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It is currently understood that gravity is not actually a force, and a fact that is often used to show this is that an object in free fall doesn't "feel" that it is accelerating and is thus an inertial frame.

However, it seems to me that Newtonian mechanics can already predict that this will be the case. Since all the parts of the object are being accelerated by the same amount simultaneously (at least approximately, like near the surface of the Earth) there won't be a tendency for this object to contract, and thus "feel" that it is accelerating. This isn't the case, for example, when I push the same object. In my understanding, the force needs to be communicated from the point I apply it to all the rest of the object, and this delay is the cause of a contraction/increase in internal forces or tension, allowing it to effectively "feel" that is accelerating.

Now, suppose that there is a uniform field that accelerates any particles with constant acceleration. Like gravity, a free object in this field will not be able to detect that it is accelerating, since all of its particles are accelerating equally at the same time. My question is: Is this object inertial, or is it only inertial if this field is a gravitational one?

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  • $\begingroup$ "It is currently understood that gravity is not actually a force" Who said this? Do you mean you can jump up on earth ans just stay in the air? or what pulls you down? $\endgroup$
    – trula
    Commented Sep 24, 2023 at 14:42
  • $\begingroup$ @trula From what I know, it is a result from general relativity. Objects that are falling aren't acted upon by a force but are simply following a path according to the curvature of space. $\endgroup$
    – WordP
    Commented Sep 24, 2023 at 14:47
  • $\begingroup$ In this you are right, but it does not apply to Newton mechanics. an object has not to be deformed to be accelerated. Still on earth gravity is a force, otherwise see your jumping up. or your weight on a scale $\endgroup$
    – trula
    Commented Sep 24, 2023 at 14:55
  • $\begingroup$ It is not necessary that you "feel"( or more appropriately "measure" ) being accelerated only if there is a compression as @trula suggests. Also consider the case of an incompressible solid in a rocket moving up in space. It is obvious that it will feel a force by the walls(floor) without being compressed. $\endgroup$
    – khaxan
    Commented Sep 24, 2023 at 16:27
  • $\begingroup$ A closely related question: What does it mean that a falling mass in space doesn't sense any force?. Look at the comments in the first answer $\endgroup$
    – khaxan
    Commented Sep 24, 2023 at 16:31

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Now, suppose that there is a uniform field that accelerates any particles with constant acceleration. Like gravity, a free object in this field will not be able to detect that it is accelerating, since all of its particles are accelerating equally at the same time.

That field isn’t like gravity, it is gravity.

My question is: Is this object inertial, or is it only inertial if this field is a gravitational one?

Unfortunately, the question cannot arise since such a field is a gravitational field.

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    $\begingroup$ I suspected this could be the case. But why can't it be, say, an electric field? $\endgroup$
    – WordP
    Commented Sep 24, 2023 at 16:18
  • $\begingroup$ Because then the accelerations would not be the same. Different objects with different charge to mass ratios accelerate differently $\endgroup$
    – Dale
    Commented Sep 24, 2023 at 16:57
  • $\begingroup$ I see. But consider a single object with uniform density and charge distribution, so that all of its parts accelerate equally. It won't be able to detect it is accelerating by the same reasons I said before, so is it inertial? $\endgroup$
    – WordP
    Commented Sep 24, 2023 at 17:04
  • $\begingroup$ Such a field does not have the specified property that the field “accelerates any particles with constant acceleration”. It is precisely that universality of free fall that is the hallmark of gravity $\endgroup$
    – Dale
    Commented Sep 25, 2023 at 1:43
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If there is a uniform field (in a small region of space, since gravity is not uniform in general) accelerating any particles with an acceleration that is different from the gravity, and you're not able to "cancel" gravity, you're ending up with a resulting field equal to the vector sum of your uniform field and the gravitational field in that region of space.

So, you're accelerating w.r.t. a ``classical mechanics'' inertial frame with acceleration $\mathbf{a}$. If you take a 3-axis accelerometer with you its measurement reads $\mathbf{g}-\mathbf{a}$.

  • You're in a "classical mechanics" (that considers gravity as a force) if you read $\mathbf{g}$ on the accelerometer, and thus $\mathbf{a} = \mathbf{0}$
  • You're in a "general relativity" (that considers gravity as a result of the curvature of the space and not an actual force) if you read $\mathbf{0}$ and thus $\mathbf{a} = \mathbf{g}$, and thus you're in free fall.

In every other situation, you need to introduce a new definition of inertial frame, considering $\mathbf{g}-\mathbf{a}$, the measurement read on the 3-axis accelerometer (torques should be zero) by every point in motion with you or in relative uniform rectilinear motion w.r.t. you as the constant ``real force'', playing the role of gravity we perceive as an example in a small region on the Earth.

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