What Do Newton's Laws Say When Carefully Analysed

[DrStupid]

Law III describes interactions, so you don't need to invoke it unless you have interactions.

Newton III not only describes interactions. It says that interactions are required for forces:

"Actioni contrariam semper et aequalem esse reactionem"
(To every action there is always opposed and equal reaction)

No interaction would not mean no force without this requirement. But that is the basis for the derivations of inertial frames from Newton I proposed in this thread. Therefore they do not work without Newton III.

 

[gmax137]

The Third Law implies Conservation of Momentum, but Conservation of Momentum does not imply the Third Law. The Third Law is not valid on an instant by instant basis because forces take time to propagate. Modern physics has given primacy to Conservation of Momentum for this very reason.

Is this idea developed anywhere? I think I understand the instant by instant idea, but when I try to work it out, my head starts spinning.

I think we did this in intro EM (Purcell); does that sound right? Moving charges, that stuff. It has been decades since I gave EM any thought.

Is instantaneous action considered to be another "built in" feature of Newton's classical mechanics (like absolute time & space) and thus, kind of outside the scope of this thread?

Thanks

 

[kith]

No interaction would not mean no force without this requirement. But that is the basis for the derivations of inertial frames from Newton I proposed in this thread.

What if @vanhees71 replaced "non-interacting particles" by "force-free particles" in his post #57? It seems to me that you would still object but what exactly would your objection be then?

 

[kith]

Is this idea developed anywhere? I think I understand the instant by instant idea, but when I try to work it out, my head starts spinning.

If you take the second law as the definition of force, it is defined as the rate of momentum change. The third law than says that for two interacting bodies, their rates of momentum change are always opposite but equal.

If our interaction is not instantaneous but mediated by a field, that's not true on an instant by instant basis anymore because the field carries non-zero momentum at intermediate times. I.e. the momentum gain of one of the bodies happens later than the momentum loss of the other.

 

[Swamp Thing]

If our interaction is not instantaneous but mediated by a field, that's not true on an instant by instant basis anymore because the field carries non-zero momentum at intermediate times. I.e. the momentum gain of one of the bodies happens later than the momentum loss of the other.

If we are willing to attribute momentum to the field or EM wave, then it does apply on an instant by instant basis, yes? We can't always assume that there is an object somewhere out there waiting to catch the field momentum (maybe a few light years away) and thus balance the books -- so we would have to count the field momentum as part of the conservation balance, right?

 

[Dale]

Instead you again start with "non-interacting objects". Which part of N1 does that come from? Citation please!

Chapter 1 of Lagrangian and Hamiltonian Mechanics by MG Calkin begins "Newton's first law deals with non-interacting bodies. It says that the velocity of an isolated body, one removed from the influence of other bodies, is constant. This law defines a set of preferred coordinate frames, inertial frames, as frames in which Newton's first law holds." https://www.worldscientific.com/doi/pdf/10.1142/9789810248154_0001

This is by no means a personal formulation of mine or @vanhees71, it is an accepted meaning of N1.

That is a consequence of N3. You are claiming not to use N3, but you failed to explain where it comes from instead.

Completely disagree. N3 doesn't even apply to non-interacting bodies. It says nothing whatsoever about them. An object is either interacting or non-interacting. If it is non-interacting then N1 defines inertial reference frames such that non-interacting objects have constant velocity. If it is interacting then N3 describes how pairs of interacting objects exert forces on each other. If it is interacting then N1 is silent and if it is non-interacting then N3 is silent.

N1 is not a consequence of N3. They are not even applicable to the same bodies, so I don't see how you can possibly think it is relevant. N1 starts with non-interacting bodies and uses them to define an inertial frame. As such, it is simply a definition and is not a consequence of anything. Definitions don't have to "come from" anything.

 

[kith]

If we are willing to attribute momentum to the field or EM wave, then it does apply on an instant by instant basis, yes?

By "it", you seem to mean momentum conservation and yes, this applies on an instant by instant basis. But we were talking about Newton's third law which implies that momentum is exchanged instantly between bodies. This is false in elaborate theories of interactions like electromagnetism where fields mediate the interaction. I consider this to be a minor point. The important thing is that there's a general principle which holds for all physical systems: momentum conservation.

We can't always assume that there is an object somewhere out there waiting to catch the field momentum (maybe a few light years away) and thus balance the books -- so we would have to count the field momentum as part of the conservation balance, right?

Sure. It's just that if there's a body which catches the momentum it can't happen at the same instant as another body gives it to the field.

 

[kith]

Griffiths also gives a general argument against the third law in relativistic theories in his "Introduction to Electrodynamics" (see chapter 12.2.4 "Relativistic dynamics"):
"Unlike the first two, Newton’s third law does not, in general, extend to the relativistic domain. Indeed, if the two objects in question are separated in space, the third law is incompatible with the relativity of simultaneity. For suppose the force of $A$ on $B$ at some instant $t$ is $F(t)$, and the force of $B$ on $A$ at the same instant is $−F(t)$; then the third law applies, in this reference frame. But a moving observer will report that these equal and opposite forces occurred at different times; in his system, therefore, the third law is violated."

 

[gmax137]

therefore

this is why i asked

Is instantaneous action considered to be another "built in" feature of Newton's classical mechanics (like absolute time & space) and thus, kind of outside the scope of this thread?

 

[atyy]

Interesting history of the concept of inertial frame: https://plato.stanford.edu/entries/spacetime-iframes/

"Neumann's definition of a time-scale directly inspired Ludwig Lange's conception of “inertial system,” introduced in 1885 . An inertial coordinate system ought to be one in which free particles move in straight lines. But any trajectory may be stipulated to be rectilinear, and a coordinate system can always be constructed in which it is rectilinear. And so, as in the case of the time-scale, we cannot adequately define an inertial system by the motion of one particle. Indeed, for any two particles moving anyhow, a coordinate system may be found in which both their trajectories are rectilinear. So far the claim that either particle, or some third particle, is moving in a straight line may be said to be a matter of convention. We must define an inertial system as one in which at least three non-collinear free particles move in noncoplanar straight lines; then we can state the law of inertia as the claim that, relative to an inertial system so defined, the motion of any fourth particle, or arbitrarily many particles, will be rectilinear. ... "

"... More simply, an inertial reference-frame is one in which Newton's second law is satisfied, so that every acceleration corresponds to an impressed force. Thomson did not reject the term “absolute rotation,” holding instead that it has to be understood as rotation relative to a reference frame that satisfies his definition. The definition does not express, as Lange's does, the degree of arbitrariness involved in the construction of an inertial system by means of free particles. Moreover, like Lange's, it leaves out a crucial condition for an inertial system as we understand it: all forces must belong to action-reaction pairs. Otherwise we could have, as on a rotating sphere, merely apparent (centrifugal) forces that are, by definition, proportional to mass and acceleration, and so the rotating sphere would satisfy Thomson's definition. Therefore the definition needs to be completed by the stipulation that to every action there is an equal and opposite reaction. (This completion was actually proposed by R.F. Muirhead in 1887.) ..."

 

[vanhees71]

Newton I and II are not limited to interactive forces without Newton III because Newton III and only Newton III excludes fictitious forces. Without this limitation to interactive forces there is no limitation to inertial frames.

Newton I is about particles moving without the action of any forces, Newton II defines forces as $\dot{\vec{p}}$ where $\vec{p}=m \vec{v}$, and Newton III is about two-body interactions. Indeed to state Newton I + II you don't need Newton III. That's all I'm saying the whole time!

The only thing, I'm now thinking about is, what about generic three-body forces (as occur in nuclear physics)? I guess these you can adequately only treat with analytical mechanics and symmetry principles to be consistent with Newtonian spacetime structure.

 

[vanhees71]

If you take the second law as the definition of force, it is defined as the rate of momentum change. The third law than says that for two interacting bodies, their rates of momentum change are always opposite but equal.

If our interaction is not instantaneous but mediated by a field, that's not true on an instant by instant basis anymore because the field carries non-zero momentum at intermediate times. I.e. the momentum gain of one of the bodies happens later than the momentum loss of the other.

Yes, and that's why fields are not belonging to Newtonian but relativistic physics ;-).

 

[vanhees71]

Griffiths also gives a general argument against the third law in relativistic theories in his "Introduction to Electrodynamics" (see chapter 12.2.4 "Relativistic dynamics"):
"Unlike the first two, Newton’s third law does not, in general, extend to the relativistic domain. Indeed, if the two objects in question are separated in space, the third law is incompatible with the relativity of simultaneity. For suppose the force of $A$ on $B$ at some instant $t$ is $F(t)$, and the force of $B$ on $A$ at the same instant is $−F(t)$; then the third law applies, in this reference frame. But a moving observer will report that these equal and opposite forces occurred at different times; in his system, therefore, the third law is violated."

Well, but in special relativity due to the homogeneity of space in any inertial frame still momentum conservation holds true and that's why the most natural description in relativistic dynamics are local field theories, and that in fact are the only hitherto successful formulation of relativistic dynamics, i.e., in some sense Newton III is still valid locally.

 

[DrStupid]

What if @vanhees71 replaced "non-interacting particles" by "force-free particles" in his post #57?

How do you know that the particle is force-free?

 

[DrStupid]

Chapter 1 of Lagrangian and Hamiltonian Mechanics by MG Calkin begins "Newton's first law deals with non-interacting bodies. It says that the velocity of an isolated body, one removed from the influence of other bodies, is constant. This law defines a set of preferred coordinate frames, inertial frames, as frames in which Newton's first law holds." https://www.worldscientific.com/doi/pdf/10.1142/9789810248154_0001

This is true because the force as used in Newton I is defined to be interactive by Newton III. Without Newton III there would be no justification for this statement.

N3 doesn't even apply to non-interacting bodies.

That depends on your definition of "apply". N3 says that non-interacting bodies are force-free. You are using this fact in your derivation of inertial frames from N1.

N1 is not a consequence of N3.

As nobody claimed something like that it seems we are talking cross-purposes. Maybe it helps when I explain it with an example:

Let's say I have a frame of reference where a single particle remains at rest no matter where I place it. That means according to

Newton I: As the particle remains ar rest, there is no force acting on it.
Newton II: As the acceleration is zero, the force acting on the particle is $F = m \cdot a = 0$

Now I switch to another frame of reference that is rotating around the origin with the angular velocity $\omega$. In this frame the particle is moving on circular paths around the rotational axis. That means according to

Newton I: As the particle doesn't remain at rest or uniform translation, there is a force acting on it.
Newton II: As the acceleration is $- \omega ^2 \cdot r$, the force acting on the particle is $F = m \cdot a = - m \cdot \omega ^2 \cdot r$

That's it. There is nothing in Newton I or II that tells me this is not allowed. That changes with

Newton III: As there is just a single particle, there is no interaction between particles and therefore no force.

The "forces" resulting from Newton I or II are violating Newton III. Or vice versa: The absense of forces resulting from Newton III violates Newton I and II. The laws of motion are not valid in the rotating frame. If inertial frames of references are defined by compliance with the laws of motion this means that it is not inertial.

The laws of motion define inertial frames of reference with (N1 or N2) and N3 but not with (N1 or N2) xor N3.

 

[DrStupid]

Interesting history of the concept of inertial frame: https://plato.stanford.edu/entries/spacetime-iframes/

"[...]Moreover, like Lange's, it leaves out a crucial condition for an inertial system as we understand it: all forces must belong to action-reaction pairs. [...] Therefore the definition needs to be completed by the stipulation that to every action there is an equal and opposite reaction. [...]

Thank you for this reference. This is exactly what I am talking about.

 

[Dale]

This is true because the force as used in Newton I is defined to be interactive by Newton III. Without Newton III there would be no justification for this statement

Sorry @DrStupid, I will stick with Dr Calkin here. His approach seems much clearer than yours.

You are using this fact in your derivation of inertial frames from N1.

I don’t derive definitions.

Newton I: As the particle doesn't remain at rest or uniform translation, there is a force acting on it.

Per the cited approach, since you have a non-interacting particle which is accelerating then your reference frame is non inertial. That seems the most direct approach to me.

Sure, you can go out of your way to use a poor formulation of Newton’s laws which requires you to use all of the laws for even the simplest cases, but you certainly don’t have to and I personally don’t choose to.

 

[vanhees71]

How do you know that the particle is force-free?

This is the crux ;-)). You have to establish an inertial reference frame first, according to the definition of Newton I. Then you assume that particles far away from any other particles are "force-free" and check whether all such particles move in rectilinear uniform motion relative to each other. You need one moving particle to establish a measure of time through the measure of distance and then at least two other particles moving in different directions to establish that you are in an inertial reference frame. Then you have the definition of force together with the definition of mass, which by assumption is a measure of "amount of matter".

 

[bhobba]

If our interaction is not instantaneous but mediated by a field, that's not true on an instant by instant basis anymore because the field carries non-zero momentum at intermediate times. I.e. the momentum gain of one of the bodies happens later than the momentum loss of the other.

Newtonian Gravity?

Thanks
Bill

 

[vanhees71]

Newtonian gravity is not a field theory but action at a distance, with the interaction potential
$$V(\vec{r}_1,\vec{r}_2)=-\frac{\gamma m_1 m_2}{|\vec{r}_1-\vec{r}_2|}.$$
Here the total momentum of a closed system of point particles is conserved, and no momentum is exchanged with any dynamical field. Newtonian Gravity is not a field theory in the narrower sense of a theory involving dynamical fields.

 

[kith]

Newtonian Gravity?

It depends on what "mediated by a field" is supposed to mean. Sure, you can formally introduce a field quantity $\vec G$ and write Newton's law of gravity as $\vec F = m \vec G$. This has the same structure as the corresponding law in electrodynamics but as @vanhees71 notes, Newtonian Gravity doesn't involve the field as a dynamical physical system which is what's usually meant by "mediated by a field".

 

[DrStupid]

Sorry @DrStupid, I will stick with Dr Calkin here. His approach seems much clearer than yours.

It seems so because you accept the limitation of Newton I to non-interactin bodies as a matter of course. You think it is just there - coming from nowhere - because it is commonly used. But that's not as self-evident as you think. It would have been possible to define forces without restriction to interactions. I already mentioned that fictitious forces are commonly used as well. You should ask yourself where the consensus, not to assume them to be forces, comes from. The answer is Newton III.

Per the cited approach, since you have a non-interacting particle which is accelerating then your reference frame is non inertial. That seems the most direct approach to me.

Is seems but it isn't. Newton I deals with particles in presence or absence of forces - not with presence or absence of interactions. That's not necessarely the same. You need to conclude from one to the other. The approach is no longer as direct as it looks like if you don't skip this step.

Sure, you can go out of your way to use a poor formulation of Newton’s laws which requires you to use all of the laws for even the simplest cases, but you certainly don’t have to and I personally don’t choose to.

On what basis are you rating Newton's formulation of the laws of motion as "poor"? What makes a formulation that needs additonal assumptions - not included in the laws of motion - better?

Even if your favorit formulation appears to be better (for what criterion ever) - why are you using it in this thread? The topic is "What Do Newton's Laws Say When Carefully Analysed". Starting with laws of motion that are not equivalent with Newton's laws of motion (resulting in different conclusions) is far from beeing careful. You shouldn't even use Newton's name for it.

And talking about poor formulations: Claiming "Newton's first law deals with non-interacting bodies." (just to take an example) is at least mistakable. Newton I says:

"Every body perseveres in its state of being at rest or of moving uniformly straight forward, except insofar as it is compelled to change its state by forces impressed."

Interacting bodies are obviously included in Newton I - even if you take the relation between interaction and forces as given. And no, this is not an outdated or unpopular version of the first law. You will find it with different wordings but same content everywhere in the scientific literature. If you claim Newton's original formulations of the laws of motion to be "poor" you should at least apply the same standards to your sources as well.

 

[DrStupid]

You have to establish an inertial reference frame first, according to the definition of Newton I.

If that means "according to the definition of Newton I only", than no I don't. I already explained why this is not even possible.

Then you assume that particles far away from any other particles are "force-free" and check whether all such particles move in rectilinear uniform motion relative to each other.

How do I assume that without using Newton III?

 

[olgerm]

An inertial frame is a reference frame where all non interacting particles travel in straight lines at constant velocity.

This only works if do not count inertial forces as forces. But if you define force by $F=a*m$, then inertial forces would be forces.
But using Newtons 3. law would not solve problem, because in uniformly accelerating frame all particles have the same inertial force.

 

[Dale]

This only works if do not count inertial forces as forces

That is why the approach of Dr Calkin is nice. Inertial forces do not come from interactions. So by focusing the definition of an inertial frame on non-interacting bodies then you automatically and naturally exclude inertial forces without even having defined forces at that point.

 

[atyy]

This only works if do not count inertial forces as forces. But if you define force by $F=a*m$, then inertial forces would be forces.
But using Newtons 3. law would not solve problem, because in uniformly accelerating frame all particles have the same inertial force.

Newton's 3rd law helps because inertial forces are not part of action-reaction pairs.

 

[Jimster41]

“Naturally” seems a stretch. Definitions are nice but... I’m having a hard time imagining a real non-interacting body I’m somehow checking for rectilinear motion...

 

[atyy]

“Naturally” seems a stretch. Definitions are nice but... I’m having a hard time imagining a real non-interacting body I’m somehow checking for rectilinear motion...

For interacting bodies (in the presence of Newtonian gravity, there is no such things as a non-interacting body), one must use all 3 laws (and maybe a bit more, I'm not sure).

If you look at posts #22 by me, #29 by @Demystifier, #45 by @DrStupid they essentially make the similar point that additional content beyond F=ma is needed. The concept of a pre-defined non-interacting particle is one way to provide the additional content. However, if there are no non-interacting particles, then one must specify the form of the forces. In fact, this is the point of the OP - if N2 is taken as a definition, and N1 is a special case of N1, then N2 and N1 alone are physically empty.

Also, in post #55 by @bhobba he points out that nowadays we define inertial frames through the symmetry of the laws, which is consistent with @DrStupid's preference to include N3, since N3 is conservation of momentum in the Newtonian framework, which is equivalent to a symmetry via Noether's theorem.

 

[Dale]

I’m having a hard time imagining a real non-interacting body I’m somehow checking for rectilinear motion.

I agree. To me, this is the weakness of Calkin’s approach. Not that it somehow implicitly uses or hides N3 in the definition of an inertial frame, but rather that it is not practical.

 

[Dale]

Newton I deals with particles in presence or absence of forces - not with presence or absence of interactions

Calkin disagrees, as do I.

The reason that you are going around in circles with me here is that you keep trying to argue based on premises that I don’t accept. I like Calkin’s formulation of Newton’s laws precisely because of how clean it is on this topic (inertial frames defined by first law alone). You keep explaining why the original formulation is not clean. That only re-convinces me to not use the original formulation.

Do you have any argument showing that Calkin’s formulation of Newton’s laws (not other formulations) requires N3 to define an inertial frame? It doesn’t seem necessary to me with his approach.

Starting with laws of motion that are not equivalent with Newton's laws of motion (resulting in different conclusions) is far from beeing careful. You shouldn't even use Newton's name for it.

It isn’t me using Newton’s name, it is Calkin. He felt that his formulation was equivalent enough to be given Newton’s name, and I agree. Complain as you will, but in science it is expected that later authors may reformulate seminal works. It is common. The seminal authors get the first word, but not the last word. Newton isn’t the pope and his words are not canonized.

Calkin has his right to reformulate Newton’s laws as he saw fit. The formulation is empirically equivalent, so the “different conclusions” are ok. Indeed, everyone mentioning the derivation from symmetry is making a similar deviation from Newton’s formulation since he explicitly included an undetectable absolute space and time as part of his original formulation. Are you objecting to their reformulation? No, nor should you; it is OK for them to do it as well as it is for Calkin.

 

[vanhees71]

It depends on what "mediated by a field" is supposed to mean. Sure, you can formally introduce a field quantity $\vec G$ and write Newton's law of gravity as $\vec F = m \vec G$. This has the same structure as the corresponding law in electrodynamics but as @vanhees71 notes, Newtonian Gravity doesn't involve the field as a dynamical physical system which is what's usually meant by "mediated by a field".

It has the same structure as the Coulomb Law of electrostatics, not electrodynamics! That's the distinction between an action-at-a-distance-approximation, which is well justified for non-relativistic motions of not too far distant charges, where radiation reaction can be neglected, and a real field theory with the fields as dynamical entities in addition to the mechanical objects (here "point particles").

 

[vanhees71]

If that means "according to the definition of Newton I only", than no I don't. I already explained why this is not even possible.
How do I assume that without using Newton III?

What has Newton III to do with that? Newton III says nothing about the range of interactions.

It's of course true that this apparently simply foundations are more subtle than one thinks. E.g., the gravitational interaction is not easy to come by, because it's long-ranged (i.e., the interaction potential goes only with $1/\text{distance}$ and it cannot be "screened", and then the asymptotic free motion is not uniform motion at all. Here, of course the equivalence principle (which applies to Newtonian gravity too) comes to a rescue: As far as tidal forces can be neglected (and their potential goes faster to 0 than with $1/\text{distance}$!) a free-falling rest frame is at least locally an inertial frame, and for this argument you indeed need Newton III (or momentum conservation and the associated center-mass law, which follows from homogeneity of space and Gailei-boost invariance, i.e., uses Newton III).

As I said, you need all three postulates (+ the tacit assumption that space for any inertial observer, and in Newton's absolute space and absolute time model that implies all observers, even accelerated ones) to make the theory complete. Nevertheless, the logic is already right in Newton's original approach: First you need to establish what inertial frames are (Newton I), then what forces are (Newton II), and finally Newton III. Of course a more systematic mathematical way is the modern one based on symmetries, and in this case it's the symmetries of the space-time model, which is a fibre bundle (identical copies of Euclidean affine manifolds along the "directed time axis").

 

[vis_insita]

I believe many of the issues discussed here can be illuminated by a covariant formulation of Newtonian Mechanics, in which all laws of motion are given independently of any reference frame. The basic idea is the following: Contrary to “velocity” which can be defined on any manifold, in order to define "acceleration" you need a method to compare velocities at different events. This requires a covariant derivative. Such a covariant derivative also defines “straightness” in space and time, which is the natural interpretation of the “uniform motion” the First Law is talking about. Thus, one needs to define acceleration as the deviation from uniform motion as $$\nabla_{\boldsymbol{u}}\boldsymbol{u}=\boldsymbol{a},$$ just like in General Relativity. The only difference is in the conditions one has to impose on $\nabla$ in order for it to be compatible with Newtonian space time.

One such requirement is that “Absolute, true and mathematical time, of itself, and from its own nature flows equably without regard to anything external.” This “equable flow” has a nice formal interpretation as $\nabla\mathrm{d}t=0$. It implies that if a vector field $\boldsymbol{x}$ is spatial i.e. $\langle\mathrm{d}t,\boldsymbol{x}\rangle=0$, then its covariant derivative is too (which will be important below), i.e. $$\langle\mathrm{d}t,\nabla_{u}\boldsymbol{x}\rangle=\nabla_{u}\langle\mathrm{d}t,\boldsymbol{x}\rangle-\langle\nabla_{u}\mathrm{d}t,\boldsymbol{x}\rangle=0.$$

Another requirement is that absolute space, defined by $t=\text{const.}$, has to be flat with a euclidean structure. (This means the curvature calculated from $\nabla$ fulfills some identities, but they are not important in the following.) With this introduction my preferred interpretation of the First Law is this:

“1st Law: There exists a covariant derivative $\nabla$ that is compatible with the space time structure in the above sense. Isolated, ie. non-interacting, particles follow geodesics in space time $\nabla_{\boldsymbol{u}}\boldsymbol{u}=0.$"

I have not defined “non-interacting”, yet. But this definition will be made independent of the Third Law below. In any case, this interpretation of the First Law enables the following definition of inertial frames, which is completely independent of forces, and only relies on $\nabla$.

An inertial frame is defined by a set of three spatial orthogonal vectors $\boldsymbol{e}_{i}$ which are parallel transported along a straight line through space time. The tangent $\boldsymbol{e}_{0}$ of that line has to fulfill the condition $\langle\mathrm{d}t,\boldsymbol{e}_{0}\rangle=1$, so that the curve described by the origin is parameterized by absolute time. This means an inertial frame is defined as a set of vectors $\boldsymbol{e}_{0},\boldsymbol{e}_{i=1,2,3}$, with

$$\langle\mathrm{d}t,\boldsymbol{e}_{0}\rangle =1$$
$$\langle\mathrm{d}t,\boldsymbol{e}_{i}\rangle =0$$
$$\boldsymbol{e}_{i}\cdot\boldsymbol{e}_{k} =\delta_{ik}$$

and the two dynamical conditions

$$\nabla_{\boldsymbol{e}_{0}}\boldsymbol{e_{0}}=0,\qquad\nabla_{\boldsymbol{e}_{0}}\boldsymbol{e}_{i}=0$$

(Because of the “equable flow of time” the spatial basis vectors remain spatial throughout, for any system.) The first of the dynamical condition is the precise meaning of the coordinate system being non-accelerating in its linear motion. The second means it is non-rotating.

Now it appears clear (I believe) that the definition of inertial frames did not rely on the notion of "interaction". We only used “straightness” in space and time. But of course in reality we cannot know if a particle actually moves along a straight line through space time, independently of knowing whether it is interacting or not. But this is not a logical or philosophical problem. It only means that in the natural sciences all theoretical postulates have to be interpreted in terms of each other for practical applications. This is due to the double nature of scientific law, as simultaneously defining its basic notions and making nontrivial statements about them. This is no peculiarity of Newtonian Mechanics.

To solve this issue practically one just has to hypothesize what force laws describe the possible interactions on the particle, to assess whether it can be judged as “non-interacting” in any particular situation. You do not need a criterion that applies to all possible situations. On the contrary, one is completely free in postualting such laws. The Second Law can thus be interpreted as

“2nd Law: For every particle (of mass $m$) there exists a spatial vector field $\boldsymbol{F}$ on space time, such that the particle's equation of motion is $m\nabla_{\boldsymbol{u}}\boldsymbol{u}=\boldsymbol{F}$.”

In this formulation the First Law is not a special case of the Second Law. The First Law postulates the existence of a covariant derivative, and the Second Law postulates the existence of a vector field. Only the second part of the First Law, concerning the motion of isolated particles, can be made a special case of the Second Law, but this is also the less important part.

The formulation of both Laws is completely covariant. Since they are indepenent of frames of reference, they are not restricted to inertial frames. In particular, this means that the vector field $\boldsymbol{F}$ only describes interactions. You may not add any “fictitious forces” on a whim. Fictitious forces only ever enter the picture when expressing the equation of motion with respect to a reference frame which is non-inertial in the above defined sense.

You also don't have to use any auxilliary criterion to distinguish real forces from fictitous forces. Any vector field is a possible real force, even those that don't satisfy the Third Law. Fictitious forces cannot be regarded as a vector fields in space time, since they are related to the coefficients of $\nabla$ in an arbitrary reference frame and thus do not transform like vectors.

To see how they come about let's just introduce an orthogonal frame S of spatial vectors $\boldsymbol{e}_{i}$ along a curve in space time, which fulfills neither of the dynamical constraints on an inertial frame, i.e. we have

$$\nabla_{\boldsymbol{e}_{0}}\boldsymbol{e}_{0}=\Gamma_{00}^{i}\boldsymbol{e}_{i}$$

(There still is no $\boldsymbol{e}_{0}$-component, because $\langle\mathrm{d}t,\boldsymbol{e}_{0}\rangle\equiv 1$, implying $\Gamma_{00}^{0}=0$.) And

$$\nabla_{\boldsymbol{e}_{0}}\boldsymbol{e}_{i}=\Gamma_{i0}^{k}\boldsymbol{e}_{k}.$$

(All other $\Gamma$ may be assumed to vanish.) The tangent $\boldsymbol{u}$ to the particle worldline and $\boldsymbol{e}_{0}$ are related by the follwing covariant version of the “Galileian velocity-addition formula”

$$\boldsymbol{u}=\boldsymbol{v}+\boldsymbol{e}_{0},$$

where $\boldsymbol{v}$ is the velocity of the particle relative to the origin, which may be equivalently defined as

$$\boldsymbol{v}=\nabla_{\boldsymbol{e}_{0}}\boldsymbol{r},$$

the derivative of the spatial position vector $\boldsymbol{r}$ from the origin to the particle. Now, if the frame rotates with anguar velocity $\boldsymbol{\omega}$ we have

$$\nabla_{\boldsymbol{e}_{0}}\boldsymbol{e}_{i}=\boldsymbol{\omega}\times\boldsymbol{e}_{i}$$

Also defining

$$\nabla_{\boldsymbol{e}_{0}}\boldsymbol{e}_{0}=\boldsymbol{b}$$

which is a spatial vector, we obtain the following form of the Second Law

$$\boldsymbol{F}=m\nabla_{\boldsymbol{u}}\boldsymbol{u}=m\nabla_{\boldsymbol{e}_{0}}(\boldsymbol{v}+\boldsymbol{e}_{0})=m\nabla_{\boldsymbol{e}_{0}}\boldsymbol{v}+m\boldsymbol{b}.$$

The first term on the right hand side may be evaluated further using

$$\boldsymbol{v} = \nabla_{\boldsymbol{e}_{0}}\boldsymbol{r}=\dot{r}^{i}\boldsymbol{e}_{i}+\boldsymbol{\omega}\times\boldsymbol{r}$$

which gives

$$\nabla_{\boldsymbol{u}}\boldsymbol{u}=\ddot{r}^{i}\boldsymbol{e}_{i}+\boldsymbol{\omega}\times(\dot{r}^{i}\boldsymbol{e}_{i})+\boldsymbol{\omega}\times(\dot{r}^{i}\boldsymbol{e}_{i}+\boldsymbol{\omega}\times\boldsymbol{r})+\nabla_{e_{0}}\boldsymbol{\omega}\times\boldsymbol{r}+\boldsymbol{b}.$$

Now, I define the following “derivatives w.r.t fixed axes of S”

$$\boldsymbol{\ddot{r}_{S}}=\ddot{r}^{i}\boldsymbol{e}_{i}$$
$$\boldsymbol{\dot{r}_{S}}=\dot{r}^{i}\boldsymbol{e}_{i}$$

The dots are only meant as suggestive notation. These are not the derivatives of any vectors, in particular $\boldsymbol{\dot{r}_{S}}$ is not equal to the relative velocity defined above $\boldsymbol{\boldsymbol{\dot{r}_{S}}}\neq\boldsymbol{v}=\boldsymbol{\boldsymbol{\dot{r}_{S}}}+\boldsymbol{\omega}\times\boldsymbol{r}$. The Second Law thus reads

$$m\boldsymbol{\ddot{r}_{S}}=\boldsymbol{F}-2m\boldsymbol{\omega}\times\boldsymbol{\dot{r}_{S}}-m\boldsymbol{\omega}\times\left(\boldsymbol{\omega}\times\boldsymbol{r}\right)-m\boldsymbol{\dot{\omega}}\times\boldsymbol{r}-m\boldsymbol{b},$$

where now the right hand side contains all the usual fictitious forces. But those are no more “fictitious” than $m\boldsymbol{\ddot{r}_{S}}$ on the left hand side. Both are just components of the absolute (covariant) acceleration of the particle. If one reviews the usual elementary treatments of “accelerated frames”, I think this is exactly what they do to derive the formulas. Only they are not talking about “covariant derivatives”, but more vaguely of “derivatives with respect to axes that are fixed in space”. But this is the same thing if “fixed in space” means $\boldsymbol{\omega}=0$ and $\boldsymbol{b}=0$.

At no point did I have to use the Third Law and still there was no logical problem defining inertial frames or formulating the First Law.

By the way it is natural to include Newtonian Gravity as curvature of space time, by stating the equivalence principle in the form "inertial motion = free fall". Then the Second Law formally holds as stated above. Only $\boldsymbol{F}$ is also free of gravitational forces, which are instead included in $\nabla$ and are in this sense more on par with inertial forces. Also all "inertial frames" become "local inertial frames".

 

[Dale]

a covariant formulation of Newtonian Mechanics, in which all laws of motion are given independently of any reference frame

Welcome to PF @vis_insita , what an excellent first post!

Do you have a reference for this approach? Also, since F is a vector field on spacetime then how do you introduce the third law?

 

[DrStupid]

Do you have any argument showing that Calkin’s formulation of Newton’s laws (not other formulations) requires N3 to define an inertial frame?

I need to know that formulation to answer this question. In the source you linked above I can see a wording for Lex III but not for I or II. There are a lot of statements about them, but I do not find the laws themselves. Maybe I just missed them. Could you please post the exact wordings?