I have the same question posted as Newton's equation under time translation except I am not seeking the physical justification of the first claim but rather the mathematical justification of the second claim. Specifically, $x(t)$ satisfies the 2nd order ODE $x''(t)=F(x(t),x'(t),t)$ known as Newton's equation. In an inertial coordinate system, this law is invariant under the galilean transformations so, for the time translation $t \mapsto t+s,$ it must be true that if $x(t)=\varphi(t)$ is a solution to the differential equation, then so is $\varphi(t+s)$ in such a system. From this, according to Arnol'd it follows that the right hand side of Newton's equation does not depend on time in such a coordinate system, i.e. $x''(t)=\phi(x(t),x'(t)).$

Why is the last sentence true? If $x(t)$ satisfies Newton's equation then so does $x(t+s)$ by the chain rule as shown in the answer of the question linked above. This does not seem to depend on the coordinate system being inertial. And, when we have $x(t+s)$ being a solution whenever $x(t)$ being one, then why does this imply that $F$ cannot depend explictly on $t?.$ The answer given in the first question shows that $F$ does not explictly depend on $t$ as one can just replace $t$ with $t+s$ and satisfy the same ODE via the chain rule.

Example: Let $F(x(t),x'(t),t)=x(t)+t.$ Then, solving Newton's equations with the appropriate initial conditions gives $x(t)=e^t+e^{-t}-t$ as a solution. Clearly, $x(t+s)=e^{t+s}+e^{-t-s}-(t+s)$ satisfies the same ODE, as we have $$\frac{d^2}{d(t+s)^2}x(t+s)=F(x(t+s),\frac{d}{d(t+s)}x(t+s),t+s).$$

P.S. I thought that the substance of this post should be a comment in the original answer but I did not have enough reputation to post the comment so I asked this question instead. Please let me know if this is bad etiquette and if I should delete this.


1 Answer 1


The assumption is that if $\varphi$ is a solution to Newtons equation then for any $s \in \mathbb{R}$ the following function $$\begin{align} \tilde{\varphi} : \mathbb{R} &\longrightarrow \mathbb{R}^3 \\ t & \longmapsto \varphi(t+s) \end{align}$$ is also a solution to Newtons equation meaning that: $$ \ddot{\tilde{\varphi}} (t) = F (\tilde{\varphi} (t), \dot{\tilde{\varphi}} (t),t) \tag{1} $$ for all $t \in \mathbb{R}$. Note that in general $\tilde{\varphi}$ does not need to be a solution to Newtons equation. Because of the chain rule we have $\dot{\tilde{\varphi}} (t) = \dot{\varphi}(t+s)$ and $\ddot{\tilde{\varphi}} (t) = \ddot{\varphi}(t+s)$. And so equation 1 becomes: $$ \ddot{\varphi}(t+s) = F (\varphi(t+s), \dot{\varphi}(t+s),t) $$ with $t=0$ $$ \ddot{\varphi}(s) = F (\varphi(s), \dot{\varphi}(s),0) \tag{2} $$ for all $s \in \mathbb{R}$. But since $\varphi$ is a solution to Newtons equation we also have $$ \ddot{\varphi}(s) = F (\varphi(s), \dot{\varphi}(s),s) $$ for all $s \in \mathbb{R}$. Combining this and equation 2 yields $$ F (\varphi(s), \dot{\varphi}(s),0) = F (\varphi(s), \dot{\varphi}(s),s). $$

Fix $s \in \mathbb{R}$ and let $x,v \in \mathbb{R}^3$. We can find a solution $\varphi$ to Newtons equation so that $\varphi(s) = x$ and $\dot{\varphi}(s) = v$ and therefore $$ F(x,v,s) = F(x,v,0) $$ for all $x,v\in \mathbb{R}^3$ and $s \in \mathbb{R}$.

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    $\begingroup$ To conclude you should also assume that $F$ is continuous, in order to have your last assumption: there is at least a solution for generic initial conditions. Uniqueness is not necessary here. $\endgroup$ Jul 12 at 15:38
  • $\begingroup$ Just to clarify, why is it true that $\tilde{\phi}$ is not always a solution to Newton's equation? Because $\ddot{\phi}(t)=F(\phi(t),\dot{\phi}(t),t) \implies \ddot{\phi}(t+s)=F(\phi(t+s),\dot{\phi}(t+s),t+s),$ correct? The example I gave seems to demonstrate that as well. $\endgroup$
    – Chordx
    Jul 12 at 16:23
  • $\begingroup$ @Chordx Yes that is correct, if $F$ depends on time we can see from what you wrote that $\tilde{\phi}$ will not be a solution in general by comparing it to the equation i wrote below equation 1. $\endgroup$
    – jd27
    Jul 12 at 17:28

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