If gravity isn't a force, then why does it "need" a boson?
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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.
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!
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.
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.
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.
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.
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.
Put an electrical field generator on a falling metal ball and it will detect changing electric field though...
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.
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.
https://physics.stackexchange.com/questions/219306/if-gravit...
https://physics.stackexchange.com/questions/61899/why-do-we-...
Second one has a highly upvoted but controversial answer by Lubos Motl.
This question is similar, I think, to others like “why is there a speed limit in the universe” and “does light slow down in a material” that are I doubly muddled for science enthusiasts because (1) they are concepts that require deep domain knowledge, (2) even for some who have (1) they’ve been biased by decades of well meaning but harmful educational aides.
This is my expectation for every single "expert" answer on Reddit. Confidence without substance.
There's lots of people out there that mean well and have strong opinions and state things with confidence that they don't necessary know for truth.
I recall thinking that I don't critically examine what appears to be expert input from redditors in a field I know little about, but those same voices in areas I am an expert in are almost always wrong, or nuanced in some way to make them useless. I finally drew the line and realized it wasn't just my field that had wrong "experts" it was every field.
I think the final straw was when I asked a really complex question about a aberration in my telescope in the astrophotography subreddits and got very authoritative sounding responses that were far far off base, I realized it wasn't just software engineering.
I wonder how much of it is teens and young adults without a lot of experience giving advice that's the problem here. Probably not, but something to think about. That person giving you advice on how to navigate your relationship with your spouse might be a 13 year old who's never dated anyone before.
I miss the forum days. People knew what they were doing. Upvotes were supposed to save us and give us top notch content, instead it's now making us dumber.
I only have an EE’s understanding of physics, so just a tiny bit of modern physics and some semiconductor stuff. But I’m glad to have had the big “it’s just a model” reveal so I can worrying too much about what physics means.
This isn't an indictment of physics and physicists, it's just an example of how advanced domain knowledge is simplified for easier consumption of non-experts across basically every field deep enough to have these sorts of complex topics.
Which is, of course, typical of Lubos Motl. :-)
What his answer actually boils down to is the quantum field theory answer: the claim that (a) we need a quantum theory of gravity, and (b) the known quantum field theory of a massless spin-2 field is at least a good low energy approximation to whatever the final quantum theory of gravity will turn out to be. Both of these claims are believed to be valid by many physicists, but neither one actually has any experimental evidence to support it at our current state of knowledge.
You say "reductionist" but don't say how. You can't just just point and say "misleading" without backing that up. Now who's being reductionist :)
Is there a reason to take your opinion as being right? I am not an expert in physics, I don't want to study for the next 20 years to be one, how can I determine if you're correct, and that the top answer is indeed "highly reductionist to the point of misleading". Maybe _you_'re wrong and the other guy is right.
Five years of studies, at minimum...
Is this understanding wrong by generally accepted science?
Yes. It's also not relevant to the discussion in this thread. The properties of gravity under discussion here have nothing whatever to do with expansion of the universe.
I was hoping for some elaboration.
> The properties of gravity under discussion here have nothing whatever to do with expansion of the universe.
At least part of the discussion is about gravity being (caused by? understood as?) the curvature of spacetime. Why would matter curve spacetime, if not for the accelerated expansion of spacetime + matter resisting acceleration due to inertia?
I'm also OOTL on the boson part (remember, high school physics). I'm asking to hope to understand the subject better.
If you mean elaboration on how the expansion of the universe works, I can give that, but as I've said, it's irrelevant to this discussion.
> At least part of the discussion is about gravity being (caused by? understood as?) the curvature of spacetime.
More precisely, the curvature of spacetime around an isolated gravitating body. Which is a very different spacetime geometry from the spacetime geometry of the universe as a whole, which is what cosmologists use to account for the observations that lead us to say that the universe is expanding.
> Why would matter curve spacetime, if not for the accelerated expansion of spacetime
You have it backwards. What you are calling "the accelerated expansion of spacetime" is an effect of spacetime curvature, not a cause of it. Plus, as I said above, that spacetime curvature is the curvature of the universe as a whole, which is very different from the curvature around an isolated gravitating body, that we are discussing here.
> matter resisting acceleration due to inertia?
This has nothing to do with spacetime curvature at all. It would be present even if spacetime were flat.
The idea behind that is that, for all other interactions (electromagnetic, weak, strong), our Standard Model of particle physics tells us that there are "gauge bosons" (for electromagnetism it's the photon, for weak it's the W+, W-, and Z bosons, and for strong it's the eight gluons) that mediate the interaction. The full details of how this works involve quantum field theory and are way too long to go into here; all we need for this discussion is the basic idea that any interaction in quantum field theory has one or more "gauge bosons".
So if we are going to try to quantize gravity, the obvious way to do it--at least if you are a Standard Model particle physicist--is to treat it the same way and have a "gauge boson" for it. This has actually been done: the quantum field theory of a massless spin-2 field, which is what a "gauge boson" for gravity would look like, was developed in the 1960s and early 1970s. Furthermore, the classical limit of this quantum field theory is known to be General Relativity, i.e., the classical theory of gravity that we already have and that we already know works. So if we want to treat gravity as if it were a Standard Model interaction and give it a "gauge boson", this would be one obvious way to do it.
The reddit questioner was asking the more basic question of why we would want to do that at all--why do we need to quantize gravity. If we don't need to, the fact that we could do it using this known mathematical model (there are other issues that come into play with that, which one of the reddit responses mentions, but leave that aside) is irrelevant. Not all mathematical models have physical realizations.
Meaning, why is the spin-2 field massless? The easiest short answer is, because gravity is a long range interaction, and only massless gauge bosons can give you a long range interaction. That's why electromagnetism is also long range (the photon is massless) but the weak interaction is not (W+, W-, and Z are massive).
The strong interaction, unfortunately, breaks this simple heuristic (the gluon is massless but the strong interaction is short range), because it has other factors involved which aren't as easy to explain. But we know that those other factors don't come into play with the spin-2 field model of gravity, so we can still safely say that the graviton should be massless.
> If I understand GR correctly, the gravitational field itself (especially in the form of a gravitational wave) has energy
Only in a certain sense, which is...
> and therefore contributes to the mass-energy tensor, further bending spacetime.
...not this sense. The stress-energy tensor does not contain any "gravity" contribution. In the Einstein Field Equation, "gravity" is on the LHS, represented by the Einstein tensor, and "mass-energy" is on the RHS, represented by the stress-energy tensor. That has to be the case in order for local conservation laws to hold.
What all that means is that there is no valid local concept of "energy stored in the gravitational field" (because there is no tensor quantity that corresponds to this). But globally, you can still look at a system that emits gravitational waves and come up with a meaningful concept of "total energy" that decreases with time as gravitational waves are emitted. (This concept is called the "Bondi energy".) So we can say that "gravitational waves carry energy" in this sense. We just can't localize where that energy is.
One of the things I love about HN: You don't just get "you're stupid not to know that" (though you sometimes get that too). You also get real explanations from people that actually know things and take the time to explain it well.
You're welcome! Thanks for the kudos.
Why not just come out and state "Someone please offer me their valuable time to explain physics exactly at my level and at my pace. Thanks!"
Perhaps because that's just how matter and spacetime work?
I mean, you could ask the same question just about anything in physics. Why is there such a thing inertia at all, for example? You can try explaining things in terms of other things, but at some point you inevitably have to arrive at "it's just how the universe works".
The expansion of the universe is not responsible for explaining gravity, it was observed and aided in determining models and constants for the theory.
Start by looking at a particles property called mass. Mass interacts with the higgs field by exchanging a particle of information called the higgs boson. The interaction strength, corresponding to mass, determines how much the space time around our particle curves. Space time curvature dictates a particles movement. Now throw another particle with mass next to the initial one, also interacting with the higgs field. This is when you get into a huge loop of interacting movements and fields.
Locally and classically we approximate it by the classical law of gravity which loosely states that two masses attract each other, depending on how close they are, i.e. GMm/r^2
The universe expanding or contracting is a consequence of mass and gravotational fields, not the other way around. And calculating the expansion and thus curvature of the universe gives you a picture of the average mass distribution.
Gravity doesn't work on things without mass (or pretend mass) so Boson is critical to gravity's behavior?
Related PBS Space Time
The Higg's Boson gives mass to many particles; this might be part of what you're thinking of.
Whether gravity is described by bosons at its deepest level is strictly speaking an open question, though many models of quantum gravity treat it this way.
Gravity changes the path light has to take, photons keep going where they were going in the first place.
Light doesn't escape a black hole because it is being pulled by gravity, it is just being given an infinite path to travel so it never gets there.
We are actually "accelerating" upwards at a constant rate that is equal to what we feel and many think of as gravity. This is because spacetime is curved by the mass of the earth.
Think of a grid being distorted or squeezed / pinched. The distortion pinching is "gravity". Everything in the grid still travels in a straight path relative tot he lines on the grid but since the [spacetime] grid is pinched, it looks as if it is not on a straight path.
Because the scale that reads 100 lb is pushing up on the weight. The same would be true if the scale was inside a rocket accelerating at 1 g in deep space, far from all gravitating bodies. (This is called the equivalence principle.) In other words, the 100 lb of force you are calling "weight" is not a "force of gravity"; there is no such thing. It's a non-gravitational force exerted by the scale on the weight.
Or stars, planets, etc. wouldn't (and couldn't) exist.
As to the exact details of how we account for it is another matter. Calling gravity 'measured spacetime curvature' or 'gravitational force', or coming up with pseudo particles, or whatever, is an accounting method. Either way, it's still there.
But it does clearly exist, as much as light (for example) does. All one needs to do to prove that is look out the window.
There is no such force. Think of the accelerating rocket: there is no "force drawing the weight and the scale together". There is just the rocket pushing. The same is true standing on the Earth: there is just the ground pushing.
One of the reasons we need "spacetime curvature" in GR is to explain how the ground can be pushing things in different directions in different parts of the Earth, while all those things still remain stationary relative to the Earth's center and each other. But the whole point is that, even with all that factored in, there is still no "force drawing the weight and the Earth together". There is just the ground pushing up, plus spacetime geometry to account for the global configuration. That's it.
So if anything, gravity wise 'pushing' is not a property of gravity at all. Merely of matter which happens to produce the spacetime distortions which result in gravity. Unless dark matter exists, which would be neat.
Or the moon wouldn't continue to be up there, instead of 'down here'.
And clouds of hydrogen atoms will eventually collect together in a vacuum barring other outside gravitational influences. Which is notably why we even have a star.
And the weight itself also has measurable (albeit extremely tiny) such effects on everything else too.
So how do you model/name/account for that force which causes that to occur? Since it does clearly exist.
Because a rocket doesn't just 'push' either. We can clearly model the chemical and physics behaviors going on there to generate that 'push', and all involve forces which we can account for. None of them meaningfully appear to be gravity.
And gravity also involve forces (spacetime distortions causing very real effects!). Which we can also account for.
So seriously, what are you even talking about?
"Moves" is frame-dependent. In the weight's own rest frame, it is the Earth that moves.
What is not frame-dependent is the fact that, if there is no ground and the weight (wrong word, as we'll see in a moment) is freely falling, there is no "weight"--the object feels no force and an accelerometer attached to it reads zero. This was the basic insight that started Einstein on the road to curved spacetime and General Relativity: if an object falls freely, it will not feel its own weight.
> how do you model/name/account for that force which causes that to occur?
Using the word "force" here is already wrong as far as General Relativity is concerned: in General Relativity gravity is not a force.
> a rocket doesn't just 'push' either. We can clearly model the chemical and physics behaviors going on there to generate that 'push', and all involve forces which we can account for. None of them meaningfully appear to be gravity.
Yes, exactly. Now apply the same principle to the Earth pushing up on the weight. We can clearly model all of the microscopic behaviors that lead to that push and the forces that account for them--and none of them are gravity.
> gravity also involve forces (spacetime distortions causing very real effects!)
The "spacetime distortions" you refer to, which do indeed cause real effects, are not "forces". That's the whole point. They are spacetime geometry. Spacetime geometry is not a force. That is what General Relativity says.
> what are you even talking about?
I am talking about standard General Relativity, as it has been understood and taught in textbooks for decades now. What are you talking about?
These are merely the aforementioned caveats.
> It does not say that the Newtonian interpretation of that equation is correct.
If the interpretation were wrong, and that's not a force, then the amount of force in that equation would be 0.
With no force, Gm₁m₂/r² = 0.
However, that's not the modification that GR applies to this, though.
With no force, Gm₁m₂/r² = 0.
Nonsense. The GR interpretation is that G m1 m2 / r^2, when it is nonzero, is describing an effect of spacetime geometry, not a force. It can't be a force in GR because it isn't felt; an object moving solely under the influence of G m1 m2 / r^2 feels no weight--an accelerometer attached to it reads zero. GR does not change the numerical value of G m1 m2 / r^2 at all. It just reinterprets what the quantity represents.
It clearly exists (as a thing).
What we decide to call it, and how we account for it, while important, doesn't make it not exist in reality.
Despite the motte and bailey argument to the contrary.
Either way you'll feel like you and the elevator floor are being pushed together, and both you and the floor will experience some compression.
That was not the question, though. The earth isn’t pushing back.
When a rocket is accelerating and you are pressed against the back and being driven forward, is the rocket "pushing" you or "resisting" you? Either one is a fine way of describing it. The net result, though, is that you are squished into the back of the rocket in a way that's completely indistinguishable from gravity.
You may say "but the ground can never push me away from the ground." Sure, but the back of the rocket can't push you away from the back of the rocket either, and yet it is clearly pushing you through space. So long as the rocket keeps accelerating, you are being pushed forward, yet you won't feel it as a push, you'll feel it as an attraction to the back of the rocket.
Yes, it is. The force you feel as weight is the ground pushing on you.
By Newton's Third Law, the ground also feels a force of equal magnitude and opposite direction from you, which can, as you say, end up compressing the ground. But that isn't what you feel as weight.
You will start to sink at various rates. Not because water is “pushing less” than earth. It’s because water is not resisting.
Also, if I am 200 pounds and the earth is pushing up on me with 200 pounds of force, what is generating that force? And where does it go if a helicopter reels me in? The 200 pounds up “force” just vanishes?
Or never existed.
Um, "is not resisting" means "pushing less". The water does not exert as much force on you as solid ground would. That's why you sink. If the water was not pushing less, you wouldn't sink.
> if I am 200 pounds and the earth is pushing up on me with 200 pounds of force, what is generating that force?
Nothing has to "generate" the force. The force is a static force that is doing no work, so no energy is being consumed. It's just static electromagnetic repulsion between your atoms and the atoms in the ground.
> where does it go if a helicopter reels me in?
If you stop making contact with the ground, there is no longer any electromagnetic repulsion between your atoms and the atoms in the ground.
> The 200 pounds up “force” just vanishes?
The "up" force from the ground vanishes, yes. See above.
But of course there is another "up" force on you now from whatever is attaching you to the helicopter.
I suggest to GP to consider the body lifted and then released. The body is moving back towards the larger body. It meets a rigid boundary and comes to rest (or bounces ..) When did the ground start "pushing" the body? It seems the smaller body was being acted on by something (we say curved space) which is motivating it towards the center of the gravity of the larger body. This motivation (spatial deformation or force, whatever) doesn't cease and the rigidity of the larger body is engendering the equal opposing 'force' pushing back against it.
Yes. You and the ground are being pushed upwards by a force. You can feel that force--it's your weight. The dropped object falls relative to you precisely because it does not feel any force--it is weightless. This is exactly the same thing that would happen in a rocket whose engine was firing in deep space, far from all gravitating bodies. This is an example of the equivalence principle, which is a central experimental fact that Newtonian gravity just had to accept as being true for no apparent reason, but which GR accounts for naturally as an aspect of spacetime geometry.
According to relativity, energy IS mass and Higgs field gives non zero energy to Higgs boson.
Radio waves are relatively low in negative mass along with photons.
Photons (a.k.a energy, including radio waves) do not interact with the Higgs field and so have zero mass.
There is no negative Higgs field, and no negative mass.
There's also no ether either as originally constructed (it was ruled out by experiment) but you could be speaking metaphorically.
When the velocity is the speed of light, solving the equation of special relativity yields m of zero.
We have a successful theory of quantum mechanics which integrates with special relativity, so that definition holds, even if you take the same approach from an Effective Field Theory perspective.
However, a photon has energy, so therefore it must have not-at-rest mass. The article doesn't really comment on not-at-rest mass. We know photons interact with gravity because of gravitational lensing.
The article doesn't have to comment on not-at-rest mass because the math is clear.
As for interacting with gravity, photons move through spacetime, and general relativity shows that gravity is the curvature of spacetime, and photons move on the shortest geodesic for that spacetime.
Fun fact from my "Modern Physics" homework: a system of two photons can have mass, if they're traveling in different directions. In parallel though, they're still massless.
To be clear, it's just a reformulation of the math involved. Both formulations of GR/SR with and without variable mass create the same predictions. It's just that it keeps the concept of mass more coherent to stick with "mass is the energy of a particle at rest".
https://en.wikipedia.org/wiki/Michelson%E2%80%93Morley_exper...
You are only behind physics community by ~140 years [1]. We know for sure that there is no "ether" so its displacement is not what causes gravity. It was a trial to explain how light would move if it is wave, which we didn't know about EM waves then.
[1] https://en.wikipedia.org/wiki/Michelson%E2%80%93Morley_exper...
"More careful reflection teaches us however, that the special theory of relativity does not compel us to deny ether. We may assume the existence of an ether; only we must give up ascribing a definite state of motion to it, i.e. we must by abstraction take from it the last mechanical characteristic which Lorentz had still left it. We shall see later that this point of view, the conceivability of which I shall at once endeavor to make more intelligible by a somewhat halting comparison, is justified by the results of the general theory of relativity ... according to the general theory of relativity space is endowed with physical qualities; in this sense, therefore, there exists an ether. According to the general theory of relativity space without ether is unthinkable."
Tidal forces are only measurable across something with size, by measuring a difference between one point and another.
Even if your body is being spaghettified, an arbitrarily small region of your body still can't tell that it's under anything but freefall.