Sabine Hossenfelder chimed in [0] on this discussion a couple days ago. I generally find her to be trustworthy on topics related to physics:
> The easiest way to see that gravity is not a force is to note that a force causes acceleration, but gravity does not.
> Acceleration is measurable with a device called an accelerometer. Acceleration is not relative (like velocity), it's absolute.
> If you are standing on the surface of Earth, an accelerometer will show that you are accelerated in the upward direction. That's because a force is acting on you from below, it's the solidity of Earth's crust (or whatever you are standing on), going back to a combination of electromagnetic forces and the Pauli principle.
> If you take away that support from Earth, eg by jumping off a plane, you are not accelerated. You are freely falling. Since you are not accelerated, there is no force acting on you. You experience gravity but no force, hence gravity is not a force.
> We can assign a pseudo-force to gravity by defining it as acceleration relative to the surface of Earth. This is how Newtonian gravity works. One can derive it from general relativity as an approximation.
> Physicists frequently do refer to gravity as a force anyway -- even I do -- because that's linguistically simpler. But it's like we say "internet" rather than "world wide web" even though we know that the two aren't the same, just because "internet" is simpler.
> So I usually don't pick on this. But strictly speaking, gravity is indeed not a force. If you have doubts about it, buy an accelerometer and do your own research...
I must be really confused about physics and reality, because that argument feels like an April Fools joke.
Stick a rocket engine to the side of the mountain, and light it up. An accelerometer glued to the mountain will show a 0. So will one glued to the rocket. Does that mean thrust is not a force now?
They will both show acceleration. The one on the mountain will show an upward acceleration from the mountain pushing against the device against the pull of gravity. The rocket would too, but presumably greater if it’s accelerating faster than gravity is pulling. The key is that the gravity acceleration vector never shows up in any context. That’s because it’s not a force and the perceived acceleration is an effect of relativity, caused by the gradient of time and space. It’s entirely counter intuitive to our understanding of force and acceleration that you can observe increasing relative motion without a force based acceleration, but there it is
In the scenario you’ve set up
m >>>>>> F, because the Earth is fucking enormous. So a will be extremely small but nonzero. A precise enough accelerometer would show you this.
Also your mountain accelerometer would show ~9.8m/s^2 plus the extremely small contribution of the rocket thrust, not 0.
The accelerometer glued to the mountain will show -9.8m/s^2, not zero. The one glued to the rocket will show the same. The force of the rocket thrust is the opposite of the force of the glue holding it to the mountain, so the sum results in zero additional acceleration.
A theoretically perfect accelerometer will show something non zero as the rocket engine is in fact shifting the entire Earth by a very very very very very very tiny amount.
Stick two identical rocket engines together in deep space so their thrust cancels out. Is thrust not a force when rockets' guidance accelerometers both show 0?
But 0 is what you expect to see in that case: you have two forces acting on the body, of equal magnitude and in opposite directions, so the net force is 0, so acceleration is 0.
But if you jump out of a plane, you will see ~0. If gravity were a force, what would be the equal and opposite force acting on you?
Both engines are exerting force in opposite directions, which cancels out, as you said. There’s no acceleration happening either from gravity or this rocket arrangement, so the accelerometer reads zero. Why is this a problem? Who is saying thrust is not a force?
Accelerometers aren't magic. They don't read out True Acceleration from the process memory of the Matrix. There's a physical process inovlved, and that process cannot tell the rocket is accelerating when that acceleration is cancelled out.
Standing on the ground, the accelerometer should show zero because gravity is cancelled out by the floor being solid. If latter can be a force, why not the former?
> Standing on the ground, the accelerometer should show zero because gravity is cancelled out by the floor being solid. If latter can be a force, why not the former?
This is the part where you are demonstrably wrong. This experiment has actually been performed. Take an accelerometer that shows 0 in space far away from the Earth, where there is no gravity. Bring it back to the earth and let it fall from your ship: it will keep showing 0 (well, because of friction with air, it will show some value, but a pretty small one). When it reaches the ground, it will "suddenly" show 9.8m/s².
This is not something you can debate: this is the measured reality. You can debate how to explain this, of course. But the reality is there: being in freefall towards the earth is equivalent from an accelerometer point of view to being in deep space with no engines running.
Another similar observation is this: the feeling of being in a gravity well is not the earth pulling you down, it is the floor pushing you up. If you're on a space station in deep space, to feel the same as on the earth, you need something pushing at your feet in the direction of your head, not something pushing on your head in the direction of your feet.
I don’t know what else to tell you, the accelerometer in your phone is reading -9.8m/s^2 right now. Go get an app that reads your accelerometer and check it.
Accelerometer on my phone is reading f(a), not a. Unlike with e.g. old-school scales, there's layers of digital electronics and software between the measuring device and measurement being displayed. For all I know, if it's saying -9.8m/s^2, it may be because there's a `value = rawMeasurement - v3(0, 0, 9.81)` line somewhere in the code, because that's the value the designers wanted to report.
I mentioned old-school scales, truth be told, they do this thing too: they're measuring weight, but reporting it as mass, with 1/9.81 factor baked into the scale label directly.
Point being, I'm less interested in what some app tells me - you should always assume apps have assumptions and massage their numbers. I'm interested in whether the measuring device can actually tell the difference, which in case of forces perfectly cancelling out, it should not.
I assure you, because I’ve I’d done it, if you talk directly to the accelerometer, it will give you an XYZ acceleration vector with magnitude 9.8m/s^2 in the direction of the Earth (whichever way you turn it). Not because there is a magic “9.8” in the code, but because that’s what the acceleration is where you’re standing.
If you take the same accelerometer up a mountain, it will read a little less. In orbit, it will read zero. On the moon, it will read about 1.6.
Internally, sure, it does measure force. It measures the force being exerted by the teeny mounting structure to keep the teeny silicon weight in place. That force has to be exerted to counteract the acceleration of the gravity field and keep the weight from flying off. So that’s a straightforward way to measure acceleration. (To be super precise, it measures capacitance between the arms and the base, which is related to distance between them, which is related to force because the arms are flexible.)
Edit: Put another way, there actually is an adjustment in the code, because most people want the reading to be zero when the phone is not moving relative to the Earth. To accomplish that, you can’t just subtract a constant vector, you have to filter the output based on the assumption that there is always a large constant acceleration in some direction. If you rotate the phone, that vector will vary wildly in direction, and you have to pick out that signal and subtract it. This means accelerometers have to have a lot of dynamic range to accurately report that large acceleration added to the tiny ones you’re more interested in. In orbit, this wouldn’t be necessary.
The point is only that the accelerometer is measuring force. If you put your phone on a centrifuge long enough so that it reads a constant force going into it, you would be able to make a situation in deep space where it reads 9.8m/s in what feels like a constant direction to any human. Right? How would the accelerometer be able to detect that?
The accelerometer will read 9.8m/s^2 along the axis of the centrifuge, because that is the acceleration that is keeping the phone moving in a circle instead of flying into space.
The difference between that and resting the phone on the Earth is that the centrifuge structure is accelerating it by exerting force on it, whereas the Earth is accelerating it by altering the spacetime metric in its vicinity. Completely different mechanisms, same reading.
You are actually wrong about the accelerometer, you're almost making the same error as the poster above.
The accelerometer shows 9.8m/s² IF you are holding it in your hand. And that acceleration is pointing up away from the earth, not down towards it. It is measuring the force of your hand/the table pushing it up to stop it from following its normal trajectory.
If you drop the accelerometer, it will show 0 until it hits the ground. If you're holding it while in a plain, it will show slightly less than 9.8, but if you drop it will still go down to 0. If you take it into space, it will show 0 even when it is touching your hand, because it's trajectory is no longer curved towards the center of the earth.
For cell phone, you don't need to trust the app much - it reports raw (X, Y, Z) vector.
Turn the phone sideways and see how it's Z went to 0 and it's now Y (or X) showing -9.8. Spin in the chair and see how the vector's magnitude is increasing. This should be a pretty effective evidence that the vector is not offset, and at most scaled.
If you built an accelerometer using only components that feel the electric force, would you be able to detect the acceleration caused by the electrical force?
I.e. part of the reason gravity is fundamentally different is that it's universal: everything obeys it, including the trajectories of radiation.
This makes it functionally a different thing than every other force, since for all other forces you can build a tool that would measure acceleration because there is always something available to you that is not affected by that other force
It's so nice to see others asking the right questions.
We can't feel gravity's pull in free-fall because it pulls on all of our accelerometers' atoms in the same way at the same time and with the same force and in the same direction. (Assuming the accelerometer is small enough to not be able to sense the tiny changes in the gravitational field due to the field being spherical and the inverse square law.)
As a theoretical physicist I would not say Sabine is very trustworthy. During some journal clubs we analyzed some papers of hers and they were really low quality. This does not invalidate everything she says though...
Agree Sabine is the rush limbaugh lite of physics ... she seems very frustrated which with our knowledge climate right now: cern struck out on susy, no serious line on dark matter, and desire to build a bigger Cern strikes her as a waste of money.
I think she was bitten by the weird state of modern academia and I just generally angry. I suppose she is right though - no new physics in recent memory. Some experimental evidence for things like the Higgs, but nothing experimentally verified that replaces or extends the standard model
That's just because gravity is also accelerating the accelerometer, and the accelerometer can only measure acceleration relative to the body of the instrument.
I've heard this argument: Because gravity is the only force that acts on all objects, including the measuring object, this makes it not a force. This is not actually an explanation, it's more of a definition, and not a useful one.
If you attach an accelerometer to a rocket and turn on the rocket, it’ll measure acceleration even in deep space. This is despite all parts of the accelerometer accelerating equally.
In fact it measures absolute acceleration, not relative.
As u/tzs explains, a rocket engine does not cause a uniform field that accelerates the rocket. It only creates a force at the engine's combustion chamber and bell nozzle. That force is then communicated to the rest of the ship via gazillions of interactions between the electrons and protons making up the atoms that make up the ship and its contents. In fact, even at the engine itself it's the same story. It is these electromagnetic interactions that we call the "normal force" that we feel and that our accelerometers measure.
It is too trivial to make the mistake you did, and then to conclude that gravity is not a force. But that's not really a correct view. A better view is that there are at least two equivalent understandings of gravity: one where it curves spacetime, and the other where it causes accelerations (therefore is a force) that cause photons and massive particles to follow the paths that the other interpretation would have them follow due to spacetime curvature.
Sorry, but no. As long as the rocket engine is producing thrust then the entire rocket will only feel the normal forces caused by that engine because those forces are electromagnetic in nature (electron clouds pushing against electron clouds). It's only because the rocket will be made of rigid materials (but still, made of atomic matter) that the forces are eventually communicated to the nose of the ship and everything in the ship. That "eventually" happens to be very fast because the speed of light is very fast, but it is not instantaneous throughout the ship, and that is why you feel the acceleration due to the engine's thrust.
Whereas when the engine is off the ship free-falls, and in the free-falling case any gravitational forces will have the same effect on every part of the ship at the same time -instantaneously-. Now if the ship is large enough then the gravitational field(s) it traverses won't be uniform enough, and then the ship will feel stresses ("tidal forces") as a result.
I might be wrong, but accelerometers measure the force between the body of an accelerometer and some inertial mass. For example, a seismometer is a big lump of heavy on springs, and when the Earth moves, there is a force between the heavy and the frame holding it. And indeed, at rest, there is a "force" pushing the heavy mass down (whatever you call it).
Drop the seismometer from space, both are falling at the same rate and there is no force and no recorded acceleration - fine.
But wouldn't that be true for any force that acts with the same strength per unit mass on both the heavy weight and its frame? If I rig it up so there's an electrical field pushing these in an empty space, and it just so happens the force is imparting equal acceleration on both the weight and its enclosure, and no acceleration will be recorded. Or indeed, set the whole thing in free space, make the heavy weight magnetic, and attach a magnet to the side. There is now a force being recorded, but no acceleration - the whole thing is a rigid body.
EDIT: in fact to simplify this even further, if I'm negatively charged, standing on some (non-gravitational) positively charged surface, I will feel a force just the same. If I charge the accelerometer, it will show an acceleration.
Objects don't fall down because of gravity, objects move along in straight lines because of inertia. It just happens that, close to large masses, "straight lines" get bent toward the center of mass, and this is what we call gravity.
But the difference is important: when you are falling down, you are in freefall, you don't experience any acceleration in spacetime. It is the surface of the earth that is experiencing acceleration towards you.
> But the difference is important: when you are falling down, you are in freefall, you don't experience any acceleration in spacetime. It is the surface of the earth that is experiencing acceleration towards you.
What if we replace the Earth in that scenario with another person? Alice and Bob are in empty space both in free fall. They each measure the relative velocity of the other and find that it is zero.
Sometime later they again measure relative velocity and now it is non-zero. Each sees that the other is now moving toward them.
Sometime later they measure again, and see that other is moving toward them even faster.
Since they are both in free fall, by your argument neither is experiencing any acceleration. But if no one is experiencing acceleration where do the velocity changes come from?
My original explanation was wrong. The correct explanation is related to the curvature of space-time: there is no acceleration, your speed and trajectory through space-time don't change.
When the two people are far away from each other, they are both moving along through space time, having a speed that is, say, 1m/s through space, each towards the other, and (c-1) m/s in the time direction, both oriented towards the future. As they get close enough to each other, spacetime gets curved by their mass-energy, such that the direction of "the future" now points towards their shared center of mass. If we project this curvature back onto the flat plain in which they were originally moving, it will look as if their direction of movement has changed and their speed through space has increased, while their speed through time has decreased.
The velocity changes are an illusion because you're looking only at the spatial metric of distance. You have to look at temporal metrics too.
The proper "a" here in F=ma is the time derivative of four-velocity or spacetime velocity, and that is constant. It's c (the speed of light). Its time derivative is zero; hence no force and no acceleration.
Yes, this gets tricky because you have to care about "the speed of time through time." Welcome to general relativity.
> It is the surface of the earth that is experiencing acceleration towards you.
This is failing to click with me. If me and my buddy on the opposite side of the world jump out of a plane at the same time the situation doesn't make sense if you say we aren't experiencing acceleration, the surface is accelerating towards us. That'd have it accelerating in two different directions.
On the other hand saying we are each accelerating towards the earth makes perfect sense with our two vectors converging on the same point.
Sorry, I was completely wrong in my explanation, and in what I thought was happening. You are completely right that the Earth of course can't simultaneously be accelerating towards both people.
The correct explanation is that there is no acceleration happening at all. The increase in speed through space is compensated by a decrease in speed through time. It can be looked at as Earth's mass bending spacetime such that "the future" for any nearby object points towards the center of the Earth. You are moving along at constant speed in a straight line towards the future, as you always do when you are not otherwise accelerated in some other direction, but because of the curvature of spacetime around the mass of the Earth, that "straight line" is pointed towards the center of the Earth (it's only a slight curvature: you're still moving much, much faster towards the future than towards the Earth).
Equivalently, we could say that there is no change in velocity: the speed increase is compensated by time dilation. The closer you are to the center of the Earth, the slower your clock ticks; if your speed is constant as measured with a clock high above the earth, it will appear to increase as your clock gets slower. Say you are moving at 1m/s as measured from outside the gravity well. Say that at some altitude inside the gravity well, when your clock shows 1 second has passed, 2 seconds passed according to the original clock. Since your speed is constant, you will have moved 2 meters in the 2 seconss, but you will experience this as moving 2m in one second. When you go deeper down, say your clock now shows one second has passed for every 3s in the original clock: now you moved 3m in 1 of your seconds, even though you're still moving at 1m/s with the original seconds. So you will think your speed is increasing, when in fact it's just your clock getting slower.
Of course, this second explanation doesn't help explain why you're moving towards the center of the Earth and not standing still relative to the earth or some other direction, so the first explanation is still better.
Suppose you were in a free falling box in a strong gravitational field. Is there an experiment that you can do within that box which would measure the acceleration? I think no. Not directly (perhaps you might get clever and do something with tidal forces). And if you can't measure it, why do you assume it's there?
By contrast, if you strap a rocket to the box, it's quite easy to measure the acceleration from within the box.
To argue that the acceleration exists because the position changes is to assume that the observer's reference frame is somehow the correct one, and Einstein has showed us that that's a problematic assumption.
Acceleration can be measured completely locally, velocity can't.
Imagine a pool table on a spaceship. If the spaceship starts to thrust, the pool balls will start to slide backwards. You can measure how fast they accelerated backwards and know how fast the ship is accelerating forwards without needing any point of reference outside the ship. There is no way to determine your velocity without having another point of reference outside the ship, and even if you do that, you still now only know the velocity between the spaceship and the reference.
I understand that. However, I do not really understand why is that so in the laws of nature.
Losely speakimg, we have F=ma. Since a second derivative of position is involved, in the integral forms with velocity and position, we have constants of integration unresolved, which leaves no absolute frame of reference.
So the spacetime somehow has it that absolute measurements are for acceleration, which implies there is an absolute recerence frame for acceleration. I wonder if this implies a higher order relativity is present or waiting to be found.
Rotational motion feels still more strange. Here, even though angular momentum is conserved, rotational velocity still seems absolute. If it weren't, then distance objects in some rotational reference frame would be moving faster than the speed of light, unless there's some relativistic modification to v = omega * r.
If you have a given velocity but no given starting position, then position is the missing constant. If you measure some acceleration and time, you can integrate that into a delta-velocity, but not an absolute velocity.
You can measure your speed against the CMB as an "absolute velocity" (in your local observable universe), but for two locations that are a significant fraction of the width of the observable universe apart, "zero movement" relative to the CMB in each of those locations, will not be zero movement relative to each other, due to the expansion of space in between.
> The easiest way to see that gravity is not a force is to note that a force causes acceleration, but gravity does not.
Maybe I'm misunderstanding as it's been a while since physics class but I'm pretty sure gravity does cause acceleration. One of the few numbers I still remember memorizing in college 9.8m/s2 -- the acceleration by gravity on Earth.
If you let go of something, it will stop accelerating because there is no longer a force on it, and start following its geodesic. You have to apply a force to make it not do that (like, by standing on the Earth so the ground can push you upward). The natural path (geodesic) of an object near the Earth plotted from a purely space-like (Newtonian) perspective accelerates toward the Earth. But the point of GR is to not have a purely space-like perspective.
(I made up the waffly term “purely space-like” to not have to explain what an inertial frame is, not sure if that’s going to help.)
That's the acceleration you feel on the surface of the earth. But that's not an inertial reference frame. In an inertial reference frame gravity doesn't cause acceleration (kind of by definition).
The argument is that gravity doesn't cause acceleration, resisting gravity does. Kind of how spinning an object doesn't cause a centrifugal force, the real force is whatever forces it to stay on a circular path instead of continuing straight
> Kind of how spinning an object doesn't cause a centrifugal force, the real force is whatever forces it to stay on a circular path instead of continuing straight
Let's say the Moon is in a circular orbit around the Earth (close enough), what's the real force that's forcing it to stay on that path? If it's not gravity, what is it?
Great question. The answer is nothing. There is no force making the moon follow a circular path. The moon "thinks" it's moving in a Newtonian, inertial "straight line." Because the spacetime around the earth and moon is curved, the moon moves in a straight line through that curved space.
Caution: The circular path we see the moon follow is not the curvature of spacetime itself. Rather it's a zero-force iso-line along that 4D spacetime. This is also called a geodesic.
The Moon moves not in circles but rather along a straight line in curved spacetime. It doesn't require any force to stay on this path — it is in free fall.
That's also true for, say, electromagnetism. The strength of the force follows the inverse square law, so the acceleration applied on an charged object depends not only on the intensity of the charges but also the distance between the two charges. The only difference between classical gravity and electromagnetism (in a mathematics sense) is that gravitational "charge" is mass, which means the mass term of the accelerated object can be canceled out since it is on both sides of the equation. For an object with a fixed mass (like the earth), that means you can say the acceleration is purely a function of distance rather than distance and mass of the other object. But I don't see how that makes the acceleration term invalid or implies that gravity doesn't actually cause acceleration.
This is because classical gravity is wrong. The easiest way to notice this is to build an accelerometer, and show that while it is in freefall towards the Earth, it measures the same reading as it does while it is in space far away from any massive body. However, when it is resting on a table on the earth, it will measure an acceleration away from the table, upwards. The same direction of acceleration it will measure if put on a rocket in the direction of movement of the rocket.
You yourself are such an accelerometer. The feeling you get while falling is the same feeling you'd get if you lived on the International Space Station. If you wanted to add a module on the ISS to make you feel just like on earth, you'd a module that generates a force on your feet up towards you head, not a force that pushes on your head up towards your feet.
So, Newton's universal force of attraction is quite clearly wrong, and there is no equivalent force in reality.
This shows that your accelerometer is broken. (or, more likely, that the vector is mislabeled)
Try this thought experiment. Assume your phone is really accelerating downwards when at rest. Then let it enter a free fall i.e. drop it. Now it's accelerating down even more so the downwards acceleration should show an increase. Actually, it goes to zero. Relative to a free-fall, being stationary is an acceleration upwards.
To mimic the effect of gravity on the surface of the earth you could stand on a platform that is accelerating. As long as that platform is accelerating you will experience something similar to gravity on the surface of the earth.
The relative direction of the acceleration would be "up". This is what she means.
I think this is a distinction between push/pull? You are interpreting the reading as it being pulled down. You could also interpret it as you having to push it up.
The example you give also doesn't make sense. The acceleration would not go to 0 when you drop something. The acceleration continues for the entire fall. It may be offset by friction if it hits terminal velocity. But, absent that, acceleration would continue the entire fall.
If the reading of an accelerometer does change in freefall, then what it is measuring isn't the acceleration, necessarily, but the difference in acceleration between components in the device. Which makes sense. Is like watching a balloon in a car when you start moving. (With the trick of whether the windows are open or not, of course.)
> If the reading of an accelerometer does change in freefall, then what it is measuring isn't the acceleration, necessarily, but the difference in acceleration between components in the device.
This would come as a surprise to the designers. :)
No, the acceleration on an object in free fall is in fact zero. It may be surprising, and may require rewriting your intuitions to fully grok, but this is indeed what GR says.
Ah, fair. I was messing myself with the knowledge that you would be accelerating the whole fall. Easy to forget that we are always measuring proxies. (For example, speedometers aren't measuring how fast cars are moving.)
Edit: (I'm actually curious what I said is fully wrong, btw? Accelerometers measure how much an inner piece drifts to other things based on differential in acceleration, no?)
Edit2: I think you thought I was saying the accelerometer would not read zero in free fall? I meant more that you would continue accelerating, despite the reading being whatever it was.
> Edit: (I'm actually curious what I said is fully wrong, btw? Accelerometers measure how much an inner piece drifts to other things based on differential in acceleration, no?)
In a steady state, all the components are accelerating equally. It measures the force between the components and uses that to compute acceleration, yeah, but the positional drift between the components is minuscule and only happens during jerk.
(This describes one type of accelerometer. I'm not up to date on every possible or currently used design.)
Apologies, referring to the drift, I felt it was safe to assume you'd measure the force between the components. Not necessarily the distances.
My point is that to an earth bound observer, something in free fall to the surface of the earth is accelerating faster to earth until they hit terminal velocity. This is despite the accelerometer on the falling body showing zero.
Note that i don't think this adds any understanding to whether or not gravity is a force. But it does feel a lot like saying that, if you are in a river and able to stay still, that you aren't fighting the force of the river. Similarly, do we say that people standing on a moving sidewalk are stationary in the same way as someone standing on the sidewalk they are passing?
Your iPhone lies to you. The acceleration is indeed upwards, as you’ll see if you look at raw accelerometer data… or the feeling from your feet, which counts.
It is already giving you one, but it's not along the normal directions you can feel: you are not moving at your normal free-fall speed that you "should" have in the Earth's curved spacetime. It's like a friction force, it's not moving you, it's actively stopping you from moving.
You're asking good questions. The reason standing on Earth gives you an upward push is that in fact this is a semantics game. What's really happening is that the upward push is what we call a "normal force", and it is entirely the result of interactions between the electrons and protons of the matter we and the Earth are made up of. The accelerations measured by accelerometers (including the ones in our ears) and by the pressure sensors in our skin are strictly electromagnetic in nature, but ultimately these are caused by gravity being a force indeed that is pulling us down, and it is the normal forces that stop us from accelerating further into the planet.
A lot of people are confused by GR. But really, gravity can be seen as any of these things:
- an effect that curves spacetime,
yielding accelerations when that
curved spacetime is mapped to
flat spacetime (think of a Mercator-
like projection of curved spacetime
to flat spacetime)
- a force that, when applied to waves
and matter (which... is standing
waves anyways) causes them to be
accelerated in such ways as to
follow paths indistinguishable from
the ones they would follow in
curved spacetime
It's really a lot simpler than some people make it out to be.
Consider the Schwarzschild metric \[ds^2 = -(1 - \frac{2M}{r}) dt^2 + (1 - \frac{2M}{r})^{-1} dr^2 + r^2 d\Omega^2\] -- its terms show you the distortions of space and time (independently, though related by the distance r to the center of the massive object) relative to flat spacetime. The first term shows you the distortion of time, and the second and third show you the distortion of space. I've yet to see a physicist put it that way, but that is exactly how it is. (This metric is a second derivative of spacetime, so it's not immediately obvious what the actual mapping between flat and curved spacetime is -- you have to do a double integral for that. However, what you get is the normal time dilation factor we know \[\sqrt{1-\frac{2M}{rc^2}] as the distortion of time, and... the same as the radial space expansion (that is, space expands, but only in the direction to/away from the massive body), + a three-dimentional angular (two angles) displacement.
It is often said that you can't tease out the space and the time curvature from spacetime curvature because they are intimately intertwined. But if you look at the Schwarzschild metric this is not really quite true. Yes, the first two terms have an r in them, so space and time curvature are very much interrelated, but they are still terms in an addition, with one term for time and one for space, so they can in fact be described separately, as indeed the Schwarzschild metric does.
> > Acceleration is measurable with a device called an accelerometer. Acceleration is not relative (like velocity), it's absolute.
The reason that you can't measure gravitation acceleration in free-fall in a uniform gravitational field[0] with an accelerometer is that all parts of the accelerometer are being accelerated together.
[0] The "uniform gravitational field" part comes from Einstein's equivalence principle, and it's pretty obvious that if the field were sufficiently non-uniform then one could expect strain gauges to indicate that the free-falling object _is_ being accelerated. Imagine you have a 10km long space ship near a massive body like Earth, or the Sun, then the gravitational field will not be uniform across that space ship, and strain gauges will a) let you know, b) will even be able to tell you the strength of the field and the heading to the massive object.
All these "gravity is not a force" arguments based on the equivalence principle are simply nonsense. The better argument is that gravity is both not a force because it only curves spacetime, and also yes equivalent to a force (especially when you do a Mercator-style projection of curved spacetime onto flat spacetime).*
Love the Hossenfelder reference (though she really needs to move to Bluesky...). That said, I'd love if someone could explain some physics to this noob:
If you jump out of a plane, at t=0 you'll not be moving (relative to the earth), at t=1 you'll be moving a bit (relative to the earth), and at t=2 you'll be moving even faster. How is that not acceleration? A quick wikipedia rabbit hole from "Accelerometer" --> "Inertial Reference Frame" --> "Fictitious Force" led me to this:
A pseudo force does not arise from any physical interaction between two objects, such as electromagnetism or contact forces. It is just a consequence of the acceleration a of the physical object the non-inertial reference frame is connected to, i.e. the vehicle in this case. From the viewpoint of the respective accelerating frame, an acceleration of the inert object appears to be present, apparently requiring a "force" for this to have happened.
They use the example of a passenger in an accelerating car, which makes sense. And there's some discussion of how the earth is rotating and thus has angular (?) acceleration, which also makes sense. I think this is what Hossenfelder is referencing by "the earth is accelerating you upwards". But the rotation discussion seems completely unrelated to gravity, no? An asteroid that isn't spinning would still exert gravitational (pseudo-)force, as all massive objects do of any size. Surely a person standing on a non-spinning asteroid in deep space isn't being accelerated upwards?
At the end of the day, I guess I'm missing why exactly "free fall" is the same thing as "no forces acting upon you". To my cynical arrogant brain, this reads like the physicists are harping on an unnecessary terminological thing, namely that a "force" is defined as "a physical interaction between two objects". In other words, its quantum bias, in the original sense of quantum; why can't objects interact with the continuous field of spacetime?
I'm commenting on your quote because her explanation especially "we have a machine with acceleration in the name, thus that's what acceleration is" set off a million alarm bells in my head, philosophically speaking!
The point is that, unlike velocity, acceleration is absolute in GR.
If we're both moving towards each other at constant speed, it's perfectly equivalent to say that I'm moving towards you and you are stationary, or to say that I'm stationary and you're moving towards me, or that we're both moving towards each other relative to some outside observer.
The same isn't true with acceleration. If we're in the same scenario and I start a rocket thruster, then I'm experiencing acceleration and you're not. Our relative velocity towards each other is increasing, but it would be wrong to say that I'm stationary and you're accelerating towards me.
So, if you fall from a plane, your relative speed towards the Earth's surface is increasing. But it's not you who is experiencing acceleration, it is the Earth, and the difference is measurable in principle.
This is similari concept to how when something is moving in a circle, it experiences an acceleration towards the center of the circle, but this is often experienced as a "centrifugal force".
Question: Two black holes that encircle each other are on geodesic orbits and thus should not feel acceleration. However, graviational waves are emitted during the orbits until they merge. How is this possible when there is no accelleration acting on the masses?
Well, with me standing where I am, and anti-me standing on the exact opposite side of the Earth, the ground must be accelerating in opposite directions at once!
It feels like taking a somewhat straightforward model and inverting it (in the x -> 1/x sense); that's how you get straight lines to split into pieces and curve away.
It's forced if you start from the perspective of "General relativity describes reality", and obviously so if you look back at the inspiration for relativity, one of which was "there is no way to differentiate between different free-falling rest frames from inside a box".
Of course it's not always the most convenient model, and there are ones in which gravity is indeed a force — the Newtonian approximation, for example — but the starting point of this article is "Here's how reality works if GR describes reality".
> The easiest way to see that gravity is not a force is to note that a force causes acceleration, but gravity does not.
> Acceleration is measurable with a device called an accelerometer. Acceleration is not relative (like velocity), it's absolute.
> If you are standing on the surface of Earth, an accelerometer will show that you are accelerated in the upward direction. That's because a force is acting on you from below, it's the solidity of Earth's crust (or whatever you are standing on), going back to a combination of electromagnetic forces and the Pauli principle.
> If you take away that support from Earth, eg by jumping off a plane, you are not accelerated. You are freely falling. Since you are not accelerated, there is no force acting on you. You experience gravity but no force, hence gravity is not a force.
> We can assign a pseudo-force to gravity by defining it as acceleration relative to the surface of Earth. This is how Newtonian gravity works. One can derive it from general relativity as an approximation.
> Physicists frequently do refer to gravity as a force anyway -- even I do -- because that's linguistically simpler. But it's like we say "internet" rather than "world wide web" even though we know that the two aren't the same, just because "internet" is simpler.
> So I usually don't pick on this. But strictly speaking, gravity is indeed not a force. If you have doubts about it, buy an accelerometer and do your own research...
[0] https://x.com/skdh/status/1850120005070799153