[proofplan] We prove the rescaling identity $\gamma_p(\lambda t, a) = \gamma_p(t, \lambda a)$ by constructing a candidate curve $\sigma(t) := \gamma_p(\lambda t, a)$ and verifying that it solves the geodesic initial-value problem with initial data $(p, \lambda a)$. Uniqueness of solutions to the geodesic ODE — which is a smooth second-order ODE on $TM$ via the geodesic spray — then forces $\sigma(t) = \gamma_p(t, \lambda a)$ on the common domain. The verification has three pieces: (i) check $\sigma(0) = p$ and $\dot\sigma(0) = \lambda a$ via the chain rule; (ii) check $\frac{\nabla \dot\sigma}{dt} = 0$ via the chain-rule for the covariant derivative along a reparametrised curve; (iii) invoke uniqueness of geodesics with given initial data. The case $\lambda = 0$ requires a brief separate treatment because $\lambda a = 0$ corresponds to the constant geodesic at $p$. [/proofplan]

[step:Set up the candidate curve $\sigma$ and verify its initial conditions] Fix $p \in M$, $a \in T_p M$, and $\lambda \in \mathbb{R}$. Let $\gamma_p(\cdot, a): I_a \to M$ denote the maximal geodesic with $\gamma_p(0, a) = p$ and $\dot\gamma_p(0, a) = a$, where $I_a \subseteq \mathbb{R}$ is the maximal open interval containing $0$ on which the geodesic is defined. For $\lambda \neq 0$, define the reparametrised curve \begin{align*} \sigma: \tfrac{1}{\lambda} I_a &\to M \\ t &\mapsto \gamma_p(\lambda t, a), \end{align*} where $\tfrac{1}{\lambda} I_a := \{t \in \mathbb{R} : \lambda t \in I_a\}$. The map $t \mapsto \lambda t$ is smooth and bijective $\mathbb{R} \to \mathbb{R}$ for $\lambda \neq 0$, so $\sigma$ is smooth on its domain (composition of smooth maps). For $\lambda = 0$, define $\sigma(t) = \gamma_p(0, a) = p$ for all $t \in \mathbb{R}$ — the constant curve at $p$, which is a geodesic by direct check ($\dot\sigma \equiv 0$, so $\frac{\nabla \dot\sigma}{dt} = 0$ by direct verification), with $\sigma(0) = p$ and $\dot\sigma(0) = 0 = 0 \cdot a$. Since $\gamma_p(t, 0)$ is the constant geodesic at $p$ by uniqueness ($\dot\gamma(0) = 0$ and the geodesic ODE has the constant solution as a fixed point), the identity $\gamma_p(0 \cdot t, a) = \gamma_p(t, 0 \cdot a)$ holds by direct substitution. Henceforth assume $\lambda \neq 0$. By the chain rule applied to the smooth map $t \mapsto \lambda t$, \begin{align*} \dot\sigma(t) = \frac{d}{dt}\bigl[\gamma_p(\lambda t, a)\bigr] = \lambda\, \dot\gamma_p(\lambda t, a). \tag{$*$} \end{align*} Evaluating at $t = 0$: \begin{align*} \sigma(0) = \gamma_p(0, a) = p, \qquad \dot\sigma(0) = \lambda \dot\gamma_p(0, a) = \lambda a. \end{align*} So $\sigma$ satisfies the initial conditions of the geodesic with data $(p, \lambda a)$. [/step]

[step:Verify the geodesic equation $\frac{\nabla \dot\sigma}{dt} = 0$]We must show that $\sigma$ satisfies the geodesic ODE on its domain. The key computational tool is the **chain rule for the covariant derivative along a reparametrised curve**: if $\beta: J \to M$ is a smooth curve and $\phi: J' \to J$ is a smooth diffeomorphism (or just smooth map), then for any smooth lift $V$ of $\beta$, \begin{align*} \frac{\nabla(V \circ \phi)}{dt}(t) &= \phi'(t)\, \frac{\nabla V}{ds}(\phi(t)). \tag{CR} \end{align*} (Here $s$ is the variable on $J$ and $t$ is the variable on $J'$.) This is a standard identity following from the local formula for $\frac{\nabla}{dt}$: in a chart with frame, both sides have components \begin{align*} \dot V_i(\phi(t))\, \phi'(t) + \sum_{j,k} \Gamma_{jk}^i(\beta(\phi(t)))\, V_j(\phi(t))\, \dot \beta_k(\phi(t))\, \phi'(t), \end{align*} where the factor $\phi'(t)$ multiplies both the $\dot V$ term (chain rule on $V \circ \phi$) and the $\dot\beta_k$ term (since $\frac{d}{dt} \beta_k(\phi(t)) = \dot\beta_k(\phi(t))\, \phi'(t)$). Apply (CR) twice. First, note that by $(*)$, $\dot\sigma(t) = \lambda\, \dot\gamma_p(\lambda t, a)$. Setting $\beta(s) = \gamma_p(s, a)$, $\phi(t) = \lambda t$ (so $\phi'(t) = \lambda$), and $V(s) = \dot\gamma_p(s, a)$, the lift $\dot\sigma$ equals $\lambda \cdot (V \circ \phi)$. By $\mathbb{R}$-linearity of $\frac{\nabla}{dt}$ (axiom 1 of the [Covariant Derivative Along a Curve](/theorems/2708)), \begin{align*} \frac{\nabla \dot\sigma}{dt}(t) = \lambda\, \frac{\nabla(V \circ \phi)}{dt}(t). \end{align*} Apply (CR): \begin{align*} \frac{\nabla(V \circ \phi)}{dt}(t) &= \phi'(t) \cdot \frac{\nabla V}{ds}(\phi(t)) = \lambda \cdot \frac{\nabla \dot\gamma_p(\cdot, a)}{ds}(\lambda t). \end{align*} Combining, \begin{align*} \frac{\nabla \dot\sigma}{dt}(t) = \lambda^2 \cdot \frac{\nabla \dot\gamma_p(\cdot, a)}{ds}(\lambda t). \end{align*} Now $\gamma_p(\cdot, a)$ is a geodesic by construction, so $\frac{\nabla \dot\gamma_p(\cdot, a)}{ds}(s) = 0$ for all $s \in I_a$. Substituting $s = \lambda t \in I_a$ (which is the condition $t \in \tfrac{1}{\lambda} I_a$), \begin{align*} \frac{\nabla \dot\sigma}{dt}(t) = \lambda^2 \cdot 0 = 0 \qquad \text{for all } t \in \tfrac{1}{\lambda} I_a. \end{align*} Hence $\sigma$ satisfies the geodesic equation on its domain.[/step]

[guided]We must show $\sigma$ satisfies the geodesic ODE on its domain. The strategy is: identify $\dot\sigma$ as a rescaling of a known geodesic velocity, apply the chain rule for the covariant derivative along a reparametrised curve, and use the geodesic equation for $\gamma_p(\cdot, a)$ to conclude. The key tool is the **chain rule for the covariant derivative along a reparametrised curve**. If $\beta: J \to M$ is a smooth curve and $\phi: J' \to J$ is a smooth map, then for any smooth lift $V$ of $\beta$, \begin{align*} \frac{\nabla(V \circ \phi)}{dt}(t) &= \phi'(t)\, \frac{\nabla V}{ds}(\phi(t)). \tag{CR} \end{align*} Why the single factor $\phi'(t)$? In a chart with frame, both sides have components \begin{align*} \dot V_i(\phi(t))\, \phi'(t) + \sum_{j,k} \Gamma_{jk}^i(\beta(\phi(t)))\, V_j(\phi(t))\, \dot \beta_k(\phi(t))\, \phi'(t). \end{align*} The factor $\phi'(t)$ multiplies *both* the time-derivative of the lift (ordinary chain rule on $V \circ \phi$) and the velocity of the underlying curve in the Christoffel-symbol term (since $\frac{d}{dt} \beta_k(\phi(t)) = \dot\beta_k(\phi(t))\, \phi'(t)$). Both scalings collapse into the clean prefactor $\phi'(t)$ on the right-hand side. Now we set up the application. By $(*)$, \begin{align*} \dot\sigma(t) = \lambda\, \dot\gamma_p(\lambda t, a), \end{align*} so $\dot\sigma$ is **not** simply $V \circ \phi$: there is an extra factor of $\lambda$ from differentiating $\gamma_p(\lambda t, a)$ in $t$. Set $\beta(s) := \gamma_p(s, a)$, $\phi(t) := \lambda t$ (so $\phi'(t) = \lambda$), and $V(s) := \dot\gamma_p(s, a)$. Then $\dot\sigma = \lambda \cdot (V \circ \phi)$. This factor of $\lambda$ is **passive** — it floats out of the covariant derivative by $\mathbb{R}$-linearity (axiom 1 of the [Covariant Derivative Along a Curve](/theorems/2708)): \begin{align*} \frac{\nabla \dot\sigma}{dt}(t) = \lambda\, \frac{\nabla(V \circ \phi)}{dt}(t). \end{align*} Now apply (CR) to the term $\frac{\nabla(V \circ \phi)}{dt}$. Here $\phi'(t) = \lambda$ is the **active** chain-rule factor — it is multiplicative because of the reparametrisation: \begin{align*} \frac{\nabla(V \circ \phi)}{dt}(t) &= \phi'(t) \cdot \frac{\nabla V}{ds}(\phi(t)) = \lambda \cdot \frac{\nabla \dot\gamma_p(\cdot, a)}{ds}(\lambda t). \end{align*} Combining the passive and active factors, \begin{align*} \frac{\nabla \dot\sigma}{dt}(t) = \lambda^2 \cdot \frac{\nabla \dot\gamma_p(\cdot, a)}{ds}(\lambda t). \end{align*} The $\lambda^2$ is morally because the geodesic equation is **second-order**: under $t \mapsto \lambda t$, velocity scales by $\lambda$, acceleration by $\lambda^2$. The right-hand side is "acceleration" — it scales by $\lambda^2$. Since the original right-hand side is zero, this scaling is harmless. Finally we use that $\gamma_p(\cdot, a)$ is a geodesic by construction, so \begin{align*} \frac{\nabla \dot\gamma_p(\cdot, a)}{ds}(s) = 0 \qquad \text{for all } s \in I_a. \end{align*} The substitution $s = \lambda t$ takes values in $I_a$ exactly when $t \in \tfrac{1}{\lambda} I_a$ — which is precisely the domain of $\sigma$. Substituting, \begin{align*} \frac{\nabla \dot\sigma}{dt}(t) = \lambda^2 \cdot 0 = 0 \qquad \text{for all } t \in \tfrac{1}{\lambda} I_a. \end{align*} Hence $\sigma$ satisfies the geodesic equation on its entire domain.[/guided]

[step:Invoke uniqueness of geodesics with given initial data to conclude]We have shown that $\sigma: \tfrac{1}{\lambda} I_a \to M$ is a smooth curve satisfying \begin{align*} \sigma(0) = p, \qquad \dot\sigma(0) = \lambda a, \qquad \frac{\nabla \dot\sigma}{dt} \equiv 0. \end{align*} The maximal geodesic $\gamma_p(\cdot, \lambda a): I_{\lambda a} \to M$ is uniquely characterised by these properties: it is the unique solution to the geodesic initial-value problem with initial data $(p, \lambda a)$. Uniqueness of solutions follows from the standard ODE theory applied to the [geodesic spray](/page/Geodesic), a smooth second-order vector field on $TM$ whose flow is the [exponential map](/page/Exponential%20Map); existence and uniqueness of integral curves of a smooth vector field with given initial conditions is the standard Picard–Lindelöf theorem. Therefore, on the intersection of their domains, \begin{align*} \sigma(t) &= \gamma_p(t, \lambda a), \end{align*} i.e. $\gamma_p(\lambda t, a) = \gamma_p(t, \lambda a)$. The intersection of the domains contains $0$ in its interior, and equality of two smooth integral curves of the same ODE with matching initial conditions extends to the **maximal** common interval — that is, to the entire intersection $\tfrac{1}{\lambda} I_a \cap I_{\lambda a}$. This is the claimed identity, valid wherever both sides are defined. The proof is complete.[/step]

[guided]The pattern of this proof — define a candidate curve via a known one, check it solves the geodesic IVP, and conclude by uniqueness — is the standard approach to symmetry identities for geodesics. The same approach handles, for instance, the fact that geodesics are preserved under isometries. We have established in the previous steps that $\sigma: \tfrac{1}{\lambda} I_a \to M$ is smooth and satisfies \begin{align*} \sigma(0) = p, \qquad \dot\sigma(0) = \lambda a, \qquad \frac{\nabla \dot\sigma}{dt} \equiv 0. \end{align*} That is, $\sigma$ is a geodesic with initial point $p$ and initial velocity $\lambda a$. The maximal geodesic $\gamma_p(\cdot, \lambda a): I_{\lambda a} \to M$ has the same defining properties — it is by definition the unique solution to the geodesic initial-value problem with data $(p, \lambda a)$. Why is this solution unique? The geodesic ODE $\frac{\nabla \dot\gamma}{dt} = 0$ with initial data $(\gamma(0), \dot\gamma(0)) = (p, \lambda a)$ corresponds to an integral curve of the [geodesic spray](/page/Geodesic), a smooth second-order vector field on $TM$ whose flow is the [exponential map](/page/Exponential%20Map). Existence and uniqueness of integral curves of a smooth vector field through a given point in $TM$ is the Picard–Lindelöf theorem. Hence any two smooth curves satisfying the geodesic IVP with matching initial conditions agree on the connected component of $0$ in the intersection of their domains. Applying this uniqueness statement to $\sigma$ and $\gamma_p(\cdot, \lambda a)$: \begin{align*} \sigma(t) &= \gamma_p(t, \lambda a) \qquad \text{for all } t \in \tfrac{1}{\lambda} I_a \cap I_{\lambda a}. \end{align*} Unpacking the definition $\sigma(t) = \gamma_p(\lambda t, a)$, this is exactly \begin{align*} \gamma_p(\lambda t, a) &= \gamma_p(t, \lambda a), \end{align*} the claimed identity, valid wherever both sides are defined. A point of fineness about domains: in fact $\tfrac{1}{\lambda} I_a = I_{\lambda a}$, so the identity holds on the entire natural domain. The inclusion $\tfrac{1}{\lambda} I_a \subseteq I_{\lambda a}$ holds by maximality of $I_{\lambda a}$ — we just exhibited a geodesic with initial velocity $\lambda a$ defined on $\tfrac{1}{\lambda} I_a$, so $I_{\lambda a}$ is at least this large. The reverse inclusion follows by applying the same construction in the other direction: rescaling $\gamma_p(\cdot, \lambda a)$ by $1/\lambda$ produces a geodesic with initial velocity $a$ defined on $\lambda I_{\lambda a}$, and maximality of $I_a$ gives $\lambda I_{\lambda a} \subseteq I_a$, i.e., $I_{\lambda a} \subseteq \tfrac{1}{\lambda} I_a$. Hence the identity holds on the full common domain. Why does the identity matter? It is the fundamental scaling property of geodesics: one can **rescale time and tangent vector interchangeably**. This is the key fact used to define the [exponential map](/page/Exponential%20Map) $\exp_p(a) := \gamma_p(1, a)$ and to reduce questions about geodesics with arbitrary initial speeds to unit-speed geodesics. It is also the engine behind the scaling identity $(d\exp_p)_0 = \operatorname{id}_{T_p M}$ in the proof of the [Differential of the Exponential Map at the Origin](/theorems/2711). The proof is complete.[/guided]