But the acceleration meter won't measure any force because gravity is acting on every part of it uniformly. 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, or would it read zero acceleration, since all parts of it were being acted on uniformly (and thus didn't "notice" any acceleration)?
> the acceleration meter won't measure any force because gravity is acting on every part of it uniformly
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.
> The acceleration meter won't measure anything because there is nothing to measure.
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.
> 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.
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'm sure you are familiar with Kalusa-Klein theory (knowing that it is incomplete and/or doesn't describe our universe), don't you think that motion in that theory is less bound to gravity alone like the parent comment suggests because the notion of proper time is different?
I tend to gravitate toward the same line of thinking because of the existence of black hole charge limit.
> Yes. Electromagnetism, weak, and strong interactions all make the acceleration meter register nonzero, even if the act "equally" on all parts of an object.
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.
> I’m genuinely curious how that acceleration meter would work.
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 reads nonzero when you are standing on Earth because of electromagnetic repulsion between your atoms and the atoms in the floor.
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.
> It works because there's an external force pushing on the surface of the phone
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).
> Which then gets transmitted by the surface of the phone to the rest of the phone.
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.
> this is a very different situation than when you have an external force acting equally on all the atoms
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.
Put more briefly, gravity does not produce a compressive stress in the phone in free fall. The external force from the table applied to the phone sitting atop it does create compressive stress. Ergo, the situations are fundamentally different.
I think the burden of explanation is on the great great grandparent post which proposed an accelerometer "entirely made out of the same magnetic field" which would let it satisfy an equivalence principle for "magnetic force".
> Electromagnetism, weak, and strong interactions all make the acceleration meter register nonzero, even if the act "equally" on all parts of an object.
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?
> 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.
But that path curvature of worldlines is in spacetime, the field of gravity. 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.
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.
> that path curvature of worldlines is in spacetime
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.
> Inertia causes the sensor in the accelerometer to read nonzero.
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.
As far as I can tell, you are claiming that there is a viable model in which, for example, a charged object moving in an electromagnetic field would have "zero proper acceleration" while an uncharged object moving solely under gravity, would have "nonzero proper acceleration". But such a model can't be right, because "proper acceleration" is a direct observable, and our direct observation is the other way around.
It is so difficult to remotely understand all this when every other term is in "quotes". Of course, I'm not a physicist, so I probably wouldn't understand it all, anyways. I am just pointing out how many dang quotes there are. Could you guys maybe define the terms first so you don't have to put two fingers around every concept?
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.
In other words, there is no gravity field, in the way there is an EM field that propagates these forces. Or, the "gravity field" is the fabric of space-time itself
Yeah, if every part of human body is accelerated equally, there's no way to feel anything (you could see it). No matter how hard you get hit or thrown around, you can't measure it internally. Each cell can only feel their neighbors, and there's no internal stress / force between cells.
Only because you feel the air rushing past, and the impact with the ground.
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.
I did sky diving some years back. I did feel acceleration immediately after the jump for a second, but then afterwards it was as if I was lying on bed. So what was that feeling I experienced (it was similar to maintain giant wheel coming down)
Going from standing on a surface where you are resisting the force of gravity (like standing on the ground, or in a stable aircraft) to being in free fall definitely has a sensation.
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.
But you don't feel the force of acceleration. Floating weightless and motionless in space feels the same as being in an elliptical orbit around a planet.
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.
If the impact somehow could stop your all cells at the same rate, you wouldn't feel it. Your body gets compressed because of different rates of acceleration.
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.
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!
