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I am studying general relativity and using the book Gravity by James Hartle. On page 170, he provides the following table: enter image description here

I don't understand what he means by "equation of motion" nor do I understand what \begin{equation} \frac{d^2 x^{\alpha}}{d \tau^2} = 0 \end{equation}

means. I assume $\frac{d^2 x^{\alpha}}{d \tau^2} = 0$ has something to do with the geodesic equation, but I am a bit lost.

Can someone

(1) explain why the stuff in the right-hand column is called "equations of motion"

(2) explain how this is relevant to understanding what the equation $\frac{d^2 x^{\alpha}}{d \tau^2} = 0$ stands for? In other words, after you explain (1), explain why this equation is an instance of an "equation of motion"

Qmechanic
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Stan Shunpike
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1 Answers1

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(1) Wikipedia

(2) The equation $$\frac{d^2 x^\alpha}{d\tau^2}=0$$ is the equation of motion of a free particle in special relativity. It is in a sense a generalization of Newton's first law. It says that free particles move in straight lines (recall that second derivatives annihilate lines). Since in special relativity the Christoffel symbols are trivially zero (in cartesian coordinates of course), the geodesic equation $$\frac{d^2x^\mu}{d\tau^2}+\Gamma^\mu_{\;\rho\sigma}\frac{dx^\rho}{d\tau}\frac{dx^\sigma}{d\tau}=0$$ reduces to the above equation. Conversely, the Equivalence Principle says that the first equation holds at precisely one point along the worldline in appropriate coordinates. It may be shown (see, e.g., Weinberg Gravitation and Cosmology p.70) that this implies the geodesic equation along the rest of the worldline.

Ryan Unger
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