When physicists teach classical mechanics, they often refer to Newton's 2nd Law as "F=ma" or more technically as:

$$\vec{F}=m\vec{a}$$

But this isn't technically what Newton wrote.What Newton actually wrote (quoted from here) was:

The alteration of motion is ever proportional to the motive force impressed; and is made in the direction of the right line in which that force is impressed.And we need to take a step back for a moment and consider how Netwon defined "motion" via Definition II:

The quantity of motion is the measure of the same, arising from the velocity and quantity of matter conjunctly.And lastly, we need Newton's Definition I, which gives us the quantity of matter as:

The quantity of matter is the measure of the same, arising from its density and bulk conjunctly.I'm not going to get into some deep discussion on what Netwon actually meant by all of this. But the way it's frequently interpreted is this.

The

*quantity of matter*is what we refer to as "mass", so his definition looks like:

\[\begin{align*}

m = \rho V \tag{definition of mass}

\end{align*}\]

where $\rho$ is density and $V$ is volume.

The

*quantity of motion*is what we refer to as "momentum" and is defined as:

\[\begin{align*}

\vec{p}=m\vec{v}\tag{definition of momentum}

\end{align*}\]

where $\vec{p}$ is the momentum and $\vec{v}$ is the velocity.

Newton's 2nd Law thus states that:

$$\vec{F} \sim \Delta \vec{p}$$

which simply says that it's proportional to the change in momentum. The change is typically defined relative to time. So what this actually looks like is this:

$$\vec{F} = c \frac{d\vec{p}}{dt}$$

So we already have a problem here. There's nothing to suppose that the constant in this equation has to be 1. But we could potentially define our measuring units so the constant ends up being 1. This leads to the following statement of Newton's 2nd Law:

\[\begin{align*}

\vec{F} = \frac{d\vec{p}}{dt} \tag{Newton's 2nd Law}

\end{align*}\]

Now there's probably an interesting question on whether or not Newton's 2nd Law

*defines*force or whether or not force is defined by something else. Newton's actual definition is:

An impressed force is an action exerted upon a body, in order to change its state, either of rest, or of moving uniformly forward in a right line.Now I'm not sure how to give that a mathematically precise definition. So I tend to think of Newton's 2nd Law as defining force. But let's leave that debate aside for the moment.

What I now want to do is derive the physicists' explication of Netwon's 2nd Law ($\vec{F}=m\vec{a}$) from the formulations I provided above. We'll actually need one more definition here: acceleration which we'll simply define as $\vec{a} = \frac{d\vec{v}}{dt}$

So let's state this a bit formally (I'll use Roman numerals to denote assumptions and definitions and Arabic numbers to denote theorems):

\[\begin{align*}

\vec{p}=m\vec{v} \tag{I: definition of momentum}\\

\vec{a} = \frac{d\vec{v}}{dt} \tag{II: definition of acceleration} \\

\vec{F}= \frac{d\vec{p}}{dt}\tag{III: Newton's 2nd Law}

\end{align*}\]

So here are the derivations:

\[\begin{align*}

\vec{F}&= \frac{d(m\vec{v})}{dt} \tag{1: from I and III} \\

\vec{F}&= m\frac{d\vec{v}}{dt} + \vec{v}\frac{dm}{dt} \tag{2: from 1 and productive rule} \\

\vec{F}&= m\vec{a} + \vec{v}\frac{dm}{dt} \tag{3: from 2 and II} \\

\end{align*}\]

So now we have a slight problem. Equation (3) is almost our result provided that $\frac{dm}{dt}=0$. So what will it take to guarantee that?

The simplest option would be to simply assume that mass doesn't change: $\frac{dm}{dt}=0$. But we have a different option at our disposal: Newton's 1st Law. Newton states it as follows:

Every body perseveres in its state of rest, or of uniform motion in a right line, unless it is compelled to change that state by forces impressed upon it.Here I think we can interpret this as stating:

\[\begin{align*}

\vec{F}=0 \iff \frac{d\vec{v}}{dt} =0 \tag{IV: Newton's 1st Law}

\end{align*}\]

This will be sufficient to derive the result since this means that acceleration is zero whenever the force is zero. That means that:

\[\begin{align*}

\vec{F}= 0 &= m\times 0 + \vec{v}\frac{dm}{dt} \tag{4: from 3 and IV} \\

&=\vec{v} \frac{dm}{dt}

\end{align*}\]

From that it follows that $\frac{dm}{dt}=0$.

So to get the statement $\vec{F}=m\vec{a}$ from Newton's 2nd Law, we ultimately needed to assume Newton's 1st Law.

Of course, we didn't have to do it that way. We could have simply assumed $\frac{dm}{dt}=0$. Newton's 1st Law would have actually followed from that assumption along with the others we made.

So there's an interesting question here: Is $\vec{F}=m\vec{a}$ equivalent to $\vec{F}=\frac{d\vec{p}}{dt}$?

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