$F = ma$ In General Relativity

Question

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)?

Answer 1

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.

Answer 2

Newton's law $F = m a$ 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" and $F$ is a "genuine force" that is applied to a particle (i.e. electromagnetic, elastic, hydrodynamic, etc...).

Newton's perspective: "$m g $ is a genuine force, so we have to include it into the total force $F$".

Einstein's perspective: "standing on the Earth's surface we have $F = m (a-g)$, so that $m g=- m a_{fr}$ is a fictitious force due to the fact that we are not falling along a geodesics".