what happened there, when GRAVITON scattering was dominant since the temperature was so hot, and the density so high, that we can not neglect gravitational interactions, usually weak when present electromagnetic or nuclear forces, or negligible only when you are not in a place where you have dense matter in a tiny volume (microscopic AND heavy black holes). That is why we need to understand better gravitons. Before the discovery of gravitational waves, that by duality imply the existence of the gravitons, some people wondered if gravity should be quantized. I think that question is not (if ever it was) relevant now. Gravitational waves do exist and then, gravitons (in some form) may exist. But, this have nothing to do with the classical existence of gravity. Before the Quantum Mechanics, physicists discussed if light was a wave or a particle. Well, light is both! Why do we need PHOTONS? We need photons since without photons (quanta of light) we could not explain wavy the photoelectric effect or the black-body radiation. Indeed, you are all embedded in a cosmic microwave background of photons emitted by the Big Bang, with temperature about 2.73 K. We believe there are also a neutrino and a graviton background as well. So, we need gravitons as well to understand the Universe! We can not understand the beginning of the Universe without understanding gravitons and the quantum nature of gravity.

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what goes wrong if you try to construct a theory of physics where gravity is not quantized?

At first glance - nothing. Present (empirically verified) theories are exactly such.

Try asking: what goes wrong if you try to construct a theory of physics where gravity is quantized?

At first glance everything. Try going deeper in: observables (operators) for the gravitational field?; redefinition of the space-time base QFTs are defined upon?; failure to describe interacting QFTs?; what is time in quantum mechanics?; how to deal with gravitation being non-linear (failure to super-pose solutions), ect.? The list is scary.

I'm not even attempting to answer these. However, it seems reasonable to assume that a quantum system should (after all, material content do curve space-time) interact with the gravitational field rather in an unknown manner, which classical limit should coincide with Einstein's equations.

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By the Heisenberg uncertainty principle, things smaller * higher energy than hbar*c must be described as waves, and this is obviously true for point singularities. Thus black holes fall not only in the domain of general relativity, but also quantum mechanics, which motivates the quest to quantize gravity. Quantum effects associated with gravity have been used for years, ie. Hawking radiation, but not in the full theoretic framework. People have tried to make gauge theories compatible with curved spacetime, for instance using the covariant derivative notation for minimal coupling similar to its use for curved space, but AFAIK, the problem with gravitons is that they are non-renormalizable. See https://arxiv.org/abs/gr-qc/0405033 for a nontraditional gauge theory based on the geometry of space.

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    $\begingroup$ Does $\hbar c$ even have units of energy? $\endgroup$ – Peter Shor Dec 25 '18 at 22:15
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    $\begingroup$ @PeterShor I believe the text means "((smaller size) $\times$ (higher energy)) than $\hbar c$", which does have the correct units. However, position uncertainty and energy uncertainty are not related in a simple way by the Heisenberg principle. The route I usually hear connecting short-distance and high-energy phenomena takes a detour through the Yukawa potential, where a massive field like the pion has an effective range $r \sim \hbar c/ mc^2$. Whether that argument works near a black hole is an interesting question whose answer doesn't fit in this comment box (but it's probably "no"). $\endgroup$ – rob Dec 26 '18 at 2:31
  • $\begingroup$ Also: welcome to Physics, new contributor! If you're my old friend and collaborator Christopher Crawford, then I hope your family is well and you're having a nice holiday and we should catch up. If you're a different Christopher Crawford, then just the first two of those. $\endgroup$ – rob Dec 26 '18 at 2:33
  • $\begingroup$ I wonder whether the number of people called Christopher Crawford who are old friends and collaborators of someone called Rob is as small as @rob thinks. Two completely different people might be catching up with each other right now. $\endgroup$ – Dawood ibn Kareem Dec 26 '18 at 7:38
  • $\begingroup$ Sorry, wrong forum. Merry Christmas, @rob! Just use d<=ct; hence hbar c. Or as you say, the usual comparison is with the Compton wavelength m/hbar c. Obsensibly so, @Dawood :) $\endgroup$ – Christopher Crawford Dec 27 '18 at 16:46
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There is no difference between waves and particles. They are identical broad concepts including any possible configuration of matter or energy.

So since gravity must be transmitted as something, whatever means through which it transmits can be called a wave or particle. Because all forces must unify, forces transmitted through quanta e.g. emag require that all other forces also be quantized.

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