0
$\begingroup$

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)?

| cite | improve this question | |
$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ Well the acceleration fictitious force for accelerated frames is identical to a gravitational force $\endgroup$ – Slereah Jan 8 '18 at 14:54
3
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

$F = m a$ 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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.