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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

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 https://x.com/skdh/status/1850120005070799153

tomp 2y ago
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.
- Filligree 2y ago
  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.
- recursive 2y ago
  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.
- dcrazy 2y ago
  An iPhone held “at rest” cannot be accelerating downward, by definition.
  - moralestapia 2y ago
    Force is not acceleration. That is extremely easy to understand but somehow you and all of these "eminent scientists" fail to grasp it.
    - Filligree 2y ago
      Force is not acceleration, and accelerometers do not measure force. We have other devices which do that, such as bathroom scales.
      - moralestapia 2y ago
        Yup. Paraphrasing, you should not measure gravity with an accelerometer, otherwise this thread happens.
        Filligree 2y ago
        You can, but you should probably take three years of physics courses first.
- AlexandrB 2y ago
  https://old.reddit.com/r/AskElectronics/comments/2ufunr/why_is_my_atrest_accelerometer_showing_1g_with/ 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.
- warble 2y ago
  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.
  - tomp 2y ago
    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
  - deleted 2y ago
    [deleted]
- taeric 2y ago
  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.)
  - Filligree 2y ago
    > 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.
    - taeric 2y ago
      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.
      - Filligree 2y ago
        > 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.)
TeMPOraL 2y ago
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?
- MichaelZuo 2y ago
  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.
  - TeMPOraL 2y ago
    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?
    - wrs 2y ago
      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?
    - tsimionescu 2y ago
      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?
    - gorkish 2y ago
      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.
CollinEMac 2y ago
> 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.
- nostrebored 2y ago
  That’s relative to the surface of the earth
  - Calavar 2y ago
    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.
    - tsimionescu 2y ago
      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.
slightwinder 2y ago
> 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?
- LaffertyDev 2y ago
  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 https://www.youtube.com/watch?v=p1W0dpunOaM
- __MatrixMan__ 2y ago
  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.
- tsimionescu 2y ago
  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.
  - Volundr 2y ago
    > 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.
gostsamo 2y ago
Some annoying student would mention that the same force is acting on the accelerometer as well so it cannot act as an independent observer.
cryptonector 2y ago
> > 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).*
ars 2y ago
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.
- Filligree 2y ago
  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.
  - tzs 2y ago
    A rocket would not accelerate all parts of the accelerometer equally.
    - Filligree 2y ago
      Not in the first couple of seconds, but by the time everything settles down it certainly will.
      - deleted 2y ago
        [deleted]
      - cryptonector 2y ago
        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.
bbor 2y ago
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!
- tsimionescu 2y ago
  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".
mrkeen 2y ago
> 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?
- cryptonector 2y ago
  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.
ithkuil 2y ago
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
- cryptonector 2y ago
  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.)
  - tsimionescu 2y ago
    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.
    - ithkuil 2y ago
      I don't understand. Why would it show an acceleration in deep space?
      - deleted 2y ago
        [deleted]
    - cryptonector 2y ago
      The accelerometers in our ears show the same acceleration when we're falling in diving airplane as when in orbit in deep space.
alok-g 2y ago
>> 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.
- hnuser123456 2y ago
  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.
  - alok-g 2y ago
    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.
    - hnuser123456 2y ago
      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.
rich_sasha 2y ago
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.
amadeoeoeo 2y ago
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...
- scrubs 2y ago
  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.
  - _kulang 2y ago
    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
gattis 2y ago
not a single one of those points is true lol

tomp 2y ago

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.

Filligree 2y ago
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.
recursive 2y ago
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.
dcrazy 2y ago
An iPhone held “at rest” cannot be accelerating downward, by definition.
- moralestapia 2y ago
  Force is not acceleration. That is extremely easy to understand but somehow you and all of these "eminent scientists" fail to grasp it.
  - Filligree 2y ago
    Force is not acceleration, and accelerometers do not measure force. We have other devices which do that, such as bathroom scales.
    - moralestapia 2y ago
      Yup. Paraphrasing, you should not measure gravity with an accelerometer, otherwise this thread happens.
      - Filligree 2y ago
        You can, but you should probably take three years of physics courses first.
AlexandrB 2y ago
https://old.reddit.com/r/AskElectronics/comments/2ufunr/why_is_my_atrest_accelerometer_showing_1g_with/ 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.
warble 2y ago
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.
- tomp 2y ago
  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
- deleted 2y ago
  [deleted]
taeric 2y ago
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.)
- Filligree 2y ago
  > 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.
  - taeric 2y ago
    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.
    - Filligree 2y ago
      > 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.)

TeMPOraL 2y ago

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?

MichaelZuo 2y ago
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.
- TeMPOraL 2y ago
  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?
  - wrs 2y ago
    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?
  - tsimionescu 2y ago
    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?
  - gorkish 2y ago
    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.

CollinEMac 2y ago

> 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.

nostrebored 2y ago
That’s relative to the surface of the earth
- Calavar 2y ago
  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.
  - tsimionescu 2y ago
    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.

slightwinder 2y ago

> 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?

LaffertyDev 2y ago
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 https://www.youtube.com/watch?v=p1W0dpunOaM
__MatrixMan__ 2y ago
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.
tsimionescu 2y ago
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.
- Volundr 2y ago
  > 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.

gostsamo 2y ago

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

cryptonector 2y ago

> > 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).*

ars 2y ago

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.

Filligree 2y ago
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.
- tzs 2y ago
  A rocket would not accelerate all parts of the accelerometer equally.
  - Filligree 2y ago
    Not in the first couple of seconds, but by the time everything settles down it certainly will.
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    - cryptonector 2y ago
      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.

bbor 2y ago

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!

tsimionescu 2y ago
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".

mrkeen 2y ago

> 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?

cryptonector 2y ago
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.

ithkuil 2y ago

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

cryptonector 2y ago
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.)
- tsimionescu 2y ago
  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.
  - ithkuil 2y ago
    I don't understand. Why would it show an acceleration in deep space?
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  - cryptonector 2y ago
    The accelerometers in our ears show the same acceleration when we're falling in diving airplane as when in orbit in deep space.

alok-g 2y ago

>> 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.

hnuser123456 2y ago
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.
- alok-g 2y ago
  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.
  - hnuser123456 2y ago
    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.

rich_sasha 2y ago

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.

amadeoeoeo 2y ago

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...

scrubs 2y ago
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.
- _kulang 2y ago
  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

gattis 2y ago

not a single one of those points is true lol