And while we’re here, Newton’s second law isn’t F=ma either

My last post on how \(V=IR\) is not Ohm’s law received some comments. @dodiscimus said: “And \(F=ma\) is not Newton’s 2^nd law. Your next post?”

I’m quoting him to:

• give him the credit for having the idea and spotting the similarity (see later)
• share the blame – otherwise people are going to think all I do is write chippy posts about how things are not the way they thought… Also, there may be a lack of appetite for what is coming next – the response of @paulmartin42 was “Please please no more Newton”. Well, we’ve all thought that at one time or another… Anyway – this post will be shorter than the Ohm’s Law one, for that very reason.

But it’s true. Newton’s 2^nd law is not \(F=ma\). Not in its modern formulation anyway. Newton’s 2^nd Law of motion states that “the rate of change of momentum of a body is directly proportional to the external force acting on the body and takes place in the direction of the force.” That’s according to Muncaster A-level Physics 3^rd edition, which sort of shows how long ago I took my A-levels…

So if we call momentum \(p\), and the external force \(F\), then

\(F=\frac{dp}{dt}\)

Now, since momentum is the product of mass (\(m\)) and velocity (\(v\)), then \(p=mv\) and

\(F=\frac{d}{dt}(mv)\)

If the mass is constant then it can sort of (with as many apologies to mathematicians as is required) ‘be taken outside’ the derivative to give

\(F=m\frac{dv}{dt}\)

And since the rate of change of velocity is acceleration (\(a\)), then

\(F=ma\)

So \(F=ma\) does follow directly from Newton’s second law, as stated by Muncaster. But only for the case of constant mass. We couldn’t take the \(m\) outside the derivative and end up with \(F=ma\) otherwise. So \(F=ma\) is only a special case of Newton’s second law – the case of unchanging mass…

This has parallels with Ohm’s law holding only for components of constant resistance (i.e. resistance not dependent on current). As @dodiscimus said, when he kickstarted this post:

“I was thinking of the similarities:

• \(a\) is only proportional to \(F\) if \(m\) is constant.
• \(I\) is only proportional to \(V\) if \(R\) is constant.”

@e=mc2andallthat then chimed in with a really interesting point (in Latin). Newton’s second law was originally stated in his Principia as in this link. The interesting part is that it’s more like \(F=ma\) than I’ve suggested above. And he (@e=mc2andallthat, not Newton) goes on to point out that \(F=ma\) first appeared sometime before 1716.  It’s not easy to tell whether the ‘rate of change of momentum’ form was kicking around at that time, because it depends on how you translate phrases of Newton’s like “the alteration of motion”. Acceleration or change in momentum? I guess people spend whole careers on history and philosophy of science, working that kind of thing out. In any case, it would be really interesting to trace that development.

By the way, the Latin version of the law reminds me of an old post of mine railing against (there I go again) stating Newton’s Third Law in terms of ‘action and reaction’. You can find it here if interested…

Now, to finish, here’s a bit the less mathematically minded can skip… So what happens if the mass is not constant? Then \(F=\frac{d}{dt}(mv)\) turns into, via the product rule:

\(F=m\frac{dv}{dt}+v\frac{dm}{dt}\)

This is a form of Newton’s second law that is more general than \(F=ma\). It contains a term dealing with the rate of change of velocity, and one dealing with the rate of change of mass. It is useful, for example, in the case of rockets, whose velocity certainly changes quite a bit, but whose mass also changes a lot as it burns up the fuel that accounts for the majority of its initial mass.

I promise my next post won’t be of the “don’t think A is B because it isn’t,” variety. Actually, I’m planning one on how weird the sievert is as a unit. Potentially niche readership I know, but hey! – it’s really interesting…