What Do Newton's Laws Say When Carefully Analysed

[vis_insita]

That is just your assumption, unless you can show it. I don't see anything in the laws of motion that implies or requires such a limitation.

The Third Law states (my emphasis)[1]: "To any action there is always an opposite and equal reaction; in other words, the actions of two bodies upon each other are always equal and always opposite in direction."

It doesn't make sense to speak of "the actions of two bodies upon each other", if the force that represents that action depends on three bodies and cannot be decomposed into a sum of two-body forces.

__________
[1] Cohen, I. Bernard, Anne Whitman, and Julia Budenz. The Principia: Mathematical Principles of Natural Philosophy. University of California Press, 1999.

Force is defined in definition IV. But that is not sufficcient. I already explained that in #120.

You said forces are defined as two-body forces and the Third Law is a "definition". What is defined by the Third Law? Are three-body forces not forces by definition?

 

[PeterEnders]

The original poster believes to having found contradictions in Newton's 3 axioms. IMHO, the contradictions stem from his personal understanding aka interpretation of Newton's text. Let me present an understanding being free of those contradictions.
The first axiom represents Galilean inertia and defines the state at rest and the state of straight uniform motion as the stationary states in free space.
The second axiom defines the linear momentum vector to be the stationary state variable and describes its change under the influence of external ("impressed motive") forces. No cause - no effect. Notice that the position vector does not enter Newton's notion of state and hence not the axioms.
The third axiom and the explanations to it postulate that to each force a counter-force of equal magnitude and opposite direction occurs. Like the first axiom it represents a conservation law, too.

 

[vanhees71]

OK, I see we have a misunderstanding here, resulting from my improper use of the term "two-body force". I just mean forces according Newton III - acting pairwise and symmetric between two bodies. I didn't want to refer to corresponding force laws. That is out of the scope of the laws of motion.

No it's not. The symmetries of spacetime constrain the types of force laws. That's the great merit of the "group-theoretical approach". It's THE achievement of 20th-century physics, without which all the other achievements wouldn't have been possible. It started with Einstein's paper of 1905 on special relativity. For Newtonian mechanics, the group-theoretical approach is of course a posteriori.

Now translation-invariance of the Galilei-Newton spacetime implies momentum conservation and thus (together with Newton I and II!) that two-body forces must obey Newton III. There are more force laws though, generic N-body forces, which fulfill the constraint, and they have even applications. In this sense Newton's concept of physics is more comprehensive than he had thought himself.

 

[DrStupid]

It doesn't make sense to speak of "the actions of two bodies upon each other", if the force that represents that action depends on three bodies and cannot be decomposed into a sum of two-body forces.

It make sense to speak of "the actions of two bodies upon each other", if the force that represents that action can be decomposed into forces acting between two bodies and that is always possible.

You said forces are defined as two-body forces and the Third Law is a "definition". What is defined by the Third Law? Are three-body forces not forces by definition?

I already explained that I actually mean forces acting between two bodies - no matter what they depend on.

 

[gmax137]

I don't have a dog in this fight, but you all seem to be talking past each other.

The Wiki cited by @ZapperZ EDIT: sorry, I mean, @vanhees71 says (emphasis added):

In general, if the behaviour of a system of more than two objects cannot be described by the two-body interactions between all possible pairs, as a first approximation, the deviation is mainly due to a three-body force.

Meanwhile (again, emphasis added)

if the force that represents that action can be decomposed into forces acting between two bodies and that is always possible.

 

[vis_insita]

It make sense to speak of "the actions of two bodies upon each other", if the force that represents that action can be decomposed into forces acting between two bodies and that is always possible.

So, you are simply going to ignore the proof sketched by Dale above, because you decided that you will just call any force a "two-body" force no matter on how many bodies it actually depends? Seems like we really do have different conceptions of "reasonable" after all.

 

[ZapperZ]

I don't have a dog in this fight, but you all seem to be talking past each other.

The Wiki cited by @ZapperZ says (emphasis added):

Er, I didn't cite any wiki article. As a matter of fact, I have not participated in this thread at all... till now.

Zz.

 

[gmax137]

Er, I didn't cite any wiki article. As a matter of fact, I have not participated in this thread at all... till now.

Zz.

Oh gosh I'm sorry... It was @vanhees71

 

[DrStupid]

because you decided that you will just call any force a "two-body" force no matter on how many bodies it actually depends?

I already admitted that this was a mistake. How often do I need to repeat it for you?

 

[vis_insita]

Your claim that any force can "be decomposed into forces acting between two bodies" seemed to repeat that same mistake. If a force depends on three bodies, it doesn't "act between two bodies".

 

[DrStupid]

If a force depends on three bodies, it doesn't "act between two bodies".

If the force and it's corresponding counterforce change the motion of two bodies only than it acts between these two bodies only - no matter what it depends on.

 

[vis_insita]

I already understood that this is what you claim. I still maintain that it makes no sense.

Any force $F_1$ on body 1 only changes the motion of body 1, and any force on body 2 that of body 2. What is meaningless is splitting that force $F_1$ arbitrarily in components $F_{12}$ and $F_{13}$, one of which is supposed to be "the action of body 2" and the other "the action of body 3", because at least one of these components (and maybe both) depend on both bodies 2 and 3.

 

[DrStupid]

Any force $F_1$ on body 1 only changes the motion of body 1, and any force on body 2 that of body 2. What is meaningless is splitting that force $F_1$ arbitrarily in components $F_{12}$ and $F_{13}$, one of which is supposed to be "the action of body 2" and the other "the action of body 3", because at least one of these components (and maybe both) depend on both bodies 2 and 3.

A rule for just two bodies that always guarantees global conservation of momentum is not meaningless but actually a brilliant idea. But that aside - wouldn't it be just as meaningless to split the net forces into n-body forces just because they depend on n bodies?

 

[vis_insita]

A rule for just two bodies that always guarantees global conservation of momentum is not meaningless but actually a brilliant idea.

Your rule does not guarantee anything like that. The property that ensures global momentum conservation in the three-body-system
$$m_i \ddot{\boldsymbol{r}}_i = \boldsymbol{F}_i(\boldsymbol{r}_1, \boldsymbol{r}_2, \boldsymbol{r}_3)$$
is the identity
$$\sum_{i=1,2,3}\boldsymbol{F}_i = 0.$$

This has nothing at all to do with how you split $\boldsymbol{F}_1$ into "action of body 2" and "action of body 3". If the above condition does not hold, then no decomposition whatsoever will help with momentum conservation. If it does hold, no arbitrary decomposition is needed for anything.

But that aside - wouldn't it be just as meaningless to split the net forces into n-body forces just because they depend on n bodies?

What split do you mean? The force $\boldsymbol{F}_1(\boldsymbol{r}_1, \boldsymbol{r}_2, \boldsymbol{r}_3)$ is defined as the force on body 1 caused by the simultaneous presence of body 2 and body 3 (not the action of either body). That's not meaningless. Now, whether a decomposition
$$\boldsymbol{F}_1(\boldsymbol{r}_1, \boldsymbol{r}_2, \boldsymbol{r}_3)=\boldsymbol{F}_{1,2}(\boldsymbol{r}_1, \boldsymbol{r}_2) + \boldsymbol{F}_{1,3}(\boldsymbol{r}_1, \boldsymbol{r}_3)$$
is possible or not, distinguishes sums of two-body forces from genuine three-body forces . Only if it is possible, does it make sense to call the first and second component the respective "action of body 2" and "action of body 3" on body 1.

 

[DrStupid]

The property that ensures global momentum conservation in the three-body-system
$$m_i \ddot{\boldsymbol{r}}_i = \boldsymbol{F}_i(\boldsymbol{r}_1, \boldsymbol{r}_2, \boldsymbol{r}_3)$$
is the identity
$$\sum_{i=1,2,3}\boldsymbol{F}_i = 0.$$

$F_{ij} = - F_{ji}$ does this job as well

 

[vis_insita]

$F_{ij} = - F_{ji}$ does this job as well

That's completely besides the point. Necessary and sufficient is the condition I gave. Yours is equivalent to mine, given an arbitrary decomposition which is physically meaningless.

 

[DrStupid]

What split do you mean? The force $\boldsymbol{F}_1(\boldsymbol{r}_1, \boldsymbol{r}_2, \boldsymbol{r}_3)$ is defined as the force on body 1 caused by the simultaneous presence of body 2 and body 3 (not the action of either body).

$F_{1,net} = F_1 \left( {r_1 ,r_2 ,r_3 } \right) + F_{1,other}$

Where $F_{1,other}$ is the sum of forces on body 1 caused by the simultaneous presence of other bodies.

 

[vis_insita]

$F_{1,net} = F_1 \left( {r_1 ,r_2 ,r_3 } \right) + F_{1,other}$

Where $F_{1,other}$ is the sum of forces on body 1 caused by the simultaneous presence of other bodies.

If $F_{1,other}$ depends, aside from body 1, only on those other bodies, then it is not meaningless.

 

[dextercioby]

The way I learned in school is that the axioms of classical Newtonian mechanics are 4: the three Newton's laws and the principle of independence of interactions (action of forces).

This last axiom is: The force with which a body acts on another body is independent of the existence of other bodies.
If the forces among particles are conservative, then mathematically (for a 3 particle system):

$U(1,2,3) = U(1,2) + U(1,3) + U(2,3)$

 

[DrStupid]

If $F_{1,other}$ depends, aside from body 1, only on those other bodies, then it is not meaningless.

The pair $F_{12}$ and $F_{21}$ affects only body 1 and 2. Why is that meaningless? Isn't the cause at least as important as the root?

 

[DrStupid]

This last axiom is: The force with which a body acts on another body is independent of the existence of other bodies.

That doesn't even work with two iron nails and a magnet. What is the benefit of such a limitation?

 

[vis_insita]

The pair $F_{12}$ and $F_{21}$ affects only body 1 and 2. Why is that meaningless? Isn't the cause at least as important as the root?

You keep saying that "a pair" of forces affects "a pair" of bodies. That's misleading. Of course $F_{12}$ only affects body 1, because it is a component of the total force on body 1. The same is true of $F_{21}$ and body 2. However, though misleading in my opinion, that's not what I called "meaningless". There's not much wrong with calling $F_{12}$ an "action on body 1". It is only meaningless to call it "the action of body 2" if it also depends on body 3.

 

[DrStupid]

Yours is equivalent to mine, given an arbitrary decomposition which is physically meaningless.

I noted that you do not see the advantage of a rule that deals with just two bodies over a common rule for all bodies in a system. We will not come to an agreement in this regard.

It is only meaningless to call it "the action of body 2" if it also depends on body 3.

It is irrelevant how it is called as long as there are symmetric pairs of forces according to Newton III. That is not the case for common actions of 2 of the bodies.

 

[Dale]

The pair F12F12F_{12} and F21F21F_{21} affects only body 1 and 2. Why is that meaningless?

It is meaningless because there is no sense in which there is a physically meaningful force between only bodies 1 and 2. You have created a meaningless label “12” which is meaningless precisely because there is no meaningful physical interaction between bodies 1 and 2 only. The physical interaction is between bodies 1, 2, and 3, so a meaningful labeling of the forces would convey that physical fact.

 

[Dale]

At this point I think this thread has run its course. I will leave it open for a bit longer to give the recent participants one last chance to respond, but then I will close it.

 

[atyy]

$F_{ij} = - F_{ji}$ does this job as well

If you look at the link that @vanhees71 gave https://www.reed.edu/physics/faculty/wheeler/documents/Classical Mechanics/Class Notes/Chapter 4.pdf, for a 3-body force Newton's 3rd law has to be generalized to $F_{i,jk}+F_{j,ki}+F_{k,ij}=0$.

 

[vis_insita]

I noted that you do not see the advantage of a rule that deals with just two bodies over a common rule for all bodies in a system.

The rule does not deal with pairs of bodies only. That's what I'm trying to tell you. It deals with artificial "pairs of forces" which, however, necessarily depend on more than two bodies. This is why your rule is entirely useless compared to $\boldsymbol{F}_{i, jk} + \boldsymbol{F}_{j, ki} + \boldsymbol{F}_{k, ij} = 0$ in a system with genuine three-body forces.

It is irrelevant how it is called as long as there are symmetric pairs of forces according to Newton III. That is not the case for common actions of 2 of the bodies.

Yes, it's not the case, which was precisely my reason for questioning your claim that the Third Law is a "definition." It is a statement about two-body forces, which is not universally valid.

 

[vanhees71]

It make sense to speak of "the actions of two bodies upon each other", if the force that represents that action can be decomposed into forces acting between two bodies and that is always possible.

Sigh. That's NOT true. As I've repeatedly said and even by providing some references, in atomic, molecular, and nuclear physics for effective (nonrelativistic) models you need interactions that are not representable by the sum over two-body forces/potentials.

 

[vanhees71]

The way I learned in school is that the axioms of classical Newtonian mechanics are 4: the three Newton's laws and the principle of independence of interactions (action of forces).

This last axiom is: The force with which a body acts on another body is independent of the existence of other bodies.
If the forces among particles are conservative, then mathematically (for a 3 particle system):

$U(1,2,3) = U(1,2) + U(1,3) + U(2,3)$

Yes, and that's not general enough to describe some more complicated systems, including atoms, molecules, atomic nuclei.

 

[dextercioby]

Yes, and that's not general enough to describe some more complicated systems, including atoms, molecules, atomic nuclei.

Yes, but that is Newtonian mechanics. It is not expected to apply to electromagnetism and quantum systems.

 

[Dale]

With that we will close the thread