Understanding Einstein Summation in the Geodesic Equation I am trying to teach myself general relativity. I believe I do not fully understand Einstein summation. I have two versions of the same question


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*Non-relativistically:   If $V^μ= ů$  (the velocity) why is $ü= 0$ equivalent to $V^μ∂_μV^ν=0$?  

*The same problem in relativity. Why is $V^μD_μV^ν= 0$ equivalent to the expanded geodesic equation? ($D_μ$ is the covariant  derivative.) I don't understand why the partial derivative in the covariant derivative is not multiplied by a factor of $V^μ$.
 A: Say you want to know the rate at which a function changes in the direction of $V$. This is just $V^\mu \partial_\mu$ by the chain rule*. The condition that $f$ does not change as we move along $V^\nu$ is the condition that $V^\nu \partial_\nu f=0$. So if $V^\nu$ itself does not change as we move in the direction of $V^\nu$, this is the condition $V^\mu \partial_\mu V^\nu=0$. So the answer to your first question is that it follows by the chain rule**. Apply the equivalence principle to get that, in the general case, we should have $V^\mu \nabla_\mu V^\nu=0$. (The equivalence principle states, more or less, "partial derivatives go to covariant derivatives")
One caveat is that we do have to imagine $V$ to be a little vector field, at least locally, so that we can actually take its partial derivatives. This is more just abuse of notation than an actual problem. 
*(Proof: imagine a path $\gamma^\mu(t)$ where $V^\mu=\frac{d}{dt}\gamma^\mu(t)$. That is, it takes in a scalar $\lambda$ and returns the coordinates of the point. Then the rate of change of a function $f$ as we move along the path is $\frac{d}{dt}f(\gamma^\mu(t))$. Apply the chain rule to get that this is equal to $V^\mu \partial_\mu f$. Interestingly: note that we did not have to invoke the metric in this discussion to this point)
**(A notational confusion: we're concerned with $\dot{u}$ where $u$ is the velocity. We don't have to consider $\ddot{u}$. It's the second derivative of position, but the first derivative of the velocity.)
