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

[octol]

I've tried reading the previous threads about Newtons second law and I've must say that I still don't understand why it's not just a definition. That is, if we want to it would be possible to just replace F with dp/dt everywhere in physics and make F=dp/dt unnecessary. For example we would instead have

\frac{dp}{dt} = G \frac{m_1 m_2}{r^2}

and for electromagnetism

\frac{d\textbf{p}}{dt} = q (\textbf{E} + \textbf{v} \times \textbf{B})

In other words, the concept of "force" is intuitive and all, but still fundamentally redundant.

 

[Cyrus]

I don't understand what the purpose of replacing F with dP/dt would solve.

 

[billiards]

depends what you're dealing with, can be handy to express force with a dt if you want to integrate the other side wrt time I guess. Other times its just superfluous to write F out like that.

 

[Hurkyl]

Why use velocity, when you could use dx/dt instead? Why use momentum when you could use mdx/dt instead? Why use mass when you could use \int_R \rho \, dV instead?

Because it's convenient, and often conceptually useful.

 

[radou]

Another major threat to the concept of classical physics.

 

[octol]

Yes I know it practical in the same way as having mass instead of an integral over the density. But F=dp/dt is usually claimed to be a fundamental law, why is that? I mean, this is never the case for m = \int \rho dV

 

[russ_watters]

I've tried reading the previous threads about Newtons second law and I've must say that I still don't understand why it's not just a definition. That is, if we want to it would be possible to just replace F with dp/dt everywhere in physics and make F=dp/dt unnecessary. For example we would instead have

\frac{dp}{dt} = G \frac{m_1 m_2}{r^2}

What if dp and dt are both zero?

 

[vanesch]

I've tried reading the previous threads about Newtons second law and I've must say that I still don't understand why it's not just a definition.

Feynman has a nice discussion about that in the beginning of his "Lectures". Essentially, F = dp/dt is an inspired definition. He compares it to G = d (mx)/dt, which he calls a "gorce" and which is a rather uninspired definition.

By defining F = dp/dt, Newton linked the interaction-between-bodies part with the kinematics in such a way that the interaction-between-bodies part took on an easy and simple form.

 

[Parlyne]

Newton's second law does not say F = \frac{dp}{dt}. What it says is \sum \vec{F} = \frac{d\vec{p}}{dt}. We can't just replace "F" in all the equations defining forces with \frac{dp}{dt} because that would only be true in the case that the force under consideration is the only force acting. If for example, electromagnetic and gravitational forces were acting, Newton's second law comes down to:

\frac{d\vec{p}}{dt} = \frac{GmM}{r^2} \hat{r} + q (\vec{E} + \vec{v} \times \vec{B})

Here, the equations for each force separately are true when written with F's, but not when written with \frac{dp}{dt}.

 

[vanesch]

Newton's second law does not say F = \frac{dp}{dt}. What it says is \sum \vec{F} = \frac{d\vec{p}}{dt}. We can't just replace "F" in all the equations defining forces with \frac{dp}{dt} because that would only be true in the case that the force under consideration is the only force acting. If for example, electromagnetic and gravitational forces were acting, Newton's second law comes down to:

\frac{d\vec{p}}{dt} = \frac{GmM}{r^2} \hat{r} + q (\vec{E} + \vec{v} \times \vec{B})

Here, the equations for each force separately are true when written with F's, but not when written with \frac{dp}{dt}.

Yes, but you could as well say that in a situation that you describe (namely where there are E and B fields as well as masses around), that:

F = \frac{GmM}{r^2} \hat{r} + q (\vec{E} + \vec{v} \times \vec{B})

simply because that's what seems to be equal to \frac{dp}{dt}

You are in fact implicitly using the superposition principle for forces when "several interactions" take place. But that was exactly the usefulness of introducing the concept of force: for a given situation, we can mentally split up the entire interaction into different 1-1 interactions (between "masses", and between "charges and field" etc...), pretend first that they act only alone (ie, considering the interaction in simplified situations, for instance, when only masses are present, and no charges, etc...), and then add them all together to find the genuine interaction.
But it is "in our minds" that this decomposition occurs.
You can just as well say that for a given setup, well, the force on a particle is given by F = \frac{dp}{dt}, and that, if you analyse the situation, that it turns out that this equals also \frac{GmM}{r^2} \hat{r} + q (\vec{E} + \vec{v} \times \vec{B})

It is just an amazing property of nature, that the different terms in there ALSO correspond to momentum changes in OTHER situations. The \frac{GmM}{r^2} term for instance, occurs in another situation, as the \frac{dp}{dt} which looks a bit like the one we're studying, but without charge.
The term q (\vec{E} + \vec{v} \times \vec{B}) occurs in still another situation as the \frac{dp}{dt}, when there are charges, but no masses present.

So in fact, what we discover here, is a specific property of nature, which says that the quantity \frac{dp}{dt} has a remarkable property in certain circumstances: for a given situation, it equals the vector sum of the \frac{dp}{dt} that occur in OTHER situations, which each seem to correspond to a certain aspect of the initial situation.
So we mentally say that in the initial situation, TWO forces work on the particle, but that's only a mental picture. Only one force works on it, which is the entire expression: F = \frac{GmM}{r^2} \hat{r} + q (\vec{E} + \vec{v} \times \vec{B}). We mentally have split it up into different terms, which correspond to the changes of momenta that would have occurred in different, simplifying situations, using this remarkable property of nature.

But there is no way to distinguish "a particle on which two equal and opposite forces act" from a "particle on which no force acts".

The two "equal and opposite" forces are just mental constructions because we think of "simplifying situations" (for instance, with only a mass left to it, and then only with a mass right to it). In each of the cases, the total force corresponds to \frac{dp}{dt}

 

[lightarrow]

I've tried reading the previous threads about Newtons second law and I've must say that I still don't understand why it's not just a definition. That is, if we want to it would be possible to just replace F with dp/dt everywhere in physics...

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

 

[arildno]

But there is no way to distinguish "a particle on which two equal and opposite forces act" from a "particle on which no force acts".

It is crucial to note that vanesch used the word "particle" here, i.e, that we are dealing with an object where we do not regard internal structure as important.

If we are dealing with an "extended object" (essentially something composed of many "particles"), then we may distinguish between an object upon two equal, opposite forces acts, and one upon which no force acts.

In the first case, the object will be in a state of stress, but in the other case, that need not be the case.

 

[Gokul43201]

But there is no way to distinguish "a particle on which two equal and opposite forces act" from a "particle on which no force acts".

Even in light of Aharonov-Bohm?

 

[octol]

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?

 

[vanesch]

It is crucial to note that vanesch used the word "particle" here, i.e, that we are dealing with an object where we do not regard internal structure as important.

If we are dealing with an "extended object" (essentially something composed of many "particles"), then we may distinguish between an object upon two equal, opposite forces acts, and one upon which no force acts.

In the first case, the object will be in a state of stress, but in the other case, that need not be the case.

This is only in the case when the points of action of the two forces are different. But if that is the case, the two forces can also be identified as the dp/dt in the two different situations where the extended body has been cut into two pieces (which is actually even the definition of the stress tensor: the change of momentum that one piece would undergo if cut away from the other part of the extended body).

So also in that case, a force is always equal to a dp/dt in a different situation (here, the body cut into two pieces).

The situation I was more thinking off, was:

Take a big mass M1, at distance R1 to the left, and a big mass M2, at distance R2 to the right.
We say that the force on a point particle, or even to a small (way smaller than R1 and R2) but extended body, equals "the gravitational force" due to M1, plus "the gravitational force" due to M2, and write:
dp/dt = F1 + F2

However, F1 is simply dp/dt of the same body, with only M1 present, and F2 is simply dp/dt of the same body, with only M2 present. It is a feat of nature that the third, and different, situation, with M1 and M2 present, has:

(dp/dt)_3 = (dp/dt)_1 + (dp/dt)_2

with 3: the situation where M1 and M2 are present
2: the situation with only M2 present
1: the situation with only M1 present.

It is rather amazing that these 3 different setups have any relationship, at first sight. Given this aspect, it becomes interesting, to give a special name to (dp/dt)_1 and (dp/dt)_2.
We call it "force".

But note that in situation 3, individually, F1 and F2 don't have any physical meaning as such, and only their sum counts. There is no internal stress in the extended body which suffers from F1 - F2.
(there will be a small tidal effect, but it is not F1 - F2).

 

[vanesch]

Even in light of Aharonov-Bohm?

I'm only talking in the frame of classical mechanics here

The only point I wanted to make, is that, strictly speaking, we don't need the concept of force in classical mechanics, but that it is a damn useful definition, and that its usefulness comes about by a peculiar property of nature (in classical mechanics), which relates momentum changes between different situations, but that we could limit ourselves, if we wanted to, to talk purely about momentum changes and their properties.

However, this would be a clumsy way of doing things, because we would then not make explicit the specific property of nature which gives us the vectorial composition of forces due to 1-1 interactions (which is nothing else but this composition property of momentum changes).

I maintain however, that for *every* use of a force, we can find a situation where it corresponds to a dp/dt. Maybe I'm wrong, but I don't think so (yet).

 

[vanesch]

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

You consider the situation where the spring is cut, and the force is the dp/dt of the weight.

 

[octol]

It is crucial to note that vanesch used the word "particle" here, i.e, that we are dealing with an object where we do not regard internal structure as important.

If we are dealing with an "extended object" (essentially something composed of many "particles"), then we may distinguish between an object upon two equal, opposite forces acts, and one upon which no force acts.

In the first case, the object will be in a state of stress, but in the other case, that need not be the case.

I fail to see how that changes anything. Fundamentally, what vanesch stated is still true. In the situation of an extended object composed of many "particles", the stress is just a consequence of the complexitivity of the constructed object. This is not in any way a fundamental principle. Take friction for example, this is not a fundamental property either, just a phenomenon that occurs for many-particle systems.

 

[octol]

I'm only talking in the frame of classical mechanics here

The only point I wanted to make, is that, strictly speaking, we don't need the concept of force in classical mechanics, but that it is a damn useful definition, and that its usefulness comes about by a peculiar property of nature (in classical mechanics), which relates momentum changes between different situations, but that we could limit ourselves, if we wanted to, to talk purely about momentum changes and their properties.

However, this would be a clumsy way of doing things, because we would then not make explicit the specific property of nature which gives us the vectorial composition of forces due to 1-1 interactions (which is nothing else but this composition property of momentum changes).

I maintain however, that for *every* use of a force, we can find a situation where it corresponds to a dp/dt. Maybe I'm wrong, but I don't think so (yet).

This is precisely my opinion as well, and so far I've never seen a convincing argument against it.

The main conseqence of this is that if we were to try to properly axiomatize classical physics, Newtons second law together with the concept of force would NOT be included as a _fundamental_ axiom/law, instead it would be a definition used in the formulation building on top of the _real_ axioms/laws.

 

[vanesch]

The main conseqence of this is that if we were to try to properly axiomatize classical physics, Newtons second law together with the concept of force would NOT be included as a _fundamental_ axiom/law, instead it would be a definition used in the formulation building on top of the _real_ axioms/laws.

Well, yes, some books do it that way (Alonso and Finn for instance, define force as dp/dt). That said, it is nevertheless a very inspired definition which captures something essential about (classical) nature. I find Feynman extremely illuminating in that respect, where he introduces his "gorce" (=d (mx) / dt), which leads him nowhere.

The definition of F = dp/dt captures the essential simplicity of the composition of the different dp/dt for different "elementary" situations, which turns F into such a useful concept.

In Landau and Lif., "force" is an entirely auxillary concept, because they start from the beginning from a variational principle as the fundamental axiom of mechanics.

 

[arildno]

Well, my post was never meant as a criticism of vanesch's post. Nor should it be possible to construe it as such.

I explicitly stated that an "extended object" was simply a collection of "particles", and that therefore, our ability to distinguish between a stressed and unstressed state (which, when not sufficient care is exercised when using the the word "particle" might seem to contradict vanesch's post) something belonging to a system of particles and not the particles themselves.

 

[rbj]

Yes I know it practical in the same way as having mass instead of an integral over the density. But F=dp/dt is usually claimed to be a fundamental law, why is that? I mean, this is never the case for m = \int \rho dV

there are also concepts of force in contexts where things are not moving, such as in a compressed spring or similarly the tension in members of a structure such as a bridge or table. but acceleration, or the time rate of change of momentum (these are not precisely propotional, only for low speeds from the pov of the observer), does not have meaning without something moving (and inertial movement is not sufficient, since relativity says that all inertial movement is equivalent). to equate this concept that also has meaning in stationary systems to something else that only has meaning in the context of movement, is what Newton's 2^nd law is about. otherwise, i might agree that it's a tautolgy.

or maybe a definition. why even have laws with momentum, p, or velocity, v, and just express all physical law in terms of relative position, time, mass, and charge? except for "charges" of other stuff, (perhaps some property of some elementary particles), everything, either measured or a variable quantity in physical law, can be expressed in terms of those four base quantities. but it's not convenient which is why we define other derived units out of the base units.

also, by setting to one the constant of proportionality in F = \frac{dp}{dt}, that is not salient to Newton's 2^nd law. it could be expressed:

F = C \frac{dp}{dt}

where momentum is defined in terms of the base units of length, time, and mass, but force is defined in terms of some units determined by deformation of some standard prototype object. then the law would have some constant of proportionality that would have to be determined by experiment, like G, c, or \hbar is.

 

[axawire]

Well I thought F was used instead of dp/dt for the lower level physics texts that are not calculus based. Then as you go up levels in physics texts and math level you learn F is really representing dp/dt. side note, rate of change of momentum with respect to time is a bit of mouth full compared to force.

 

[vanesch]

to equate this concept that also has meaning in stationary systems to something else that only has meaning in the context of movement, is what Newton's 2^nd law is about. otherwise, i might agree that it's a tautolgy.

Well, that's because historically, people knew first about statics, and there was already a concept of force there. But what I tried to point out is that in all these static cases, there is an equivalent dynamical case where the force is dp/dt of something. In all things with internal stress, this equivalent dynamical situation can be obtained by taking away (cutting, breaking...) some part of the system and see how the remaining part now accelerates.

also, by setting to one the constant of proportionality in F = \frac{dp}{dt}, that is not salient to Newton's 2^nd law. it could be expressed:

F = C \frac{dp}{dt}

where momentum is defined in terms of the base units of length, time, and mass, but force is defined in terms of some units determined by deformation of some standard prototype object. then the law would have some constant of proportionality that would have to be determined by experiment, like G, c, or \hbar is.

You can do that. The constant would then just determine the unit of "force" which you desire to use.

The whole point of the definition of force and Newton's second law and all that, is simply that there's a property of nature that we try to capture: the fact that the change in momentum dp/dt of different situations (with objects removed, cut away etc...) is related in a rather simple way (vectorial sum).
As such it is highly useful to give an "independent existence" and a name to this dp/dt thing.

 

[vanesch]

Well I thought F was used instead of dp/dt for the lower level physics texts that are not calculus based. Then as you go up levels in physics texts and math level you learn F is really representing dp/dt. side note, rate of change of momentum with respect to time is a bit of mouth full compared to force.

Yes, if you follow the development of landau and lifsh*tz, then at no point force is a fundamental concept, and they derive "Newton's second law" as some peculiar form of the equations of motion when the lagrangian is written with cartesian coordinates in an inertial frame.

 

[Andrew Mason]

This is precisely my opinion as well, and so far I've never seen a convincing argument against it.

The main conseqence of this is that if we were to try to properly axiomatize classical physics, Newtons second law together with the concept of force would NOT be included as a _fundamental_ axiom/law, instead it would be a definition used in the formulation building on top of the _real_ axioms/laws.

F = dp/dt is a conclusion based on observation. Unlike "density=m/V", Newton's second law is subject to empirical testing.

Force can be understood as an interaction (eg. pushing or pulling) between bodies. The magnitude of that interaction can be measured independently of dp/dt (eg. the force exerted on a body by a spring with a certain displacement can be doubled by adding another identical spring with the same displacement; the force exerted by two identical hanging masses can be halved by removing one of the masses).

AM

 

[rbj]

Well, that's because historically, people knew first about statics, and there was already a concept of force there.

yup. but it's not just historical. tension or pressure has physical meaning also even in contexts where nothing is getting accelerated.

You can do that. The constant would then just determine the unit of "force" which you desire to use.

actually, the unit of force was already determined (by some definition that has something to do with the stress and strain of some anthropometric prototype object). the constant would be related to the Young's modulus of the material of and the geometry of that prototype object. this unit of force would be defined completely independently of motion (other than displacement involved in squishing the prototype object, but no acceleration).

The whole point of the definition of force and Newton's second law and all that, is simply that there's a property of nature that we try to capture: the fact that the change in momentum dp/dt of different situations (with objects removed, cut away etc...) is related in a rather simple way (vectorial sum).
As such it is highly useful to give an "independent existence" and a name to this dp/dt thing.

well, if Newton's 2^nd law simply says that the vector sums of individual dp/dt vectors is the total dp/dt vector, that's almost saying that momentum is a vector that adds linearly. maybe that's not a tautology, but it feels like one. but if it relates this notion of force that exists as pressure or tension in static objects (which, until something snaps, is only detected as some deformation of the object) to how fast some mass accelerates (when the balancing component is removed), then that is clearly not a tautology. it actually says something. what if the time rate of acceleration was proportional to the square (rather than directly proportional) of this measure of force as perceived by the stretch or deformation of some object? that would say something different.

 

[vanesch]

well, if Newton's 2^nd law simply says that the vector sums of individual dp/dt vectors is the total dp/dt vector, that's almost saying that momentum is a vector that adds linearly. maybe that's not a tautology, but it feels like one.

Well, it surely doesn't work for d(mx)/dt !

Consider a situation with 3 masses, M1, M2 and m. The x1,x2 and x3 will correspond each time to the position of m in 3 different situations.
We look at the position of mass m, when there is only M1 present, and this gives us a certain d(mx1)/dt = m v1
We look at a different situation where there is only mass M2 present, and this gives us a certain d(m x2)/dt = m v2

Now, look at the third situation where there are both masses present, now we have d (m x3)/dt = m v3.

There is no relationship between d(mx1)/dt, d(mx2)/dt and d(mx3)/dt. So there's no point in giving a special name to d(mx1)/dt = G1 and so on.

However, if you do the same with d(mv1)/dt, d(mv2)/dt and d(mv3)/dt, you will see that d(mv3)/dt = d(mv1)/dt + d(mv2)/dt

So there's something special and "transitive" between different situations for the quantity d(mv)/dt.
We give it therefor a special name, "force".
So our observation is here that in the situation with only M1 and m present, m accelerates as d(mv1)/dt = F1 ;
that in the situation with only m and M2 present, m accelerates as:
d(mv2)/dt = F2; and, now comes the special property:
that if M1 and M2 are both present, that m accelerates through:
d(mv3)/dt = F3 = F1 + F2.
This didn't need to be. It is far from tautological, and I think it is this property which makes the definition of F = d(mv)/dt useful.

 

[vanesch]

the constant would be related to the Young's modulus of the material of and the geometry of that prototype object. this unit of force would be defined completely independently of motion (other than displacement involved in squishing the prototype object, but no acceleration).

Although I surely understand the intuitive appeal to that (and this was the historical approach), the problem I have with it is that in order to define a quantity of fundamental importance, we need to rely on a highly convolved property of certain mass configurations: namely the elasticity of solid matter, which is, by itself, a resulting property of miriads of binding forces inside the material, the crystaline structure and so on. If the world were made out of butter, we would not be able to define force that way, if you see what I mean.

 

[rbj]

However, if you do the same with d(mv1)/dt, d(mv2)/dt and d(mv3)/dt, you will see that d(mv3)/dt = d(mv1)/dt + d(mv2)/dt

i think you need to be more clear about what "the situation" is. or, at least, i need you to be more specific.

So there's something special and "transitive" between different situations for the quantity d(mv)/dt.
We give it therefor a special name, "force".
So our observation is here that in the situation with only M1 and m present, m accelerates as d(mv1)/dt = F1 ;
that in the situation with only m and M2 present, m accelerates as:
d(mv2)/dt = F2; and, now comes the special property:
that if M1 and M2 are both present, that m accelerates through:
d(mv3)/dt = F3 = F1 + F2.
This didn't need to be. It is far from tautological, and I think it is this property which makes the definition of F = d(mv)/dt useful.

are you talking about three collisions?:

1. between m and M1
2. between m and M2
3. and lastly between m and M1+M2 stuck together

wouldn't this law simply be that of the vector conservation of momentum in a closed system? is that not what this is? it defines momentum and says it's conserved (and i s'pose that's not a tautology; momentum defined as \mathbf{p} \equiv m \mathbf{v} wouldn't necessarily have to be conserved). otherwise, i am not sure i understand your "situation".

but Newton's 2^nd law (along with Hooke's law which, for the sake of argument, is just the pre-existing working definition of "force") says that when i take a mass M, mount it on my bow string and pull back 1 cm and release, it accelerates at some rate A. when i take a two of those same masses stuck together and pull back 2 cm, it acclerates at A again. and a single mass M pulled back 2 cm and released accelerates at rate 2A. that's where i can give Newton's 2^nd law meaning more than the vector conservation of momentum in closed systems.

i can kinda see some of the OP's question. it's a good naive POV that helps us rethink what we believe.

 

[rbj]

F = dp/dt is a conclusion based on observation. Unlike "density=m/V", Newton's second law is subject to empirical testing.

Force can be understood as an interaction (eg. pushing or pulling) between bodies. The magnitude of that interaction can be measured independently of dp/dt (eg. the force exerted on a body by a spring with a certain displacement can be doubled by adding another identical spring with the same displacement; the force exerted by two identical hanging masses can be halved by removing one of the masses).

i think, Andrew, that is what i was trying to say. to keep it from being circular, or a tautology, the meaning of this "pushing or pulling" has to be quantified in an independent manner than measuring dp/dt. otherwise

\mathbf{F} = \frac{d \mathbf{p}}{dt}

is a tautology if force is defined in terms of momentum to begin with.

 

[vanesch]

i think you need to be more clear about what "the situation" is. or, at least, i need you to be more specific.

Three different "Newtonian worlds".
One is a world with two particles, one with mass M1 at position u(t) and one with mass m at position x1(t), where u and x1 are two world lines. Consider u(t) given, while x1 is the "solution to the dynamical problem".

A second one is a world with again two particles, one with mass M2 at position v(t) and one with mass m at position x2(t). Again, consider v(t) given.

We can assume x1(t=0) = x2(t=0) = x0, and x1'(t=0) = x2'(t=0)= v0.

The third one is a world with 3 particles. One with mass M1 with world line u(t) (the same u), one with mass M2 with world line v(t) (the same v), and our third particle with position x3(t), such that x3(t=0) = x0 and x3'(t=0) = v0.

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.

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. As such, the interactions resolve into simple situations, and in these simple situations the law relating, say position and velocity to the famous dp/dt may become a relatively simple expression (thanks to symmetry arguments and all that).

Imagine for instance a gravitational interaction between 7 particles (A,B,C,D,E,F,G). We consider one of these particles, A.
Now, we consider the "sub-situations" where A is only there in presence of particle B. We could now try to find the most general expression that gives us a vector dp/dt of A as a function of positions, velocities and "constants determining the particle" such as mass, charge etc... and then, thanks to empirical observation, fill in the free functions.

We'd find something like dp/dt = f(r_A, v_A,r_B,v_B,c1,c2,c3,...), but in such a way that the function is invariant under rotations or translations of space: it needs hence to be a vectorial expression, which reduces in the case of gravitational interaction to something like dp/dt = -G.mA.mB r/r^3
So this two-particle situation is manageable, and the peculiar property of addition of dp/dt for subsituations makes that it has a sense to consider only 1-1 interactions, and then analyse a more complicated situation as a combination of 1-1 interactions. This is so useful that we give a name to the dp/dt for the 1-1 interaction, and call it the "force excerted by one particle onto the other", and then when there are many particles present, we simply sum over all the 1-1 "contributions".

are you talking about three collisions?:

No, I'm talking about 3 interaction situations (say, gravitational pull, or electrical coulombic pull or something of the kind).

(along with Hooke's law which, for the sake of argument, is just the pre-existing working definition of "force") says that when i take a mass M, mount it on my bow string and pull back 1 cm and release, it accelerates at some rate A.

But Hooke's law is a very complicated phenomenon ! It doesn't work with water or with butter !
If the ambient temperature is 20 000 degrees, there is no Hooke's law for instance. So one would need a very involved phenomenon in order to define a fundamental quantity in nature ?

While relating dp/dt for different situations is elementary and universal. You don't need a special material like a solid body to define it.

 

[octol]

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.

 

[vanesch]

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

This is the way I also understand things. However, it is a very inspired definition, by an empirically observed phenomenon which is not obvious at all, and which relates the dp/dt in a given situation to the different dp/dt in various "simplified" situations, by simple vectorial sum. This property is highly suggestive of an "independent reality" of these dp/dt, and so it is very beneficial to give it a name of its own.
If it weren't for this "combination" property, the definition of force wouldn't have much utility (imagine that we could not write the "total force" as the "combination of 1-1 interaction forces", but that we have a different force law if there are 2 particles, or 3 particles, or 4 particles, or... with no relationship between them).

Moreover, as rbj pointed out, it is again this same quantity which is also responsible for what we intuitively call "pushing or pulling" in statics which is nothing else but a manifestation of a combination of Hooke's law, the property mentionned above, and another property which is conservation of momentum. It is true that historically, and intuitively, this approach is easier to grasp, but I find it less "fundamental".

 

[Andrew Mason]

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.

If it was just a definition, it need not have any physical significance or meaning. The fact is that F = dp/dt has a physical significance in the real world. It provides the means of quantifying matter interactions.

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.

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.

Newton postulated, but did not prove that inertial and gravitational mass were equivalent when he postulated his Law of Universal Gravitation. It has since been shown to be true to within an infinitessimal margin of error. F = GmM/R^2 is not "defined" any more than F = dp/dt is defined.

AM

 

[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