What Do Newton's Laws Say When Carefully Analysed

[DrStupid]

How can you apply N3 to a single non-interacting particle at all?

See #85 for an example.

 

[Dale]

In reality lines and circles are defined as geometric objects with coordinate-independent properties.

Yes, and using curved coordinates to define a straight line doesn't make the straight line curved nor does it make the curved coordinate lines straight.

As I mentioned above, I don't think that Newton's original intentions are really important to modern science, but he didn't explicitly describe things in terms of coordinates or frames (both of which were developed later). So his statements could just as easily be taken as coordinate- and frame-independent geometric statements rather than frame- or coordinate-based statements. Those would then be formalized in modern geometric notation as you posted earlier. (Except for N3)

 

[DrStupid]

The question I was answering is whether you have demonstrated in #85 that Newton's axioms are violated in a rotating frame. As I have shown, your conclusion becomes false if acceleration is defined as second derivative

$$\boldsymbol{a} = \frac{\mathrm{d}^2\boldsymbol{r}}{\mathrm{d}t^2}$$
of the position vector $\boldsymbol{r}$ relative to the common origin, which seems very natural.

The example in #85 is based on the results of measurements using the different frames of reference. These measurements result in the coordinate acceleration and I just used them unchanged. You are right that the resulting violation of the laws of motion can be avoided using proper acceleration. But that is not the primary result of the measurement. At least one measurement doesn’t give you the proper acceleration and you need to decide which one it is and how to turn it into the correct value. Thus I cannot agree that you solved the problem. You just changed it.

He simply argues that Earth's gravity is affecting the satellite in a way that causes precisely that acceleration,

$$m\boldsymbol{a} = \boldsymbol{F}.$$

In order to do that he need to know “precisely that acceleration”. Where does it come from? The satellite is at rest from his point of view. What does he need to do in order to know that it is actually accelerated and how?

 

[DrStupid]

N3 is silent on non-interacting bodies.

No, it says that non-interacting bodies are free of forces.

 

[atyy]

Not any more than the axioms of Euclidean geometry are violated in curvilinear coordinates. In geometry a line can be defined by $\ddot x^i = 0$, but only if you are using a Cartesian or at least affine coordinates. In spherical coordinates the same condition can describe a circle. Of course, you could say everything depends on how you define "line", but that would be a very strange position to take in this regard. In reality lines and circles are defined as geometric objects with coordinate-independent properties. No one makes the difference between them a matter of definition, because that would utterly confuse every discussion of geometry.

My point is that discussions of Newtonian physics become equally confused -- and for essentially the same reasons -- if you interpret the "a" in F=ma as "coordinate acceleration in arbitrary reference frames". Such confusion should be avoided by defining inertial motion as straight lines through spacetime independent of reference frames.

I don't agree with your insistence on your preferred interpretation, when both interpretations can be made physically and mathematically equivalent.

Anyway, as you have given it, how does your preferred formulation distinguish between gravity being formulated as a force or not a force, ie. do test particles in gravity follow geodesic or non-geodesic paths?

 

[olgerm]

If energy and momentum are defined in all frames of references as same functions of velocities as in inertial frame of reference.
Definition is: Inertial frame of of reference is frame of reference where momentum and energy of any system(aka setup) is conserved.

If energy and momentum are defined in all frames of references so that energy and momentum is conserved:
Definition is: Inertial frame of reference is frame of reference where momentum of any system(aka setup) is $\sum(\vec{v(i)}*m(i))$ and energy of any system(aka setup) is $\sum(\frac{\vec{v^2(i)}*m(i)}{2})$.

I do not know which of these definitions of energy and momentum is more correct/standard.

 

[vis_insita]

I don't agree with your insistence on your preferred interpretation, when both interpretations can be made physically and mathematically equivalent.

I'm not insisting on anything. I'm just explaining where I see the advantage of a covariant approach. My motivation is a desire for clarity and consistency. If someone understands both approaches and still prefers the non-covariant one, I won't have an argument with them.

Anyway, as you have given it, how does your preferred formulation distinguish between gravity being formulated as a force or not a force, ie. do test particles in gravity follow geodesic or non-geodesic paths?

In Newton-Cartan-Theory free falling particles follow geodesics, i.e. the same trajectories as do force-free particles according to my formulation of the 1st Axiom. Thus, the Equivalence Principle can be interpreted as "free fall = inertial motion." This means that spacetime curvature and gravity are the same thing. The only non-vanishing component of the Ricci tensor is related to mass density, via the field equation
$$R_{00} \propto \rho.$$
(Absolute space is still completely flat.) This eliminates the gravitational potential as well as its gradient, both of which are observer-dependent quantities and not uniquely definable as fields on spacetime.

Loosely speaking one could say that fictitious forces are everything that can be included in the definition of $\nabla$, while real forces can not and must appear on the right hand side of $m\nabla_u u = F$. The reason will of course be, that those forces depend on instrinsic properties of the particle other than mass.

 

[olgerm]

I do not know which of these definitions of energy and momentum is more correct/standard.

an someone tell me?

 

[atyy]

In Newton-Cartan-Theory free falling particles follow geodesics, i.e. the same trajectories as do force-free particles according to my formulation of the 1st Axiom. Thus, the Equivalence Principle can be interpreted as "free fall = inertial motion." This means that spacetime curvature and gravity are the same thing. The only non-vanishing component of the Ricci tensor is related to mass density, via the field equation
$$R_{00} \propto \rho.$$
(Absolute space is still completely flat.) This eliminates the gravitational potential as well as its gradient, both of which are observer-dependent quantities and not uniquely definable as fields on spacetime.

Loosely speaking one could say that fictitious forces are everything that can be included in the definition of $\nabla$, while real forces can not and must appear on the right hand side of $m\nabla_u u = F$. The reason will of course be, that those forces depend on instrinsic properties of the particle other than mass.

I see, would it be right to say that there is freedom in the definition of $\nabla$?

In the Newtonian system, for one definition of $\nabla$, there are global inertial frames and particles follow non-geodesic paths in gravity.

In the Newton-Cartan system, a different definition of $\nabla$ is used, and there are no global inertial frames, and test particles follow geodesic paths in gravity.

 

[vis_insita]

The example in #85 is based on the results of measurements using the different frames of reference. These measurements result in the coordinate acceleration and I just used them unchanged. You are right that the resulting violation of the laws of motion can be avoided using proper acceleration. But that is not the primary result of the measurement.

It is true that proper acceleration is not the result of the same measurement as $\ddot r^i$. But still it can be measured. You just have to measure the rotation of your frame of reference too. In practice this will be done by observing the apparent movement of other systems, e.g. a Foucault pendulum or a gyroscope.

Probably at this point already the first suspicion of "circularity" will come up. But it only seems circular under the assumption that we would have to infer everything from a single fact about a single particle. If we only had a single observation about the whole universe, then of course we would be hopelessly uncertain about almost everything. But in reality we are not nearly as ignorant.

Thanks to Newton, we already now the universal law of gravity. From its very universality, we also know that gravity affects the stellite from your previous example too. And from the presumed universality of the 2nd Law we also know that force results in acceleration, our observation that $\ddot r^i = 0$ notwithstanding.

At least one measurement doesn’t give you the proper acceleration and you need to decide which one it is and how to turn it into the correct value.

Yes, but the decision was made in my calculation in a completely unambiguous way: proper acceleration equals coordinate acceleration only with respect to inertial frames. In that example I knew that $\boldsymbol{a}=0$, because you assumed that $\dot r^i \equiv 0$ w.r.t an inertial frame.

In order to do that he need to know “precisely that acceleration”.

He only needs to know the force. From this he immediatley knows the proper acceleration from the 2nd law. Remember F=ma is valid independently of reference frames if a is interpreted as proper acceleration. I don't infer proper acceleration from coordinate acceleration alone, because I don't have to.

 

[vis_insita]

I see, would it be right to say that there is freedom in the definition of $\nabla$?

Yes, mathematically this is always the case. You have to decide what trajectories to call "geodesics of spacetime" on physical grounds. Free falling particles provide a natural choice because you cannot shield the influence of gravity from those particles anyway.

In the Newtonian system, for one definition of $\nabla$, there are global inertial frames and particles follow non-geodesic paths in gravity.

In the Newton-Cartan system, a different definition of $\nabla$ is used, and there are no global inertial frames, and test particles follow geodesic paths in gravity.

That's true, but in general there is no natural choice for global inertial frames, with respect to which you can define a gravitational force.

Under special circumstances, however, this is possible. If mass is confined to a bounded region, you could impose the boundary condition $\Phi\to 0$ if $r\to\infty$ on the potential to eliminate the ambiguity $\Phi' = \Phi + \boldsymbol{b}\cdot \boldsymbol{r}$ caused by the relative acceleration $\boldsymbol{b}$ of two observers. This singles out a preferred class of local inertial frames, which map onto each other by means of the usual Galilean transformations.

 

[atyy]

That's true, but in general there is no natural choice for global inertial frames, with respect to which you can define a gravitational force.

Under special circumstances, however, this is possible. If mass is confined to a bounded region, you could impose the boundary condition $\Phi\to 0$ if $r\to\infty$ on the potential to eliminate the ambiguity $\Phi' = \Phi + \boldsymbol{b}\cdot \boldsymbol{r}$ caused by the relative acceleration $\boldsymbol{b}$ of two observers. This singles out a preferred class of local inertial frames, which map onto each other by means of the usual Galilean transformations.

So we still need additional conditions to define a global inertial frame (and not just local inertial frames)? Could Newton's third law (or equivalently momentum conservation or spatial translation invariance) provide or help to provide the additional conditions?

 

[Dale]

No, it says that non-interacting bodies are free of forces.

Not according to Calkin

 

[vanhees71]

It's a bit strange to claim that non-interacting bodies are not free of forces. Are you claiming that there are forces which are not due to interactions (and here I don't consider inertial forces which are just reinterpretations of parts of the covariant time-derivative in non-inertial frames brought to the other side of the equation as forces).

 

[Dale]

It's a bit strange to claim that non-interacting bodies are not free of forces.

Nobody is claiming that. We all agree that non-interacting bodies are free of forces. The question is if it is a claim made in the third law or elsewhere.

 

[vis_insita]

So we still need additional conditions to define a global inertial frame (and not just local inertial frames)? Could Newton's third law (or equivalently momentum conservation or spatial translation invariance) provide or help to provide the additional conditions?

I'm not sure how. Fundamentally the question is how to distinguish physically a curved spacetime from a flat one. As far as I understand it, the logic of Newton-Cartan-Theory is quite the same as in General Relativity: Equivalence principle -> free falling particles follow geodesics -> spacetime is curved -> there are no global inertial frames. The only step that has physical content is the first one, identifying test particle trajectories with geodesics. This must be false if there are supposed to be global inertial frames. But I don't see any relation to the Third Law or momentum conservation.

 

[vanhees71]

The 3rd law is simply saying that two-body interactions are equal in magnitude and antiparallel, i.e., $\vec{F}_{12}=-\vec{F}_{21}$ and thus that (for this special case) of interactions momentum conservation holds. In standard Newtonian theory by assumption there exists a global inertial frame (and thus infinitely many related by Galilei transformations).

 

[DrStupid]

You just have to measure the rotation of your frame of reference too. In practice this will be done by observing the apparent movement of other systems, e.g. a Foucault pendulum or a gyroscope.

And how do you know that these measurements show you the rotation of the frame of reference and not just forces acting on the pendulum or gyroscope? In my example in #85 I used Newton III to conclude that there are no such forces because there can't be corresponding counterforces (due to the lack of another body). That tells me that the "forces" are not real. How do you do that?

Thanks to Newton, we already now the universal law of gravity. From its very universality, we also know that gravity affects the stellite from your previous example too. And from the presumed universality of the 2nd Law we also know that force results in acceleration, our observation that $\ddot r^i = 0$ notwithstanding.

Without Newton III that could mean that there is another force that cancels gravity out.

In that example I knew that $\boldsymbol{a}=0$, because you assumed that $\dot r^i \equiv 0$ w.r.t an inertial frame.

No, I carefully avoided such an assumption. I didn't even conclude it. I just demonstrated that the rotating frame is not inertial.

Remember F=ma is valid independently of reference frames if a is interpreted as proper acceleration.

Yes, but that just shifts the problem from the force to the acceleration. In my example in #85 I conclude that the forces (and therefore the accelerations) in the rotating frame are not real and you say that the accelerations (and therefore the forces) in the rotating frame are not real. That's pretty much the same.

 

[DrStupid]

The 3rd law is simply saying that two-body interactions are equal in magnitude and antiparallel, i.e., →F12=−→F21F→12=−F→21 and thus that (for this special case) of interactions momentum conservation holds.

Newton's original version of the 3rd law also says that every force results from such two-body interactions. The use of different versions of the laws of motion is a major problem in this thread. As we obviously can't agree about a single version, we will need to indicate what we refer to - for every single statement - if we don't want to keep on talking at cross-purposes.

 

[atyy]

I'm not sure how. Fundamentally the question is how to distinguish physically a curved spacetime from a flat one. As far as I understand it, the logic of Newton-Cartan-Theory is quite the same as in General Relativity: Equivalence principle -> free falling particles follow geodesics -> spacetime is curved -> there are no global inertial frames. The only step that has physical content is the first one, identifying test particle trajectories with geodesics. This must be false if there are supposed to be global inertial frames. But I don't see any relation to the Third Law or momentum conservation.

I don't think I'm asking a "physical" question, since the two formalisms are identical (I think) in physical content. It's just that the connection is defined differently in each, such that gravity is a force in one formalism but not the other.

I guess my question is: is there global conservation of momentum in the Newton-Cartan formalism? Heuristically from GR, where energy momentum is not globally conserved, I would guess that global conservation of momentum is also absent in Newton-Cartan.

If global conservation of momentum is absent in Newton-Cartan, then the third law or global momentum conservation could be one criterion that helps to differentiate thee Newton connection from the Newton-Cartan connection.

 

[vanhees71]

Newton's original version of the 3rd law also says that every force results from such two-body interactions. The use of different versions of the laws of motion is a major problem in this thread. As we obviously can't agree about a single version, we will need to indicate what we refer to - for every single statement - if we don't want to keep on talking at cross-purposes.

May be, but that's not true (e.g., for the forces acting within nucleons, i.e., protons and neutrons in the non-relativistic approximation). Newtonian mechanics is more general than Newton's formulation of it ;-).

 

[vis_insita]

And how do you know that these measurements show you the rotation of the frame of reference and not just forces acting on the pendulum or gyroscope? In my example in #85 I used Newton III to conclude that there are no such forces because there can't be corresponding counterforces (due to the lack of another body). That tells me that the "forces" are not real. How do you do that?

I told you how I know: I am starting from specific hypotheses about all relevant interactions, e.g. that there are no interactions other than
$$\boldsymbol{F}=-\frac{mM}{r^3}\boldsymbol{r},$$
and then I examine how they work out in many situations.

You are starting from a general statement about all possible interactions, namely the Third Law, and then just postulate that it must be true in every situation.

But there is no advantage to your approach. Of course you can ask how I know that a specific gyroscope is free of torque. That is an assumption I make that can be false and is always open to reasonable doubt (but not to unreasonable doubt). But I can ask the same thing about your analysis. How do you know the Third Law is correct? By the authority of Newton? This is bad science. (In fact the Third Law doesn't seem correct for all known interactions.)

Without Newton III that could mean that there is another force that cancels gravity out.

Yes, of course there is always the possibility that there is an entirely unknown force that invalidates my whole analysis of the situation. This remaining uncertainty is just an unavoidable situation in science. Also there is the (very real) possibility that the Third Law is false.

But there is absolutely no reason to assume that there is any force other than gravitation relevant to your satellite example. So, suspecting the effects of such a force seems unreasonable.

No, I carefully avoided such an assumption. I didn't even conclude it. I just demonstrated that the rotating frame is not inertial.

Right, you didn't conclude it, but I did. In your example there was just a single non-interacting particle at rest in one frame. This means that frame is an inertial frame.

Also, I didn't object to the rotating frame being non-inertial. I regard that as a trivial statement that doesn't need any demonstration. I objected to the claim that any of Newton's axioms is violated in any of both frames.

Yes, but that just shifts the problem from the force to the acceleration.

No, I'd say the opposite is true. The problem is and always has been to know all force laws, i.e. interactions. Everything else follows from that.

In my example in #85 I conclude that the forces (and therefore the accelerations) in the rotating frame are not real and you say that the accelerations (and therefore the forces) in the rotating frame are not real.

That's not exactly what I was saying. I was saying that $\boldsymbol{a} = \frac{d\boldsymbol{ v}}{d t}$ is the same in both frames (because it is independent of any frame). Therfore there is no violation of Newtons laws (formulated in terms of $\boldsymbol{a}$) in any of the frames.

 

[atyy]

May be, but that's not true (e.g., for the forces acting within nucleons, i.e., protons and neutrons in the non-relativistic approximation). Newtonian mechanics is more general than Newton's formulation of it ;-).

How do the nucleons violate the third law?

 

[DrStupid]

How do you know the Third Law is correct?

It's a definition.

 

[vis_insita]

I don't think I'm asking a "physical" question, since the two formalisms are identical (I think) in physical content. It's just that the connection is defined differently in each, such that gravity is a force in one formalism but not the other.

That's correct. But the torsion free part of that connection is determined by the geodesics, isn't it? So if you make the physical assumption that free falling particles are geodesics, then you end up with a connection that implies curvature. If you want to avoid this, then your geodesics necessarily must be some other sets of curves in spacetime. I would argue that those curves are not physically significant. They don't correspond to particle trajectories in an arbitrary region of space, since those particles are always affected by gravity.

I guess my question is: is there global conservation of momentum in the Newton-Cartan formalism? Heuristically from GR, where energy momentum is not globally conserved, I would guess that global conservation of momentum is also absent in Newton-Cartan.

If global conservation of momentum is absent in Newton-Cartan, then the third law or global momentum conservation could be one criterion that helps to differentiate thee Newton connection from the Newton-Cartan connection.

I think momentum conservation holds in Newton-Cartan-Theory and I'm not sure why you suspect this might not be the case. Problems defining a conserved energy-momentum vector in GR only exist in spacetimes that aren't asymptotically flat. But in Newton-Cartan-Theory space is globally flat, so you can always sum the local momenta and forces to meaningful global quantities. And if the total force vanishes, then the total momentum is conserved.

 

[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