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I'm no expert in general relativity, so please bear with any misconceptions in my understanding :)

In general relativity, Einstein showed that we experience gravity because standing on earth is actually being in a non-inertial (accelerating) frame of reference in a curved space-time.

Only free falling along a geodesic contoured by the curvature of the local space-time is considered an inertial frame of reference.

On the other hand, we are led to believe that Newton's second law: $F=ma$ is valid only when one is in an inertial frame of reference.

So shouldn't $F=ma$ be invalid in most use cases classical mechanics (obviously it is valid, but what am I missing)?

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    $\begingroup$ Yes, $F = ma$ is not accurate in most frames in classical mechanics. You will usually get coordinate forces such as the acceleration force, centrifugal force, Coriolis force or Euler force in addition to $F$ in a general coordinate frame. $\endgroup$ – Slereah Jan 8 '18 at 8:03
  • $\begingroup$ @Stereah. I do not think this is related to the Q. We are aweare of fictitious forces even in classical mechanics. The answer is when a "classical" gravity force appear as fictitious force in a non- inertial frame in GR as in the answer below. $\endgroup$ – Alchimista Jan 8 '18 at 12:24
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    $\begingroup$ Well the acceleration fictitious force for accelerated frames is identical to a gravitational force $\endgroup$ – Slereah Jan 8 '18 at 14:54
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Classically, gravity appears in a force diagram as a regular force (albeit one that depends on the mass of the object). This is necessary when we assume the surface of the earth represents a (nearly) inertial frame.

Because the same frame in GR is non-inertial, we can expect fictitious forces to appear. The classical gravitational force appears this way and makes the force diagram sum up as expected.

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$F = m a$ is valid only when one is in an inertial frame of reference. In a non-inertial frame you have $F = m (a+a_{fr})$, where $-m a_{fr}$ is a "fictitious force". $F$ is a "genuine force" that is applied to the particle (electromagnetic, elastic, hydrodynamic, etc...).

Newton: $m g $ is a genuine force, so include it into the total force $F$.

Einstein: standing on the surface of our planet we have $F = m (a-g)$, so that from the General Relativity point of view $m g=- m a_{fr}$, a fictitious force due to the fact that we are not falling along a geodesics.

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