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First, I would like to apology for my lousy English used here, as being a non-native english speaker,

I am currently using Classical Mechanics as a tool to learn how to construct a theory as an end. Therefore, I have a few questions, and a few reasonings behind them (I only consider reality, not imagined world), and I would like your suggestions or corrections :D

To make things simple, I first consider when mass is a constant, so F=dp/dt can be written as F=ma.

Why F=ma? Instead of, say, I define new constants or variables, and claim

1.) $I=N \frac{d^3x}{dt^3} $ ?

Or using the same F both literally and physically, and claim 2.) $F=N(x,\dot x,\ddot x,...x^{(n)};t) \frac{d^3x}{dt^3} $ or

3.) $F=Q \frac{dx}{dt} $?

I mean, force is not something that you can touch nor see, either is mass; so, we pretty much define them via indirect observations, recording the data, and then comparing with other objects, in order to give operational definitions on mass and force (in case of "mass", we will have to define a unit mass first, and then use F= ma as operational definition for mass.).

Under this line of reasoning, So why can't we work things in a simple way - $I=Q \dot x $? Yes, we can indeed.

for question 1.) given reality: A falling object a=g:

$I=Q \frac{dx}{dt}$

$\dot x=gt$



So if I define such way $I=Q \dot x $, it is just the law of falling. So my answer to my 1st question is: Yes, we can define things such way, but it has nothing to do with the subject we are dealing right now - inertia. Lesson learned: Know what you want to deal with.

For question 2.) $F=N(x,\dot x,\ddot x,...x^{(n)};t) \frac{d^3x}{dt^3} $

Since I only consider our reality, that means F=ma is truth. So I will comparing F=Na' with F=ma with a few cases (constant a, and SHM) And it turns out that F=Na' is far more complicated (often a variable) and N is going to be ill defined given constant a.

Reality: A falling object a=g:

$F=N\dot a=0 =>N\dot a=mg =>N=mg/\dot a$.

as $\dot a=0$, N(x,t) is ill defined.

Reality: SHM:

$ m\ddot x=-Ax =>\dddot x= -A\dot x/m $. Given $ Na'=ma => N= ma/a'=-m^2\ddot x/\dot x A=(-m^2/A)(d\dot x/dx) $

Too complicated. (comparing with the fact that m is just a constant.)

Lesson learned: 1.) best work for most of the cases. 2.)When a constant will do, don't use a function. 3.)Theory is about productivity.

For question 3.) $F=Q \frac{dx}{dt} $ or any other orders, actually, I am just asking:

for $ F= \frac {d^nR(W,\dot W,...,W^{(n)},x,\dot x,...,x^{(n)};t)}{dt^n}= \frac {dP(m,\dot m,x,\dot x;t)}{dt} $

4.) is there exist a n, such that there exist a R makes things simpler than $F=\dot P$ ? Define $R(W,x;t)=W(x,\dot x,\ddot x,...x^{(n)};t) x$, (R, W new variables, x is position function of time)

if there doesn't exist such an n, then 5.) why F= ma turns out to be the simplest way to work things out?

And thanks for your reading. Please give me some criticisms to my reasonings.

p.s. I think I am making the reinvent wheel mistake. I hope these fancy mathematical philosophy will end up some educational purpose on physics.

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Related: physics.stackexchange.com/q/4471/2451 and links therein. –  Qmechanic Aug 29 '13 at 19:32
Because it works. –  ja72 Aug 29 '13 at 23:57
$F$ is more correctly defined as $\frac{dP}{dt}$ where $P$ is the momentum. You are assuming mass $m$ to be a contant, but do your equations work well for variable mass systems(when $m$ is not constant, eg. rocket engines)? What about forces on massless objects which have momentum, like light? –  udiboy1209 Aug 30 '13 at 5:21
@udiboy, are you suggesting generality as a importance factor for constructing a equation? I am just self learning how to construct a theory, and I picked newton mechanics simply because it is the simplest physics theory that I know,,and sure the new equations don't apply well, whenever the Newton Mechanics doesn't. –  Shing Lau Aug 30 '13 at 9:01
Yes, generality is an important factor for a law. By the way, newton's laws work for variable mass systems, but I doubt your equation will. –  udiboy1209 Aug 30 '13 at 15:33

2 Answers 2

I mean, force is not a observable, either is mass

If this were the case, then it would indeed be completely arbitrary how to write Newton's second law, and we wouldn't have such a law of physics at all.

Force and mass are both observable.

I can measure mass like this. Start with some arbitrary object to be my basic unit, and then making another object that balances against it on a double-pan balance. In this way I can make up as many of these unit masses as I need. Now if 7 of these balance against my dog, I know that my dog's mass is 7 units.

Force is something we measure directly using spring scales, bathroom scales, etc.

It is possible to consider F=ma as a definition of force, or a definition of mass, but you certainly can't consider it as defining both force and mass. Even if you do consider it to be the fundamental definition of force, you are constrained by the fact that force can be measured in other ways, such as by spring scales.

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Thanks, at first I had thought that as long as we need some device to measure force or mass, then it is probably not a observable, but then I have realized that our eye balls are device too. And speaking on Physics, an observable is something physical of Physical content, am I right? –  Shing Lau Aug 30 '13 at 8:49
In Newton II law, force is how you measure interaction quantifiably. How about mass? Also, would you elaborate a bit on why F=ma can't be seen as defining both mass and force? –  Shing Lau Aug 30 '13 at 8:57
You can't use F=ma to define both force and mass because a single equation can't be solved to find two unknowns. –  Ben Crowell Aug 30 '13 at 14:31

It is just convenient to define force from perspective of modern physics,it is not necessary,but if we had excluded the force from our theory,we would get a little bit more complex theory,you may do it to get what happens,and since we are used to it ,it is not harmful to omit it from modern theory.

Force is not even like an idea in different but equivalent axiomizated system,all idea is necessary or at least independent,force is not independent,and I don't think it is observable since what we observe are other quantity of object.

share|improve this answer
Thanks, yeah, it is true in Hamilton and Lagrangian Mechanics. It had bee quite strange to me until I realized that Newton Mechanics is just a way to model or approximate nature, and surely all roads lead to Rome. –  Shing Lau Aug 30 '13 at 9:05

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