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In 4-acceleration,

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

I wonder whether the term $\frac{dU^\lambda}{d\tau}$ (for indexes $\lambda=0,1,2,3$) should be called coordinate acceleration or not. In one answer, John Rennie called it coordinate acceleration and I got confused.

Based on my knowledge, coordinate acceleration of an object is defined as $\boldsymbol{a}=\frac{d^2\boldsymbol{x}}{dt^2}$ where $\boldsymbol{a}$ is spatial part of 4-acceleration $A$, i.e. $\boldsymbol{a}=(A_1, A_2, A_3)$, $\boldsymbol{x}$ is spatial part of 4-position $X$, i.e. $\boldsymbol{x}=(X_1, X_2, X_3)$ (I think positions are recorded by some observer observing the object in some reference frame), and $t$ is time measured probably by the same observer (one who records $\boldsymbol{x}$).

What's the universally agreed definition for coordinate acceleration?

Qmechanic
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Jimmy Yang
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1 Answers1

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Let's take the example of a freely falling particle near the surface of the earth. Since such a particle never gets really close to relativistic speeds, we can take $\gamma \approx 1$. This also implies we can take $t\approx \tau$, and so the four velocity of such a free falling particle is: $$ U^\lambda = \gamma(c,\vec{v})\approx(c,-gt\ \hat{z}) $$ Now, let's try to apply this to the equation you wrote for the proper acceleration, $A^\lambda$:

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

Since this particle is in geodesic motion, it experiences no proper acceleration. An accelerometer attached to it will give a reading of $0$. Now since $t\approx\tau$ we also see that $\mathrm{d}U^3/\mathrm{d}\tau\approx -g$. In this approximation the only non vanishing component of the equation is for $\lambda=3$ (the $z$-direction) and it reads:

$$ \Gamma^3 {}_{\mu \nu}U^\mu U^\nu \approx g $$

Now, to come back to the main question, why call $\mathrm{d}U^3/\mathrm{d}\tau$ a coordinate acceleration? Because, as we can see it is not an acceleration that an observer freely falling with the particle's frame measures. Rather, in setting up the problem we have implicitly in fact assumed that the components of $U^\lambda$ will be written with respect to an external coordinate system, of some other observer on the surface of the earth. For example, we made the arbitrary choice that negative $\hat{z}$ will stand for the downward vertical direction.

In fact, both the $U^\lambda$ and the $\Gamma^\lambda{}_{\mu\nu}$ take on coordinate dependent values. Nonetheless the combination of them always produces the proper acceleration vector. The magnitude of this vector is invariant and frame independent. In particular, if it vanishes in one frame, it will vanish in every frame. This is in fact expressed in what is known as the geodesic equation:

$$\frac{dU^\lambda }{d\tau } + \Gamma^\lambda {}_{\mu \nu}U^\mu U^\nu = 0 $$

Which describes inertial motion on a geodesic worldline, in the absence of external forces. The equation holds good regardless of the coordinate system we pick.

Amit
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