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Comment by marcus_holmes

4 days ago

I am in a gravitational field. I have no idea what my acceleration is, I just know that I feel 1G (I could be falling in a stronger gravity and only feel 1G, or I could be climbing in a weaker gravity and feel 1G). The only way of determining it is to see if I'm moving relative to the stuff around me. Even then, that's not definite - I could be in an elevator and everything around me is also accelerating.

I'm not disagreeing with you, I'm just pointing out that there are circumstances where "you can determine your acceleration without any external reference" isn't true. You might even say that this is relative to your circumstances ;)

19 comments

marcus_holmes

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adrian_b  2 days ago

If you measure acceleration in a single point, without other information you cannot know whether it is caused by gravitational attraction or by movement that is not rectilinear and uniform.

However, if you measure acceleration in many points, you typically can discriminate the 2 cases, because the spatial variation of the 2 kinds of acceleration fields is normally very different.

If you also move relatively to the local system of reference while measuring acceleration, you have additional distinguishing information from Coriolis forces.

So with enough measurements, gravitational forces and inertial forces can always be separated.

p1mrx  4 days ago

According to general relativity, you (and the ground) are accelerating at 1g, and feel weight because your inertia resists that acceleration. If you jump off a cliff, you'll stop accelerating for a bit, until the ground hits you.

Edit to reply:

> I am standing on the ground. I feel 1G acceleration. My speed is not changing. How much am I accelerating?

You are accelerating at 1g through curved spacetime. Newtonian "speed" behaves strangely in curved spacetime.

  • JumpCrisscross  4 days ago

    > According to general relativity, you (and the ground) are accelerating at 1g

    I don't believe this is correct. If I lock two rockets in opposition to each other, they aren't accelerating. They're pushing at each other. And their propellant is accelarating away. But their displacement and orientation are unchanging, which means their velocity is zero which means acceleration isn't happening.

    Similarly, the normal force resists your gravitational force to produce zero net acceleration. (An object at rest in a gravity well is its own local frame.)

    > If you jump off a cliff, you'll stop accelerating for a bit, until the ground hits you

    I don't believe this is correct. In GR, free fall is still inertial motion. You're just free of fictitious forces and thus following the curvature of spacetime.

    • ben_w  4 days ago

      As I understand it, in GR acceleration is indistinguishable* from gravity, so while you're on the ground feeling 1 gee, you're being accelerated up at 1 gee, and so is the ground.

      When you're in free-fall, that's when you're in a non-accelerating frame, even though a non-relativistic description** would say that you are, in fact, accelerating.

      Caveat: I only do physics as a hobby, neither academically nor professionally, so take with appropriate degree of doubt.

      * for point-like observers at least

      ** ignoring rotation and curved orbits

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    • tsimionescu  4 days ago

      It is correct, and you're also right that two rockets tethered to each other would not feel acceleration. The acceleration we feel in Earth's gravitational field is affecting our speed, though - it's slowing down the speed at which we move towards the future.

      1 reply →

    • adrian_b  2 days ago

      Declaring some forces as "fictitious" is just a convention and in my opinion it is a useless convention that brings more complexities than simplifications.

      In my opinion, a much healthier way of thinking is to not apply any such conventional labels to some forces and to just treat all forces equally, which is simpler, and it also matches the real world, where there is no practical difference between "fictitious" and "non-fictitious" forces.

      If you treat all forces equally, then the rule (a.k.a. Newton's 2nd law) is that the resultant of all forces that act in a point is always null. From this, the conservation of energy is an obvious consequence, because whenever that point moves the total mechanical work is null.

      The mechanical work of the "fictitious" forces is the variation of the kinetic energy, in the same way as the mechanical work of other forces is equal to the variation of some corresponding kinds of energy. For instance, the mechanical work of an elastic force is the variation of a potential elastic energy.

      A "fictitious" force can squish you like a bug exactly like a "non-fictitious" force, and when that happens you would not be saved by the thought that the force is "fictitious".

      There is a baseless claim that "fictitious" forces can be distinguished from "non-fictitious" forces because they depend on the system of reference. However the same is true for some of the so-called "non-fictitious" forces, which are functions of velocities and accelerations of bodies, in the same way as the inertial forces. Moreover, what actually varies between systems of reference is how forces are distributed into various kinds of components, not the resultant forces.

      If you are pressed on a wall by a so-called "fictitious" inertial force, no change in the system of reference will change the compression force that you feel, but it may change the interpretation of the kind of forces that result in the compression that you feel.

      In general relativity the "fictitious" forces do not disappear, but like in classic mechanics where you can define the "fictitious" forces using the variation of the kinetic energy, in relativistic mechanics there is an analogous definition based on the variation of the momentum-energy quadri-vector. In the gravitation theory of Einstein you can compute this inertial 4-force using the gravity force that is derived from the "curvature" of the space (which is in turn determined by the spatial distribution of the momentum-energy 4-vector of matter). The rule of a null resultant 4-force remains true in general relativity, so for a body that moves freely, like a satellite or a body in free fall, the inertial 4-force is equal in magnitude and opposite in sign with the gravitational 4-force.

      In some problems of general relativity, you do not need to compute the inertial forces, because e.g. the trajectory of some body might be along a geodesic and that is all that is of interest for you. This is the same like when you use kinematics to determine the possible movement of a mechanism, where you want to know the path on which something moves, but you do not care which forces are exerted on the parts of the mechanism. However, in other problems of general relativity, you may want to know the forces that act upon a body, for instance for computing the required strength of materials, and then you may need to compute inertial forces, exactly like in classic mechanics.

      The physics as taught in schools is full of such useless conventions caused by historical accidents. Moreover, besides conventions that just make things more complex than they should be, the standard textbooks contain definite mistakes that have been perpetuated for generations, like wrong definitions for all physical quantities related to rotation motions. In conclusion, trusting the "authority" of the school textbooks is a mistake and people must attempt to verify with their minds all that they are taught, instead of trusting. As a schoolboy I was more skeptical than most, so some of the less competent teachers feared my questions, but I still believed much of what I was being taught and only years later I realized that I was duped.

      The fact that the AI models are trained on scientific literature that contains widespread inefficient methods or even mistakes, guarantees that they will provide wrong answers in comparison with really competent humans.

  • marcus_holmes  4 days ago

    I am standing on the ground. I feel 1G acceleration. My speed is not changing. How much am I accelerating?

    • brabel  4 days ago

      You say later that you think gravity and acceleration look the same but cannot be the same , which is funny since that’s exactly what relativity says: if two things are indistinguishable from each other even in principle, then they must be the same. Which is what led Einstein to realize that gravity really is just a curvature in space time. Hard to wrap your head around that! But if you study relativity, you eventually understand what being relative actually means.

    • layer8  4 days ago

      Your speed relative to what? There is no absolute speed. Relative to an inertial rest frame, you're accelerating upwards at 1G, which is what you are feeling and what an accelerometer is measuring. Of course, relative to the non-inertial reference frame of the ground, your speed doesn't change.

    • tsimionescu  4 days ago

      You need to take into account your entire 4-vector for speed. You don't just have a speed in the 3 spatial coordinates, you're also moving thorough the "time" coordinate, and that is happening at a slower pace near a large mass like the Earth than it would of you were far away from here.

    • hcs  4 days ago

      You are more quickly being carried by the ground further from where you would otherwise be. Hope that clears it up.

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qubitcoder  4 days ago

You can always hold an accelerometer in your hand. If you did so now, assuming you're on Earth's surface, it'd register approximately 9.8m/s/s pointing in the upward direction.

You could also perform one of many historical experiments, such as dropping an object from an elevated height with careful timing, or rolling a round ball down a gently sloped track, and so on.

BDPW  4 days ago

You're conflating coordinate and proper acceleration.