# Inertial coordinate systems being invariant under time translation in Newton's Principle of Detrimancy

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.

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 $$s \in \mathbb{R}$$.
• 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. Jul 12 at 15:38
• 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. Jul 12 at 16:23
• @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.