# Can Newton's First Law be treated as a form of bias? [closed]

This is a quirky question, so bear with me. I'm fairly certain it is relevant. Also, I am using the term "bias" here very loosely. By that term, I basically mean any way in which a conclusion is affected by how data is collected, how it is modeled, or any other decision by an analyst (rather than by a feature of the actual underlying phenomenon). Now here's the question:

One way to formulate the basic problem of mechanics is to say that we want to model the position $$S$$ of an object or system of objects. Presumably $$S\in\mathbb{R}^{3}$$ for a single object or $$S\in\mathbb{R}^{3n}$$ for a system of $$n$$ objects. If $$H\in\mathbb{R}^{m}$$ represents the set of variables which are believed to affect the motion of our (collection of) object(s) then we want a model that looks something like

$$S(t) = F(H;t)$$

where the form of $$F$$ will not be specified. This could work, but $$S$$ depends on choice of origin. So no one should object if we instead model displacement

$$S(t)-S(0) = F(H;t)$$

which eliminates translational dependence on the coordinate frame. Now, maybe we run an A/B test where $$H=H_{A}$$ for the "A" test and $$H=H_{B}$$ for the "B" test. Note that we will see no effect when

$$F(H_{A};t) = F(H_{B};t)$$

which amounts to

$$S_{A}(t)-S_{A}(0) = S_{B}(t)-S_{B}(0)$$

and dividing by $$t$$ gives

$$\cfrac{S_{A}(t)-S_{A}(0)}{t} = \cfrac{S_{B}(t)-S_{B}(0)}{t}$$

All well and good. But what if we only have observational data to work with? Unable to experiment, we will likely just have to compare different periods of time where different $$H$$ conditions are observed. Suppose $$H_{A}$$ was observed for the interval $$(t_{A},t_{A}+\delta t)$$ whereas $$H_{B}$$ was observed for the interval $$(t_{B},t_{B}+\delta t)$$. Well, now the equation is

$$\cfrac{S(t_{A}+\delta t)-S(t_{A})}{\delta t} = \cfrac{S(t_{B}+\delta t)-S(t_{B})}{\delta t}$$

which says that the two intervals should have the same average velocity. From this alone we already get the sense that no effect will be detected which does not change velocity. We can make this more explicit though. Set $$t=t_{A}$$ and $$t_{B}=t+\delta\tau$$. We get

$$\cfrac{S(t+\delta t)-S(t)}{\delta t} = \cfrac{S(t+\delta t+\delta\tau )-S(t+\delta\tau)}{\delta t}$$

Doing a little more simple math

$$\cfrac{S(t+\delta t+ \delta\tau )-S(t+\delta\tau)} {(\delta t)(\delta\tau )} -\cfrac{S(t+\delta t)-S(t)} {(\delta t)(\delta\tau )} = 0$$

in the limit $$\delta t,\delta\tau\rightarrow 0$$ this is

$$\text{acceleration} = 0$$

so, to recap, no effect will be detected if it does not accelerate the object or at least one of the objects in the collection, and this followed straight-forwardly from the choice to model displacement. Now, it is not clear to me whether physicists have a single clear definition of force, but it seems like "something that affects motion" is a pretty general description thereof. And, from this simple argument, it would seem that no force should be detected when modeling constant-rate displacement. This looks suspiciously like Newton's First Law, but it results simply from the choice to model displacement rather than from any underlying quality of nature.

That's not a criticism or a dismissal. It's just another perspective. My question is: is it valid?

I've done a little write-up that expands on my thoughts, which can be found here: https://docdro.id/dnlLWvk. But it does not provide any critical context for the question at hand.

• This looks a lot like the traditional derivation of the laws of classical mechanics from the invariance over space and time translations (and space rotations, for a more general case). In other words, the Galilean group of isometries. If it is, then it's a very common derivation from the 19th century that got a powerful generalization at the beginning of the 20th century when physicists discovered Noether's theorem. Oct 10, 2022 at 23:14
• @Miyase Interesting. I have heard of Noether's theorem but never got around to looking into it. Would that lead me to an affirmative answer? Oct 10, 2022 at 23:16
• On a slightly humoristic note... I don't believe Newton and Kepler were ever given the budget to experiment with the solar system.... so the need to "experiment" is certainly not a pre-requisite of science. Slightly closer to your point, Galilean symmetry follows from nothing else than the existence of a homogeneous and isotropic metric space, time, the conventional definition of velocity and the requirement that all physics be relative. The slight problem with it is... that's only one of the two possible solutions and it's the wrong one. The other, correct one, is the Lorentz group. Oct 10, 2022 at 23:29
• @FlatterMann That was actually the origin of the question. Newton's first two laws look almost derivable from the formula for differences-in-differences estimation -- which is a tool for simulating experimental controls in observational settings. It seems like such a tool would be necessary in lieu of the ability to experiment. My attempt to explain this is in the linked doc (not that I'd expect you to go in read it, I just wanted to make this connection). The rest of your comment I don't entirely follow, but it sounds like something I should look in to. Oct 10, 2022 at 23:35
• I don't fully follow your derivation, but your conclusion is too strong given your assumption. An equation like $\dot{x}=F$ is fully consistent with time translation invariance. Newton's first law, $F=0$, would become $\dot{x}=0$, or "an object will remain at it's same position unless acted on by an external force", which is not correct. You need an extra assumption to derive Newton's first law, in particular the assumption that the laws of physics are the same in different inertial frames. I suspect that one of your steps doesn't logically follow, but maybe you slipped that assumption in. Oct 11, 2022 at 1:24

Newton's first law is actually a very special statement about dynamics, for which we can imagine dynamical systems that violate it. The law says that under an absence of interaction, an object will either be at rest, or else moving in a straight line with a uniform velocity. Or better, we can find a coordinate system - in both space and time - such that in every physical situation, if we have a collection of non-interacting objects, they must move with uniform velocity (including velocity $$\mathbf{0}$$). That is, it has a universal quantifier in it - even better, it is an axiom schema, with one instance for each natural number $$n$$ of objects we may consider, involving a predicate $$I_j$$ which means its arguments, of which there are $$j$$, are engaged in a $$j$$-body interaction.

The instance for $$n = 2$$, for example, looks like

$$\forall O_1 \forall O_2 \left[\neg I_2(O_1, O_2) \rightarrow \left(\frac{d\mathbf{v}_{O_1}}{dt} = \mathbf{0} \wedge \frac{d\mathbf{v}_{O_2}}{dt} = \mathbf{0}\right)\right]$$

where the quantifiers quantify over all physically possible scenarios for the objects $$O_1$$ and $$O_2$$, i.e. all possible shapes, compositions, sizes, and positions and velocities at time zero.

(There is no instance for $$n = 1$$, because a "one-body interaction" would go against the spirit or, we could say, there is only one such interaction, and that is the autokinetic interaction which is what causes objects to move.)

The rigorous part is the concept of interaction, which objects are either engaged in or not, and the best way to see this is to get imaginative and consider a fictional counterexample: We could imagine a universe where that non-interacting objects instead always ran in little circles of a certain radius about a suitable nearest grid point. This would be a perfectly valid dynamics. It would break (continuous) translational symmetry and thus also violate Newton's third law, but it would still work in that it is mathematically cogent as a way things could be for someone. It's not our universe, but maybe it's someone else's - who knows? But now consider what happens in your scenario in this fictitious universe. The noninteracting objects are clearly accelerating, but your method would still say acceleration is zero ... or does it?

And there we see that your last equation is not actually a statement about the accelerations observed individually. It's a statement that the accelerations observed in the second scenario equal those in the first:

$$\frac{S(t + \delta t + \delta \tau) - S(t + \delta \tau)}{(\delta t)(\delta \tau)} = \frac{S(t + \delta t) - S(t)}{(\delta t)(\delta \tau)}$$

and indeed in our hypothetical world that would be the case - they are both undergoing uniform circular acceleration. But that's not Newton's first law. Newton's first law, or this specific instance of it, specializes to that

$$\frac{S(t + \delta t + \delta \tau) - S(t + \delta \tau)}{(\delta t)(\delta \tau)} = 0$$

and

$$\frac{S(t + \delta t) - S(t)}{(\delta t)(\delta \tau)} = 0$$

individually (remember the quantifier - it loops transfinitely over each potential pair, triple, etc. of objects and considers their various interactions) for each scenario, and clearly that is false in this world. Your statement is that the two scenarios show no difference between each other, i.e. that the laws of physics do not change with respect to time. That's actually a useful point - it tells us energy is conserved, but that is not the same as Newton's first law.

So no, it's not a form of bias.