Quantum gravity vs. general relativity

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Let's suppose gravitons exist, and you have a machine that is 100% effective at detecting them. If you were in a room with no windows, and there is an apparent gravitational field, then would using this machine let you tell if you were in a gravitational field or accelerating?
 

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  • #2
tom.stoer
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I don't know, because I don't know what a graviton really is.

Counterquestion: Let's suppose photons exist, and you have a machine that is 100% effective at detecting them. If you were in a room with no windows, and there is an apparent electromagnetic field, then would using this machine let you tell if you were in a electromagnetic field or accelerating?

This has been answered by the so-called Unruh-effect which states that an accelerating observer measures thermal photons (and other particles).
 
  • #3
marcus
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https://www.physicsforums.com/showthread.php?t=242145

Steady acceleration would correspond to a constant gravitational field and
that means virtual gravitons, not detectable as particles.

===============

As Tom.S says, same thing with photons. If you just have a static electric field in the room, say the ceiling is more negative and the floor more positive (to get something approximately analogous to gravity) then no matter how good your detector is it will not see photons.

You see photons when there is an electromagnetic wave---a change in the field.
 
  • #4
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https://www.physicsforums.com/showthread.php?t=242145

Steady acceleration would correspond to a constant gravitational field and
that means virtual gravitons, not detectable as particles.

So what you're saying is that gravitons only exist for gravitational waves not in a static gravitational field, right? That doesn't sound like a complete description of gravity, does it?
 
  • #5
marcus
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So what you're saying is that gravitons only exist for gravitational waves not in a static gravitational field, right? That doesn't sound like a complete description of gravity, does it?

In my post I distinguished between gravitons and virtual gravitons, friend.

The original poster, kashiark, imagined having a perfect graviton detector. My claim is that such a detector, no matter how perfect, would not pick up virtual gravitons. No more does a photon detector register the virtual photons which mediate a static electric field. This was by way of responding to kashiark's question.

Could you detect the putative virtual gravitons mediating a static gravitational field? I believe even in a thought experiment you could not.
=====================

The issue of giving a complete description of gravity is a different topic, I think.

I believe that the graviton is a useful mental construct which appears in the mathematics when the background space is nearly flat, or has some other suitable fixed geometry. I don't think one would expect to use gravitons to analyze a highly dynamic, high curvature situation. Like the collapse of a star to form a black hole. I would not advise anyone to consider gravitons as providing a fundamental description of gravity's reality.

That's just my personal viewpoint. I think of them as small ripples in a nearly flat geometry. The quanta associated with models of grav waves. Limited applicability but very useful for special purposes and valid within the appropriate context.

No reason to suppose that individual gravitons will ever be detected, as Freeman Dyson has famously pointed out. But that's a completely separate issue.

A fundamental description of gravity (according to a common view which I share) has to be a fundamental description of spacetime geometry. Useful as they might be in certain situations, the graviton does not offer a complete description of geometry. So, as you suggest, it cannot provide a complete description of gravity.

==========

But this wasn't what kashiark was asking about. He was talking about using a "graviton detector" to distinguish between grav and acceleration. (so they wouldn't really be equivalent!) Now he has been answered by various people. Let's see what he says, and what else, if anything, he wants to talk about :biggrin:

BTW did you read what Janus said in that thread I linked to?
https://www.physicsforums.com/showthread.php?t=242145
 
  • #6
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ok so let's say it doesnt detect virtual gravitons then you could tell if you were accelerating or in a "real" gravitational field disproving the little adage of general relativity "no local experiment can differentiate between acceleration and a gravitational field"
 
  • #7
gel
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ok so let's say it doesnt detect virtual gravitons then you could tell if you were accelerating or in a "real" gravitational field disproving the little adage of general relativity "no local experiment can differentiate between acceleration and a gravitational field"

I think you misunderstood the previous posts. Gravitions making up a static gravitational field would be virtual, so no, it wouldn't detect the difference. Real, non-virtual, gravitons would be present if you have a gravitational wave.
 
  • #8
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ah yes you're right i knew i must have been missing something thanks :D
 
  • #9
tom.stoer
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marcus, this is a dangerous concept. It seems to me that you are mixing the idea of gravitons as small ripples on spacetime with the detector's ability to detect these small ripples but not static fields. If your detector is not able to detect static fields that does not necessarily mean that static fields are not "gravitons" in some sense!

Let's start with QED: QED in an Hamiltonian, gauge-fixed formalism contains a differential operator D. Inverting this operator leads to the Coulomb potential between charges and currents; that means the interaction term in the Hamiltonian density contains something like h(x) = j(x) V(x,y) j(y) which is formally V = 1/D; 1/D is a nonlocal object.

Now let's continue with QCD. The formalism is much more involved, but the central step is the same. You have a differential operator D; inverting this operator gives you the "static" chromo-electric field in h(x). In QCD other interaction terms will appear, of course. Why QCD? Because due to its non-abelian nature D = D[A], that means D[A] depends on the gauge fields!!! That's why it is tricky to calculate 1/D[A], but formally this operator exists. This shows that the chromo-electric field consists of gluons, a fact that is hidden in QED due to its abelian nature.

So what I am saying is that even static fields are to be derived from the dynamic gauge fields.

Of course your restriction to small ripples leads to gauge fields as "plane waves plus interaction terms" and therefore this difficulty disappears. Nevertheless it does not explain what gravitons are. The perturbative concept of gravitons fails because of non-renormalizibility of GR. Non-perturbative concepts like LQG (you know this much better than me :-) are hardly able to tell you what gravitons are. So what are gravitons?
 
  • #10
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Just a quick question regarding gravitons.

If they do in fact exist, exactly how big would one be? It seems that a gravitational field is a steady state wave, unlike the point-to-point nature of something like a photon.

Wouldn't the wavelength of the sun's gravitons be something like twice the size of the solar system?
 
  • #11
tom.stoer
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As the sun's gravitational field is static, there are no waves at all.

Regarding frequencies and wavelength I found the following data (w/o calculation):
- two merging stellar black holes: ~1mHz
- two circulating stellar black holes: ~10mHz
- core collaps SN = SN II: ~10kHz

You should check the GEO600, LISA and LIGO websites
 
  • #12
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My claim is that such a detector, no matter how perfect, would not pick up virtual gravitons. No more does a photon detector register the virtual photons which mediate a static electric field.

Are they not measurable because they are not moving? How could you measure particles if they don't move through your detector machine, right? We know that gravitons from graviational waves are moving with the wave. But static gravitons are not moving and so are not measurable and thus considered virtual, right?

Like virtual electrons and positrons, they pop into existence and then almost immediately they disappear in the same place. So how would you detect them, they don't exist long enough to move through any machine that would register them. That's why we can not directly measure the zero point energy, right?

But positrons and electrons annihilate each other as they pop into and then out of existence. Would virtual gravitons also pop into and then out of existence? What would they annihilate with to pop back out of existence?
 
  • #13
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As the sun's gravitational field is static, there are no waves at all.

Regarding frequencies and wavelength I found the following data (w/o calculation):
- two merging stellar black holes: ~1mHz
- two circulating stellar black holes: ~10mHz
- core collaps SN = SN II: ~10kHz

You should check the GEO600, LISA and LIGO websites

That's what I was trying to say. The Sun's gravity isn't propagating outward; it's just there. The curvature of space describes a waveform, but it's not going anywhere.

Why should we expect to find gravitons at all? It seems like we would have a better chance at detecting "Magnetons" from refrigerator magnets.
 
  • #14
tom.stoer
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You still have problems with the concept of gravitons.

If you think of gravitons in terms of waves and fields, then the sun's "gravitons" are static and do not propagate; I doubt that many peoplewould call this "graviton".
If you think of gravitons in terms of particles described by propagation and annihilation operators (second quantized gravity), then they are always propagating with the speed of light; unfortunately the theory becomes inconsistent!

So everybody should specify what is meant by gravitons.

Regarding the detector: it is made of atoms; the sun is made of atoms, too; two atoms are interchanging a graviton; this is a "virtual graviton" according to standard concepts of QFT; so the detector will detect virtual gravitons ONLY. What I am saying is that by construction asymptotic states are the real states, whereas exchanged particles are virtual (they are off-shell). As soon as a particle is absorbed and detected, it is no longer a real particle but a virtual particle.

=> I think this distinction is fine for certain mathematical concepts but missleading for the description of many physical processes, especially if static fields are involved.

If you want to compare gravitons with other particles I would not use electrons. Photons and gluons are much more similar to gravitons, except for the fact that gravitons lead to inconsistent quantum theories.
 
  • #15
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You still have problems with the concept of gravitons.

I do. I recognize that there is a need for something to fill in the gap of understanding gravity, but gravitons seem like a placeholder at best.

Photons; they do something. Object A radiates energy outward, and when it hits object B, that quantized energy is registered as a photon. That's an idiot level simplification, but it is possible to talk about what photons actually do and how they behave in plain english.

Gravitons sound like magic. What is the commonsense explanation for how object A emits a graviton towards object B, and when it gets there it says, "Hey, come over here." Especially when gravitational fields are for the most part stationary.

Does anyone have a reasonable explanation of "how" gravitons mediate the gravitational force, or are they simply self-defined?
 
  • #16
nrqed
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I do. I recognize that there is a need for something to fill in the gap of understanding gravity, but gravitons seem like a placeholder at best.

Photons; they do something. Object A radiates energy outward, and when it hits object B, that quantized energy is registered as a photon. That's an idiot level simplification, but it is possible to talk about what photons actually do and how they behave in plain english.

Gravitons sound like magic. What is the commonsense explanation for how object A emits a graviton towards object B, and when it gets there it says, "Hey, come over here." Especially when gravitational fields are for the most part stationary.
why is this more magical than a photon going from a positive charge to a positive charge and saying "hey, move away" or going from a positive charge to a negative charge ans saying "hey! come over here" ?
 
  • #17
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why is this more magical than a photon going from a positive charge to a positive charge and saying "hey, move away" or going from a positive charge to a negative charge ans saying "hey! come over here" ?

Well, photons are affected by gravity, as demonstrated by gravitational lensing and black holes. Do photons exchange gravitons with whatever gravitational source they find themselves affected by? That sounds unlikely.

Do black holes emit gravitons? If so, how? Light cannot escape. What is the mechanism for graviton emission from a black hole?
 
  • #18
tom.stoer
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nrqed is right. The main problem with gravitons is not that they are more mystical than photons. The two main problems are simply:
- they have not been detected (no gravitational wave has been detected directly, so far)
- if used for quantization they lead an inconsistent theory

If those two issues were absent gravitons and photons would be quite similar, except for the fact that a raviton is spin-2 and therefore the force is always attractive.

You can compare "plane wave photons" with "plane wave gravitons" and you can compare "static photons", e.g. the Coulomb potential with "static gravitons". More or less the same idea.
 
  • #19
nrqed
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Well, photons are affected by gravity, as demonstrated by gravitational lensing and black holes. Do photons exchange gravitons with whatever gravitational source they find themselves affected by? That sounds unlikely.
Why does it sound unlikely?
According to GR, any stress energy momentum tensor couples with gravity, so we would simply have the corresponding tensor of any theory (whether it is QED or anything else) coupled to gravitons.

Do black holes emit gravitons? If so, how? Light cannot escape. What is the mechanism for graviton emission from a black hole?
They don't emit gravitons anymore than an electron at rest emits photons (even though there is a static electric field). We have to wiggle the electron to get photons produced. We have to wiggle a black hole to get gravitational waves emitted (and not any wiggling will do)
 
  • #20
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nrqed is right. The main problem with gravitons is not that they are more mystical than photons. The two main problems are simply:
- they have not been detected (no gravitational wave has been detected directly, so far)
- if used for quantization they lead an inconsistent theory

If those two issues were absent gravitons and photons would be quite similar, except for the fact that a raviton is spin-2 and therefore the force is always attractive.

You can compare "plane wave photons" with "plane wave gravitons" and you can compare "static photons", e.g. the Coulomb potential with "static gravitons". More or less the same idea.

Okay, so how do gravitons reconcile with black holes again? How does the matter inside the event horizon of a black hole exert a gravitational influence on matter outside the event horizon of the black hole if gravitons are the mediating particle? How would such a particle escape? <--- The recursive nature of this question makes my brain hurt.
 
  • #21
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They don't emit gravitons anymore than an electron at rest emits photons (even though there is a static electric field). We have to wiggle the electron to get photons produced. We have to wiggle a black hole to get gravitational waves emitted (and not any wiggling will do)

I understand this, and it's exactly why I don't see how gravitons usefully explain anything about the nature of gravity. Sure, if you move a gravitational field relative to something else you get a waveform that can be quantized into a particle at some level.

That still doesn't explain how gravity actually works. Isn't that the point of speculating the existence of these particles? Are we trying to figure out how the engine works, or just what color the paint happens to be?
 
  • #22
nrqed
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I understand this, and it's exactly why I don't see how gravitons usefully explain anything about the nature of gravity.

I am not sure what you have in mind by "explaining the nature of gravity".
In your language, do photon "explain" the nature of electromagnetism?

What do you mean by "explaining how a force works"? What is your criterion?

Gravitons are just a byproduct of quantizing the gravitational force using quantum field theory.

Sure, if you move a gravitational field relative to something else you get a waveform that can be quantized into a particle at some level.

That still doesn't explain how gravity actually works. Isn't that the point of speculating the existence of these particles? Are we trying to figure out how the engine works, or just what color the paint happens to be?


Gravitons can be used to calculate quantum corrections to classical GR, that's all.
 
  • #23
tom.stoer
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I understand this, and it's exactly why I don't see how gravitons usefully explain anything about the nature of gravity. ...
That still doesn't explain how gravity actually works.

I agree. Gravitons do not really explain how gravity works. As we just figured out, gravitons are something like a concept taken from a perturbative treatment of conventional quantum field theory.

This concept has failed. Nobody is able to recover full non-perturbative physics from perturbative concepts only, neither in gravity nor in QCD. In gravity it's even worse because the introduction of gravitons (perturbation theory) breaks background independence; that means the geometry is no longer able to evolve dynamically; only small ripples are able to propagate.

If you would have a full understanding of QG, you might be able to go to a certain limit of that theory where gravitons turn out to be a useful approximation. The full theory may be LQG or string theory; neither starts with gravitons as fundamental objects.
 
  • #24
marcus
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In line with what you said, in LQG spinfoam papers where they derive and study the graviton propagator they must first restrict to the case where they have an approximately flat background. This is difficult because the theory is essentially backgroundless, so it has taken some ingenuity and effort to force restriction to a flat background case---and so get gravitons. Gravitons are somewhat artificial in the theory. Or they emerge only in the appropriate circumstances. I think this agrees with what you said.

I will get some links to a sampling relevant papers. Rovelli co-authored of most, or if not he then Simone Speziale. I should be able to find a few just by listing articles by those authors since 2005. Here are a few:

http://arxiv.org/abs/gr-qc/0604044
Graviton propagator in loop quantum gravity
Eugenio Bianchi, Leonardo Modesto, Carlo Rovelli, Simone Speziale
(Submitted on 10 Apr 2006)
We compute some components of the graviton propagator in loop quantum gravity, using the spinfoam formalism, up to some second order terms in the expansion parameter.
Comments: 41 pages, 6 figures

http://arxiv.org/abs/gr-qc/0608131
Group Integral Techniques for the Spinfoam Graviton Propagator
Etera R. Livine, Simone Speziale
(Submitted on 30 Aug 2006)
We consider the proposal of gr-qc/0508124 for the extraction of the graviton propagator from the spinfoam formalism. We propose a new ansatz for the boundary state, using which we can write the propagator as an integral over SU(2). The perturbative expansion in the Planck length can be recast into the saddle point expansion of this integral. We compute the leading order and recover the behavior expected from low-energy physics. In particular, we prove that the degenerate spinfoam configurations are suppressed.
Comments: 16 pages

http://arxiv.org/abs/gr-qc/0508124
Graviton propagator from background-independent quantum gravity
Carlo Rovelli
 
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  • #25
tom.stoer
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Gravity is certainly the most complex theory, especially because of diffeomorphism invariance the the missing PT expansion. Even in well-understood theoris like QCD the plane waves are a very poor approximation or essentially useless in studying non-perturbative effects like confinement and chiral symmetry breaking.

The main benefit of plane wave states in conventional QFT was that calculation of scattering amplitudes was "simple" (ever tried to calculate a two-loop amplitude in QCD?). This benefit is lost in gravity as the result of the calculation is meaningless. The reason to study this approximation is lost, too, simply because of the missing graviton colider experiments.
 
  • #26
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I am not sure what you have in mind by "explaining the nature of gravity".
In your language, do photon "explain" the nature of electromagnetism?

No, photons don't explain the nature of electromagnetism, but they demonstrate the properties of electromagnetism. By empirically observing them and their behavior, we can infer quite a bit about the nature of spacetime.

What do you mean by "explaining how a force works"? What is your criterion?

Isn't this the ultimate goal of studying physics? I consider the fundamental forces to be the emergent properties of some even more fundamental underlying structure. Once that underlying structure can be accurately described, I'll have a satisfactory answer. Of course, then I'll want to understand what makes that tick.

Gravitons can be used to calculate quantum corrections to classical GR, that's all.

Thank you.
 
  • #27
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Gravity is certainly the most complex theory, especially because of diffeomorphism invariance the the missing PT expansion. Even in well-understood theories like QCD the plane waves are a very poor approximation or essentially useless in studying non-perturbative effects like confinement and chiral symmetry breaking.

Many years ago (published in 1985) we demonstrated how one can estimate the ground state energy (meson masses, for example) starting from a free motion and considering the binding quark potential as a perturbation in case of confinement. If you are interested, I can give details.

Bob_for_short.
 
  • #28
tom.stoer
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Tell me more!
 
  • #29
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Tell me more!

Ok, here it is. Let us consider a 3D non singular potential that can model the confinement phenomenon. For example, a 3D oscillator potential V(r)=mω2r2/2. The Schrodinger equation has exact solutions in this case. The lowest energy is equal to (3/2)ω.

Let us consider the Green's function G(r,t|r',t'). Its spectral representation is known:

G(r,t|r',t') = ∑ψn(rn(r')e-iEn(t-t') (0)

I will consider a particular case of G(r,-it|0,0) that I denote as G(r,τ). Its equation is the following (kind of a diffusion rather than a wave equation):

{∂/∂τ - ∇2/2m + V(r)}G(r,τ) = δ(r)δ(τ) (1)

G(r,τ) does not oscillates but decays when τ increases.

Let us now introduce the Green's function of free motion equation G0(r,τ):

{∂/∂τ - ∇2/2m }G0(r,τ) = δ(r)δ(τ) (2)

It is equal to G0(r,τ) = (m/2πτ)3/2e-mr2/2τ

From equation (1) one can obtain G(r,τ) by the perturbation theory where a perturbation is the non singular potential:

G(r,τ) ≈ G0(r,τ){1 - u(r) + w(r) - ...} (3)

where u(r) and w(r) are some integrals.

These integrals are simplified if r=0. Then u(0)=u=(ωτ/2)2, w(0)=α⋅u2, α=19/30 for the 3D oscillator. The functions u, w, and the coefficient α can be calculated exactly for many non singular potentials.

However the expansion (3), especially if truncated, is good only for small τ. At large times τ the series {1 - u + α⋅u2 - ...} diverges as the highest degree of time in it.

In order to extrapolate G(0,τ) from (3) to finite and large times, we can apply a non-linear series summation. It is kind of summation of divergent series. We can use a positive Padé approximant [0/2]:

{1 - u + α⋅u2} ≈ 1/{1 + u + (1-α)⋅u2} (4)

This approximation represents well the function even at large τ. But at large times the Green's function is also well approximated with only one spectral term - the lowest level contribution:

G(0,τ) ≈ |ψ0(0)|2e-E0⋅τ (5)

Taking the logarithmic derivative of G(0,τ) with help of (4) at big times, one obtains an estimation of E0. For the 3D oscillator we obtained E0 ≈ (2.62/2)ω.

We used also a Sommerfeld approximation:

{1 - u + α⋅u2} ≈ (1+A⋅u)B, A=2α-1, B=1/(1-2α) (6)

and we obtained E0 ≈ (3/2)ω.

Of course, one should not choose too large time τ for calculating the logarithmic derivative of G(0,τ): any approximation gets worse when τ→∞. The calculation point should be finite: just sufficient to use the one spectral term approximation (5) of the sum (0).

It is a very brief explanation of how one can do it.

It was published in the Soviet Journal of Nuclear Physics in 1985, V. 41, N.2, under the title:

"On sum rules for non-singular potentials in quantum mechanics."

Bob_for_short.
 
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  • #30
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Of course, such a program can be only realized for non singular potentials for which the Green's function exists and converges well to one term (5) at large times.

Bob_for_short.
 
  • #31
tom.stoer
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That seems to be rather interseristing for special cases in QM. But how does it affect my statement "even in well-understood theories like QCD the plane waves are a very poor approximation ..."?
 
  • #32
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Plane waves are the initial approximations for PT calculations. My simple QM example may serve to estimate the meson masses from PT calculations. Concerning three-quark baryons, it is probably harder but I think it is still possible.

Of course, the plane waves without PT corrections are poor approximations.

Bob.
 
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  • #33
Buckethead
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Let's suppose gravitons exist, and you have a machine that is 100% effective at detecting them. If you were in a room with no windows, and there is an apparent gravitational field, then would using this machine let you tell if you were in a gravitational field or accelerating?

I'm being (somewhat) whimsical here, but a simple bathroom scale can be used to detect gravitons. If there is a reading on the scale, you have gravitons, if not, you don't. The interesting thing about this (whimsical) situation is that if the scale is accelerating, it turns out to display a reading, which would imply that acceleration generates gravitons!
 
  • #34
tom.stoer
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I think we are mixing different things:
a) are gravitons only small quantum exitations of the gravitational field?
b) or is a static field composed of gravitons?
c) if b) does acceleration generate gravitons?

I think that gravitons in the sense of a) are a very special concept that has been shown not to produce a viable theory; therefore I think b) is the rigth answer. And that means that I have to admit that c) is correct, too.

Neverthelesee, I still think we that don't agree what gravitons really are.
 
  • #35
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I think we are mixing different things:
a) are gravitons only small quantum exitations of the gravitational field?
b) or is a static field composed of gravitons?
c) if b) does acceleration generate gravitons?

I think that gravitons in the sense of a) are a very special concept that has been shown not to produce a viable theory; therefore I think b) is the rigth answer. And that means that I have to admit that c) is correct, too.

Neverthelesee, I still think we that don't agree what gravitons really are.

I forgot the name of it, but there is a theory that proposes acceleration is quantized. It is used to explain dark matter, and the Pioneer annomoly. So if gravity is equivalent to acceleration, then are gravitons a form of quantized acceleration in general?
 

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