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I wonder when a particle undergoes free fall onto Earth's surface, does all components of its proper acceleration vanish so that the magnitude of proper acceleration is zero? If a charged particle undergoes "free fall" (uniform acceleration) with acceleration of $g$ in electric field, does all components of its proper acceleration also vanish in this case? If not, why?

The components of 4-acceleration $A$ are

$$A^\lambda = \frac{dU^\lambda }{d\tau } + \Gamma^\lambda {}_{\mu \nu}U^\mu U^\nu$$

where $U$ is 4-velocity. Based on my understanding, the magnitude of $A$ gives proper acceleration: $\alpha=||A||$. So I must show that $||A||=0$ during free fall. I suspect that during free fall, it must be the case that $\frac{dU^\lambda }{d\tau } = -\Gamma^\lambda {}_{\mu \nu}U^\mu U^\nu$. I am not sure why it is true.

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A particle in free fall performs geodesic motion, so that its equation of motion is given by the geodesic equation, which is (in local coordinates): $$ \frac{d^2X^\mu}{d\tau^2} + \Gamma^\mu{}_{\rho\sigma}\frac{dX^\rho}{ds}\frac{dX^\sigma}{ds} = 0 $$ If we now call the worldline tangent $\frac{dX^\rho}{ds} = U^\rho$, we immediately end up at your statement. The point is that in general relativity, proper acceleration denotes the deviation of a particle's worldline from geodesic motion - this is the general relativistic analog of Newton's first law, which assumes the flat-space case in which all geodesics are straight lines. "Free fall" in this case means exactly that: The particle is "free" in the sense that no external forces act on it, so that it is in an inertial frame.

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  • $\begingroup$ So the geodesic equation tells that the 4-acceleration vector of a particle in free fall is $A=(0,0,0,0)$ right? I wonder what the zero time component $A^0=0$ signify. $\endgroup$
    – John Davies
    Commented Sep 25 at 18:16
  • $\begingroup$ @JohnDavies Note that one can also write the geodesic equation as $$ \frac{dU^\lambda }{d\tau }=A^\lambda - \Gamma^\lambda {}_{\mu \nu}U^\mu U^\nu$$ So that now $\dfrac{dU^\lambda }{d\tau }$ is also a four acceleration, but not the proper four acceleration. Its zeroth component will in general not vanish when the object is in motion in curved spacetime. This means that $A^0=0$ signifies that any observed temporal effects related to the motion of the body (like changes in relativistic mass for example too) are an artefact of the spacetime geometry and not a result of any real force $\endgroup$
    – Amit
    Commented Sep 25 at 18:52
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    $\begingroup$ @JohnDavies Near the surface of the earth, yes, exactly $\endgroup$
    – Amit
    Commented Sep 25 at 19:49
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    $\begingroup$ @JohnDavies To nitpick a bit, it's right for the radial direction. For the time component it is a bit different but overall, the time dilation effect from free falling near the surface of the earth will be negligible $\endgroup$
    – Amit
    Commented Sep 25 at 19:55
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    $\begingroup$ @JohnDavies Not really but in terms of keywords you can search something like "Newtonian limit" with "General relativity"... also see if this post here at least partially helps $\endgroup$
    – Amit
    Commented Sep 25 at 22:50

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