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In Sean Carroll's GR book pg. 109, it was said that the geodesic equation for timelike paths can be written in terms of the four momentum $p^\mu=mU^\mu=m\frac{dx^\mu}{d\tau}$: $$p^\lambda \nabla_\lambda p^\mu=0$$ It was then said that this relation expresses the idea that freely falling particles keep moving in the direction in which their momenta are falling. I don't understand why this equation implies that. Is there some kind of dot product involved here?

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    $\begingroup$ Yes, there is a dot product at $p^{\lambda} \nabla_{\lambda}$. If you'd like you can instead view it as $p^{\lambda} ( \nabla_{\lambda} p^{\mu})$ which is just another way to say $p\cdot (\nabla p^{\mu})$ (technically $(p\cdot \nabla)p^{\mu}$). $\endgroup$
    – Triatticus
    Sep 5, 2021 at 15:19

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If the $\nabla$ were $\partial$, the equation would simplify by the chain rule to $m\frac{d}{dt}p^\mu=0$, so the momentum is "conserved" in the sense of being unchanged by a change in $\tau$. This notion of free motion is just the intuition of Newton's first law, in relativistic language. But since a geodesic can be described with any choice of affine parameter, not just the proper time $\tau$, we generalize the $\partial$ to $\nabla$.

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Along time-like or null geodesic we can associate "rate of change" wrt some affine parameter $\tau$ as $$\frac{d}{d\tau}=P^a\nabla_a$$ where $P^a$ is tangent vector along geodesic. Then the geodesic equation can be stated as $\frac{dP^a}{d\tau}=0$. The intuition is that, if we work in a flat space-time and in non-relativistic limit (assume $P^a=mU^a$ with $U^a\approx (1,\vec{v})$), we see that $$\frac{d}{d\tau}=\partial_t+\vec{v}.\vec{\nabla}$$ which is our usual time derivative (negelecting an overall factor of $m$). Thus geodesic equation in this limit will boil down to simple free particle equation:$$\frac{d\vec{v}}{d\tau}=0$$.

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A geodesic, when parametrized by the arc lenght, can be expressed as the parallel transport of the tangent vector being zero. $U^\mu=\frac{dx^\mu}{d\tau}$ is exactly the tangent vector with such parametrization.

$$U^\lambda \nabla_\lambda U^\mu=0$$

Multiplying by $m$ the $U$'s we get the expression for the momentum.

For a visualisation, we could change from the spacetime to the surface of a spherical planet perfectly smooth without friction. The momentum of a free moving particle along the surface would follow a great circle, the geodesic in this case.

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