Coordinate-independent metric implies constant metric along geodesic

Question

I am struggling to understand a problem I was given. The problem is as follows:

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Show that if the metric does not explicitly depend upon a coordinate (e.g. $x^1$) then the term $g(\dot{x}, \partial_1)$ is constant along every geodesic.

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The thing is, I am not sure how to interpret $g(\dot{x}, \partial_1)$, since $\dot{x}$ is a vector, whereas $\partial_1$ is not (since $\partial_j$ would be the gradient, i suppose $\partial_1$ is just one component).

Furthermore, how do I prove this? What I have so far and seems promising but the last step is missing:

Considering Euler-Lagrange Equations for the Lagrange function $L = \tfrac{1}{2} g_{ab} \dot{x}^a \dot{x}^b$ we can conclude:

$\partial_1 L = 0$

and therefore

$\frac{d}{dt}\frac{\partial L}{\partial \dot{x}^1} = 0 \quad \Longrightarrow \quad \frac{\partial L}{\partial \dot{x}^1} = \text{const.} = g_{i1} \dot{x}^i$

Any ideas on how to complete this?

Answer 1

In differential geometry $\partial_i$ is usually tangent vector (or vector field) to the coordinate curve of $i$-th coordinate. The curve is given by the parametric equations: $$x^j(t)=x_0^j+\delta^j_i t$$ so the vector $\partial_1$ has components $\delta^j_1.$

The completion is by using the fact that $$ g(\dot{x},\partial_1)=g_{ij}\dot{x}^i\delta_1^j=g_{i1}\dot{x}^i $$