[proofplan] The strategy is to show that the squared speed $f(t) := g_{\gamma(t)}(\dot\gamma(t), \dot\gamma(t))$ has $f'(t) \equiv 0$ on $I$, hence is constant; constancy of $|\dot\gamma| = \sqrt{f}$ follows since $f \ge 0$. The derivative $f'(t)$ is computed by differentiating along $\gamma$ using the covariant derivative along the curve $\frac{\nabla}{dt}$ (which is the natural derivative for sections along $\gamma$, since $\dot\gamma$ is a section of $\gamma^* TM$, not a vector field on $M$). Two ingredients are needed: the metric-compatibility of the Levi-Civita connection in its **covariant-derivative-along-a-curve form** ($\frac{d}{dt} g(V, W) = g(\frac{\nabla V}{dt}, W) + g(V, \frac{\nabla W}{dt})$ for any lifts $V, W$ along $\gamma$), and the geodesic equation $\frac{\nabla \dot\gamma}{dt} = 0$. Combining the two yields $f'(t) = 0$ at every $t \in I$ in two lines. [/proofplan] [step:Reduce constancy of $|\dot\gamma|$ to constancy of $f := g(\dot\gamma, \dot\gamma)$] Define the squared-speed function \begin{align*} f: I &\to \mathbb{R}_{\ge 0} \\ t &\mapsto g_{\gamma(t)}\bigl(\dot\gamma(t), \dot\gamma(t)\bigr). \end{align*} Since $\dot\gamma(t) \in T_{\gamma(t)} M$ and $g_{\gamma(t)}$ is the inner product on $T_{\gamma(t)} M$, the value $f(t) \ge 0$ is well defined. Smoothness of $f$ follows from smoothness of $\gamma$ and $g$. We have $|\dot\gamma(t)|_g = \sqrt{f(t)}$. The square root is smooth on $\{t : f(t) > 0\}$, and $f$ is constant on $I$ if and only if $\sqrt{f}$ is constant on $I$ (taking square roots preserves and is preserved by constancy on the non-negative reals). Thus it suffices to prove $f'(t) = 0$ for all $t \in I$. [/step] [step:Differentiate $f$ using metric compatibility of $\frac{\nabla}{dt}$] The Levi-Civita connection $\nabla$ is metric-compatible: $\nabla g = 0$. The same holds for its associated covariant derivative along a curve. Concretely, for any pair of smooth lifts $V, W: I \to TM$ of $\gamma$ (i.e., $V(t), W(t) \in T_{\gamma(t)} M$), the function $t \mapsto g_{\gamma(t)}(V(t), W(t))$ is smooth and satisfies \begin{align*} \frac{d}{dt} g_{\gamma(t)}(V(t), W(t)) &= g_{\gamma(t)}\!\left(\frac{\nabla V}{dt}(t),\, W(t)\right) + g_{\gamma(t)}\!\left(V(t),\, \frac{\nabla W}{dt}(t)\right). \tag{MC} \end{align*} This is the [metric-compatibility identity for the covariant derivative along a curve](/page/Levi-Civita%20Connection), which follows from $\nabla g = 0$ by combining axiom (3) of the [Covariant Derivative Along a Curve](/theorems/2708) with the general identity $X(g(Y, Z)) = g(\nabla_X Y, Z) + g(Y, \nabla_X Z)$ for vector fields $X, Y, Z \in \mathfrak{X}(M)$ — applied locally with $X$ extending $\dot\gamma$ and $Y, Z$ extending $V, W$. (We verify (MC) carefully in the `[guided]` block, since it is the main analytic input.) Apply (MC) with $V = W = \dot\gamma$ (which is a smooth lift of $\gamma$ to $TM$ by smoothness of $\gamma$): \begin{align*} f'(t) &= \frac{d}{dt} g_{\gamma(t)}\bigl(\dot\gamma(t), \dot\gamma(t)\bigr) = 2\, g_{\gamma(t)}\!\left(\frac{\nabla \dot\gamma}{dt}(t),\, \dot\gamma(t)\right). \end{align*} We used the symmetry $g(V, W) = g(W, V)$ to combine the two terms on the right of (MC) into a factor of $2$. [guided] The metric-compatibility identity (MC) is the form of $\nabla g = 0$ adapted to sections along curves. Let us verify it from scratch. Fix $t_0 \in I$ and choose a chart-and-frame neighbourhood as in the proof of the [Covariant Derivative Along a Curve](/theorems/2708). Write $V(t) = \sum_i V_i(t) e_i(\gamma(t))$ and $W(t) = \sum_i W_i(t) e_i(\gamma(t))$ in a local orthonormal frame for $TM$ — alternatively, use a coordinate frame and the metric components $g_{ij}$. We use a coordinate frame $\{\partial_{x_k}\}$ for clarity. Then $g_{\gamma(t)}(V(t), W(t)) = \sum_{i,j} g_{ij}(\gamma(t)) V_i(t) W_j(t)$. Differentiating in $t$ with the product and chain rules, \begin{align*} \frac{d}{dt}\bigl(g_{ij}(\gamma(t)) V_i(t) W_j(t)\bigr) = \sum_k \dot x_k(t) (\partial_{x_k} g_{ij})(\gamma(t)) V_i W_j + g_{ij}(\gamma(t)) \dot V_i W_j + g_{ij}(\gamma(t)) V_i \dot W_j. \end{align*} Now invoke metric compatibility on $M$: $\partial_{x_k} g_{ij} = g(\nabla_{\partial_{x_k}} \partial_{x_i}, \partial_{x_j}) + g(\partial_{x_i}, \nabla_{\partial_{x_k}} \partial_{x_j}) = \sum_\ell \Gamma_{ik}^\ell g_{\ell j} + \sum_\ell \Gamma_{jk}^\ell g_{i\ell}$. Substituting and rearranging, the three terms regroup into $g(\frac{\nabla V}{dt}, W) + g(V, \frac{\nabla W}{dt})$ using the local formula $(\frac{\nabla V}{dt})_i = \dot V_i + \sum_{j,k} \Gamma_{jk}^i V_j \dot x_k$ from the previous theorem. The mechanical computation is straightforward; the structural reason the identity holds is that $\nabla g = 0$ holds on $M$ and the covariant derivative along $\gamma$ is constructed to preserve all algebraic identities of $\nabla$ that involve only directional differentiation along $\dot\gamma$. Why apply (MC) to $V = W = \dot\gamma$? Because $\dot\gamma$ is itself a smooth lift of $\gamma$ to $TM$, and the squared-speed function $f$ is exactly $g(\dot\gamma, \dot\gamma)$ — so (MC) gives $f'$ in terms of $\frac{\nabla \dot\gamma}{dt}$, which is what the geodesic equation controls. Note that $\dot\gamma$ is **not** in general the restriction to $\gamma$ of any vector field on $M$ — it is intrinsically a section of $\gamma^* TM$, which is why the covariant derivative *along the curve* is the right object here, not $\nabla_X Y$ for vector fields $X, Y$ on $M$. [/guided] [/step] [step:Apply the geodesic equation $\frac{\nabla \dot\gamma}{dt} = 0$ to conclude $f' \equiv 0$] By the definition of a [geodesic](/page/Geodesic), $\gamma$ satisfies \begin{align*} \frac{\nabla \dot\gamma}{dt}(t) &= 0 \qquad \text{for all } t \in I. \end{align*} Substituting into the expression for $f'$ from Step 2, \begin{align*} f'(t) &= 2\, g_{\gamma(t)}\!\left(\frac{\nabla \dot\gamma}{dt}(t),\, \dot\gamma(t)\right) = 2\, g_{\gamma(t)}(0,\, \dot\gamma(t)) = 0 \end{align*} for every $t \in I$. Hence $f$ is constant on the connected interval $I$. By Step 1, $|\dot\gamma(t)|_g = \sqrt{f(t)}$ is also constant on $I$. This completes the proof. [/step]