Geodesic- the derivation of thei scalar product

Question

I would like to know how follows the $1$st $=$ in the very last formula in the snippet below (i.e. one on the l.h.s. below) $$\frac{d}{dt}\langle\frac{dx}{dt},\frac{dx}{dt}\rangle=2\langle\frac{dx}{dt},\frac{\nabla}{dt}(\frac{dx}{dt})\rangle =0$$

Answer 1

It seems like this boils down to understanding what is meant in your source by $\frac{\nabla}{\mathrm{d}t}$. In the comments you described the gradient, a sort of derivative of a function, but here $\frac{\nabla}{\mathrm{d}t}$ is a gadget for differentiating vector fields. For now, let's work in $\mathbb{R}^n$. Let $x=x(t)$ be a parameterized curve and let $v=(v^1,\dots,v^n)$ be a vector field. If we restrict $v$ to the curve $x(t)$, we can differentiate $v(x(t))$ with respect to $t$, obtaining a new vector field $$ \frac{\nabla}{\mathrm{d}t} v = \left( \frac{\mathrm{d} v^1}{\mathrm{d}t}, \dots , \frac{\mathrm{d} v^n}{\mathrm{d}t} \right) $$ defined along $x(t)$.

Now let $w=(w^1,\dots,w^n)$ be another vector field on $\mathbb{R}^n$. You can check that $$\frac{\mathrm{d}}{\mathrm{d}t} \langle v,w \rangle_{x(t)} = \left\langle \frac{\nabla}{\mathrm{d}t} v, w \right\rangle + \left\langle v, \frac{\nabla}{\mathrm{d}t} w \right\rangle .$$ This equation says that the covariant derivative along a curve preserves the inner product, and it's the reason we call $\nabla$ orthogonal. (You don't have to worry about what a connection is and what it means to say a connection is orthogonal at this point. Just note that $\nabla$ has this special property involving the metric.) All Riemannian manifolds have a unique torsion-free orthogonal affine connection, called the Levi-Civita connection; it's a gadget that let's us differentiate vector fields along curves.

I don't know what book you are using, but hopefully if you take a look at the section about covariant differentiation it should help. My suggestion would be to first make sure you understand the case for $\mathbb{R}^n$, then for submanifolds of $\mathbb{R}^n$, then for covariant differentiation on a Riemannian manifold intrinsically. The calculation you are asking about $$\frac{\mathrm{d}}{\mathrm{d}t} \left\langle \frac{\mathrm{d}x}{\mathrm{d}t},\frac{\mathrm{d}x}{\mathrm{d}t} \right\rangle_{x(t)} = \left\langle \frac{\nabla}{\mathrm{d}t} \frac{\mathrm{d}x}{\mathrm{d}t}, \frac{\mathrm{d}x}{\mathrm{d}t} \right\rangle + \left\langle \frac{\mathrm{d}x}{\mathrm{d}t}, \frac{\nabla}{\mathrm{d}t} \frac{\mathrm{d}x}{\mathrm{d}t} \right\rangle = 2 \left\langle \frac{\mathrm{d}x}{\mathrm{d}t},\frac{\nabla}{\mathrm{d}t} \frac{\mathrm{d}x}{\mathrm{d}t} \right\rangle$$ is just a special case of the property that $\nabla$ is orthogonal, plus the fact that the metric is symmetric. This equation implies that geodesics have constant speed.