Still don't understand why we need F=dp/dt

[vanesch]

The fact is that F = dp/dt has a physical significance in the real world. It provides the means of quantifying matter interactions.

Well, what I'm trying to point out in this thread is that it is "dp/dt" that is ultimately related to what we call interactions. And we can call them interactions, because we can separate a complicated situation into simpler ones, usually 1-1 situations, and the dp/dt are additive for the different 1-1 situations into the more complicated one, so it is very attractive to consider these 1-1 observations of dp/dt as resulting from the interaction of the two said bodies. As such, what has a physical significance is the "resolution of dp/dt into a sum of simple contributions, one from each 1-1 situation that can be extracted from the more complicated situation". THIS was the discovery of Newton. And it is this which makes dp/dt a special quantity, which deserves hence a special name.

The magnitude of the interaction between matter objects is not measured by the speed of the object (an object can have any speed and no interaction at all with other matter). It is not measured by the rate of change of speed (the same interaction will cause different rates of change of speed to objects with different quantities of matter). It is not measured by the amount of mass that is interacting (an interaction with a small mass at high speed can have the same effect as a large mass at slow speed). It is measured by the rate of change of mass x speed. That was the great discovery of Newton.

Yes, exactly. But what allows us to say this, is that this "interaction" is a sum of "simple" 1-1 terms of more elementary situations. It is the fact that, for instance, in Newtonian gravity, we have a contribution for each pair of particles which is the same whether the others are present or not, so the "simplified situations" here are simply the setups were only each time one pair is present. We can calculate the dp/dt for each of these situations, and the dp/dt of the total situation is the sum of the dp/dt. This makes it extremely useful to give dp/dt a name.

So Newton's Second Law represents a deduction from empirical observation, not a definition. Because the law is so perfect and universal, it is used as if it were a definition.

I don't think so. You could just as well *define* Feynman's "gorce" g = d(mx)/dt. It would be a perfectly legal definition. But it is not useful, because there is not this property where the "total gorce working on an object is the sum of gorces of simplified situations". It is the fact that we can have this simplification into 1-1 situations, and then simply make the vectorial sum, that makes the dp/dt quantity so useful.

 

[lightarrow]

"You have a stationary weight on a contracted spring. You substitute the force made from the spring with what?"

Well with dp/dt of course, why wouldn't that work?

In this case dp/dt = 0. You would conclude there are no forces on it?

 

[vanesch]

"You have a stationary weight on a contracted spring. You substitute the force made from the spring with what?"
In this case dp/dt = 0. You would conclude there are no forces on it?

You cut the spring, and you see the weight accelerate, and you see the spring accelerate. The two dp/dt are equal and opposite.

And yes, you would conclude that, in the original situation, there is no force acting on the *point mass*. And if it is an extended mass, we already went through this.

 

[rbj]

Well, it turns out that d(m x1'(t))/dt + d(m x2'(t))/dt = d(m x3'(t))/dt at t=0. That is the fundamental property of forces.

but vanesch, that plus saying: x1'(t=0) = x2'(t=0) = x3'(t=0) = v0, is just another way of saying m x1'(t) + m x2'(t) = m x3'(t) . in vector form, it's just saying that momentum is a linearly additive vector quantity and that it is conserved in a closed system. that is an empirical fact and can be called a law. it's also true that you can have some convenience to call d(m x'(t))/dt "force". but that's still just a definition.

to say something like Newton's 2^nd law as something not tautological, you have to relate force as defined as the time rate of change of momentum to something else that is not the time rate of change of momentum. all of these other physical laws (such as gravitation and electrostatic or electromagnetic interaction) can be expressed in a form that has no force in it.

saying (assume mass $M$ is at the origin)

$$\mathbf{F} = -G \frac{M m}{|\mathbf{r}|^3}\mathbf{r}$$

is the same as saying (assuming no relativistic effects)

$$\frac{d^2 \mathbf{r}}{dt^2} = -G \frac{M}{|\mathbf{r}|^3}\mathbf{r}$$

that's the empirical law that governs motion. $\mathbf{F} = d \mathbf{p}/dt$ is just a definition that transforms the latter equation to the previous.

The reason is that, if we are allowed to "pull apart" a situation, when there are 7 particles interacting, that we can consider one as our "test particle", and then put it in 6 different situations (different toy worlds), each time with ONE partner.

that's just saying that momentum is a vector quantity that can be added linearly and that momentum in a closed system is conserved.

You don't need a special material like a solid body to define it.

no, but if you define it in terms of time rate of change of momentum, that's what it is: a definition, not an empirical law. now if you relate that definition of force (in terms of time rate of change of momentum) to another, completely different definition of force, and say they are equal (or proportional, since these two definitions might define different units for it) that is an empirical law.

you haven't persuaded me.

 

[rbj]

Ok reading the posts I'll try to summarize what I've understood so far:

1) Strain and pressure does not enter the discussion as they are not fundamental properties, i.e systems showing strain and pressure can always be decomposed to subsystems without them.

2) The situation of a point particle being subject to two opposing forces of equal magnitude is not possible to distinguish from the situation of the same particle where no forces are applied. Hence the two situations are identical.

3) So far no examples has been provided of a situation of where the forces involved cannot be replaced by rates of change of momentum.

Conclusion: the relation F=dp/dt does not provide any new information and hence is a definition.

The sitation is the same with velocity, where because it is convenient we introduce the letter v and the concept of velocity to denote dx/dt. If we wanted to we could just as well get rid of all velocities and only speak of derivatives, hence
$$v = \frac{dx}{dt}$$
is not a fundamental relation, just a definition. "Force" is no different, it is a name for dp/dt.

i'm with you except for assumption 1) which is not true ("does not enter the discussion") just because you say so and that leads to a different conclusion.

if they only define force in terms of time rate of change of momentum and do nothing to relate that quantity they call "force" to another independent definition of force, then Newton's 2^nd is a definition and not a law. what makes it a law is from equating that definition of force (dp/dt) to some other definition of force and saying that these two quantities are proportional to each other.

but, even taking force out of the picture (by undefining the quantity), there is a law that can still be emprically determined (that would otherwise be a consequence of Newton's 2^nd law) from observation that is that this vector quantity we define as "momentum" as $\mathbf{p} = m \, d \mathbf{r}/dt$ is linearly additive and, in closed systems, conserved. that's a law that is a degenerate case of Newton's 2^nd law.

 

[vanesch]

but vanesch, that plus saying: x1'(t=0) = x2'(t=0) = x3'(t=0) = v0, is just another way of saying m x1'(t) + m x2'(t) = m x3'(t)

No, it isn't !
If x1' = v0, and x2' = v0 and x3' = v0, then x1' + x2' = 2v0 is NOT x3' !

(don't forget that x1', x2' and x3' are each time the SAME initial condition of the SAME particle in 3 different setups...)

saying (assume mass $M$ is at the origin)

$$\mathbf{F} = -G \frac{M m}{|\mathbf{r}|^3}\mathbf{r}$$

is the same as saying (assuming no relativistic effects)

$$\frac{d^2 \mathbf{r}}{dt^2} = -G \frac{M}{|\mathbf{r}|^3}\mathbf{r}$$

that's the empirical law that governs motion.

That's the law that governs motion in a 2-particle setup. In principle, for a 3-particle setup, we could have something entirely different, and for a 4-particle setup, still something entirely different. The "miracle" is that the 4-particle setup can be analysed as a combination of 2-particle setups. It is this property which makes force a useful concept.

 

[rbj]

No, it isn't !
If x1' = v0, and x2' = v0 and x3' = v0, then x1' + x2' = 2v0 is NOT x3' !

(don't forget that x1', x2' and x3' are each time the SAME initial condition of the SAME particle in 3 different setups...)

and you're right (that'll teach me for not paying attention), but the issue is simply that of an arbitrary constant of integration (which is non-zero because x1'(0)+x2'(0) does not equal x3'(0)). my basic point remains. it's still the case that the momentums add vectorily (with an offset) and when you differentiate w.r.t. time, the derivatives also add (and that constant offset goes away). that momentums can do that is an emperical law (conservation of momentum in a closed system), but to replace the derivative of momentum with some humanly derived quantity we'll call "scurk" and then say that these scurks also add linearly, is just a definition, not a law.

That's the law that governs motion in a 2-particle setup. In principle, for a 3-particle setup, we could have something entirely different, and for a 4-particle setup, still something entirely different. The "miracle" is that the 4-particle setup can be analysed as a combination of 2-particle setups. It is this property which makes force a useful concept.

the empirical law is that we can superimpose the vectors. without an independent concept or definition of force (that is not dp/dt), Newton's 2^nd law is just a definition. to make it say more is a tautology that is essentially saying dp/dt = dp/dt. it's true but doesn't say anything new. i understand octal's point and agree with him/her up to the point where he/she says that alternative concepts of force (such as we find in static structures) are not germane. there are different manners in which we can define the concept of "push".

 

[Andrew Mason]

now if you relate that definition of force (in terms of time rate of change of momentum) to another, completely different definition of force, and say they are equal (or proportional, since these two definitions might define different units for it) that is an empirical law.

ok:

The rate of increase of momentum of a canal boat is proportional to the number of (same size and breed) of horses pulling on it.

The rate of increase of speed of a fixed mass cart on a table is proportional to the number of 1 kg. masses pulling on it (ie tied to it by a string and hanging over a pulley).

The rate of increase of momentum of a car is proportional to the number of (identical) springs with extension = x attached to it.

AM

 

[vanesch]

that momentums can do that is an emperical law (conservation of momentum in a closed system), but to replace the derivative of momentum with some humanly derived quantity we'll call "scurk" and then say that these scurks also add linearly, is just a definition, not a law.

Exactly, that's the point. There is an empirical observation which is the fact that derivatives of momentum in different situations are additive (and this time, there is no extra choice in an integration constant!). It are only the derivatives of momentum which have this property in all generality.
This makes it very useful to give them a name, by definition. It is the additive property of these quantities (additive between different situations, not within one single situation) which makes the definition useful.

As such, one can say that the definition captures the essence of an empirical observation and that it is not a random happening. True. But it is nevertheless a definition.

i understand octal's point and agree with him/her up to the point where he/she says that alternative concepts of force (such as we find in static structures) are not germane. there are different manners in which we can define the concept of "push".

I think they can all be reduced to some form or other of dp/dt for a "sub-situation". I still have to encounter a case where this is not true. Now, I know that historically, intuitively and in most intro courses, one starts with the intuitive static "force" which is the "push" you're referring to. However, what I don't like about it is that you need very specific and complicated systems (solid material systems) in order to define something quite fundamental, while dp/dt is way more close to fundamental concepts without having to use a complex system such as a solid.

 

[vanesch]

ok:

The rate of increase of momentum of a canal boat is proportional to the number of (same size and breed) of horses pulling on it.

Yes, but it would be inconvenient to have to have, as an international standard, a kind of breed of horse, if you see what I mean. It would be simpler to refer to the dp/dt that the horse undergoes the very moment you cut its link.

The rate of increase of speed of a fixed mass cart on a table is proportional to the number of 1 kg. masses pulling on it (ie tied to it by a string and hanging over a pulley).

But here you are already using the gravitational force to define the concept of force. A more clever way would be to put all that stuff in a rocket, and have the rocket accelerate at 1 g. But then we have again, your number of masses of 1 kg times the acceleration of the rocket, which is again dp/dt...

The rate of increase of momentum of a car is proportional to the number of (identical) springs with extension = x attached to it.

Here, we are using springs, which are very specific complex structures (solid material bodies with a Hooke's law of elasticity) and we are using the definition of the stress tensor (the dp/dt that a small element will undergo when cut away along a surface normal to a given direction).

So in all these cases, we can reduce the forces to a situation where the force is given by a dp/dt.

 

[Andrew Mason]

Yes, but it would be inconvenient to have to have, as an international standard, a kind of breed of horse, if you see what I mean. It would be simpler to refer to the dp/dt that the horse undergoes the very moment you cut its link.

The point is that you need not know anything about dp/dt of the horses. All you know is that the pulling "force" provided by one horse is 1/2 the pulling force provided by 2.

But here you are already using the gravitational force to define the concept of force.

Not the concept of GMm/R^2. It is using the concept that weights pull with a force, that is all. If I double the weight, I double the force. No concept of dp/dt is required at all. This is why high school physics labs use weights and variable mass carts to prove F=ma.

Here, we are using springs, which are very specific complex structures (solid material bodies with a Hooke's law of elasticity) and we are using the definition of the stress tensor (the dp/dt that a small element will undergo when cut away along a surface normal to a given direction).

All I have to know is that a spring produces a mechanical motive force. I don't have to know anything about dp/dt.

So in all these cases, we can reduce the forces to a situation where the force is given by a dp/dt.

Of course we can:because F = dp/dt = ma. But we can also prove this by simply observing that forces (ie. number of horses, bricks, springs) are proportional to dp/dt.

AM

 

[vanesch]

The point is that you need not know anything about dp/dt of the horses. All you know is that the pulling "force" provided by one horse is 1/2 the pulling force provided by 2.

But this is the essential non-trivial contents ! It doesn't work for "velocities" for instance: the velocity of 2 identical horses is not twice the velocity of a single horse ; the position of 2 horses is not twice the position of a single horse. But the "force" is. Why ? It is an empirical observation. There's nothing a priori evident about it. The *only* thing that we can observe, is that the situation with two horses gives us a dp/dt of the boat which is such that it is twice that of the situation of a single horse. Hence, we say that both give us the same force. This is by itself a highly remarkable fact. There is nothing evident about the fact that two identical horses "pull with twice the force" of a single horse. This is only so simply because they double the dp/dt of the boat.

Not the concept of GMm/R^2. It is using the concept that weights pull with a force, that is all. If I double the weight, I double the force. No concept of dp/dt is required at all.

And how do you establish that when you double the weight, you double the force ? Why would doubling the weight double the force a priori ? It doesn't double the position or the velocity for instance.

This is why high school physics labs use weights and variable mass carts to prove F=ma.

Yes, that is because there has been introduced a priori a notion of force in statics, with some a priori given, postulated, specific force values (for instance, that the force acting upon a mass in the Earth gravitational field at its surface, equals its weight which is g.m). But this is based upon a very earth-bound, intuitive notion, which is difficult to put forward as a fundamental, abstract concept. One uses a lot of intuitive day-to-day "common knowledge" to arrive at it, and it is very difficult to formalise it entirely independent of daily knowledge. For kids, such a non-abstract approach is fine. But in order to set up a formal system it is not very appropriate to make reference to "daily objects" such as solid matter, Earth surface etc...

Of course we can:because F = dp/dt = ma. But we can also prove this by simply observing that forces (ie. number of horses, bricks, springs) are proportional to dp/dt.

Yes, but in doing so, you need to refer to daily known objects, and some intuition. It is difficult to set up an entirely formal notion of force that way. And the problem is, for instance, that it is difficult to understand things such as "inertial forces".

 

[rbj]

Exactly, that's the point. There is an empirical observation which is the fact that derivatives of momentum in different situations are additive (and this time, there is no extra choice in an integration constant!). It are only the derivatives of momentum which have this property in all generality.
This makes it very useful to give them a name, by definition. It is the additive property of these quantities (additive between different situations, not within one single situation) which makes the definition useful.

As such, one can say that the definition captures the essence of an empirical observation and that it is not a random happening. True. But it is nevertheless a definition.

a definition is not the same as an emperical observation. and a law is not a definition, it says something more (a conclusion or generalization that comes from emprical observation). so, if we define force as the time rate of change of momentum, what is the law in Newton's 2^nd law?

I think they can all be reduced to some form or other of dp/dt for a "sub-situation". I still have to encounter a case where this is not true.

what is the sub-situation in the tension of static members of a structure or a compressed or static spring that is some form or other of dp/dt?

 

[vanesch]

so, if we define force as the time rate of change of momentum, what is the law in Newton's 2^nd law?

It's empty, of course, in this approach. It is often recognized that Newton's second law is or empty, or tautological, or circular in a certain way. On the other hand it is not useless. Newton's second law tells us that it is a good idea to define force that way.

what is the sub-situation in the tension of static members of a structure or a compressed or static spring that is some form or other of dp/dt?

Well, from the moment that you consider solid bodies, you have to define what you mean by the stress tensor, and the usual definition is: the dp/dt that a small piece of the solid would undergo, if you were to cut away the material next to it, perpendicular to a given vector and with a surface proportional to that vector, is given by the product of the stress tensor and this vector (with a minus sign).
If you consider gases for instance, then this is clearly so: the pressure is the dp/dt the "element of gas" undergoes when it would be facing a small hole with vacuum on the other side, all the rest equal. Before the "vacuum cut", there were particles going in both directions and the average momentum of the "gas element" (a statistical average over a set of particles) was 0. After the cut, the gas particles go only in one direction, and hence now the "gas element" has a finite momentum.

In other words, stress situations are related to a dp/dt by cutting away a part of the solid.

 

[Andrew Mason]

But this is the essential non-trivial contents ! It doesn't work for "velocities" for instance: the velocity of 2 identical horses is not twice the velocity of a single horse ; the position of 2 horses is not twice the position of a single horse. But the "force" is. Why ? It is an empirical observation.

You simply observe that if the horses are indistinguishable there is no reason to believe that they pull any differently. Do the experiment with 10 horses and a large ship. Differences between horses will be statistically much less significant. If the boat is massive enough the horses won't move much, initially, and you should be able to establish the linear relationship between dp/dt and number of horses.

And how do you establish that when you double the weight, you double the force ? Why would doubling the weight double the force a priori ?

It is very easy to show that the force produced by weights is additive. Take a 1kg, 2 500 g, and 10 100g. weights. Have them pull against each other by hanging them over a pulley. 10 100g. weights balance the 1 kg. weight, 5 100 g weights balance the 500g and 2 500 g balance the 10 100g and 1kg weights. You quickly conclude exactly what your muscles tell you: weights are additive.

AM

 

[vanesch]

You simply observe that if the horses are indistinguishable there is no reason to believe that they pull any differently. Do the experiment with 10 horses and a large ship. Differences between horses will be statistically much less significant. If the boat is massive enough the horses won't move much, initially, and you should be able to establish the linear relationship between dp/dt and number of horses.

Yes, but you don't seem to get the gist of what I'm trying to say. In order to even be able to say that "identical horses pull in the same way" you first even have to establish that horses do something such as "pulling", hence you already have intuitively introduced a notion of "force". This is what is a good idea in intro courses, but it doesn't allow you to set up a totally formal definition. Why should "horses pull" on a boat, and why should this "pulling" be an additive quantity ? Why doesn't the "pull" by the horses go, say, as the square of their number or something ? This already pre-supposes that there is some thing like a vectorial quantity, that only comes about from the single "horse-boat" relationship, and that this vectorial quantity is additive.

It is very easy to show that the force produced by weights is additive. Take a 1kg, 2 500 g, and 10 100g. weights. Have them pull against each other by hanging them over a pulley. 10 100g. weights balance the 1 kg. weight, 5 100 g weights balance the 500g and 2 500 g balance the 10 100g and 1kg weights. You quickly conclude exactly what your muscles tell you: weights are additive.

Well, that assumes already that "to balance" the rope over a pulley, they have to "pull in same amounts". It is then simpler to say that the dp/dt must be zero, and that this dp/dt is the sum of the dp/dt that the rope would undergo by each weight individually!

Again, I'm not arguing against all these approaches (which are often used in intro courses and don't pose any intuitive problems). I'm only saying that they are at a certain level dependent on some intuitive notions based upon everyday experience, and that you can hence not build a formal system around that (but of course you can do practical stuff that way). And that the simplest way to define force purely formally, is to say that it is nothing else but another name for dp/dt, and that dp/dt has a very special property (namely its additiveness in different situations: from "sub-situations" up to a more involved situation). If you do so, you can get rid of the intuitive notion behind a force, that it is some "kind of pushing or pulling". In fact, you can start to understand our intuitive notion of "pushing and pulling" as a consequence of this special additive property of dp/dt.

 

[Andrew Mason]

Yes, but you don't seem to get the gist of what I'm trying to say. In order to even be able to say that "identical horses pull in the same way" you first even have to establish that horses do something such as "pulling", hence you already have intuitively introduced a notion of "force". This is what is a good idea in intro courses, but it doesn't allow you to set up a totally formal definition. Why should "horses pull" on a boat, and why should this "pulling" be an additive quantity ? Why doesn't the "pull" by the horses go, say, as the square of their number or something ? This already pre-supposes that there is some thing like a vectorial quantity, that only comes about from the single "horse-boat" relationship, and that this vectorial quantity is additive.

The simplest assumption is that they are additive, which seems to accord with our observations. There is no reason to assume it is proportional to the square of their number or some other non-obvious relationship. You then test that assumption against what is observed and see if there is a reason to change your assumption.

You would agree that empirically $dp/dt \propto \sum{F}$. Now you could assume that the pulls are not additive. You would then conclude that the force of their pulls is proportional to the square of their number but you would then observe that $dp/dt \propto \sqrt{F}$. That is not the simplest relationship and in the absence of some evidence requiring a more complicated relationship, you use the simplest explanation.

AM

 

[vanesch]

You would agree that empirically $dp/dt \propto \sum{F}$. Now you could assume that the pulls are not additive. You would then conclude that the force of their pulls is proportional to the square of their number but you would then observe that $dp/dt \propto \sqrt{F}$. That is not the simplest relationship and in the absence of some evidence requiring a more complicated relationship, you use the simplest explanation.

Yes, but in doing so, you needed to refer to dp/dt in order to define what it means, the "total force of seven horses equals the sum of the forces of each horse". This is all what I want to point out: that the deep, fundamental meaning of force, is not some "pulling or pushing" or "stressing materials" or the like, but is simply "dp/dt". Although we intuitively see it the other way around, and although in intro courses, one first does it that way (by first going to statics, and then seeing dynamics as "statics out of equilibrium"). It was also the historical way (the "d'Alembert force" = -dp/dt in order to have "equilibrium" again).

 

[Andrew Mason]

This is all what I want to point out: that the deep, fundamental meaning of force, is not some "pulling or pushing" or "stressing materials" or the like, but is simply "dp/dt".

And all I wanted to point out that this realization was the great "discovery" of Newton. It was not about insight into creating a "definition" of force.

Newton did this by showing that a force - which we intuitively understand and feel - is measured by the change in motion that it produces in a given time interval: $\Delta v \propto F$ and that these changes in velocity are additive:

"Law II: The alteration of motion is ever proportional to the motive force impressed; and is made in the direction of the right line in which that force is impressed.

If any force generates a motion, a double force will generate double the motion, a triple force triple the motion, whether that force be impressed altogether and at once, or gradually and successively. And this motion (being always directed the same way with the generating force), if the body moved before, is added to or subtracted from the former motion, according as they directly conspire with or are directly contrary to each other; or obliquely joined, when they are oblique, so as to produce a new motion compounded from the determination of both."

AM

 

[vanesch]

And all I wanted to point out that this realization was the great "discovery" of Newton. It was not about insight into creating a "definition" of force.

Newton did this by showing that a force - which we intuitively understand and feel - is measured by the change in motion that it produces in a given time interval: $\Delta v \propto F$ and that these changes in velocity are additive

Sure, and it is also taught that way, initially. But it is based on intuitive, and hence ill-defined concepts which make it difficult to be entirely rigorous when giving a more formal definition.

 

[Aero]

This all depends on what you consider obvious.

Newton considered it obvious that F_total = F1 + F2 + ...

And therefore for him it definitely was a law, to say that F = dp/dt.

But the modern view is that it is silly to think something like the fact that forces are additive obvious before you have actually defined force. For Newton, defining force was easy: force is the amount by which something is pushed or pulled.

But this is not rigorous enough for modern physicists: they want to define it in terms of what they already know i.e. measuring things. And so the only way to define force is to define F = dp/dt.

And so while Newton considered it a law relating two quantities that he already knew about, nowadays it is just a definition in preparation for the modern law F_total = vector sum of individual forces.

One can either be intuitive and concrete, and explain informally what a force is, and then uncover the law F = dp/dt.
Or one can be formal and rigorous, and define a quantity called F = dp/dt, and show that it has this remarkable additive-when-split-up-into-simpler-situations property.

 

[vanesch]

Yes, aero, that sums it up nicely

 

[Andrew Mason]

Sure, and it is also taught that way, initially. But it is based on intuitive, and hence ill-defined concepts which make it difficult to be entirely rigorous when giving a more formal definition.

I am not so sure they were ill-defined concepts. Force, including gravity, was well understood as a concept long before Newton related force to dp/dt. But until one realizes that force is proportional to the change in momentum it produces in an object, the only way to measure it is by how many bricks/springs etc. are used to produce the force.

Newton considered it obvious that F_total = F1 + F2 + ...

And therefore for him it definitely was a law, to say that F = dp/dt.

I don't think he just considered it obvious that forces were additive, he observed it.

Newton observed that if you apply a given (constant) force to a object with mass m, the quantity p = mv of the object will increase linearly with time as: $\Delta p = constant\Delta t$. He observed also that if you double that force (by applying two of the things that gave rise to the intial force), the proportionality constant between $\Delta p$ and $\Delta t$ doubled. So he concluded that the force is proportional to dp/dt.

AM

 

[vanesch]

Imagine you live in a Newtonian universe, and there are particles around you.
However, you don't know their interaction laws at all. How do you define force ?
Remember, everything bites the tail of everything else: first of all, you have to establish what is an inertial frame, because you know somehow some must exist somewhere. But in order to do so, you need to have particles on which there are no forces acting, to take them as reference points on which to build an inertial frame. You don't know which ones (how do you determine whether a force is acting on a particle or not ?). You don't know what force laws (if any) are valid in your universe. How do you go about to define such a thing as force ?

On earth, we are somehow lucky: a stupid frame attached to the surface of the Earth is already not such a bad inertial frame, without thinking. Also, there's a simple law of gravity which gives you already a nice measure of mass. But you can't count on that in every Newtonian universe, where other force laws may exist. So how do you go about finding a definition of force which holds in any Newtonian universe ?

I claim that the "trick" is, to find a coordinate system (which is a transformation of a "feasible" coordinate system based upon existing particles, which may, however, undergo certain forces), such that the dp/dt of all particles can be written as relatively simple sums of contributions from expressions containing the coordinates and velocities of other particles, in that transformed frame. This is a kind of complicated fit, in which the masses of particles, the transformation to the "trial" coordinate frame, and the expressions in the dp/dt are all "free fit variables/functions".

In most frames this will be a complicated affair, and in a certain set, this will vastly simplify (or so we hope). We take it that *that* is then an inertial frame. We might even observe that for some particles, dp/dt in that frame is 0 or nearly so. We take it that that are then particles on which no forces act (free particles) and we can use them in the future as references to build directly an operational reference frame which is inertial. The expressions occurring in the dp/dt are then the "forces" due to the interactions with other particles, and we have to hope to find some general rule for them.