Derivation of Newton's Second Law of Motion [duplicate]

Question

How does one derive the equation that $F=Δp/Δt$ without using $F=ma$? I reviewed Deriving $F = ma$ - Newton's Second Law of Motion, but it just assumed $F=Δp/Δt$.

Answer 1

Depends the framework in which you are working, in other words, in your system of axioms and your definitions. To give an example from pure Mathematics, in Euclidean space $\mathbb{R}^d$ a compact set can be defined as a subset which is closed and bounded or either as a subset with the property that any open cover admits a finite subcover. Now if you decide to define a compact set as one which is closed and bounded, then you can prove that any open cover has a finite subcover. Otherwise, if you decide to define a compact set by means of open coverings you can then prove that it is closed and bounded.

This is sort of what is going on here. In Newtonian Mechanics, Newton's second law is one axiom of the theory. So it is something you assume to be true and derive results from it. Since we are talking about Physics, these results are then later compared to experiment thereby validatings several tests of that underlying basic axiom.

But this need not be the case. Indeed you could work in another framework in which you can actually derive Newton's second law. To give a concrete example of what I mean, in Lagrangian Mechanics the basic axiom is another one. It is the variational principle which states that the path of evolution of a system is the one which extremizes the action $$S=\int_{a}^b L(q(t),\dot q(t),t)dt.\tag{1}$$

Here the evolution of the configuration of the system in time is described by one or more generalized coordinates $q(t)$ and the associated derivatives $\dot{q}(t)$. The equations of motion that follow from the variational principle are the Euler-Lagrange equations $$\dfrac{d}{dt}\dfrac{\partial L}{\partial \dot q}=\dfrac{\partial L}{\partial q}\tag{2}.$$

If one then defines the generalized force $F$ and the generalized momentum $p$ by the equations $$F\equiv \dfrac{\partial L}{\partial q},\quad p\equiv \dfrac{\partial L}{\partial \dot{q}}\tag{3}$$ then the Euler-Lagrange equation reads $$F=\dfrac{dp}{dt}\tag{4}.$$

In other words, in the framework of Lagrangian Mechanics, with suitable definitions of force and momentum, the Euler-Lagrange equation takes the form of Newton's second law.

In particular, in this theoretical framework the derivation of the Euler-Lagrange equation from the action by means of the variational principle is a derivation of Newton's second law.

So in conclusion your answer depends on the framework you are working on. In Newtonian Mechanics there is no derivation since it is one basic axiom. In other frameworks like Lagrangian Mechanics you may derive it from whatever other basic axiom is in place.

Answer 2

How does one derive the equation that F=Δp/Δt...

You can't derive it, it is a postulate of Newtonian mechanics. It is proven by comparison to experimental results.

The other form of the second law you wrote ($\vec F = m\vec a$) can be obtained from the basic postulate $\vec F = \frac{d\vec p}{dt}$ when the mass is constant.

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Update, to expand a bit more:

The above answer is pithy, but I think it is an appropriate answer to the question at the level of an introductory high-school or college mechanics/physics course.

Of course, Newton did not write his second law in the same way, nor did he arrive at his second law by divine inspiration (as we seem to have). But rather he conducted, and observed, and knew of, various mechanics experiments such as: (1) a single small particle moving near the surface of the earth under earth's gravity (in approximate vacuum or in air); (2) two small masses connected by a relatively inextensible string also moving under earth's gravity; (3) observations of the motion of the planets and other astronomical bodies; (4) etc.

Based on the results of such experiments, we don't usually jump right to momentum, but rather start off by considering acceleration, which is defined as the second time derivative of position, as well as mass, which is defined below and is fixed/unchanging in the above-mentioned observations/experiments.

A general statement in English that summarizes and generalizes the above-mentioned observations/experiments can be found in Whittaker's mechanics textbook:

"If any set of mutually connected particles are in motion, the acceleration with which any one particle moves is the resultant of the acceleration with which it would move if perfectly free, and accelerations directed along the lines joining it to the other particles which constrain its motion. Moreover, to the several particles A, B, C, ..., numbers $m_A$, $m_B$, $m_C$, ... can be assigned, such that the acceleration along AB due to the influence of B on A is to the acceleration along BA due to the influence of A on B in the ratio $m_B$:$m_A$. The ratios of these numbers $m_A$, $m_B$, ... are invariable physical constants of the particles."

The above statement, together with the definition of a "unit mass" (e.g., a kilogram weight), provides the working definition of mass.

Clearly, when the masses are fixed (and given the definition of momentum $\vec p = m \vec v$, where $\vec v$ is the first time derivative of the position), the two forms of Newton's second law ($F=ma$ and $F=dp/dt$) are equivalent.

Answer 3

The expression $$ \boldsymbol{F} = \tfrac{ {\rm d}}{{\rm d}t} \boldsymbol{p} $$

is the definition for momentum $\boldsymbol{p}$.

Momentum is the result of a net force $\boldsymbol{F}$ acting over a finite time

$$ \Delta \boldsymbol{p} = \int_{\Delta t} \boldsymbol{F}\,{\rm d}t $$

Momentum is measured with $\boldsymbol{p}=m \boldsymbol{v}$, but it is not defined by it. Momentum is the result of a force acting over time.

This means that if you assume a constant force $\boldsymbol{F}$ over a time period $\Delta t$ then

$$ \Delta \boldsymbol{p} = m \Delta \boldsymbol{v} = \boldsymbol{F} \Delta t $$

or

$$ \boldsymbol{F} = \frac{ \Delta \boldsymbol{p}}{\Delta t} = m \frac{ \Delta \boldsymbol{v}}{ \Delta t} $$

and for an infinitesimal time frame

$$ \boldsymbol{F} = \tfrac{\rm d}{{\rm d}t} \boldsymbol{p} = m \boldsymbol{a} $$

Answer 4

I'll answer in general. Not every law of Physics is "derivable". It's not pure math. In mathematics you have a set of axioms and then derive some conclusions from it. In Physics, derivations have no added value, unless it helps to build a new theory. Instead, added value is in verification of laws, principles, assumptions and what not, in experimental or observational ways.

Answer 5

You haven't specified what we are supposed to take as axioms in deriving this equation. One possibility is the following. Suppose a particle is moving in a one-dimensional potential $V(x)$. We assume that the particle obeys the conservation of energy, given as $$ \frac{d}{dt}\left(\frac{p^2}{2m}+V(x)\right)=0, $$ where we define momentum as $$ p = m\frac{dx}{dt}. $$ We also define the force as $$ F=-\frac{dV}{dx}. $$ Then differentiating the energy conservation equation gives $$ \frac{d}{dt}\left(\frac{p^2}{2m}+V(x)\right)=\frac{p}{m}\frac{dp}{dt} + \frac{dx}{dt}\frac{dV}{dx}=0 $$ Then using the definition of momentum and the definition of force, we have $$ \frac{dx}{dt}\frac{dp}{dt} - \frac{dx}{dt}F=0. $$ Whenever $dx/dt\neq 0$, it follows that $$ \frac{dp}{dt}=F. $$ However, conservation of energy alone is not sufficient to prove the statement when $dx/dt=0$.