# Nonlinear time transformation and Newton's law

This question is based on Astrophysicist Thanu Padmanabhan's online lecture here, time around 13.00-14.00.

Consider a map where the absolute time $t$ is mapped to another absolute time $f(t)$ i.e., $$t\to \bar{t}=f(t)\tag{1}$$ where $f(t)$ is a single-valued, monotonically increasing function of time. For example, $f(t)=at^3$. Under this transformation, the Newton's law $$m\frac{d^2\textbf{r}}{dt^2}=\textbf{F}$$ changes to $$m\frac{d^2\textbf{r}}{d\bar{t}^2}=\alpha(t)\textbf{F}+\beta(t)\frac{d\textbf{r}}{dt}\tag{2}$$ where $\alpha(t),\beta(t)$ can be expressed in terms of $f(t)$ and/or its derivatives.

If I understand it correct, then he says such a transformation is not allowed because even if one starts with $\textbf{F}=0$, after one makes the time-transformation (1), the particle is no longer force-free due to the second term in (2) on the RHS. It looks like a "viscous drag".

But the real reason such a transformation is not allowed, I think, is that it makes the flow of time non-uniform. As time passes, the time flows faster and faster. Why should we focus at what happens to Newton's law under $t\to f(t)$ when such a time transformation is also not possible in special relativity. What is the significance of his mathematical exercise?

Transformations like that are allowed in general relativity, and in fact they are responsible for gravitational acceleration. In effect the transformation is creating a curved time coordinate $\bar{t}$ and the result is that the acceleration $d^2\mathbf x/d\bar{t}$ is no longer zero even when there is no applied force. This is exactly what happens in curved spacetime. The equation of motion of a freely falling body (i.e. $\mathbf F = 0$) is the geodesic equation:
$${d^2 x^\mu \over d\tau^2} = - \Gamma^\mu_{\alpha\beta} {dx^\alpha \over d\tau} {dx^\beta \over d\tau}$$
In fact transformations like that are allowed in special relativity as well, but the associated frame will be a non-inertial frame. For example consider the rest frame for an observer with constant acceleration (constant proper acceleration). The coordinates for that observer are the Rindler coordinates and the transformation to the time of the accelerated observer $\tau$ is given by:
$$\tau = \frac{c}{a} \sinh^{-1}\left(\frac{at}{c}\right)$$
where $a$ is the constant (proper) acceleration. This is precisely the sort of mapping function discussed in the video, and of course in an accelerated frame dropped objects accelerate away from you so Newton's law doesn't hold.