Gravity is not a force | Hacker News Reader

Gravity is not a force

superposer.substack.com

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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...

[0] https://x.com/skdh/status/1850120005070799153

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

F = ma so a = F/m.

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.

> An accelerometer glued to the mountain will show a 0.

Incorrect. An accelerometer glued to the mountain will show -9.8m/s^2 because it's being accelerated by said mountain.

No?

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?

If you are accelerating, the accelerometers will not "show 0". I have no idea where you are getting this notion from; it's completely wrong.

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.

My point was the accelerometer would not be enough to expose what is happening. Imagine a centrifuge large enough that you can't determine the angle.

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.)

If this were true, then an accelerometer that shows 0 in free fall on Earth would show some value when it was in deep space.

The accelerometers in our ears show the same acceleration when we're falling in diving airplane as when in orbit in deep space.

I don't understand. Why would it show an acceleration in deep space?

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.

A rocket would not accelerate all parts of the accelerometer equally.

Not in the first couple of seconds, but by the time everything settles down it certainly will.

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.

> The easiest way to see that gravity is not a force is to note that a force causes acceleration, but gravity does not.

When did objects stop falling down from gravity? Or is this some hairsplitting about how gravity takes effect by pulling instead of pushing?

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.

The answer here goes into more detail:

https://physics.stackexchange.com/questions/102910/why-would...

> 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.

When Einstein literally discovered General Relativity.

Gravity is the curvature of Space Time. From that perspective, objects under gravity are traveling in a straight line.

Fun Veritasium video on the topic: https://www.youtube.com/watch?v=p1W0dpunOaM

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 is not relative (like velocity), it's absolute.

May be a naive question, but why is this so? Is there a different type of 'ether' when we consider acceleration?

Edit: Actually I have the same question for rotational motion too.

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.

I do not know if I understand these things right.

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.

It took me a while to understand your second para.

I found a lot more on the broader topic here.

https://en.wikipedia.org/wiki/Mach%27s_principle

It mentions that rotational motion does seem to have an absolute reference frame.

Some annoying student would mention that the same force is acting on the accelerometer as well so it cannot act as an independent observer.

> 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 relative to the surface of the earth

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.

I don't understand her comment.

This is really easy to verify.

You drop an iPhone, the gravity "acceleration" really goes to 0 - so far so good.

But at rest (i.e. holding the iPhone in your hand), the acceleration is pointing downwards, not upward as she claims.

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.

ok this clears it up, thanks!

this is obviously a very confusing topic, as evidenced by the other replies that either don't get my post, or the post above :D

https://old.reddit.com/r/AskElectronics/comments/2ufunr/why_...

I think there's some sign inversion happening there. But don't quote me, all the details are fuzzy to me.

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.

An iPhone held “at rest” cannot be accelerating downward, by definition.

Force is not acceleration.

That is extremely easy to understand but somehow you and all of these "eminent scientists" fail to grasp it.

Force is not acceleration, and accelerometers do not measure force. We have other devices which do that, such as bathroom scales.

Yup. Paraphrasing, you should not measure gravity with an accelerometer, otherwise this thread happens.

You can, but you should probably take three years of physics courses first.

> If you are standing on the surface of Earth, an accelerometer will show that you are accelerated in the upward direction.

How long until this upward a gives me a little upward v? What's the dt here?

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?

There is acceleration. It just isn’t on you.

The ground is accelerating upwards, which keeps it in place. You are not accelerating, and therefore you move downwards.

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".

not a single one of those points is true lol

First off, the author doesn't appear to be a crank.

Here's the Wikipedia entry: https://en.wikipedia.org/wiki/Jonathan_Oppenheim

Second, we should note that even Einstein himself cautioned against believing spacetime was actually curved. His writings inform us he didn't believe it. I don't want to appeal to authority, that's just to say smart people, including the main developer of general relativity, didn't believe it. But he didn't believe in the non-local nature of quantum theory either, which we have now, since Einstein's death, proven to be true.

Third, the claim that only gravity can be described using geometry is false, which the author himself notes later in this article. The stress-energy-momentum tensor simply makes gravity universal, unlike the other forces. I don't see any reason why that universality confers something special to gravity with regards to interpreting it as geometry. Just because we can model gravity as geometry, doesn't make gravity a result of geometry, and the author notes that modeling gravity that way makes it so we can't unify the forces.

Finally, as the old saying goes, if you think gravity isn't a force, drop a brick on your toe! :)

I'll also point out that singularities are generally considered to be a sign of issues with a model. GR has singularities. Maybe that should tell us something.

> we should note that even Einstein himself cautioned against believing spacetime was actually curved. [...] I don't want to appeal to authority

Another example that comes to mind is Max Planck believing that light being absorbed in discrete packets of energy was only a neat mathematical hack he came up with. It took Albert Einstein to say "but what if light is discrete packets". And then as you say Einstein having major reservations against the field of quantum physics that he himself spawned.

> if you think gravity isn't a force, drop a brick on your toe

By that logic the centrifugal force has to be a force. If you don't believe it, just drive a vehicle around a curve, or whirl a rock on a string.

> By that logic the centrifugal force has to be a force.

Are you trying to claim it's not?

Not the OP, but I think I know what wongarsu is referring to.

In order to make an object turn, it needs to experience an centripetal acceleration (towards the centre of rotation). This is the force causing objects to change trajectory.

If there is another object inside the turning object (like clothes inside a washer, or a person inside a car) they will "feel" like they are being flug out as if a centrifugal force existed, but actually that is just the effect of Newton's first law: the natural tendency of every moving body is to continue to move in a straight line, so when the containing object is changing direction (due to the centripetal force), Newton's first law tends to push you outwards.

All of the above is from the static (world) frame of reference.

It is also possible to put a coordinate system on the rotating object, in which case something like a centrifugal force will exit, but we kind of created it by choosing an accelerating reference frame, so it's not real. Sometimes called a pseudoforce.

"There is no centrifugal force" is often found together with "centrifugal forces appear as a term when using the frame of reference of the object going around", and this implicitly says that some reference frames are more privileged than others which -in a world that accepts the principle of Relativity- is just not acceptable.

So of course gravity is a force, and of course centrifugal forces are real. These dogmas serve to do nothing more than to scare away students, and to make the dogmatic seem like geniuses because only they can understand these things.

Relativity, classical Galilean relativity, special relativity, and general relativity, all say that all inertial frames of reference are equivalent. Accelerated frames of reference are not equivalent to inertial ones, nor to each other.

That is, there is a difference in all of these theories between an observer who is at rest with a train accelerating towards them, and the same observer accelerating towards a train that is at rest.

This doesn't mean that you can't do coordinate transforms to look at the world from the point of view of the accelerated observer, and get apparent forces. But different accelerated observers will come up with different apparent forces, while all intertial observers agree on the same forces experienced by any object.

> But he didn't believe in the non-local nature of quantum theory either, which we have now, since Einstein's death, proven to be true.

We have not proven quantum theory to be non-local. We’ve only proven that it can’t be both local and contain particles, as opposed to particles being emergent from the wavefunction’s interactions.

MWI chooses the latter, and is therefore a local theory.

(Alternately, collapse. But collapse theories are largely nonsensical.)

> First off, the author doesn't appear to be a crank.

Is this the first most people are hearing about this? In the 90's, I had a physics textbook for the layperson (i.e. non-STEM fields). It had a fantastic chapter on general relativity, and it also went with "not a force but a spacetime curvature".

Popular scientists have been saying it for decades and always use the flexible rubber sheet as the learning aid. Serious physicists tended to stay out of the fray, preferring to just "shut up and compute", as they do for quantum mechanics.

It's a relatively recent development that physicists have entered the fray and say maybe curved spacetime isn't a model, maybe it's reality. I urge caution to all as we have no evidence that the model of curved spacetime is indeed reality. I think the caution is warranted considering that GR contains singularities, which is a sign the theory has issues.

GR is a theory of the bulk. It’s like fluid mechanics, which treats fluids as infinitely divisible and doesn’t acknowledge the existence of molecules: There’s no way it’s accurate at small enough scales that molecules matter, which is what would be happening inside a black hole.

At larger scales, however, fluid mechanics is extremely reliable.

A fun fact about black holes that isn't often mentioned is that a black hole has two singularities. One is in the center — the one people always talk about — and the other one is at the event horizon. My impression is that the second one is eliminated by choosing to represent the calculations with a coordinate system to eliminates it at the geometrical level, but it's unclear if this is a "hack" or not. But I find it interesting (as a layman) that one can be handwaved away as irrelevant, while the other cannot.

What if singularities do in fact exist? What issues does that create?

They might not be observable if superextremal black holes don't exist and thus naked singularities do not exist.

We’d have no way to predict what they do, since the math breaks down. This is fine if you treat the math as only a model, and you’re willing to accept that it can’t predict the behaviour of a thing that can’t affect you anyway, but given your wording — “do in fact exist” — I assume that’s not what you mean.

A theory that purports to describe how the universe actually functions, can’t have places where the theory says “and now the universe bluescreens”.

At a minimum it would need additional postulates about computational order that allows the inside of the black hole to be NaN without that causing external reality to seize up.

> A theory that purports to describe how the universe actually functions, can’t have places where the theory says “and now the universe bluescreens”.

A theory that provides useful predictions in some conditions and doesn't provide useful predictions in other conditions is a useful theory if you can determine which conditions are which. My lay understanding is general relativity provides useful predictions as long as mass isn't too dense (black hole) and as long as nothing breaks the speed limit; which is pretty general for everyday use, even if it doesn't cover the whole universe.

> “and now the universe bluescreens”.

I love it!

Singularities mean that the math stops working, so we don’t know what’s really happening there, the mathematical model is failing. We’d like to have a working model.

It would be fascinating if the universe actually has a “NaN” phenomenon that conveniently can never be observed or directly interacted with because it’s always inside an event horizon.

Event horizons aren't physical objects, they're mathematical constructs.

It is possible to unify the electromagnetic field into a geometric framework via the Kaluza-Klein theory which is just rewriting GR + ED. Then electromagnetic forces are also of geometric origin. How do you interpret this? Is electromagnetism then also not a force? I am not a physicist, so I am asking the experts.

https://en.wikipedia.org/wiki/Kaluza%E2%80%93Klein_theory

I found this paper which details how a charged particle falling into a black hole radiates. How is this possible when there is no acceleration on the particle?

https://adsabs.harvard.edu/full/1971PASP...83..633R

Title: Radiation from Particles Falling into Black-Holes

Authors: Ross, D. K.

Journal: Publications of the Astronomical Society of the Pacific, Vol. 83, No. 495, p.633

Bibliographic Code: 1971PASP...83..633R

I'm going to have to read that one. Thanks for the link!

> First off, the author doesn't appear to be a crank.

Perhaps, but the author is being dogmatic. See commentary elsewhere in this post.

I like the model of an insulated tunnel bored through the Earth from the north to the south pole and filled with a vacuum (technologically implausible, yes). If we drop a steel ball into the tunnel at the north pole, what forces does it experience?

From Newton's perspective, F = ma and the ball accelerates towards the center of the Earth. The value of g diminishes to zero at the center, the ball is at its maximum velocity, and then enters the negative acceleration regime until it just reaches the surface of the Earth at the south pole. This will continue indefinitely in harmonic motion. It's not a perpetual motion machine because machines do work and we're not doing any work on the ball; it's similar to an orbiting sphere. (ChatGPT-o1 claims the period is 84.4 minutes, assuming uniform density)

The general relativity perspective seems to be the ball is just rolling up and down a bowl of spacetime, without any friction or drag, which isn't all that satisfying a picture, since it implies a restorative force being involved to keep the ball from escaping the bowl. It helps to consider the state of the ball right before it is kicked into the tunnel - it is being prevented from following its natural geodesic trajectory by the electromagnetic forces of the rocks of the Earth's crust upon which it is being held up - that's the only relevant force in this picture.

Accelerometers don't measure acceleration, they measure the difference between the force on a mass on springs and the body of your phone. If you drilled a tiny hole and pulled the mass up, the sensor would think the phone was being pulled down.

Acceleration due to gravity has an important property of acceleration in general: radiation. Spiraling black holes emit gravitational waves, in the same way that an accelerating electric charge emits light. There are free falling reference frames where uniform gravitational fields can go away, but the gravitation of a massive body isn't uniform and can't be eliminated by changing the coordinates.

The relationship between gravity and fictitious forces is an important stepping stone, but it does not have all the properties of a fictious force, only some of them.

He lost me there:

> Is this purely a semantic difference? You could argue that it doesn't really matter whether we describe gravity as a force or through geometry, and we should conflate these two concepts. But I think this distinction is important to make because it has predictive power. If you believe that gravity is manifest through spacetime bending, then you will never find two different test particles that follow different geodesics.

Don't we get the same conclusion if we believe gravity is a force _and_ equivalence principle is true?

We do. The argument here is that there are two equivalent interpretations but that one is more fundamental (or closer to actual reality anyways) than the other. But this is mostly a result of the starting point taken by Einstein and then everyone understandably accepting that as the more fundamental thing.

Even if no one ever finds a way to mathematically and equivalently derive GR from flat spacetime + gravity is a force first principles, it is certainly a lot easier for humans to understand gravity as being a force, and the curvature of spacetime as being other distortions on flat spacetime, not unlike the various 3D->2D projections we use for maps of Earth. Pedagogy matters.

[Notice that I'm not describing the distortions that one gets when mapping GR to flat spacetime. I want to leave that to the reader, though I might pop up and reply with a list later if someone asks.]

This is all pretty far up the semantic creek anyways because forces are only defined in the Newtonian framework.

Question to the physicists out there: When an electron gets accelerated it emits "Bremsstrahlung" because it radiates away photons when it changes its velocity vector.

So for an electron on a circular path in a magnetic field, we know that it emits this radiation because this is the synchrotron radiation.

Now what happens to an electron on a circular path around a black hole? Does it emit synchrotron radiation or not?

Bremsstrahlung radiation involves accelerating electric charges through non-zero electromagnetic fields. When all charged particles are accelerated by gravity in the same way by a [close to] uniform gravitational field then they are not being accelerated through non-zero electromagnetic fields because the causes of those fields are also accelerating in the same way. However, if an electron were falling into a charged, rotating black hole then I'd expect some Bremsstrahlung radiation indeed.

That said, IANAP. And when I say "accelerated by gravity" I am taking the interpretation that gravity is a force, which is a valid interpretation (see other commentary above).

“If you accept that we live in spacetime, and it can be curved, then I think you should accept that gravity cannot be a force.”

I am on the opposite time of thinking — where spacetime itself emerges from particle interaction events. Pairwise distance (i.e. metric) is just yet another interaction parameter in the world graph.

The metrics in GR relate the effects on spacetime of massive bodies, but... the effects relative to... what? Well, relative to flat spacetime. Those metrics are really projections. We need them to understand the effects on waves and matter (which is standing waves anyways, so it's all waves, all the time). But we don't have to take the view that curved spacetime is more fundamental than flat spacetime . There is an equivalence between them, therefore we can say that neither is more fundamental. That means that we can use the interpretation that is easiest to understand.

There is no such thing as “flat spacetime”, or, above all, “absolute” background spacetime. This is what relativity explicitly denies. Spacetime appears as pairwise metrics between objects (i.e. quanta/particles), and doesn’t exist beside objects.

There are no absolute coordinates. But things like the Schwarzschild metric very much map between flat and curved spacetime.

You should search for "conformal mappings to flat spacetime". This is not a controversial thing.

"is to say that when no force acts on a test particle in curved space, it should move along a geodesic"

Every single author who calls gravity not a force, just hand waves right past this: why should the particle move at all?

Sure if the particle is moving it will follow a curved path thinking the path is straight.

But if the particle just sits there, okay the path is curved, but it just sits there.

I've heard explanations having to do with the fact that particles move through time, that doesn't really answer the question because it can continue moving through time while just sitting there.

"in the absence of being pushed or pulled, test particles in a curved spacetime will free fall"

Really? Why exactly will they free fall? Why can't they just stay exactly where they are?

I've also never heard anyone explain what the issue is with calling gravity a force. One person said it can't be a force because it makes light move, and light is massless.

However gravity does not act on mass it acts on energy, and mass is just a form of energy. Since photons have energy obviously they gravitate.

I think the missing piece in your puzzle is adding the "time" to space-time. A particle moving through time is "falling" in the time dimension. The curvature of the 4d space makes this movement through time also move through space due to the warping effect of a gravity well.

There is also the asymmetry of time dilation. Someone moving in a rocket ship will be younger than the person staying behind. Isn't their relative acceleration the same though? Why is the rocket man privileged in staying young? This kinda works if you imagine a grid of sand particles being acted upon by either gravity or traversed by the rocket. The rocket traverses many more grids than the person standing on Earth. The person floating in naked space will traverse the fewest grids.

The kicker is that there is no place in space with zero grid traversal. Accounting for all grid traversals becomes a nightmare.

what do you mean by "just sits there"?

Sits where?

In the same position relative to the sun? Relative to the center of the Galaxy? Relative to the local galaxy supercluster?

A particle does not just sit still!

Site there relative to that other mass that wants to accelerate it toward each other.

You’re moving in the time direction — and when spacetime is curved, you end up moving in the space direction as well, to follow the geodesic.

The river model of general relativity.

https://www.youtube.com/watch?v=hFlzQvAyH7g&list=PL__fY7tXwo...

Not arguing for or against this explanation, but it's hard enough (impossible for me) to visualize a curved 3d space, once you try to think of what curved 4d space might be like any intuitive sense of what should happen goes out the window.

Classic diagrams of curved spacetime portray space as a 2d sheet with divots in it. I wonder what a similar visualization of curved 1d time + 1d space would look like. Long valleys for gravitational wells and everything moving along the sheet in the direction of positive time?

> and when spacetime is curved, you end up moving

You did the same as everyone else: You hard waved right past "end up moving". Why do I end up moving instead of just sitting there?

And if I "end up moving" that means I exchange momentum, if it's just bent space that magically makes me move, what did I exchange momentum with, if not via a force toward another particle?

Gravity is a force that acts on energy. If you disagree please give me a counter example.

You do just sit there. In the GR-realism view, the only thing particles _ever_ do is self-propagate forwards in time.

The problem is, this means the particle’s definition of “time”. Spacetime is 4D, and this can be any timelike direction. That’s what causes it to appear to move, from the perspective of someone whose time vector is pointed differently.

Then time can be curved, which reorients said vector as the particle propagates through it. That’s gravity.

> Really? Why exactly will they free fall? Why can't they just stay exactly where they are?

Absent any forces, they can only stay on a geodesic. It’s like the arrow that springs from the bow. No hesitation, no doubts. The path is clear. There’s only ever one geodesic, one path through spacetime that takes no energy to follow.

This is true even if you ignore general relativity and just focus on classical mechanics, as formulated by Lagrange. In Lagrangian mechanics, you must compute a number called the “action” for every path through space that your system could possibly take. Then your system will take the path that minimizes the action. If your system is a particle in space, then it will stay still if that has the minimum action, or it will follow some path through space if that has the minimum action. It can’t ever do both. Only one of them will minimize the action.

> one path through spacetime that takes no energy to follow.

Again, this if fine for an orbit for example, it does not explain how a particle will accelerate, it also does not explain momentum, because when that particle accelerate it exchanges momentum with another particle - it does not give momentum to the "geodesic".

And explain please what information is not captured by saying "Gravity is a force that acts on energy". What nuance is lost by saying that?

Particles in orbit do accelerate. They accelerate continuously, from our perspective, as they go around the orbit. But seen from the perspective of GR, that orbit is really just a straight line through a curved spacetime. It might be easier to visualize though if you think of it as the bottom of a valley. The walls of the valley rise away from the floor as you get closer to the planet you are orbiting, or further away. You can’t climb those walls without energy, so if you have no energy will you stay in the valley forever.

> it also does not explain momentum, because when that particle accelerate it exchanges momentum with another particle - it does not give momentum to the "geodesic".

No one has said that it should. The geodesic is merely the straight path that the particle must follow when it _isn’t_ exchanging momentum with other particles.

> And explain please what information is not captured by saying "Gravity is a force that acts on energy". What nuance is lost by saying that?

You can’t explain time dilation that way, or length contraction.

A charged particle on a straight line does not emit synchrotron radiation. What happens if it is on a geodesic? Do electrons that circle around a black hole emit synchrotron radiation?

That’s beyond my level of expertise, but yes.

Quoting from the abstract of The Motion of Point Particles in Curved Spacetime <https://web.archive.org/web/20151203142700/http://relativity...>

> This review is concerned with the motion of a point scalar charge, a point electric charge, and a point mass in a specified background spacetime. In each of the three cases the particle produces a field that behaves as outgoing radiation in the wave zone, and therefore removes energy from the particle. In the near zone the field acts on the particle and gives rise to a self-force that prevents the particle from moving on a geodesic of the background spacetime. The self-force contains both conservative and dissipative terms, and the latter are responsible for the radiation reaction. The work done by the self-force matches the energy radiated away by the particle.

Note especially that the charged particle won’t actually follow the geodesic. Because the synchrotron radiation carries away some energy, the particle will be continually pushed off of the geodesic; it will spiral inwards instead of orbiting forever.

Does this mean that when the geodesic is a straight line then the particle stays on the geodesic, but when the geodesic is curved the particle interacts with its own e.m. field and deviates from the geodesic because it feels a deflecting force?

Then there would be a fundamental difference between curved and straight geodesics which would be a contradiction to the Einstein Equivalence Principle. How do physicists explain that contradiction?

The equivalence principle only applies to uniform fields and uniform motion. A charged particle in orbit is not in a uniform field and is not experiencing uniform motion.

Hmm, if you look at another comment here: https://news.ycombinator.com/item?id=41971973

It seems like the point is that the particle is actually not accelerating when it follows the one path through spacetime. It only accelerates when it's stopped by something (e.g. the surface of the earth).

Another angle that might help: Newton was correct when he said a particle with no forces on it would keep going in a straight line. He was just missing a dimension in his coordinate system. The geodesic is the straight line in spacetime. And mass distorts the spacetime metric (mostly time) in a nonlinear way, so when you project that straight line back into 3D space, it’s not linear, it’s a curve. An “acceleration” curve.

The author is 50% correct. As John Wheeler stated "Spacetime tells matter how to move; matter tells spacetime how to curve."

The author is unfortunately forgetting the second part. If matter is quantized spacetime curvature must also be quantized, because matter defines our spacetime.

In principle one can even shorten the sentence and say: matter tells matter how to move. Spacetime appears only to be a convenient calculation tool. And in that sense spacetime=gravity isn't a force. It actually doesn't even exist.

I am sympathetic to the author's thesis. I favor the idea that gravity is a different thing from the other fundamental forces, and possibly an emergent phenomenon rather than a fundamental thing in its own right.

But, I don't buy the argument made here:

> To call gravity a force, is to privilege flat space as somehow being special.

Flat space is special, and we didn't make it special.

This is taking an important aspect of known physics---that there exist various symmetries and all elements of the corresponding group are equal players (there is no privileged reference frame, positive charge and negative charge are indistinguishable save for their oppositeness, etc.)---and attempting to apply this principle to spacetime curvature. But the zero curvature state is a unique one that is differentiatable from the others. It's the only one where a circle is perfect, having circumference 2 * pi * r. And pi is a fundamental invariant of geometry, curved or otherwise. The mathematics privileges flat space. Further, experiments can be constructed to detected whether we are in flat space or not [1]. That wouldn't be possible if the whole concept of flat were only relative to an arbitrary frame.

[1] https://en.m.wikipedia.org/wiki/BOOMERanG_experiment

I tend to think about this in via a simple 0-1-many heuristic - there's infinitely many ways to have a curved space, but there is exactly one way to have a flat space. That by itself makes it special.

Is this heuristic wrong?

> Is this heuristic wrong?

I would say it is, but in a subtle way. There is only one way in which gravity curves spacetime, and also one set of effects it has when seen from the lens of flat spacetime.

> There is only one way in which gravity curves spacetime, and also one set of effects it has when seen from the lens of flat spacetime.

Is there though? Isn't that the holy grail of science - expressing all of physics, particularly all of "fundamental constants", in a formula where there is one clearly preferred answer? One function with a single global minimum in the parameter space, that defines our universe?

Perhaps let me put my heuristic differently: when there exists multiple (especially infinitely many) solutions, you still need to explain why a particular solution would be the solution, and not any of the other ones. A single solution is naturally privileged, because there is only one and there can be no other.

> But, I don't buy the argument made here:

> > To call gravity a force, is to privilege flat space as somehow being special.

> Flat space is special, and we didn't make it special.

Neither is curved spacetime. Metrics like the Schwarzschild metric merely let us map between the two. They inherently do this, so flat spacetime seems like inherently as real as -no more, and no less than- curved spacetime. And in flat spacetime gravity is very much a force.

We should really not teach that gravity curves spacetime, but that there are equivalent representations of what massive objects do to their surroundings. The reason we should teach both alternatives is that humans have a hard time wrapping their minds around "curved spacetime". Giving a person two different ways to think about the same effects will increase the likelihood that they will understand, and only at a very small cost of increased cognitive load (but since understanding eases the cognitive load, the overall effect should be to reduce cognitive load).

Pedagogy matters.

This is not article-content-related but article-presentation-related: the trouble with 3-pixel-wide scrollbars is that you have to be very exact when you try to use them. Yes, there are scroll wheels and keyboard buttons, but forcing the mobile experience (on a mobile I probably wouldn't notice) on non-mobile setups is at least annoying.

Actually even Newtonian gravity is no force, because Newtonian spacetime is curved:

https://youtu.be/IBlCu1zgD4Y?feature=shared

It must be so, because it is just a approximation to general relativity.

Whatever comes out of this kerfuffle, the based dial will be set to 11.

It’s unfortunate that we can’t monetize this stuff in the same vein as crypto and ChatGPT because I’d love to hear YouTube grifters telling you how you can make money overnight from quarks and neutrinos.

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