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.

To make things simple, I first consider when mass is a constant, so $F=\frac{\mathrm{d}p}{\mathrm{d}t}$ can be written as $F=ma$.

Why $F = m a$? Instead of, say, I define new constants or variables, and claim:

$$ I~=~N \frac{\mathrm{d}^3x}{\mathrm{d}t^3} \,. \tag{1} $$

Or using the same $F$ both literally and physically, and claim

$$ F~=~N\left(x,\,\dot{x},\,\ddot{x},\,{\dots},\,x^{\left(n\right)};\,t\right) \frac{\mathrm{d}^3x}{\mathrm{d}t^3} \,, \tag{2} $$or$$ F~=~Q \frac{\mathrm{d}x}{\mathrm{d}t} \,. \tag{3} $$

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, why can't we work things in a simple way, e.g.$$ I~=~Q \dot{x} \,? $$


  1. Assuming a reality in which a falling object $a=g$ [?]:$$ \begin{align} I & ~=~ Q \frac{\mathrm{d}x}{\mathrm{d}t} \\ \dot{x} & ~=~ gt \\ I & ~=~t \\ Q & ~=~ \frac{1}{g} \,, \end{align} $$

    I define such way $I=Q \dot{x}$, it is just the law of falling.

    I assume that, 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.

  2. Noting $\operatorname{Eq}{\left(2\right)}$, 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 $a=g$:

    $$ \begin{array}{ccccc} F & = & N\dot{a} & = & 0 \\ & \to & N\dot{a} & = & mg \\ & \to & N\phantom{\dot{a}} & = & \frac{mg}{\dot{a}} \\ \end{array} $$

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

    Reality: SHM:

    $$m\ddot x=-Ax$$ $$\to \dddot x= -A\dot x/m$$

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

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

    Lessons 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.

  3. $F=Q \frac{\mathrm{d}x}{\mathrm{d}t}$, 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?

    Please give me some criticisms to my reasonings.

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    $\begingroup$ Related: physics.stackexchange.com/q/4471/2451 and links therein. $\endgroup$ – Qmechanic Aug 29 '13 at 19:32
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    $\begingroup$ Because it works. $\endgroup$ – John Alexiou Aug 29 '13 at 23:57
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    $\begingroup$ $F$ 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? $\endgroup$ – udiboy1209 Aug 30 '13 at 5:21
  • $\begingroup$ @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. $\endgroup$ – Shing Aug 30 '13 at 9:01
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    $\begingroup$ 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. $\endgroup$ – udiboy1209 Aug 30 '13 at 15:33

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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  • $\begingroup$ 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? $\endgroup$ – Shing Aug 30 '13 at 8:49
  • $\begingroup$ 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? $\endgroup$ – Shing Aug 30 '13 at 8:57
  • $\begingroup$ You can't use F=ma to define both force and mass because a single equation can't be solved to find two unknowns. $\endgroup$ – user4552 Aug 30 '13 at 14:31
  • $\begingroup$ You're using a theory to measure the very concept of "mass". If my theory does not contain any such concept, you'll have hard time giving it a value. Don't be so naive, linking abstraction to perception is never trivial and involves some kind of (non canonical) choice. People are "cheating" when they mix both, but that's what physics do. $\endgroup$ – sure May 30 '15 at 15:30

From a philosophical perspective, one may prefer the original $F=\dot{p}$ over your formulism simply by Occam's razor. $F=\dot{p}$ is sufficient to describe all classical dynamics, and by adding another derivative you have only added another layer of complexity. First, your equation needs to be integrated to recover Newton's Law, and in the process a constant of integration will appear which will need to be fine tuned in order to agree with Newton's law and therefore classical reality. From a Popperian perspective, Newton's theory would be prefered over yours because, while they both account for classical mechanics, yours adds another layer of complexity (for which you haven't provided a satisfactory motivation, and which solves no new problem).

From a mathematical perspective, if you're willing to begin classical mechanics beginning from a configuration space, then $F=\dot{p}$ arises naturally as a consequence of some differential geometric machinery on that space. But I don't see why this would make it preferred over your method except by the Popperian criteria I outlined above.

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  • $\begingroup$ Recently, I am thinking if we should not make up math out of philosophical idea/ mathematics, but out of reality. which means, if interaction (which can be defined as never occuring by one object alone) is causing speed in stead of acceleration, then we have a law F proportional to speed (the Greek idea?). Anyway, I need some time to think it through, recent school work is too much. $\endgroup$ – Shing Jun 3 '15 at 8:27
  • $\begingroup$ Interactions cause both, and wondering if we should prefer the velocity interaction is a philosophical inquiry. You've just named your own philosophy "reality" in the comment above. The reality of the situation corresponds to classical mechanics, which is well established, and replacing newton's law is again a philosophy question. I have answered it with Occam's razor. $\endgroup$ – danielsmw Jun 3 '15 at 12:41

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.

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  • $\begingroup$ 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. $\endgroup$ – Shing Aug 30 '13 at 9:05

A theory is based on some primitive concepts you never define. What you define is the "equation of motion" describing the (algebraic) relations between all these primitive concepts. This is what a theory does, and it does not evaluate or explicit a form of the primitive concepts: this is what a model does.

Ex: In Newtonian mechanics, the primitive concepts are the ones of (absolute) space, time, inertial mass, and force. Now, Newton tells us that absolute change is equivalent to acceleration, and that acceleration is induced by an interaction process (ie, a force acting on some object and reacting back). The force has to respect some symmetries (basically, be invariant under Galilean transformation and maybe more if you want). This tells us that $F = ma$.

Now, a model is either

1) giving an explicit form of $a$ knowing $m$, and computing back the form of $F$ thanks to 2nd law,

2) giving an explicit form of $F$ (usually as a function of relative space-time position, and relative velocity of at least two bodies) knowing $m$, and computing back the acceleration of such a force on a given object. In particular, note that if you do things rigorously, 2nd law is a system of equations (each for each physical system you consider, because of 3rd law).

PS: you might find the first five part of that http://www.sure.zhln.eu/wp/?p=126 interesting. You'll notice that what you claim as being "too complicated" actually suffers from much more philosophical problems than just complication. It is yet only an unfinished draft, but you'll find all answers to what you ask there.

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