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# Why Does $$F = m \cdot a$$?

Our question this week was:

Newton's laws (in 1-d) state F=ma, where a is the second time derivative of the position. Why does this equation have a second time derivative and not, say, 1,3,4, etc. time derivatives? In particular, what are the physical reasons behind this?

We first need to figure out the options we want to compare F=ma to. I'll start from a general construction that contains some possibilities and show what's wrong with each possibility besides F=ma. Let's first take as our set of possibilities that

F=\sumn cn \frac{d^n x}{dt^n}

where $$n=1...\infty$$ and $$c_n \neq 0$$ for only one value of $$n$$. F=ma then corresponds to $$c_2=m$$, all other $$c_n=0$$. This provides the scope of our inquiry. The scope is non-trivial in that it allows for different possibilities and it encompasses the original question, but it also is tractable, in that I have neglected anything goes type approaches (I've restricted to the position of an object being an analytic function for example). I have also assumed the definition of force remains unchanged. This is one of the hardest lessons to learn as a student - when is the scope too big or too small. We always want to ask the big questions, but asking things like "what is time?" will often get you nowhere without being more specific. Conversely, making only a small change can get you somewhere, but where is often not all that interesting.

We now look at the various values of $$n$$. Many of you had great answers about $$n=0$$ ($$F=c_0 x$$) and $$n=1$$ ($$F=c_1 v$$). Here's mine.

n=0: In nature, we must be able to describe physics of extended objects. Therefore there must exist distinct points $$x,y$$ in space. What would happen if we used $$F=mx$$? Well, if an object had no forces on it, then it must be at $$x=0$$. However, we have all sorts of objects in our world that are extended and are not moving, and so have no net force. Therefore just by the requirement that our physics must describe a world where things can be not moving, yet here or there have meaning, we can get rid of $$F=mx$$.

n=1: In this case, we usually apply the notion of relativity - that the physics in different inertial reference frames should be the same. Let's assume the principle of relativity holds here and choose an observer O, moving at a speed $$v$$ with respect to an object at some position $$x=0$$. Since our notion of force is the same, the force on the object is unchanged (or invariant) no matter what $$v$$ is. But, $$v$$ is a totally arbitrary quantity in that it depends on the choice of observer. Hence the n=1 case is mathematically inconsistent if we maintain the principle of relativity.

$$n\geq 2$$: This is where it gets interesting (at least to me). We usually have Newton's first law, part of which is that an object at rest tends to remain at rest unless acted on by a force. So, let's consider the F=0 case and ask what happens. If we had $$F=ma$$, then the $$F=0$$ solution is just $$x=v_0t + x_0$$. If we still assume the principle of relativity, then we can always transform to an inertial frame where $$x=0$$ for all time. This is good, as it means that Newton's first law holds. If we have the $$n>2$$ case, however, Newton's first law breaks down. For example, let $$n=3$$. Then the $$F=0$$ solution is $$x=A t^2+v_0t + x_0$$ where $$A$$ is a constant. Using relativity, we can still transform to a reference frame where $$v_0=x_0=0$$. Therefore the solution is $$x=At^2$$ in this frame. This is a problem though! Particles that experience no force can just fly off to infinity, thereby violating Newton's first law, not to mention things like the conservation of energy. A similar argument holds for all higher derivatives. In fact, higher derivative differential equations generally have solutions that are "unstable", they just fly off to infinity, violate energy conservation, and do other nasty things. This instability goes by the name of the Ostrogradski instability, and it generates a very important restriction on the number of time derivatives we can have in our theories.

In sum, we find that having a standard notion of force, the need to have here vs. there for static objects be meaningful, the principle of relativity, and the requirement that our solutions don't run off to infinity and violate energy conservation, constrain our equations for the motion of particles to be $$F=ma$$ and nothing else. These are the physical reasons behind this equation.

BONUS: For the adventurous types, there was an idea a decade or so ago called MoND (Modified Newtonian Dynamics) which postulated that $$F=ma + a^2/a_0$$, where $$a_0$$ is an acceleration scale. This theory was proposed to explain the motion of stars at the edges of galaxies, and it actually worked well for spiral galaxies. Dark matter has gotten rid of the need for MoND, but it was an area of active interest not all that long ago. You can read about it here. Which physical principle did MoND give up?

Note by David Mattingly
3 years, 11 months ago

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So, as an answer to the question asked in the end, did MoND give up the principle of inertia? Cuz, the second law of motion is the underlying explanation to inertia.

- 2 years, 1 month ago