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By definition, under a unit conversion of time $$t\to\bar{t}=\alpha t\tag{1}$$ (i.e., measuring time in minutes rather than in seconds) the physics should not change, and therefore, the form of the equations too, don't change. Newton's law for a time-independent force $\textbf{F}$ transforms under unit conversion as $$m\frac{d^2\textbf{r}}{dt^2}=\textbf{F}\Rightarrow m\frac{d^2\textbf{r}}{d\bar{t}^2}=\frac{1}{\alpha^2 }\textbf{F}=\frac{1}{\alpha^2 }\alpha^2\bar{\textbf{F}}=\bar{\textbf{F}}$$ where the force $\bar{\textbf{F}}$ is measured not in Newton but $N/\alpha^2$. This is easy to understand why the numerical values of the accelerations are different by a factor of $\alpha^2$. This is because they are measured in two different units.

Now consider the equation $$m\frac{d^2\textbf{r}}{dt^2}+\gamma\frac{d\textbf{r}}{dt}+\omega^2\textbf{r}=\textbf{F}\tag{2}$$ which under the unit conversion (1) becomes $$m\alpha^2\frac{d^2\textbf{r}}{d\bar{t}^2}+\alpha^2\bar{\gamma}\frac{d\textbf{r}}{d\bar{t}}+\alpha^2\bar{\omega}^2\textbf{r}=\textbf{F}=\alpha^2\bar{\textbf{F}}$$ because under unit conversion $\omega\to\alpha\bar{\omega}$ and $\gamma\to\alpha\bar{\gamma}$. Therefore, the equation is again invariant as expected: $$m\frac{d^2\textbf{r}}{d\bar{t}^2}+\bar{\gamma}\frac{d\textbf{r}}{d\bar{t}}+\bar{\omega}^2\textbf{r}=\bar{\textbf{F}}.$$

Now, a scale transformation on time $$t\rightarrow\bar{t}=\lambda t\tag{3}$$ is fundamentally different from a trivial change of units. Under (3), the physics can change if the form of the equations change. For example, under (3), Eqn. (2) becomes $$m\lambda^2\frac{d^2\textbf{r}}{d\bar{t}^2}+\gamma\lambda\frac{d\textbf{r}}{d\bar{t}}+\omega^2\textbf{r}=\textbf{F}.$$ This is of different form than (2). This equation is not invariant under scale transformation of time.

A unit conversion is easy to understand and easy to implement in practice. Either use seconds or use minutes to measure i.e., either you count the time elapsed in seconds or in minutes. But how can I understand and implement scale transformations in practice? What kind of physical operation would correspond to making a scale transformation on time$\dagger$?

$\dagger$ However, scale transformation on space is easy to understand: to implement $\textbf{r}\to\lambda\textbf{r}$, one just need to zoom in by using a magnifying glass or higher and higher resolution (which is different from a unit conversion i.e., making measurements in centimeter and metre). I cannot think of what is the analogous way of understanding scale transformation of time.

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A unit conversion is not a transformation $t \mapsto \lambda t$. A unit conversion, is, rather by definition, a trivial transformation of $t$, since $1\ \mathrm{s} = 1/60\ \mathrm{min}$. Both sides are equal, there is no transformation of the equation happening. The choice of units is something external to the physical equations under consideration. When we write $F = ma$, this isn't a statement bound to a particular unit system. It's the statement that force is equal to mass times acceleration, regardless of the units used.

You might be confused by the sloppy wording of physicists who often "work in units of $c=1$", and then write e.g. $E=m$ instead of $E=mc^2$. What you should think here is that the $m$ in the first equation is really a new quantity $m' = m/c^2$ and reads $E=m'$, and in a unit system where $c = 1$, $m'$ and $m$ are numerically equal so in that unit system you can use $E=m$ to compute things.

A scale transformation of time is physically a transformation that makes time go slower/faster by the scaling factor. Its importance usually arises because such scale transformations are part of the conformal transformations.

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  • $\begingroup$ Dear @ACuriousMind I understand but there is a confusion. How can you make time go slower or faster in non-relativistic physics? Is it meaningful? Isn't the flow of time absolute in non-relativistic physics? $\endgroup$ – SRS Jan 21 '17 at 5:00
  • $\begingroup$ @SRS We cannot "make" time go slower or faster. I cannot spatially rotate the solar system either, yet nobody objects to making rotations in the Kepler problem. Transformations of time simply are possible transformations of classical mechanics, albeit usually better grasped in the Lagrangian or Hamiltonian formalisms. $\endgroup$ – ACuriousMind Jan 21 '17 at 14:19

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