# $F = ma$ In General Relativity

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)?

• 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. Jan 8, 2018 at 8:03
• @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. Jan 8, 2018 at 12:24
• Well the acceleration fictitious force for accelerated frames is identical to a gravitational force Jan 8, 2018 at 14:54

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".