Can Newton's Second Law be absolutely proven mathematically?

[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!

 

[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.

 

[atyy]

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.

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

 

[Pengwuino]

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

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.

 

[atyy]

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.

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

 

[Pengwuino]

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

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.

 

[atyy]

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.

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.

 

[prosteve037]

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.

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)

 

[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.

 

[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

 

[Pengwuino]

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

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.

 

[prosteve037]

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.

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?

 

[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.

 

[Andrew Mason]

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?

\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

 

[prosteve037]

\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

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? :/

 

[Superstring]

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? :/

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

 

[Andrew Mason]

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? :/

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

 

[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.

 

[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.

 

[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)

















9

 

[Claude Bile]

Newtons laws are postulates, whose predictions happen to agree (for the most part) what we observe.

Claude.

 

[Andrew Mason]

Newtons laws are postulates, whose predictions happen to agree (for the most part) what we observe.

Claude.

That is an interesting point of view. I am not saying it is incorrect, but I am not sure that Newton would have agreed. I think he would have said the laws of motion are based on observation on earth. The generalization that the laws of motion apply to all masses in all reference frames everywhere in the universe would be a postulate.

AM

 

[prosteve037]

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

As far as demonstrating that \vec{F}\propto m\vec{a} though, does that count as a derivation? Or is that the fundamental/the only basis that the formula has for mathematical validity?

 

[my_wan]

As far as demonstrating that \vec{F}\propto m\vec{a} though, does that count as a derivation? Or is that the fundamental/the only basis that the formula has for mathematical validity?

I see the proportionality as all the proof that is needed to count as a derivation. A proportionality between X and Y merely indicates that X = cY, where c is a constant. Since force itself can take whatever scaling you choose it can be defined such that its definition contains whatever constant is needed. Hence there is no need of specifying any proportionality constant to give the equality. Just scale the force accordingly. Thus, in this case, \vec{F}\propto m\vec{a} is equivalent to \vec{F}= m\vec{a}.

Even if mass is ultimately found to contain a combination of variables, such that it is a product of differing variables, it would still remain true as that product remains the definition of mass. Even in Relativity, Special and General, where differing observers disagree on the mass of an object, the same observers disagree by the same amount on the energy expended to boost that mass. Hence the equality holds even when specific values of the equality differ between observers of the same system. It is equivalent to multiplying both side of the equality by the same factor.

To me that is proof enough.