Can Newton's Second Law be absolutely proven mathematically?

1. Jun 19, 2011

prosteve037

So if you've seen/posted in any of my other threads you probably noticed that I've been trying to figure out how the modern form of Newton's Second Law ($\textit{F = ma}$ or $\textit{F = m}$$\frac{dv}{dt}$) came to be formulated.

After reading many different websites and sources, I think I've developed a pretty good handle on how $\textit{F= ma}$ was formed, why it's the way it is, and how it's an equation that is scientifically viable.

However, I'm at a loss when it comes to explaining how the MATHEMATICAL formulation can be proven from the ground up. I can get to explaining how $\textit{a}$$\propto$$\frac{x}{m}$ ($\textit{x}$ being the constant of proportionality), but I can't explain why "$\textit{x}$" turns to "$\textit{F}$". From my understanding, $\textit{F = ma}$ is merely the result of that proportionality statement which is backed up directly by experiment. But how does "$\textit{F}$" take the place of "$\textit{x}$"?

Is it merely because Force, "$\textit{F}$", is the only variable left out of the equation? Or is there a mathematical reason behind it? If so, what are the steps that show how "$\textit{x = F}$"?

Thanks!

2. Jun 19, 2011

Pengwuino

F=ma is a postulate, it can not be derived from any mathematical principles.

There's no reason why Newton's 2nd law can't be $F = ma^2$ mathematically. Experiments show that that's not the case, however.

3. Jun 19, 2011

atyy

Could I make it true, by defining the force of gravity to be $F=GMm/((r^2)a)$?

4. Jun 19, 2011

Pengwuino

I assume you mean the acceleration is on top? I don't see why not; the problem is the physical interpretation/consequences would be non-sense.

5. Jun 19, 2011

atyy

Yes, that's what I meant. Why would it be non-sense? At least in 1D it seems to be mathematically equivalent.

6. Jun 19, 2011

Pengwuino

Hmm maybe not, I'm trying to imagine everything that would follow if you actually had forces that increased simply because the object is accelerating. I suppose you could simply modify all of classical mechanics and the Lagrangians in the same manner and everything would still work.

7. Jun 19, 2011

atyy

I've often heard that Newton's second law was a definition of force, which I think means something like F=ma^2 being ok. I don't know how that would work out with vectors though.

8. Jun 21, 2011

prosteve037

Wait, so since it's a postulate can it take any mathematical form? (Within the validity of experimental data)

Could it be for example, written as "F = $\pm$|(m + a)|? ("$\pm$" for direction) Of course, this wouldn't be mathematically correct since the units wouldn't work out, and $\textit{a}$ and $\textit{m}$ must both exist for the formula to be in accordance with experiment.

But since it was 'postulated', couldn't this formula stand as applicable in instances where there exist a mass and nonzero acceleration? (While also yielding a dimensionless quantity)

9. Jun 21, 2011

arildno

Not if "m" and "a" are to retain their current meanings, no.
Alternatively, your "F" is not the same as our "F".

For example, it is an observed fact that if you use the same force-generating mechanism (say, stretching a spring by the same amount) then a twice as large mass object will get only half the acceleration the first got.

And that would be generally untrue with your equation.

10. Jun 21, 2011

prosteve037

Alrighty. But as far as "How does 'x' turn into 'F'?", there is no answer to that? The equation is just the way it is by manner of definition?

That just seems too... uncomplicated haha

11. Jun 21, 2011

Pengwuino

It almost sounds like you're saying if 3 = x/2, how do we know x becomes 6? It's how we postulated multiplication to work. You could say that x = 9, but then you'd have to re-arrange the whole system to get it to work, but you could get it to work I suppose.

12. Jun 21, 2011

prosteve037

But isn't that kind of a different situation? In that situation, x HAS to equal 6.

I'm just having trouble understanding why "F" is the only thing that can fill in the blank: __ = ma

Is it just ASSUMED that "F" belongs there?

13. Jun 21, 2011

Nabeshin

Here's my take on the issue:

It seems to me your way of thinking about it is fundamentally backwards! I can imagine saying, "OK, so the acceleration a body feels is proportional to the force with which I push on it". That makes good sense, and is probably what most people think.

So now you have $a \propto F$ or $a=x F$

The question should be why is x the inverse of the mass! Of course, that's precisely how the inertial mass is defined, so it's really again a matter of definitions.

14. Jun 21, 2011

Andrew Mason

$\vec{F} = \frac{d\vec{p}}{dt}$ is essentially the definition of Force. Force has no independent definition. One can think of force as being the measure of a pushing or pulling on an object - where the measurement is determined by the time rate of change of momentum that it causes.

But it is not an arbitrary definition. If you define units of force by something that exerts a constant pull (eg a 1 kg weight suspended by a string mounted over a pulley) you can show that the number of units of force is directly proportional to the acceleration of a constant mass on a frictionless horizontal surface AND is directly proportional the mass at constant acceleration.

AM

15. Jun 22, 2011

prosteve037

But couldn't you combine those proportionality statements together to get "F = ma"? Though that would mean that "F = ma" ISN'T a postulate wouldn't it? :/

16. Jun 22, 2011

Superstring

F=dp/dt is the definition of a force. You can derive F=ma from that by holding mass constant.

17. Jun 22, 2011

Andrew Mason

F=ma is an empirically derived law. It is derived from observations of how different numbers of units of "pulling" or "pushing" affect motion of masses. The postulate may be simply that forces are additive: that if I apply 2 units of force F to a given mass, the total force that is applied is 2F. If you accept that postulate, it is a relatively simple matter to demonstrate that $\vec{F}\propto m\vec{a}$.

AM

18. Jun 23, 2011

Nabeshin

I really don't think force is 'defined' by $\frac{dp}{dt}$. Force is supposed to be an intuitive concept, almost without need for definition. I push on something, that's a force. Then you have the empirical statement that if I push twice as hard, it accelerates twice as fast. So then you empirically arrive at athe fact that $F \propto a$.

Then of course you extend what you mean by 'push' to be things other than me mechanically shoving the object.

At least, if I was in the 17th century thinking about these things this is how I believe I'd go about it.

19. Jun 23, 2011

daveb

Newton's Second Law is a Law. As such, from a philosophy of science perspective, it is a mathematical representation of a pattern that has been obseved in nature, and is always such (hence being a Law). In science, there are patterns and processes. Patterns are given the name "Law" since they never seem to deviate from that pattern. It says nothing about why it is that way, and thus can't really be proven, other than than that it accurately represents the pattern.

Processes, on the other hand, are explained by theory (as are Laws). Theories are attempts to explain "Why" the pattern is the way it is. Thus, we have the Law of Natural Selection, and the Theory of Evolution. We notice the pattern of natural selection, and the theory explains the pattern. Newton's Laws are the pattern, and Einstein's Theory explains the pattern.

This is why theory is so much more powerful than law. Laws don't usually predict anything other than what they describe. Theories can predict. This is why, if creationists say, "Evolution is only a theory", you say "Exactly!" and then explain to them why theory such a powerful scientific domain.

20. Jun 24, 2011

nceba

to prove that F=ma ,
consider a particle with a piosition vector ,r donated by <f(t),g(t),h(t)> and mass m
if YOU compute r' =<f'(t),g'(t),h'(t)> =V(t) which is the velocity of the particle in the direction of the force and r" =<f''(t),g''(t),h''(t)> = a(t) which is the acceleration of the partice.
Then Force is = to the dot product of m(scalar) and r"(t) the second derevative of the position vector
henec F= m.(r")=m. r" =<f''(t),g''(t),h''(t)> =m. a(t)

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