# Why can’t gravitons distinguish gravity and inertial acceleration?

If gravitons mediate the gravitational force, couldn’t the detection of gravitons by an observer be used to distinguish whether they are experiencing gravitational acceleration vs. inertial acceleration, contradictory to general relativity? If this is not the case, and detection of gravitons can not be used to distinguish gravity from other acceleration, shouldn’t acceleration affect the way objects interact with the gravitational field?

Obviously, this can not be correct, so what am I missing?

Gravitons do not mediate the gravitational force and you cannot detect gravitons flashing to and fro between objects interacting gravitationally. Since you cannot detect the gravitons you cannot use said gravitons to find out whether acceleration is inertial or gravitational.

It is often said that forces are due to the exchange of virtual particles, for example the EM force is due to the exchange of virtual photons while the gravitational force is due to the exchange of virtual gravitons. But virtual particles are a computational device and do not actually exist. Those Feynman diagrams you have seen showing the exchange of a virtual particle are just a graphical representation of an integral called a propagator and do not show a physical process. I cannot emphasise this strongly enough:

Real gravitons are the quanta of gravitational waves, just as real photons are the quanta of light waves, but real gravitons do not transmit the gravitational force any more than real photons transmit the EM force.

When we write the four-acceleration of some observer we write it as a sum of the inertial and gravitational terms:

$$A^\alpha = \frac{\mathrm d^2x^\alpha}{\mathrm d\tau^2} + \Gamma^\alpha{}_{\mu\nu}U^\mu U^\nu$$

where the first term on the right hand side is the inertial part and the second term is the gravitational part. However neither of the terms on the right hand side are tensors so both are changed when we change the coordinate system. It is a fundamental principle in general relativity that we cannot distinguish between the two terms since either can be made zero just by choosing appropriate coordinates. In fact this is the equivalence principle stated mathematically.

• "Virtual particles do not exist!" What about Hawking radiation? – nick012000 Oct 24 at 2:46
• @nick012000 Here's a great answer by John on that very topic: physics.stackexchange.com/a/252236/123208 Also see math.ucr.edu/home/baez/physics/Relativity/BlackHoles/… – PM 2Ring Oct 24 at 3:05
• As pointed out, IMHO it is at least not straightforward to say that virtual particles do not exist. In fact, all the particles we ever actually observe are necessarily "slightly off-shell". For example, Ron Maimon's answer here: physics.stackexchange.com/a/17097/20427. – Dvij D.C. Oct 24 at 3:34
• True, but for an accelerating asymmetric mass one could in principle detect real gravitons – anna v Oct 24 at 8:09
• "But virtual particles are a computational device and do not actually exist." This statement is nonsense and confuses models and reality. Everything in physics is a "computational device". – Jannik Pitt Oct 24 at 21:25

Firstly, pure General Relativity theory doesn't have gravitons, it just has spacetime curvature. Gravitons are a quantum particle, and GR isn't a quantum theory. Hopefully, some kind of Quantum Gravity theory will unite GR & quantum field theory, but we don't have a successful QG theory yet. So we don't know if gravitons even exist, but considering how successful quantum theories have been, it's not too much of a stretch to say that they probably do exist.

However, even if a QG theory says that gravitons do exist, we may not be able to detect single gravitons directly because (as Anna V mentions) they interact so weakly with other particles (and with each other). It's hard enough just detecting powerful gravitational waves. If we have trouble seeing a veritable gravity tsunami, what hope do we have of catching a single droplet from that tsunami?

On the other hand, you don't need to go looking for gravitons to distinguish between uniform acceleration and a gravitational field. The equivalence principle is local: it essentially says that a sufficiently small chunk of curved spacetime looks approximately like flat spacetime. It's not possible for a real gravitational field to be perfectly uniform everywhere. So if you're permitted to make measurements over an extended region of space you can easily detect that the field isn't uniform.

For example, if you're in a 1 km tall spaceship in deep space, accelerating perfectly smoothly at 1 g, then the equivalence principle says that it feels just the same as if you're at rest on the Earth. However, on Earth if you ascend 1 km you will measure a slightly smaller gravitational acceleration, whereas in the ship, the acceleration will be the same up in the top section as it is in the lower section.

Actually, that example may be a little misleading, since in both cases a clock in the top section runs slower than a clock in the lower section, in accordance with the equivalence principle. Alfred Centauri explains this effect nicely in this answer. This was famously tested on Earth in 1960 in the Pound-Rebka experiment. A better illustration is when we drop 2 masses. On the rocket they both fall straight down. On Earth, they both fall towards the centre of the Earth, so their paths converge slightly, as illustrated in this anim, courtesy of Markus Pössel:

• Overall a great answer, thank you! However, I think your last example is wrong (although I may be wrong). From my understanding, the equivalence principle dictates that the changing acceleration felt by moving up away from Earth is equivalent to experiencing jerk in the deep space spaceship. Essentially, any gravitational experience can be indistinguishably “recreated” with inertial motion. Please correct me if I am wrong though! – Jack Edwards Oct 23 at 22:31
• @JackEdwards Your second to last sentence is true only if an observer is allowed to observe one location at a time. If you put measuring instruments at the top and bottom of a 1km high building, they will record slightly different data. With a strong enough tidal force, the top and bottom of your body would have different accelerations. – Acccumulation Oct 23 at 23:28
• @JackEdwards There are various subtle issues here. Yes, you can get the same effect as the Earth's changing g by changing the acceleration of your ship. To detect that you're sitting on a planet & not in deep space your non-local measurements need to be over a large enough region (& of sufficient duration) that you can detect the tidal effects of spacetime curvature. But I guess my last paragraph above could be a little misleading, since even with uniform gravity / acceleration there are vertical time dilation effects that the equivalence principle says are equal. – PM 2Ring Oct 24 at 2:45
• @JackEdwards You may find this helpful: The elevator, the rocket, and gravity: the equivalence principle. Also see Clarifying possible misconceptions in the foundations of general relativity. – PM 2Ring Oct 24 at 2:54
• This. Gravitons are a feature of string theory. We have never detected a graviton, let alone a virtual one, so maybe they would violate equivalence. Either way it would be interesting but people make too big a deal out of principle theories. Relativity does not hinge on principle theories. They are just convenient starting points from which one happens to be able to derive the rest of the hitertho known consequences. M91 isnt going to disappear if you find your virtual graviton, any more than time will stop dilating if you do manage to triangulate a preferred frame. – Eelco Hoogendoorn Oct 24 at 8:53

In the theory of general relativity, the equivalence principle is the equivalence of gravitational and inertial mass, and Albert Einstein's observation that the gravitational "force" as experienced locally while standing on a massive body (such as the Earth) is the same as the pseudo-force experienced by an observer in a non-inertial (accelerated) frame of reference

Take string theories, which allow for the quantization of gravity, and have a graviton, and at the same time, can show that they are compatible with the classical (non quantized) theory of General Relativity , which is emergent, a reviw . These are two different mathematical models, each valid in its variable domains. For example thermodynamics is emergent from statistical mechanics, one does not expect or mix up laws/principles from one framework with the other. It is only conservation laws that are expected to hold for all frameworks.

• Ok, I think I follow but I want to clarify because it seems that if an observer knew they were undergoing acceleration, in inertial frames of reference gravitons could be detected while in non-inertial frames gravitons would not be detected. Doesn't that distinguish gravitational force from inertial forces? Or does the problem lie in the detection of gravitons? – Jack Edwards Oct 23 at 16:09
• The probability of detecting single gravitons, given the coupling constants is infinitessimally small. If one could detect them as we detect photons, yes, it would be a confirmations of the quantization of gravity. But there are other , more hopeful experiments in bulk that could show whether gravitons exist, as the LIGO experiment hopes. At the quantum level there is an in principle a detectable difference between gravity and accelerating inertial frame. – anna v Oct 23 at 16:25
• Also BICEP experiments are looking for the footprint of real gravitons bicepkeck.org – anna v Oct 24 at 6:44

I like John Rennie's and anna v's answer, and, I am just adding a little clarification in this case to emphasize one topic.

It is very important to distinguish between a static (GW or EM) field and a (GW or EM) wave. I do believe that this distinction is the answer to your question. Detecting gravitons will not help us to falsify the equivalence principle because:

1. the effects of gravity in your case are caused by the static gravitational field, which we describe by using virtual gravitons. These virtual gravitons are a mathematical model. Not real particles (Please note that there is a debate on this site whether these virtual particles can become real or not, but in this case, we use them as a mathematical model). They are used to describe the phenomenon of the effects of the static gravitational field, which includes the one (acceleration and the equivalence principle) in your example. The same goes for static EM fields, where we use virtual photons that describe the phenomenon of the effects of this static EM field.

https://en.wikipedia.org/wiki/Electromagnetic_field

https://en.wikipedia.org/wiki/Gravitational_field

1. There exist EM and gravitational waves, that are made up of respectively photons and gravitons. These are not virtual photons or gravitons. We already detected gravitational waves, though, this does not qualify as an experimental evidence for the existence of single gravitons.

https://en.wikipedia.org/wiki/Gravitational_wave

So the answer to your question is, that even if we would be able to detect single gravitons, this could not help to prove/disprove the equivalence principle, because this principle describes the effects of the static gravitational field.

In general terms, gravitational waves are radiated by objects whose motion involves acceleration and its change, provided that the motion is not perfectly spherically symmetric (like an expanding or contracting sphere) or rotationally symmetric (like a spinning disk or sphere). A simple example of this principle is a spinning dumbbell. If the dumbbell spins around its axis of symmetry, it will not radiate gravitational waves; if it tumbles end over end, as in the case of two planets orbiting each other, it will radiate gravitational waves. More technically, the second time derivative of the quadrupole moment (or the l-th time derivative of the l-th multipole moment) of an isolated system's stress–energy tensor must be non-zero in order for it to emit gravitational radiation.

Now please note that there is one more distinction to be made. This is a crucial point to your question. To create real gravitons, to create gravitational waves, you need a time dependent quadrupole moment (the second time derivative of the stress-energy tensor needs to be nonzero). If, and only if your spaceship can do this, only then it could emit gravitons, and you might detect them. Now if you can make a spaceship, and it can move as a dumbbell, that tumbles end over end, it will emit gravitational waves, which are made up of gravitons, and you might detect them.

• As I say in my answer's comments,I think if one could measure gravitons the way one meauses photons, then accelerating masses would radiate gravitons but an accelerating inerial frames would not at the same rate. For example the pseudogravity of a rotating spaceship, even if given a shape for a quadrupole, would not give the same distribution of gravitons as an equivalent gravitational field from a mass. imo. – anna v Oct 24 at 18:06
• @annav thank you so much! – Árpád Szendrei Oct 24 at 19:19
• My last sentence should read :"would have a distribution of gravitons but an equivalent gravitational field from a mass would not be be radiating real gravitons. – anna v Oct 25 at 3:42

In the language of quantum field theory, the gravitational force is self-coupling, which means that gravitons carry gravitational charge and can feel the gravitational field.

In the language of general relativity, sufficiently small gravitational waves are perturbations of spacetime travelling in a curved background spacetime, and will follow null geodesics in that spacetime, which means that they will curve in that spacetime in exactly the same way light does.

• I understand that gravity is self-coupling, but I am still confused as to why detecting gravitons could not be used to distinguish gravitational acceleration from non-gravitational acceleration. Can you clarify, because there is clearly something I am still not understanding? Thank you! – Jack Edwards Oct 23 at 14:31
• @JackEdwards: because gravitational waves move in a gravitational field just like anything else, in exactly the same way? – Jerry Schirmer Oct 23 at 14:52

In a classical field theory, the motion of particles changes the field, as the field changes the motion of particles. In a quantum theory, the two sides are treated as the same phenomenon. Particles like photons and gravitons represent changes in the field due to the motion of 'matter' particles, and when you make a change in one bit of the field it propagates elsewhere, carrying momentum, until it is able to wiggle another material particle elsewhere. So really, a force-carrying particle is just another way of saying the field is in a configuration that applies a force to matter passing through it, as if it was a particle you could collide with.

Think of it like a water wave causing a cork to bob. The essence of the wave is that the water surface is not stationary, so a force is applied to anything floating on it. A sinusoidally moving water waves pushes on anything in it. But you can get the same sort of effect with a 'static wave', where the water is moving constantly in the same direction like a river. The water pushes the cork in the same way, but it's no longer an oscillating effect. You can think of the river as the limit of something like an infinitely slow changing, infinitely long wavelength wave. It's like a 'virtual' wave. In one sense, it's a mathematical fiction. In another, it's a perfectly valid description of wave physics taken to a particular extreme.

But the motion of a river is relative. You can think of it as either the water is moving past you, or you can be moving past the water. The physics is the same. In the same sort of way, a field can look different depending on your state of motion.

So in the case of gravity, a large mass bends spacetime so as to apply a force on matter floating on the surface of it, like water flowing down a drain. In the case of rotation, spacetime appears bent, like you're moving past the water instead of the water flowing past you.

To the extent that the curvature of spacetime is describable in terms of gravitons, both gravity from a mass and centrifugal force from rotation both constitute a 'graviton', a disturbance in the apparent flatness of spacetime, in the same sort of way. You can't distinguish them.

(Caveats - this is a very loose analogy, gravity doesn't actually work like a flowing fluid. And gravity isn't quantisable yet. I understand there was an early attempt to model it as the field of a spin 2 quantum particle dubbed a 'graviton', but it didn't quite work.)

• The question wasn't about the classical field theory, though? And you might want to expand on the part where an accelerating observer detects gravitons while an inertially moving one doesn't detect any at the same location. – user253751 Oct 26 at 10:02

This second entry in answer to:

Why can’t gravitons distinguish gravity and inertial acceleration?

I have found this article "Gravitational Waves: Sources, Detectors and Searches" which gives data (from a reference) for gravitational waves from a man made dumbbell, , given a rotation.

imagine a dumbbell consisting of two 1-ton compact masses with their centers separated by 2 meters and spinning at 1 kHz abouta line bisecting and orthogonal to their symmetry axis, as shown in figure 2.

Detection of the waves is not possible due to their very small amplitude.

When the dumbbell is not rotating there are no gravitational waves, just the newtonian gravity from its masses.

Detecting gravitational waves would tell if there was a rotation or not. One would be able to tell if the masses were in a non-inertial frame.

Thus, if gravitons exist and it were possible to detect single gravitons, if the two masses were part of a space station, there would be a way for outside observers of detecting whether there was rotation or not. Of course it would be much simpler to see this with photons, as a change of the location in space as a function of time.

Stop mixing concepts from general relativity and QFT. There is 0 experimental evidence for gravitons.

In a feynmann diagram gravitons are particles which when they are exchanged between two fermions or two bosons the result is gravitational attraction.