Is it just making up BS regarding the attributes imparted by various spin values, or is that a reasonable explanation of why gravitons are presumed to be spin-2 particles?
It's disappointing here, for sure.
No. The acceleration meter won't measure anything because there is nothing to measure. An object in free fall is in free fall; there is no "gravity" acting on it at all. It's just as if the object were floating out in deep space, far from all gravitating bodies. That's the point of the equivalence principle.
> If you had an acceleration meter entirely made out of the same magnetic substance, and you brought a magnet near it, would the acceleration meter register anything
Yes. Electromagnetism, weak, and strong interactions all make the acceleration meter register nonzero, even if the act "equally" on all parts of an object.
Put this way, isn't it almost begging the question? In GR the definition of acceleration is movement in contrast with the movement of gravity. If course gravity will never meet this criteria - all movement due to gravity will be aligned with movement due to gravity.
If instead we had a universe where instead of all matter having a gravitational effect, it was that all matter had a magnetic effect the we'd see no acceleration due to the magnetic effect and gravity would "produce a field" and cause acceleration in your above examples.
You can't use a gravitational biased tool to proclaim gravity is a neutral actor and everything else is a field.
It seems like more accurately, everything is "gravitationally charged", so instead we say it warps spacetime, but really is no different.
No, it isn't. You have it backwards. The definition of acceleration in GR is proper acceleration, i.e., what an accelerometer reads. The "movement" property is then a consequence of this plus picking an appropriate frame of reference.
> If instead we had a universe where instead of all matter having a gravitational effect, it was that all matter had a magnetic effect the we'd see no acceleration due to the magnetic effect
Yes, you would, because unlike gravity, magnetism does not obey the equivalence principle, so differently charged objects in the same magnetic field with the same initial conditions can have different motions. With gravity, all objects in the same field with the same initial conditions have the same motion, regardless of their mass. That is why it is possible to model gravity using spacetime curvature, and that property is unique to gravity.
I tend to gravitate toward the same line of thinking because of the existence of black hole charge limit.
I’m genuinely curious how that acceleration meter would work. There won’t be any internal forces as a consequence of the external field and no relative motion.
Look up how the one in your phone works. It reads nonzero when you are standing on Earth because of electromagnetic repulsion between your atoms and the atoms in the floor.
> There won’t be any internal forces
Yes, there will, because the object's internal state (including its shape, size, and internal stresses) when it is accelerated is different from its shape when it is in free fall. Why? Because the acceleration sets up internal forces in the object that result in a different equilibrium from the one it was in while it was freely falling.
It works because there's an external force pushing on the surface of the phone, and not equally on all its parts, which is the scenario we are discussing.
> Because the acceleration sets up internal forces in the object that result in a different equilibrium from the one it was in while it was freely falling.
And the cause of this is what you fail to explain.
Which then gets transmitted by the surface of the phone to the rest of the phone.
> not equally on all its parts
Yes, equally on all its parts, once you take into account that the parts of the phone can exert forces on each other.
> the cause of this is what you fail to explain
Cause of what? The internal forces? That's obvious: the distances between adjacent atoms change, and the electromagnetic forces between adjacent atoms are distance dependent. In other words, as I have already pointed out, the size and shape of the object changes when it is accelerated (in the sense of proper acceleration), compared to when it is not. Or, to put it another way, the equilibrium state of the object is different when it is accelerated than when it is not, with different inter-atomic distances and therefore different internal forces (as in, nonzero internal forces when accelerated, compared to zero when not).
But this is a very different situation than when you have an external force acting equally on all the atoms equally, and you will, as you correctly point out, have forces between the atoms. Of course in this case it is trivial, because you can easily measure the tension between different parts (for example).
But that is not the situation we are discussing. The situation we are discussing is when were are an external force on all parts equally. So there will be no tension that you can measure.
Gravity is not an "external force" in GR. In GR, an object moving solely under gravity, i.e., in free fall, feels no force (no internal stresses, zero reading on an accelerometer) because there is no force--not because there is "an external force acting equally on all the atoms".
If you want to say that the Newtonian interpretation, where gravity is "an external force acting equally on all the atoms", is indistinguishable experimentally from the GR interpretation, where gravity isn't a force at all, I suppose that's true. But then the "external force acting equally on all the atoms" is just like Carl Sagan's undetectable dragon in his garage. We have a model that works just as well without it, so it gets scraped right off by Occam's Razor. That is the GR position.
Also, no matter how you want to resolve the above issue, it still remains true that non-gravitational forces do not obey the equivalence principle, so there is no "spacetime geometry" interpretation for them that works. And those are also the cases where you do have internal stresses in the objects and an accelerometer reads nonzero. So again GR's interpretation--that these cases are forces while gravity is not, and that explains the difference in accelerometer readings--is simpler than the Newtonian one, where you have to argue that gravity is a "force" but doesn't work like other "forces" work.
I'm struggling to wrap my head around this assertion. If all parts of the object are acted upon "equally" (why is this in quotes?) where would this acceleration come from?
It is proper acceleration, not coordinate acceleration. An object can have nonzero proper acceleration even if none of its parts are in relative motion. Geometrically, proper acceleration corresponds to path curvature of the worldlines of the atoms in the object. "No relative motion" means all the worldlines of the object's atoms have the same path curvature (modulo some technicalities that don't really matter here). It does not require that that path curvature be zero.
Physically, a typical accelerometer works by measuring the internal stresses that are set up in an object when it is accelerated. These stresses put the object into a different equilibrium state than it was in when it was freely falling: the object's size and shape can change. For typical solid objects at typical Earthbound accelerations these changes are too small for us to see with our unaided senses--but sensitive instruments like accelerometers can detect them.
And to develop internal stresses, there needs to be some difference in the forces acting upon different parts of the body. Again, you are arbitrarily using free fall as your choice of a default state, against which you are comparing. If you instead chose being stationary on the ground, which would correspond to when you are in electromagnetic free fall, you would register an acceleration when in gravitational free fall.
It's the path curvature of worldlines in spacetime.
> the field of gravity
Only if the spacetime is curved. But worldlines can have path curvature even in flat spacetime.
> If you chose your worldline along the electromagnetic field instead, you'd see no "proper" acceleration when being pulled by the electromagnetic field and you would see "proper" acceleration due to gravity.
There is no such thing as "worldline along the electromagnetic field" as you appear to be using the term. A charged object in an electromagnetic field will have nonzero proper acceleration, as measured by an accelerometer. That's an invariant prediction; there is no alternate model in which it's any different.
> to develop internal stresses, there needs to be some difference in the forces acting upon different parts of the body
No, there doesn't. There just needs to be a difference in the shape and size of the body, from its state when under no external forces.
It is true that, if the body is large enough, the distribution of internal stresses in the body might not be uniform when the body reaches its equilibrium state under some externally applied force. For example, if the object is tall enough, the internal pressure at the top will be measurably less than the internal pressure at the bottom. But this is not due to any difference in external force being applied to the object. It's due to how the object adjusts itself to be in equilibrium under the applied external force.
> If you instead chose being stationary on the ground, which would correspond to when you are in electromagnetic free fall, you would register an acceleration when in gravitational free fall.
Again, this is just wrong. There is neither a valid theoretical model, nor any experimental data, to support this claim.
Meta: let's stop arguing what might happen, and do the experiment!
This can't be right, because an object moving solely under gravity is moving solely under its own inertia, but an accelerometer attached to it reads zero.
> do the experiment!
Do what experiment? Experiments showing that accelerometers attached to objects moving solely under gravity read zero, while objects subjected to non-gravitational forces read nonzero, have been done.
BTW I made this "comment" by "clicking" on the reply button and then I "typed" it using my "keyboard". No, I don't mean "keyboard" when I say "keyboard", hence the quotes.
I guess one answer to this is that particles which are (supposedly?) massless, like the photon, are affected by the space-time warping effects of gravity. A parallel construction wouldn't be true of magnetism or an electric field. Furthermore, when we detect gravity waves, they come at the same time as corresponding gamma ray bursts; since the gamma rays are affected by the space-time warping effects of gravity, this means that the gravity waves themselves are affected by the space-time warping effects of gravity.
So gravity probably is something else. But who knows!
If you do this in a capsule where the air is moving with you, and avoid hitting the ground, we call the sensation "weightlessness" or "zero g", like is experienced by astronauts in orbit.
Gravity is absolutely acting on astronauts orbiting the earth, at nearly the same strength as if they were standing on the ground. Depending on the shape of the orbit their linear speed may be increasing or decreasing, and they are definitely experiencing directional acceleration as their path bends in a circle around the Earth. But internally there is no bodily sensation of acceleration. It feels the same as floating, or free fall without the air rushing past.
https://www.npl.washington.edu/eotwash/equivalence-principle
This has to be true, because if you can't (internally) detect the difference between nothing and a planet being nearby, you obviously also can't detect how massive the planet is, so you can't know if your acceleration relative to the nearby environment is due to gravity or something else, or even how much "absolutel" acceleration you have.
Standing on earth (or the floor of your rocket) feels different due to electromagnetic effects of the nearby "touching" external objects.
It's a sensation that actually causes a decent percentage of people to feel physically ill for a period after launching to orbit on a rocket.
It's like going over the top of a hill on a roller coaster where you go from being pulled down into your seat to floating up against the restraints.
But once you're falling, or or in orbit (which is also free fall just outside an atmosphere), you don't feel the changes in velocity (acceleration) due to gravity.
If aliens have tech which can apply this type of force field / acceleration, then they won't get squished in their spaceships no matter how hard they accelerate. You basically need a large force field like gravity, instead of transferring force via small intermolecular force fields.
A feather falling the same speed as an elephant in the Earth reference frame is an example.
It is the same as other apperant forces like centrifugal force.
Even Newton himself said 'Hypotheses non fingo,' or 'I feign no hypotheses. '
It was a conjecture.
Put an electrical field generator on a falling metal ball and it will detect changing electric field though...
The reason the "g-meter" (e.g. a weight on a scale) doesn't move in the gravity case is that the weight is affected by the same field. The weight is a weight and feels gravity just like you and everything else in your environment does.
But by construction, you're imagining that the scale you have holds a different electrical charge than the object to which it's attached. Which is "normal" according to our everyday experience, but just an artifact of the way charges work on large objects (they distribute themselves on the "outside" of a conductive environment and everything inside tends to have a neutral distribution).
But that's just arbitrary. You could equally demand (in your gedankenexperiment, though doing this in practice would be very difficult) that your electrical charge be distributed just like the mass is, in which case the force measured would be zero too.
No, they don't. Gravimeters of the type you describe measure the coordinate acceleration of a freely falling test object in the accelerated frame of the gravimeter. In other words, it's the gravimeter (the part that isn't the freely falling test object) that has a force acting on it, which makes it accelerate upward (proper acceleration--an accelerometer attached to the gravimeter reads nonzero), and a freely falling test mass therefore appears to accelerate downward (coordinate acceleration in the frame of the gravimeter), just as if the gravimeter were inside an accelerating rocket out in deep space far from all gravitating bodies.
In other words, gravimeters of this type rely on the equivalence principle, which is the same principle that GR uses to justify the statement that gravity is not a force.
That really depends on construction of your g-meter. If instead of mass (ie. gravitational charge) you use electric charge in your accelerometer then that electric charge on the falling ball - ie. moving with acceleration - will generate EM wave thus providing clear detection of acceleration.
Wrt. the "boson" - gravity effects propagate with finite speed, i.e. wave, and the neutron in gravitational potential experiment shows that the gravitational potential/energy is quantized, and thus we have wave and quantized nature -> boson (wave packet/quant mediating interaction of a charge with the field).
https://en.wikipedia.org/wiki/Paradox_of_radiation_of_charge...
Curious interpretation, but beware this bit wasn't substantiated that well.
You can't.
Quick question, how would I even measure speed itself internally? I thought motion and speed was always measured relative to something else, why would an increase of these properties then have to be internal?
Not trying to be a dick, I’m genuinely curious. And yes, I obviously do not know anything about physics. Any explanation or link to a source to help me understand?
Yes, somehow actually. KK theories introduce an extra spatial dimension beyond our usual (3+1) which is postulated to be "compactified". This just means that it's curled up on itself at such a tiny scale that we don't directly perceive it. The way this extra dimension is curled and shaped affects the geometry of the overall 5-D spacetime. How it is connected to EM is that now these geometrical variations in the 5-D spacetime, when viewed from our 4-D perspective, manifest as the EM force and its associated field.
So you can say like in GR which have gravity arises as consequence of geometry, it is in KK that EM is consequence of geometry. However, the geometry and details are different.
> How does this relate to QFT?
Not much in the sense that they can provide useful information to each other. QFT describes forces in terms of interactions mediated by particles (e.g., photons for EM). KK, while primarily geometrical, give hints that perhaps these force-carrying particles can be associated with specific vibrational modes of the extra dimension. Of course KK theory only include EM and gravity. So we know for sure that we need to go beyond KK. This was the actual motivation for people to think about string theory to expand the original KK work.
You've just explained "compactified" in terms of being "curled up", but that doesn't really help (for me, at least). What does it mean for a dimension to be "curled up"?
To be honest, it is hard to visualize, as it is counterintuitive of what we think of space. While I know many people would disagree, I really like the garden hose analogy [1]. The idea is to simply imagine a very long garden hose. From a great distance, it looks like a one-dimensional line. Now you get closer, and realize it has a second dimension, which is its circumference curled around that seemingly 1-D line. An ant walking on the hose can move along its length, but also in a circle around it. The extra dimension in KK is like this circumference. It is tiny, curled up so we don't directly notice it, but still potentially there.
[1] https://www.preposterousuniverse.com/blog/2004/06/30/extra-d...
Both GR and QFT are insanely accurate models, but they are just models.
The N-body problem is undecidable, and Gödel, Turing, Church and other s proved that is the best we can do.
https://philsci-archive.pitt.edu/13175/1/parker2003.pdf
Western reductionism or Laplacian determinism is a good framework for practical, computable models. QFT actually is actually one of the counterexamples to Western reductionism.
But models are reductive and scientific models are just models. Don't confuse the map for the territory.
All models are wrong, some are useful; is another way of saying the same thing.
Superposition being inseparable is a large part on why the many words concept is popular with some people. It is about regaining a form of determinism.
Looking at gravity, we can compare what General Relativity and Quantum Mechanics say about the center of a black hole. Remember, both GR and QM have both been able to accurately model the way the world works at every scale we have been able to measure and test them. But they are incompatible with each other in certain points, such as the singularity in the black hole. GR says the center of the black hole is an infinitely dense point. QM says this can't be true because everything is made up of waves in a field, which requires things be spread out over some amount of an area. These can't both be true, yet GR and QM have both stood up to every single test and observation we can throw at them. Every prediction they make that we can verify has been verified and lines up with the theories. And this is not the only place they disagree, of course, but it is one example.
And that's really, from my understanding, the more fundamental answer to the question asked in the reddit post. It's not that unifying the two requires a gauge boson like the graviton, though that it is one possible outcome when quantizing gravity, but that we have two very useful and very tested models of how things work that are incompatible with each other in certain ways. Maybe gravitons exist, though it currently seems impossible for us to reach the point where we can detect them - Dyson calculated that using an Earth sized detector we'd be able to detect about one graviton from the sun per billion years, if they exist - and maybe they don't.
As for predicted things not expecting to exist in reality, this is really just par for the course for models. It's not like tensors are some real physical thing either, for example.
Approximately: consider gravity as small deviations from inertiality, so it's approximately linear, like the other forces we know and love.
A. When you incessantly hunt for every boson you can find anywhere, you end up with more of them.
Nor does the basic rationale for why we think gravity needs to be quantized involve a "gauge boson". It involves simple reasoning about how QM works. Say we have an experiment which puts an object with non-negligible stress-energy into a superposition of being in two different positions (for example, we make its position depend on the outcome of a spin measurement on a qubit). QM would say that spacetime would then need to also be in a superposition of two different geometries. But GR, as a classical theory, has no way to handle that. We would need a quantum theory of gravity, i.e., a quantum theory that can handle superpositions of different spacetime geometries.
The energy to move the object into a given position is an additional element here unaccounted for in your model. 2 different positions to move object into - 2 different energies (more specifically 2 different changes to the starting, before the experiment, stress energy distribution of the Universe). When corresponding moving energies (ie. their GR effects) are accounted for in those 2 cases it may as well be that those 2 cases are indistinguishable from the GR point of view, ie. those 2 supposedly different spacetime geometries happen to be the same. The superposition of 2 indistinguishable cases - it doesn't really matter is it superposition or not.
You can set the experiment up so the energy is the same in both cases (for example, both positions at the same height, just horizontally separated). If you don't, then yes, you have to include the effects of the different energies in your model.
> When corresponding moving energies (ie. their GR effects) are accounted for in those 2 cases it may as well be that those 2 cases are indistinguishable from the GR point of view, ie. those 2 supposedly different spacetime geometries happen to be the same.
I'm not sure how this would work if the energies were different, since "different" means a different source for the spacetime geometry.
But in any case, yes, for such an experiment to be relevant at all to the question I was discussing, the spacetime geometries being superposed would have to be different.
The action of placing them, say with your hands for simplicity, into different horizontal positions means differently pushing the Earth with your legs. For more cleaner illustration - let's say in our experiment a space ship is placed into orbit clockwise or anti clockwise. We can't just teleport the ship, so let's say we move it by rocket engines. So the ship goes in one direction, rocket engine exhaust goes in the other. The exhaust does have mass and speed. Even if it wouldn't eliminate the superposition gap, it will definitely decrease it, and decreasing the superposition gap increases the chances that some other unaccounted for factor(s) (for example gravitational waves caused by all these movements) will eliminate it or decrease further. Even if ultimately we still can't fully eliminate the gap, significantly decreasing it may eliminate various divergencies arising from quantization or make them very smallscale/localized (an observer from Alfa Centauri wouldn't care about the ship's orbit direction like we don't care about the spin of a given particle in the air around us) and average-able out on larger scales.
This might change the momentum, but not the energy if the heights are the same. But if your point is that there will always be some difference in a conserved quantity, yes, that's a fair point.
But it also means that there will always be some difference in the spacetime geometry. None of the other factors you talk about would eliminate the "superposition gap", because none of them cancel out any changes in the spacetime geometry; they just add more changes to it.
> average-able out on larger scales
But if you don't have a theory that can represent the variations you're going to average out, you can't do the averaging. That's the problem: classical GR cannot represent "variation in spacetime geometry" at all. It can only represent one spacetime geometry. It can't represent a superposition of them, not even to do an average.
But the forces and factors that work on the smallest particle should be the same forces that work on the largest of galaxies. If they are not, that's a completely different mystery and means our entire view of the universe is missing something substantial.
That is one key principle that drives the effort, yes. However, that doesn't mean things will always work out that way. Freeman Dyson, for one, published at least one paper making arguments for why gravity didn't need to be quantized.
> the forces and factors that work on the smallest particle should be the same forces that work on the largest of galaxies.
If you mean fundamental forces, then this is true (that's the definition of "fundamental"), but it also means that you have to adopt many levels of indirection between those fundamental forces and what actually happens with macroscopic objects. Or, to put it another way, the models we actually use to make predictions can have "forces and factors" in them that are not any of the fundamental ones, and that's fine, as long as we have some chain of reasoning that connects those models to the fundamental forces and factors. For example, our models of macroscopic objects can have dissipative forces like friction and viscosity in them; those aren't fundamental forces. But we have a chain of reasoning that connects them to fundamental forces (electromagnetic forces between electrons in atoms).
That's correct. However...
> That would imply that fermions are therefore the force carriers for gravity, not a hypothetical new boson.
That's wrong. Energy/mass-momentum (which can, as hughesjj points out, be bosons or fermions or both) is the source of gravity. The source is not the same as the "force carrier". (For example, in electromagnetism the source is charge/current, but the force carrier is the photon.)
In GR, gravity has no "force carrier" because it is not a force. In the simplest quantum model that has a "force carrier" for gravity, the quantum field theory of a massless spin-2 field, the "force carrier" is the massless spin-2 graviton, which is not the same as any source that occurs in ordinary matter.
No, it can't, because the EM interaction does not obey the equivalence principle, as gravity does. The geometric interpretation of gravity relies on the equivalence principle.
To state this another way: if I put two objects with different masses at a given point in spacetime and give them both the same initial velocity in the same direction, their paths through spacetime under gravity will be the same. But if I put two objects with different charges at a given point in the same electromagnetic field and give them both the same initial velocity in the same direction, their paths through spacetime will not be the same. And this remains true even if I add "extra dimensions" to "spacetime" along the lines of Kaluza-Klein theory, to represent the EM field.
Gravity is extremely non linear, when you look at the expression for Ricci tensor, it is much more complicated than other forces.