Galilean transformations are said to have 10 degrees of freedom. Four for translation in space and time, three for rotation, and three for direction of the uniform motion.

If I scale space axis by $\alpha$ and do the same with time axis, you can see that Newton's second law remains the same.

So why don't we consider scaling (of time and space) another type of Galilean transformation?

  • $\begingroup$ I agree that "scale" is a badly neglected degree of physical freedom. $\endgroup$ – Steve Jun 1 '19 at 23:50
  • $\begingroup$ @Steve, it is not a degree of freedom in classical mechanics. $\endgroup$ – Akerai Jun 2 '19 at 0:18
  • $\begingroup$ @Akerai, why exactly is that? Is it simply because no apparent practical means of changing the scale of physical things currently exists (and thus the theoretical possibility didn't need to be considered by the classical physicists)? $\endgroup$ – Steve Jun 2 '19 at 0:38
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    $\begingroup$ @Steve It’s because, as Akerai states in an answer here, the laws of classical mechanic are in fact generally not scale-invariant as OP claims. $\endgroup$ – Thatpotatoisaspy Jun 2 '19 at 3:10
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    $\begingroup$ Interestingly enough, the Schrödinger equation is both invariant under galilean transformations and dilations, even if it has a length scale (mass). It is even conformally invariant! cf. mathoverflow.net/a/270122/106114 $\endgroup$ – AccidentalFourierTransform Jun 2 '19 at 14:13

Imagine you take the transformation you mentioned above: $$x^i \rightarrow x'^i = \alpha x^i,\\ t \rightarrow t' = \alpha t,$$ where $\alpha \in \mathbb{R}$.

Then assuming the Newton's law holds in the new coordinates, it will be of the form $$F^i = m \frac{d^2x'^i}{dt' ^2} = m \frac{d^2 (\alpha x^i)}{ dt^2} \left(\frac{dt}{dt'} \right)^2.$$

As you can see, the derivative $dt/dt' = 1/ \alpha$, and therefore the equality becomes

$$m \frac{d^2x'^i}{dt' ^2} = m \frac{d^2 (\alpha x^i)}{ dt^2}\frac{1}{\alpha^2} = \frac{m}{\alpha} \frac{d^2 (x^i)}{ dt^2}.$$

Therefore the Newton's law is actually not invariant under this transformation as you claimed before, the transformation actually transforms an object of mass $m/\alpha$ to an object of mass $m$.

  • $\begingroup$ Thanks, I made a mistake in my calculation. Thank you so much for clearing up. $\endgroup$ – Shuheng Zheng Jun 2 '19 at 3:40
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    $\begingroup$ This make sense. If it is truly scale invariant, we probably won't need units on acceleration forces etc. $\endgroup$ – Shuheng Zheng Jun 2 '19 at 3:41

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