# Does the Newton's law break scale invariance?

Under a scale transformation $$t\rightarrow \bar{t}=\mu t\hspace{0.3cm}\text{and}\hspace{0.3cm}\textbf{r}\to\bar{\textbf{r}}=\lambda\textbf{r},\tag{1}$$ Newton's law take the form $$m\frac{d^2\textbf{r}}{dt^2}=\textbf{F}\Rightarrow m\frac{d^2\bar{\textbf{r}}}{d\bar{t}^2}=\frac{\lambda}{\mu^2}\textbf{F}.\tag{2}$$ which shows that Newton's law is not scale-invariant for a time-independent $\textbf{F}$.

This looks surprising to me because scaling investigates whether the physics is same at all scales (of magnification), and scale invariance is broken/spoiled if there is a built-in length scale or time scale in the problem. Now, Newton's law for a particle of mass $m$ is not scale invariant as I've shown in (2).

What is the reason for this? There is no built-in length scale or time scale in the problem that one can construct from the $\textbf{F}$ and $m$. Therefore, physically it is surprising to me. Does it mean that breakdown of scale invariance has nothing to do with intrinsic length scale or time-scale?

• Your statement is rather obviously correct. What's the real question you want to ask? Why are you investigating time scaling symmetry for Newton's laws? You sound surprised to find it is not invariant - why would you have assumed it would be? Jan 20 '17 at 22:03
• @ACuriousMind I've throughly edited the question and the title, and made it clearer. Does it now convey the question clearly?
– SRS
Jan 21 '17 at 13:17

What you have shown is that Newton's law is not scale-invariant for a force $F(x,\dot{x},t)$ that is scale-invariant, since you implicitly assumed that $F$ transforms as a scalar under the dilation1. This is kind of a trivial statement: If the l.h.s. of an equation transforms non-trivially and you assume that the r.h.s. transforms trivially, the equation as a whole cannot be in- or covariant.

The point is that it is a priori undetermined how $F$ transforms under such a transformation. It is the precise functional form of $F$ that determines whether or not the equation of motion is invariant under any transformation, in particular the scale transformation.

Your confusion seems to be that you expect "Newtonian mechanics" to exhibit scale symmetry. But symmetries are properties of physical systems, not of physical theoretical frameworks. Since many Newtonian systems have equivalent Lagrangian descriptions in which we can apply Noether's theorem, expecting all Newtonian systems to have scale invariance is patently absurd, since this would expect all of them to have a corresponding conserved quantity. Your "explicit length/time scales" are simply hidden from you because you haven't picked a particular system and therefore an explicit expression for $F$.

1Time-independence is not enough to guarantee that Newton's law is not scale-invariant, consider the force $F = \frac{\dot{r}^2}{r}$ as a counter-example.

• 1. I get your point. Since the scale transformation property of $\textbf{F}$ is not known a priori, because the form of $\textbf{F}$ is not known, Eq.(2) does not prove that Newton's law is not scale invariant. 2. But if $\textbf{F}$ is a constant force then Eq.(2) does break the scale invariance. 3. Isn't that physically surprising? Surprising because there is no intrinsic time scale or length scale in the problem that can be constructed from the constant $F$ and $m$. Then why is the scale invariance broken with constant $\textbf{F}$ (physically)? @ACuriousMind
– SRS
Jul 8 '17 at 6:34

Newton unit is $kg⋅m/s^2$. The unit tells you that if you scale the time by $t \to \lambda t$, the result won't be invariant, it will be divided by $\lambda^2$. It tells you too that there are infinitely many ways of scaling the other units (meter and mass) to obtain an invariant force. For example, it you scale the length by $r \to λ^2r$ and leave the mass unchanged, then the force will be invariant. You could also scale the length by $r \to λr$ and the mass by $m \to λm$...

If a given quantity is invariant under time scaling (with no other scaling), it simply means that the unit of this quantity does not involve time.