# Physical meaning of geodesic equation $p^\lambda \nabla_\lambda p^\mu=0$

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?

• 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 $(p\cdot \nabla)p^{\mu}$). Sep 5, 2021 at 15:19

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 $$\nabla$$.
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:$$\frac{d\vec{v}}{d\tau}=0$$.
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.