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

  • Thread starter octol
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In summary, the concept of force is redundant and not intuitive. Feynman discusses this in greater detail in his lectures. The concept of force is based on an inspired definition- G=d (mx)/dt- which is a much simpler and easier to understand definition than F=dp/dt. 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
  • #1
octol
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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

[tex] \frac{dp}{dt} = G \frac{m_1 m_2}{r^2} [/tex]

and for electromagnetism

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

In other words, the concept of "force" is intuitive and all, but still fundamentally redundant.
 
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  • #2
I don't understand what the purpose of replacing F with dP/dt would solve.
 
  • #3
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.
 
  • #4
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 [itex]\int_R \rho \, dV[/itex] instead?

Because it's convenient, and often conceptually useful.
 
  • #5
Another major threat to the concept of classical physics. :tongue2:
 
  • #6
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 [tex] m = \int \rho dV [/tex]
 
  • #7
octol said:
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

[tex] \frac{dp}{dt} = G \frac{m_1 m_2}{r^2} [/tex]
What if dp and dt are both zero?
 
  • #8
octol said:
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.
 
  • #9
Newton's second law does not say [tex]F = \frac{dp}{dt}[/tex]. What it says is [tex]\sum \vec{F} = \frac{d\vec{p}}{dt}[/tex]. We can't just replace "F" in all the equations defining forces with [tex]\frac{dp}{dt}[/tex] 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:

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

Here, the equations for each force separately are true when written with F's, but not when written with [tex]\frac{dp}{dt}[/tex].
 
  • #10
Parlyne said:
Newton's second law does not say [tex]F = \frac{dp}{dt}[/tex]. What it says is [tex]\sum \vec{F} = \frac{d\vec{p}}{dt}[/tex]. We can't just replace "F" in all the equations defining forces with [tex]\frac{dp}{dt}[/tex] 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:

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

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

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:

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

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

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 [tex] F = \frac{dp}{dt}[/tex], and that, if you analyse the situation, that it turns out that this equals also [tex]\frac{GmM}{r^2} \hat{r} + q (\vec{E} + \vec{v} \times \vec{B})[/tex]

It is just an amazing property of nature, that the different terms in there ALSO correspond to momentum changes in OTHER situations. The [tex]\frac{GmM}{r^2}[/tex] term for instance, occurs in another situation, as the [tex]\frac{dp}{dt}[/tex] which looks a bit like the one we're studying, but without charge.
The term [tex] q (\vec{E} + \vec{v} \times \vec{B})[/tex] occurs in still another situation as the [tex]\frac{dp}{dt}[/tex], 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 [tex]\frac{dp}{dt}[/tex] has a remarkable property in certain circumstances: for a given situation, it equals the vector sum of the [tex]\frac{dp}{dt}[/tex] 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: [tex]F = \frac{GmM}{r^2} \hat{r} + q (\vec{E} + \vec{v} \times \vec{B})[/tex]. 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 [tex]\frac{dp}{dt}[/tex]
 
  • #11
octol said:
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?
 
  • #12
vanesch said:
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.
 
  • #13
vanesch said:
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?
 
  • #14
lightarrow said:
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?
 
  • #15
arildno said:
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).
 
  • #16
Gokul43201 said:
Even in light of Aharonov-Bohm?

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

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).
 
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  • #17
lightarrow said:
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.
 
  • #18
arildno said:
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.
 
  • #19
vanesch said:
I'm only talking in the frame of classical mechanics here :cool:

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.
 
  • #20
octol said:
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.
 
  • #21
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.
 
  • #22
octol said:
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 [tex] m = \int \rho dV [/tex]

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 2nd 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 [itex] F = \frac{dp}{dt} [/itex], that is not salient to Newton's 2nd law. it could be expressed:

[tex] F = C \frac{dp}{dt} [/tex]

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 [itex]\hbar[/itex] is.
 
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  • #23
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.
 
  • #24
rbj said:
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 2nd 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 [itex] F = \frac{dp}{dt} [/itex], that is not salient to Newton's 2nd law. it could be expressed:

[tex] F = C \frac{dp}{dt} [/tex]

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 [itex]\hbar[/itex] 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.
 
  • #25
axawire said:
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.
 
  • #26
octol said:
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
 
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  • #27
vanesch said:
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 2nd 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.
 
  • #28
rbj said:
well, if Newton's 2nd 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.
 
  • #29
rbj said:
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.
 
  • #30
vanesch said:
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 [itex] \mathbf{p} \equiv m \mathbf{v} [/itex] wouldn't necessarily have to be conserved). otherwise, i am not sure i understand your "situation".

but Newton's 2nd 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 2nd 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.
 
  • #31
Andrew Mason said:
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

[tex] \mathbf{F} = \frac{d \mathbf{p}}{dt} [/tex]

is a tautology if force is defined in terms of momentum to begin with.
 
  • #32
rbj said:
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.
 
  • #33
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
[tex] v = \frac{dx}{dt} [/tex]
is not a fundamental relation, just a definition. "Force" is no different, it is a name for dp/dt.
 
  • #34
octol said:
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".
 
  • #35
octol said:
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
[tex] v = \frac{dx}{dt} [/tex]
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
 

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