What Do Newton's Laws Say When Carefully Analysed

[vanhees71]

How do the nucleons violate the third law?

It's well known that effective theories describing nuclei as bound states of nucleons need three- and even more n-body forces. E.g., the fact that He-3 is stablecannot be understood using two-body forces only.

Nothing violates the 3rd law though since translation invariance is a fundamental symmetry of Newtonian spacetime. The only thing that's violated is the claim by @DrStupid that any forces are necessarily sums of two-body forces only. There's nothing in Newtonian mechanics not allowing for three- and even more general n-body forces. Most simply you just describe them by a potential $V^{(n)}(\vec{x}_1,\ldots,\vec{x}_n)$. The only thing you need to ensure the 3rd law (i.e., momentum conservation) is translation invariance, as we know from Noether's theorems, i.e., there's a symmetry constraint on the potential
$$V^{(n)}(\vec{x}_1+\vec{a},\vec{x}_2+\vec{a},\ldots,\vec{x}_n+\vec{a})=V^{(n)}(\vec{x}_1,\ldots,\vec{x}_n).$$
making $\vec{a}$ very small and expanding leads to the conclusion that the total momentum is conserved, i.e.,
$$\sum_{j=1}^n \vec{\nabla}_j V=-\sum_{j=1}^{n} \vec{F}_j=0.$$

 

[DrStupid]

The only thing that's violated is the claim by @DrStupid that any forces are necessarily sums of two-body forces only.

Citation please!

 

[vis_insita]

It's a definition.

No, it's not. It says that $\boldsymbol{F}_1(x, y) \neq - \boldsymbol{F}_2(y, x)$ can never describe the dynamics of a real two-particle-system. That is a hypothesis with empirical content, not a definition.

 

[DrStupid]

It says that $\boldsymbol{F}_1(x, y) \neq - \boldsymbol{F}_2(y, x)$ can never describe the dynamics of a real two-particle-system.

That means that fictitious forces are not considered to be forces. That is a definition. You may use it or not. You may like it or not. But you cannot test if it is correct or not.

 

[atyy]

It's well known that effective theories describing nuclei as bound states of nucleons need three- and even more n-body forces. E.g., the fact that He-3 is stablecannot be understood using two-body forces only.

Nothing violates the 3rd law though since translation invariance is a fundamental symmetry of Newtonian spacetime. The only thing that's violated is the claim by @DrStupid that any forces are necessarily sums of two-body forces only. There's nothing in Newtonian mechanics not allowing for three- and even more general n-body forces. Most simply you just describe them by a potential $V^{(n)}(\vec{x}_1,\ldots,\vec{x}_n)$. The only thing you need to ensure the 3rd law (i.e., momentum conservation) is translation invariance, as we know from Noether's theorems, i.e., there's a symmetry constraint on the potential
$$V^{(n)}(\vec{x}_1+\vec{a},\vec{x}_2+\vec{a},\ldots,\vec{x}_n+\vec{a})=V^{(n)}(\vec{x}_1,\ldots,\vec{x}_n).$$
making $\vec{a}$ very small and expanding leads to the conclusion that the total momentum is conserved, i.e.,
$$\sum_{j=1}^n \vec{\nabla}_j V=-\sum_{j=1}^{n} \vec{F}_j=0.$$

Here (in saying that 3 body forces are consistent with Newton's third) are you still formulating Newton's third as saying that if body 1 exerts a force on body 2, then body 2 exerts an equal and opposite force on body 1?

 

[vis_insita]

That means that fictitious forces are not considered to be forces.

The Third Law is a statement about interactions (more precisely the relation between "action" and "reaction"), not about fictitious forces. That's the only reasonable interpretation. (I would provide a quote, but I think you already know the words.)

 

[DrStupid]

The Third Law is a statement about interactions (more precisely the relation between "action" and "reaction"), not about fictitious forces.

The 3rd laws says that to every action there is always a reaction. That is a clear statement about fictitious fources: they are no forces because there is no reaction.

That's the only reasonable interpretation.

Maybe we just have a different concept of "reasonable".

 

[vis_insita]

The 3rd laws says that to every action there is always a reaction.

Or "in other words, the actions of two bodies upon each other are always equal and always opposite in direction."

This is a statement about all possible interactions between two bodies, which is precisely the part that is not a definition.

That is a clear statement about fictitious fources: they are no forces because there is no reaction.

According to that logic the Third Law is a clear statement about everything that is not an interaction, like, say, temperature.

 

[DrStupid]

Or "in other words, the actions of two bodies upon each other are always equal and always opposite in direction."

This part of Newton III isn't violated by fictitious forces but the first part is.

This is a statement about all possible interactions between two bodies, which is precisely the part that is not a definition.

Really? How are you going to test it?

According to that logic the Third Law is a clear statement about everything that is not an interaction, like, say, temperature.

And that means that everything that is not an interaction cannot be a force - including fictitious forces or temperature.

 

[vanhees71]

Citation please!

Google for many-body forces. You get some hits in atomic and particularly nuclear physics. In the latter topic it's even subject of current research!

 

[Dale]

Google for many-body forces. You get some hits in atomic and particularly nuclear physics. In the latter topic it's even subject of current research!

@vanhees71 please actually provide any requested citations, including ones that you think should be obvious or common knowledge. Providing sources on request is an important part of what we do here at PF that makes this forum different. It ensures that what is said is consistent with mainstream science and it also gives others a place to go for further study.

 

[vanhees71]

$\newcommand{\uvec}[1]{\underline{#1}}$

And that means that everything that is not an interaction cannot be a force - including fictitious forces or temperature.

That makes sense now. Forces are caused by interactions, and the Newtonian space-time structure dictates translation invariance and thus energy, momentum, and angular-momentum conservation for closed systems. That determines the action of Newtonian point-particle systems. It does however not restrict everything to two-body interactions only. There can be generic n-body interactions (see #176). I've no clue what anything of this has to do with temperature (at least not within non-relativistic Newtonian mechanics) at all.

"Fictitious forces" (I prefer to call them "inertial forces") are terms that from a principle point of view do not belong to the right-hand side of the equation of motion. They occur from bringing parts of $m \ddot{\vec{r}}$, where $\vec{r}$ is a vector within an inertial reference frame to the other side in the equation, when expressing the equations of motion in terms of the vector components (sic!) wrt. a non-inertial reference frame. The latter is defined by the origin $O'$ of the non-inertial reference frame, which is moving in an arbitrary way against the origin of the inertial frame, and a right-handed Cartesian basis $\vec{e}_k'(t)=D_{jk}(t) \vec{e}_j$ which can arbitrarily rotate against the inertial Cartesian basis $\vec{e}_j$, where $(D_{jk})=\hat{D} \in \mathrm{SO}(3)$. This is the most general form of an accelerated reference frame (accelerated against an inertial frame of course).

Now take the most simple example of a particle moving in some external field (e.g., a gravitational field of a very heavy body, e.g., the Earth's gravitational field acting on a stone or a charged particle in an electrostatic field, neglecting radiation reaction, which is a relativistic effect anyway). Then in the inertial frame of reference you have
$$m \ddot{\vec{r}}=\vec{F}(\vec{r})$$
or in components
$$m \frac{\mathrm{d}^2}{\mathrm{d} t^2} r_j \vec{e}_j=F_j(\vec{r}) \vec{e}_j \; \Rightarrow \; m \ddot{r}_j = F_j(\vec{r}).$$
Now you have
$$\vec{r}=\vec{R}+\vec{x},$$
where $\vec{R}=\overrightarrow{OO'}$ is the position vector of the accelerated reference frame's origin wrt. the inertial frame's origin, and $\vec{x}$ is the position vector wrt. $O'$.

To get the equations of motion for the components $\uvec{r}'=(r_1',r_2',r_3')^{\text{T}}$ of the position vector $\vec{r}=\overrightarrow{RP}$, we need the time derivative of an arbitrary vector $\vec{V}(t)$ in terms of it's non-inertial components, i.e.,
$$\dot{\vec{V}}=\mathrm{d}_t (V_k' \vec{e}_k')=\dot{V}_k' \vec{e}_k + V_k' \dot{\vec{e}}_k'.$$
Now we have
$$\vec{e}_k' = D_{jk} \vec{e}_j \; \Rightarrow \; \dot{\vec{e}}_k' = \dot{D}_{jk} \vec{e}_j = D_{jl} \dot{D}_{jk} \vec{e}_l'=\Omega'_{lk} \vec{e}_l'.$$
Next we show that $\Omega'_{lk}=-\Omega'_{kl}$. This follows from the orthogonality of the rotation matrix $\hat{D}$, i.e., $D_{jl} D_{jk}=\delta_{lk}$ and thus
$$\Omega_{lk}'=\dot{D}_{jl} D_{jk}=-\dot{D}_{jl} D_{jk}=-\Omega_{kl}'.$$
So we can write
$$\Omega_{lk}'=\epsilon_{klm} \omega_m'=-\epsilon_{lkm} \omega_m'.$$
Then we find
$$\dot{\vec{V}}=\dot{V}_k' \vec{e}_k + V_{k}' \epsilon_{klm} \omega_m' \vec{e}_l'=(\dot{V}_l'+\epsilon_{lmk} \omega_m' V_k') \vec{e}_l'.$$
For simplicity we can now introduce a covariant time-derivative for vector components by
$$\dot{\uvec{V}}'=\dot{\uvec{V}}'+\uvec{\omega}' \times \uvec{V}'.$$
Now the equation of motion for $\vec{r}$ are easily expressed in terms of the non-inertial components. We only need the 2nd covariant time derivative. First we have, as just derived,
$$\mathrm{D}_t{\uvec{r}}'=\dot{\uvec{r}}'+\uvec{\omega}' \times \uvec{r}',$$
and then after some algebra
$$m \mathrm{D}_t{\uvec{r}}'=m[\ddot{\uvec{r}}' + 2 \uvec{\omega} \times \dot{\uvec{r}}' + \dot{\uvec{\omega}}' \times \uvec{r}'+\uvec{\omega}' \times (\uvec{\omega}' \times \uvec{r}')]=\uvec{F}'(\uvec{r}').$$
As you see everything is form-invariant when using the proper time derivatives $\mathrm{D}_t$ for components in the intertial frame. Then there are just the "real forces", i.e., in our case just $\uvec{F}'$ on the right-hand side of the equation, but now it is customary to single out $m \ddot{\uvec{r}}'$ and write the equation of motion as
$$m \ddot{\uvec{r}}' =\uvec{F}'-m[2 \uvec{\omega} \times \dot{\uvec{r}}' + \dot{\uvec{\omega}}' \times \uvec{r}'+\uvec{\omega}' \times (\uvec{\omega}' \times \uvec{r}')]$$
and calls the additional pieces on the right-hand side "fictitious forces" or "inertial forces" and gives them fancy names like "Coriolis force", "centrifugal force", and the part with $\dot{\uvec{\omega}}'$, for which I don't know whether there's also a specific name.

It's just written the usual equations of motion, which are well-defined in the intertial frame in a non-covariant way in terms of the vector components with respect to the basis fixed in the non-inertial reference frame which can time-dependently rotate against the basis fixed in the inertial frame.

 

[vis_insita]

I've no clue what anything of this has to do with temperature (at least not within non-relativistic Newtonian mechanics) at all.

Nothing, of course, which was exactly the point. "Anything" in this context was the Third Law, which is a statement about forces, not about non-forces.

and calls the additional pieces on the right-hand side "fictitious forces" or "inertial forces" and gives them fancy names like "Coriolis force", "centrifugal force", and the part with $\dot{\uvec{\omega}}'$, for which I don't know whether there's also a specific name.

I think I have seen the $\dot \omega$-term being called "Euler force" before. But I don't remember where.

 

[vanhees71]

@vanhees71 please actually provide any requested citations, including ones that you think should be obvious or common knowledge. Providing sources on request is an important part of what we do here at PF that makes this forum different. It ensures that what is said is consistent with mainstream science and it also gives others a place to go for further study.

Ok, it's almost never discussed in general mechanics textbooks. There's a nice manuscript, where it's mentioned:

https://www.reed.edu/physics/faculty/wheeler/documents/Classical Mechanics/Class Notes/Chapter 4.pdf
which is a chapter from

https://www.reed.edu/physics/faculty/wheeler/documents/

in the folder classical mechanics->class notes->Chapter 4.pdf

About three-body forces there's even a short Wikipedia article:

https://en.wikipedia.org/wiki/Three-body_force

The current state of the art of three-body nuclear forces from effective chiral theory, can be found in the following RMP article (it's of course not classical mechanics anymore but QFT)

https://arxiv.org/abs/1210.4273

 

[Dale]

Thanks @vanhees71 that is perfect!

 

[DrStupid]

Google for many-body forces. You get some hits in atomic and particularly nuclear physics. In the latter topic it's even subject of current research!

I am asking for a reference for "the claim by @DrStupid that any forces are necessarily sums of two-body forces only."

 

[vanhees71]

That's what I understood you claimed in several postings in this thread. Maybe I misunderstood you.

 

[Dale]

That's what I understood you claimed in several postings in this thread. Maybe I misunderstood you.

Also, “Citation please” sounds like a request for a scientific reference rather than a claim that he personally never said it.

 

[Dale]

I am asking for a reference for "the claim by @DrStupid that any forces are necessarily sums of two-body forces only."

It's a definition.

If the 3rd law is a definition of force then only two-body forces are forces by definition. I agree with @vanhees71 characterization of your position. That is also how I understood your intent from what you wrote.

Furthermore, I highly recommend that you make an effort to communicate more clearly, both of us came to the same incorrect conclusion about your intention because of your excessively brief posts. "It's a definition" and "Citation please" clearly did not adequately communicate your intention, neither regarding your claim nor your objection.

 

[DrStupid]

That's what I understood you claimed in several postings in this thread. Maybe I misunderstood you.

I didn’t say that forces are necessarily sums of two-body forces only. I say that they are defined to be two-body forces. That’s a difference. The limitation to two-body forces is far from being obvious. There are infinite different possibilities to express net forces as sums of binary, ternary or higher order forces (given a sufficient number of bodies) or even non-interactive forces and they are all equivalent because they are indistinguishable by experiments. It was Newton’s decision to use binary forces only and this decision has been commonly accepted.

 

[Dale]

they are all equivalent because they are indistinguishable by experiments.

They are distinguishable by experiment. A three body force cannot be written as a sum of two body forces. There is an additional term that cannot exist in two body forces.

 

[DrStupid]

They are distinguishable by experiment. A three body force cannot be written as a sum of two body forces. There is an additional term that cannot exist in two body forces.

Can you show that with an example?

 

[Dale]

Can you show that with an example?

Yes, the ones that @vanhees71 posted earlier.

 

[DrStupid]

Yes, the ones that @vanhees71 posted earlier.

Can you be a bit more specific please?

 

[Dale]

Can you be a bit more specific please?

Is it really that hard to scroll up and click on the links he already posted? Here again is the Wikipedia link he posted.

https://en.m.wikipedia.org/wiki/Three-body_force

 

[DrStupid]

Is it really that hard to scroll up and click on the links he already posted? Here again is the Wikipedia link he posted.

https://en.m.wikipedia.org/wiki/Three-body_force

That's the example, but where is the demonstration I asked for?

 

[Dale]

That's the example, but where is the demonstration I asked for?

In the links. Please read them.

 

[DrStupid]

In the links. Please read them.

The link doesn't show why such a force cannot be expressed as a sum of forces acting between each two bodies according to Newton III. I try to unterstand what you are talking about. At least in classical mechanics it is always possible to express the three net forces $F_1$, $F_2$ and $F_3$ acting on three bodies as a sums of six binary forces:

$F_1 = F_{12} + F_{13}$
$F_2 = F_{21} + F_{23}$
$F_3 = F_{31} + F_{32}$

with

$F_{21} = - F_{12}$
$F_{13} = - F_{31} = F_1 - F_{12}$
$F_{23} = - F_{32} = F_2 + F_{12}$

There is even a degree of freedom left. And that doesn't work for three bodies only. For n bodies I get a system of ${\textstyle{1 \over 2}}n \cdot \left( {n + 1} \right) - 1$ linear equations and $n·(n-1)$ unknown variables. Such a system has at least one solution if the number of linear independent equations is less or equal the number of unknowns. That is the case for $n \ge 2$, which means always.

What prevents me from doing that with the three-body force mentinoned in the link?

 

[Dale]

What prevents me from doing that with the three-body force mentinoned in the link?

The problem comes when you try to write down an equation for $F_{1,2}$. For it to be a two body force then it can only depend on properties of bodies 1 and 2. I.e. $F_{1,2}=f(\theta_1,\theta_2)$ where $\theta_i$ is the set of all physical properties of body $i$.

For a three body force that cannot be done. The force on object 1 would be $F_{1,23}=f(\theta_1,\theta_2)+f(\theta_1,\theta_3)+g(\theta_1,\theta_2,\theta_3)$

In the first link that @vanhees71 posted it showed how requiring translational and rotational invariance (which leads to a generalized 3rd law) further restricts the possible form of the law. Table 1 shows that for two bodies there is only one argument while for three bodies there are six in general. So for translationally and rotationally invariant three body forces there are in general three more arguments than can be accounted for by a sum of two body forces.

So again, please read the links already posted.

 

[vanhees71]

Of course, Hamilton's principle is always your best friend. The ingenious principia is not very popular anymore for some good reason ;-)).

 

[DrStupid]

For it to be a two body force then it can only depend on properties of bodies 1 and 2.

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.

 

[vis_insita]

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.

It can't be out of scope for the Third Law. It makes no sense to talk of a force acting "pairwise between two bodies" if the corresponding force law must depend on the state and properties of three or more bodies. According to your interpretation of the Third Law is completely void of content as you have shown yourself in #203 above.

 

[DrStupid]

It makes no sense to talk of a force acting "pairwise between two bodies" if the corresponding force law must depend on the state and properties of three or more bodies.

The 3rd law doesn't say that the corresponding force law must depend on the state and properties of three or more bodies. It doesn't say anything about force laws.

According to your interpretation of the Third Law is completely void of content as you have shown yourself in #203 above.

1. The idea to apply conservation of momentum not only to a total system of n bodies but also to every pair of bodies within this system is not void of content. It is is a very useful tool if you want to divide a complex problem into many simple problems.

2. This is not the only content of Newton III. Even more important is the requirement of interaction for forces. In #85 I demonstrated how to use it to identify non-inertial systems.

 

[vis_insita]

The 3rd law doesn't say that the corresponding force law must depend on the state and properties of three or more bodies.

I know. I never claimed it did. I consistently assumed that the Third Law presupposes that all force laws depend only on the state of two particles, not more. That's why I don't think the Third Law qualifies as "definition of force", which was one puzzling claim of yours. Of course, the forces important to Newton's analysis in the Principia have that property.

 

[DrStupid]

I consistently assumed that the Third Law presupposes that all force laws depend only on the state of two particles, not more.

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.

That's why I don't think the Third Law qualifies as "definition of force", which was one puzzling claim of yours.

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