What Do Newton's Laws Say When Carefully Analysed

[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?

 

[DrStupid]

Newton III says nothing about the range of interactions.

That's not the point. The range of interactions doesn't matter if forces would need no interactions at all. How do you exclude that without Newton III?

 

[Jimster41]

That was clear. #103. In fact I rather enjoyed that math.

It reinforces what @vanhees71 said about Newton's assumption re Euclidean background space-time.

It also clarifies what bothers me... in what space-time does this co-variant derivative exists? It feels like an "arbiter of all Newtonian dilemmas" but I don't get what gives it reference, how does it avoid being accused of being a "preferred frame" or in some sense occupying one, like some background geometry?

Unless the idea of the third law is to say that the covariant derivative is a benign and strictly helpful artifice because it always treats any two frames it connects as equals (forces are always equal and opposite). IOW, it's reference is only physically defined via interactions - for which no frame is considered preferred?

 

[Jimster41]

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 FF\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.

I guess that's kind of what @vis_insita said?

 

[vis_insita]

Thank you, @Dale, for the friendly welcome.

A reference I intended, but forgot, to include in my first post is Misner, Thorne, Wheeler, Gravitation, which contains a chapter dedicated to the covariant formulation of Newtonian Gravity. This approach goes back to Cartan and is also known as Newton-Cartan-Theory. I am only aware of one other book that also discusses it, but it is in German: Dirschmid, "Tensoren und Felder".

To be honest, I haven't really thought about how the Third Law fits in, since it always appeared somewhat unimportant to me. It may be true or false without affecting much of the rest of the theory as far as I can tell. Because the formulation of the first two laws I have given is completely local, I don't expect that the Third Law can be included in a natural way, since it appears inherently nonlocal to me. (But I am attempting one formulation below.) Still I believe that even in this local formulation the theory is general enough to incorporate all cases of physical interest. As long as the influence of the other particle can be described by a field equation, that has to be solved simultaneously with the equations of motion of the particles I don't see a fundamental problem.

Consider gravity which is a paradigmatic two-body-force obeying the Third Law. Yet, in Newton-Cartan-Theory there is a field equation (very similar to the Einstein equation)

$$(\text{Ricci-Curvature}) = (\text{mass density})\mathrm{d}t\otimes\mathrm{d}t,$$

wich makes the equation of motion of every particle completely local. Of course the fields thus defined will in general not be independent degrees of freedom, since the field equations will not contain time derivatives in the characteristic ways, as is necessary in Relativity. But this doen't invalidate their appearance in Newtonian Mechanics. Also, I think in view of the equivalence principle, gravity is more naturally included among the inertial forces, since it shows up only implicitely in $\nabla$, not in the force law $\boldsymbol{F}$. So here the Third Law used as a criterion for the distinction between real force and fictitious force is somewhat in the way of a covaraint formulation of the equivalence principle.

Now, if one wanted to include non-local interactions, I think one must rephrase the 2nd law. The n-particle force $F_i$ on particle i is not a vector field on space time anymore, but a more complicated object that assigns a vector field to each absolute time t and n-1 events happening at that time. So for every events $x_1, x_2, ...,$ with $t(x_1)=...=t(x_{n-1})$ (all happening at the same absolute time) there is a vector field: $x \mapsto \boldsymbol{F}_i(x, x_1, ...)$ which is tangent to x and spatial. But, since I just made all that up, I'm not sure if it makes sense.

In any case, I believe the Third Law must be regarded as a condition on two body forces which reads "action equals reaction at each point in absolute time". This can be formulated on Newtonian space time as

$$\boldsymbol{F}_1(x(t), y(t)) = -\boldsymbol{F}_2(y(t), x(t))$$

where $x$ and $y$ are the world lines of both particles. This condition makes sense since all vectors are taken at the same absolute time and are thus contained in the same absolute Euclidean space.

 

[vis_insita]

It also clarifies what bothers me... in what space-time does this co-variant derivative exists? It feels like an "arbiter of all Newtonian dilemmas" but I don't get what gives it reference, how does it avoid being accused of being a "preferred frame" or in some sense occupying one, like some background geometry?

A covariant derivative does not define a preferred frame.

The principle of relativity is very elegantly incorporated in the structure of space time. Note that the usual approach to Newtonian Mechanics gives space time a structure $\mathbb{R}\times\mathbb{E}^3$. This somewhat contradicts the principle of relativity since it allows to define absoluet rest as "staying at the same point in absolute space $\mathbb{E}^3$ for all times."

In the covariant formulation one potulates a more general structure: Spacetime is a 4-dimensional manifold $\mathbb{N}^4$, say, on which a scalar field $t: \mathbb{N}^4 \rightarrow \mathbb{R}$ is definied, which assigns each event the absolute time at which it happens. (Of course there are some requirement on $t$, like $\mathrm{d}t\neq 0$ without which the following wouldn't make much sense.) Now you define absolute space at time $\tau$ as the set of events $x$, that satisfy $t(x)=\tau$. The crucial thing is that each time t has its own absolute space. It makes no sense to talk about two events at different times happening in the same place. Thus, you cannot define absolute rest here. The covariant derivative doesn't change anything about that fact. You can easily verify this by showing that for each inertial frame with origin $O(t)$ there is a different one, whose origin $O'(t)$ has a nonvanishing constant relative velocity to the first one, i.e.

$$\nabla_{e_0}(O'(t) -O(t)) \neq 0.$$

I think there is an interesting pattern to recognize here

1) The elementary formulations of Newtonian Mechanics involve a global split of space and time. This makes the Principle of Relativity a pure accident of the dynamics, which only shows up as a symmetry of the force laws that is in no way reflected in the fundamental structure of space time. To remedy this you have to unify space and time, even in Newtonian Mechanics.

2) The elementary formulations of Newtonian mechanics treat gravity as a normal interaction force. This makes the Principle of Equivalence a pure accident of the equality of "inertial" and "gravitational" mass. To adequately represent that principle in the Laws of Motion as well as in the spacetime structure you have to make spacetime curved, even in Newtonian Mechanics.

 

[atyy]

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".

This is important, and shows a great difference between your presentation and the usual Newton's laws that we have been discussing. While the overall physics is the same, the definition of force is not the same in both formalisms.

 

[vis_insita]

This is important, and shows a great difference between your presentation and the usual Newton's laws that we have been discussing. While the overall physics is the same, the definition of force is not the same in both formalisms.

Since I am not using the Third Law anywhere I believe the definition of force is a little more general. But I think this is only a minor difference, not a great one. The point was more that even absent the Thrid Law there is no conceptual problem in formulating the other two laws of motion, or in treating inertial forces, or in defining inertial frames. The Third Law doesn't seem necessary for that.

The treatment of the Equivalence Principle, however, is independent of the treatment of all non-gravitational forces. It is possible to incorporate gravity as spacetime curvature in a covariant formulation, as well as in a non-covariant one. But it is not necessary in either. You can also postulate the existence of a field $\Phi$ on spacetime, which satisfies $\Delta\Phi = \rho$ on each slice of constant absolute time, and couples to the particles via

$$m\nabla_{u}u = -m\nabla\Phi + \boldsymbol{F}.$$

But the gravitational force $-m\nabla\Phi$ is a little elusive and cannot be unambiguously (i.e. observer-independently) assigned to a particular event without further assumptions. Since $\partial_i \Phi$ can always be absorbed in the redefinition

$\tilde{\Gamma}^i_{00} = \Gamma^i_{00} + \partial_i\Phi,$

mechanics provides no basis to distinguish a locally vanishing gravitational field observed by a non-inertial observer with linear acceleration $b^i = \Gamma^i_{00}$ from a non-vanishing field of strength $\Gamma^i_{00} = \partial_i\Phi$ observerd by an inertial oberver.

This is a real physically important difference to other forces and can justifiably influence the way we see and define forces.

 

[atyy]

Since I am not using the Third Law anywhere I believe the definition of force is a little more general. But I think this is only a minor difference, not a great one. The point was more that even absent the Thrid Law there is no conceptual problem in formulating the other two laws of motion, or in treating inertial forces, or in defining inertial frames. The Third Law doesn't seem necessary for that.

The treatment of the Equivalence Principle, however, is independent of the treatment of all non-gravitational forces. It is possible to incorporate gravity as spacetime curvature in a covariant formulation, as well as in a non-covariant one. But it is not necessary in either. You can also postulate the existence of a field $\Phi$ on spacetime, which satisfies $\Delta\Phi = \rho$ on each slice of constant absolute time, and couples to the particles via

$$m\nabla_{u}u = -m\nabla\Phi + \boldsymbol{F}.$$

But the gravitational force $-m\nabla\Phi$ is a little elusive and cannot be unambiguously (i.e. observer-independently) assigned to a particular event without further assumptions. Since $\partial_i \Phi$ can always be absorbed in the redefinition

$\tilde{\Gamma}^i_{00} = \Gamma^i_{00} + \partial_i\Phi,$

mechanics provides no basis to distinguish a locally vanishing gravitational field observed by a non-inertial observer with linear acceleration $b^i = \Gamma^i_{00}$ from a non-vanishing field of strength $\Gamma^i_{00} = \partial_i\Phi$ observerd by an inertial oberver.

This is a real physically important difference to other forces and can justifiably influence the way we see and define forces.

Would it be correct to think that what you have added beyond the traditional Newton's first and second, is that you have specified their transformation properties in arbitrary coordinates?

 

[vanhees71]

That's not the point. The range of interactions doesn't matter if forces would need no interactions at all. How do you exclude that without Newton III?

Now I'm completely lost. If the forces are not interactions, Newton III doesn't make any sense.

It's of course clear that a physical theory is not simply a system of axioms as in mathematics but you have to operationally define it. Again, I can only repeat the logic of Newton's arguments:

Newton I: Assumes the existence of an absolute time and space, where the law of inertia holds, i.e., bodies which are not "influenced" by other bodies (and for Newton there's nothing else than material bodies of course, because fields haven't been discovered yet) move in rectilinear uniform motion in such an "inertial frame".

Newton II: Forces acting on a body change the state of motion within an inertial frame, being proportional to the acceleration of a body. The proportionality constant is a measure for the inertia of the body and also a meausure for "amount of substance".

Newton III: Forces act pair-wise and if $\vec{F}_{12}$ is the force on body 1 through interaction with body 2, the force on body 2 through interaction with body 1 is $\vec{F}_{21}=-\vec{F}_{12}$.

As we know today, that's not even complete since also within Newtonian physics there are generic $N$-body forces with $N>2$ (e.g., the nucleon-nucleon interaction in relativistic models).

It's anyway a very complicated issue, and from a modern point of view it's simpler to just state the space-time model and analyze its symmetries, but that we've also discussed already above.

 

[vis_insita]

Would it be correct to think that what you have added beyond the traditional Newton's first and second, is that you have specified their transformation properties in arbitrary coordinates?

I would say that the transformation properties to arbitrary coordinates are made especially transparent in this approach. But really the only thing that is added beyond traditional approaches is, I think, clarity. It is perfectly valid to base Newtonian Mechanics on the notion of inertial frames and derive everything, including the form of the equations of motion, in non-inertial frames or arbitrary coordinate systems from that.

But personally I always found that a little confusing. And I encountered many statements that I suspect are the result of similar confusions. Some such statements are plainly false, e.g. the very common claim that Newton's axioms are only valid in inertial frames. Against such claims a generally covariant formulation of those axioms should provide a decisive argument.

One of the subtleties that can in the traditional approach become the more confusing the more you think about it, is the derivation of inertial forces. It usually involves a discussion of the time-dependence of basis vectors which goes like this:

In an inertial frame S we have
$$\frac{d v}{d t} = \dot{v}^i e_i \qquad (1).$$
But in a non-inertial frame S' the basis vectors move, so we have to take that into account too. Thus
$$\frac{d v}{d t} = \dot{v}^ie'_i + v^i\dot{e}'_i.$$
But what justifies the asymmetry? The basis vectors of S' clearly appear to move if viewed from S. But so do the basis vectors of S appear to move when viewed from S'. I once had a discussion where I pointed out that condition (1) is precisely one of the characteristics of an inertial frame. My opponent replied that this cannot be true, since from the point of view of an observer at rest in a rotating frame, the basis vectors don't move. So for that observer the same condition holds. I said that the time derivative of a vector is again a vector, which is a geometric object. So it either is or isn't zero. But, in any case, that statement is independent of any reference frames. I believe he remained unconvinced, but I wonder how he would derive inertial forces.

I think the cause for this confusion is that an important part of mathematical structure, namely the covariant derivative, is completely absent from the discussion. It is never really made clear what the inertial motion of particles has to do with time derivatives of basis vectors. The covariant approach makes that perfectly clear: $\nabla$ defines straightness, i.e. inertial motion, as well as vector derivatives.

 

[atyy]

I would say that the transformation properties to arbitrary coordinates are made especially transparent in this approach. But really the only thing that is added beyond traditional approaches is, I think, clarity. It is perfectly valid to base Newtonian Mechanics on the notion of inertial frames and derive everything, including the form of the equations of motion, in non-inertial frames or arbitrary coordinate systems from that.

In practice the covariant first law is still useless if there are no non-interacting particles, so that seems the same as the elementary version.

The covariant second law is still a force definition, as in the elementary version, since it is empty unless F is specified.

The covariant second law seems able to distinguish non-inertial forces from real forces, but similarly to the elementary version, you add something. In the elementary version, one of the things one can add is Newton's 3rd law, but that doesn't have to be the extra ingredient, it can also be the transformation properties of the forces.

However, in the elementary version, adding transformation properties does not distinguish between an accelerated frame and a uniform gravitational field. Thus an accelerated frame can cancel a uniform gravitational field not only locally, but over a large region of space and time. Does the covariant version have any advantage in distinguishing an accelerated frame from a uniform gravitational field?

 

[vis_insita]

In practice the covariant first law is still useless if there are no non-interacting particles, so that seems the same as the elementary version.

The covariant second law is still a force definition, as in the elementary version, since it is empty unless F is specified.

It is true that both the covariant and the non-covariant version have the same content. This is so by intention. But we have to be clear about what that content is.

The First Law both defines what "straight uniform motion" means, and states a physical condition under which it occurs. The latter condition would also follow from the Second Law. But the definition of "straight uniform motion" is not useless in the sense that it could be discarded without substitution. From a theoretical point of view I think it is most important.

Also it seems to me that traditional expositions are usually not very consistent in applying that definition even if they carefully formulate it. E.g. a non-covariant formulation that would be (absent gravitation) equivalent to the one I have given is

1st Law: "There exists a frame of reference in which all straight uniform trajectories have the form
$$\left(\frac{\mathrm{d}^2 x^i}{\mathrm{d} t^2}\right)\boldsymbol{e}_i = 0. \qquad (1)$$
Isolated, non-interacting particles follow those straight uniform trajectories."

Now, changing to an arbitrary reference frame will in general violate condition (1) in a well-defined way, but it will not change anything about the straightness or the uniformity of the motion, because both properties have absolute, frame-independent meaning within the theory. One cannot neglect defining them in a frame-independent way without confusing everything. Yet, some authors seem to say that condition (1) defines straightness in any frame which absolutely makes no sense. Thus inertial forces are introduced to "explain" the departure from straight uniform motion in light of the Second Law, which is completely backwards. Inertial forces occur even if the motion is straight and uniform merely as a consequence of chosing a non-inertial reference frame.

The covariant second law seems able to distinguish non-inertial forces from real forces, but similarly to the elementary version, you add something. In the elementary version, one of the things one can add is Newton's 3rd law, but that doesn't have to be the extra ingredient, it can also be the transformation properties of the forces.

I think here you are saying that inertial forces arise as a consequence of the transformation properties of forces. They do not; at least not in the way they are usually introduced in textbooks. They arise because the time derivative that defines acceleration is caluclated in a time-dependent basis. Thus, the only transformation property involved is that of the time derivative
$$\left(\frac{\mathrm{d}}{\mathrm{d}t}\right)_{S}[\cdot] = \left(\frac{\mathrm{d}}{\mathrm{d}t}\right)_{S'}[\cdot] + \boldsymbol{\omega}\times[\cdot]$$
which is of fundamental importance to the theory and is, as the covariant formulation makes clear, nothing but a special case of the usual "transformation law" of an affine connection.

(Maybe this equation is that something you say must be "added" to make a clear distinction between real and fictitious forces. I think I would agree with that.)

However, in the elementary version, adding transformation properties does not distinguish between an accelerated frame and a uniform gravitational field. Thus an accelerated frame can cancel a uniform gravitational field not only locally, but over a large region of space and time. Does the covariant version have any advantage in distinguishing an accelerated frame from a uniform gravitational field?

Why would that be an advantage? I think it is precisely the content of the Equivalence Principle that you cannot distinguish both situations. In the Newton-Cartan-Theory both the uniform acceleration and the "local gravitational field" originate from
$$\nabla_{\boldsymbol{e}_0}\boldsymbol{e}_0 = \Gamma^i_{00}\boldsymbol{e}_i$$
As in General Relativity that term can always be made to vanish at the origin, by changing to a local inertial frame. If there is only a uniform "gravitational force" (which really isn't a gravitational force at all) it vanishes everywhere and the local inertial frame is a global one.

 

[Dale]

There exists a frame of reference in which all straight uniform trajectories

Would it be better to just say “geodesic” here?

 

[vis_insita]

Would it be better to just say “geodesic” here?

In my opinion this is exactly the meaning that should be attributed to "straight uniform motion" also in Newtonian Mechanics, yes.

 

[DrStupid]

If the forces are not interactions, Newton III doesn't make any sense.

I'm not sure what you mean. Newton III defines forces to be interactive. Thus, I would reather say that non-interactive forces make no sense (due to Newton III).

Again, I can only repeat the logic of Newton's arguments:

As we can't ask Newton anymore, nobody knows for certain what the logic of his arguments really is. You just repeat your opinion about them. Here is my opinion, based on the original wordings^1,2:

Newton’s laws of motion are about forces. That’s why it makes sense to start with Newton’s definition of force:

"Def. IV: Vis impressa est actio in corpus exercita, ad mutandum ejus sta-tum vel quiescendi vel movendi uniformiter in directum."
(An impressed force is an action exerted upon a body, in order to change its state, either of rest, or of moving uniformly forward in a right line.)

With this definition we have a general impression what forces are (something that changes the state of motion of bodies) and what they not are (something that preserves the state of motion [e.g. in contrast to the Aristotelian concept of force]). But it is not ready to use. There are some questions open:

1. The definition of force says that it is an action that changes the state of motion of a body, but are there other reasons for a body to change its motion, in particular, can it spontaneously change its motion without an action exerted upon it? That is answered by

"Lex I: Corpus omne perseverare in statu suo quiescendi vel movendi uniformiter in directum, nisi quatenus a viribus impressis cogitur statum illum mutare."
(Law I. Every body perseveres in its state of rest, or of uniform motion in a right line, unless it is compelled to change that state by forces impressed thereon.)

That means that forces are the only reason for a change of the state motion.
Now we know that every force changes the state of motion of a body (according to Def. IV) and every change of motion is caused by a force (according to Law I). Or

$F = 0 \Leftrightarrow a = 0$

which is logical equivalent to

$F \ne 0 \Leftrightarrow a \ne 0$

2. Now we have a full qualitative relation between forces and changes in the state of motion but what how to quantify it? That is answered by

"Lex II: Mutationem motus proportionalem esse vi motrici impressae, & fieri secundum lineam rectam qua vis illa imprimitur."
(Law II: The alteration of motion is ever proportional to the motive force impressed; and is made in the direction of the right line in which that force is impressed.)

In modern notation and with a unified system of units this means

$F = \dot p$

3. Together with the definition of momentum (Def. II) we now have a full qualitative and quantitative relation between forces and changes in the state of motion. But where do forces come from? That is answered by

"Lex III: Actioni contrariam semper & aequalem esse reactionem: sive corporum duorum actiones in se mutuo semper esse aequales & in partes contrarias dirigi."
(Law III: To every action there is always opposed an equal reaction: or the mutual actions of two bodies upon each other are always equal, and directed to contrary parts.)

That implies (but is not limited to)

$action \Rightarrow reaction$

which is logical equivalent to

$\neg reaction \Rightarrow \neg action$

Forces are limited to interactions. Without interaction there is no force.

Law III says furthermore that the interactions need to be symmetrical. For two bodies A and B the force $F_{AB}$ exerted by A upon B and the counter-force $F_{BA}$ exerted from B upon A must comply with

$F_{AB} = - F_{BA}$

That also implies that the interactions are binary only. For more than two bodies the additivity of forces (resulting from the additivity of momentum and also described in Corollary I) would for infinite different linear combinations. Newton III tells us that the combination of symmetric pairs of forces shall be used only.

That’s what Newton’s laws of motion say to my understanding.

What has that to do with inertial frames? The laws of motion (with the wording above) are not valid in all frames of reference. In #85 I demonstrated how rotating frames violate them. They can either comply with Law I and II or with Law III but not with all of them at once. Frames of references that comply with all laws of motion are called inertial. Of course there are other possibilities to define inertial frames, but that’s how it works with the laws of motion as given above.

If and how the laws of motion define inertial frames strongly depends on the exact wordings. There are a lot of different versions out there, resulting in different conclusions. In #34 I already mentioned a version that has been introduces by Newton himself. By deleting the first part of Newton III ("To every action there is always opposed an equal reaction") he allowed fictitious forces and therefore extended the laws of motion to non-inertial frames. There are a lot of cases where something like this is very useful. But with this modification the laws of motion alone do not define inertial frames anymore.

That’s why it is very important to clarify what we actually mean with “Newton’s laws of motion”. I refer to the original or equivalent wordings. If you refer to another version that please post the wording.¹Latin original: http://cudl.lib.cam.ac.uk/view/PR-ADV-B-00039-00001/1
² Eglish translation: https://en.wikisource.org/wiki/The_Mathematical_Principles_of_Natural_Philosophy_(1846)

 

[vanhees71]

Of course, you can describe everything in any frame you like, but to establish the laws you need "absolute time" and "absolute space", i.e., (the equivalence class of) inertial frames. There's no problem with accelerated reference frames though, and the laws can be formulated covariantly, introducing covariant time derivatives acting on the components of vectors (and general tensors) wrt. any basis. There's also no problem to introduce genereric 3- and N-body forces/interactions in addition to pair interactions.

 

[atyy]

(Maybe this equation is that something you say must be "added" to make a clear distinction between real and fictitious forces. I think I would agree with that.)

Yes, that is what I am saying. To distinguish between real and inertial forces, one could add N3, but one could also do as you did and add the transformation criteria.

Why would that be an advantage? I think it is precisely the content of the Equivalence Principle that you cannot distinguish both situations. In the Newton-Cartan-Theory both the uniform acceleration and the "local gravitational field" originate from
$$\nabla_{\boldsymbol{e}_0}\boldsymbol{e}_0 = \Gamma^i_{00}\boldsymbol{e}_i$$
As in General Relativity that term can always be made to vanish at the origin, by changing to a local inertial frame. If there is only a uniform "gravitational force" (which really isn't a gravitational force at all) it vanishes everywhere and the local inertial frame is a global one.

Yes, if the gravitational field is truly uniform over all space, then one cannot distinguish it from an accelerated frame. However, suppose we have an infinite plane mass that produces a uniform gravitational field over the half space. N3 can distinguish between a real uniform gravitational field and an accelerated frame, since N3 would predict that the real gravitational field is produced by matter.

 

[olgerm]

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

If the noninertial frame of reference is bounded to a body that is aceelereating with constant acceleration and is not spinning in inertial frame of reference,
then Newton 3. law would still be valid in it.

If you have no information (for examle force as function of coordinates and speeds) about (noninertial) force, then Newton 3. law is same as conservation of momentum.

 

[DrStupid]

Of course, you can describe everything in any frame you like, but to establish the laws you need "absolute time" and "absolute space"

Newton introduced them in Scholium I + II [https://en.wikisource.org/wiki/The_Mathematical_Principles_of_Natural_Philosophy_(1846)/Definitions]

 

[DrStupid]

If the noninertial frame of reference is bounded to a body that is aceelereating with constant acceleration and is not spinning in inertial frame of reference, then Newton 3. law would still be valid in it.

That only works as long as the reason for the acceleration of the body remains unknown. It is no surprise that such a lack of information may lead to wrong conclusions.

If you have no information (for examle force as function of coordinates and speeds) about (noninertial) force, then Newton 3. law is same as conservation of momentum.

Newton 3 implies conservation of momentum but it is not the same.

 

[olgerm]

That only works as long as the reason for the acceleration of the body remains unknown.

Lets say than that the it not bounded to a body, but is just accelerating like it were bounded to a body that was constantly accelerating in inertial frame of reference. aka a frame of reference with homogeneous and constant inertial force.

 

[DrStupid]

Lets say than that the it not bounded to a body, but is just accelerating like it were bounded to a body that was constantly accelerating in inertial frame of reference. aka a frame of reference with homogeneous and constant inertial force.

Momentum is not conserved in such a frame of reference. That violates Newton III.

 

[olgerm]

Momentum is not conserved in such a frame of reference. That violates Newton III.

You are right. But how would you define inertial frame of reference WITH Newton's 3. law?

 

[vis_insita]

What has that to do with inertial frames? The laws of motion (with the wording above) are not valid in all frames of reference. In #85 I demonstrated how rotating frames violate them. They can either comply with Law I and II or with Law III but not with all of them at once.

From post #85:

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 $F = m \cdot a = - m \cdot \omega ^2 \cdot r$

But there is a catch. I think nothing in your (or Newton's) formulation of the axioms implies that switching to a rotating frame changes the particle's acceleration to $-\omega^2 r$. You probably think that you are just using standard definitions here, but you are not. It is your definition of "acceleration" that only applies to inertial frames, not any of Newton's axioms.

I suppose we agree that acceleration is just the time derivative of velocity, both of which are represented as vectors. The velocity can be calculated as time derivative of position, which may also be represented as vector relative to the common origin. Now we just have to carefully apply those definitions: Since both the inertial frame S, $\boldsymbol{e}_i$, and the rotating frame S', $\boldsymbol{\tilde e}_i$ always share the same origin, the position vector $\boldsymbol{r}$ at every instant of time is given by (left hand side is always S, right hand side is S' in the following)
$$ r^i\boldsymbol{e}_i = \boldsymbol{r} = \tilde r^i\boldsymbol{\tilde e}_i.$$
Its time derivative gives the velocity
$$\dot{r}^i \boldsymbol{e}_i = \boldsymbol{v} = \dot{\tilde r}^i\boldsymbol{\tilde e}_i + \boldsymbol{\omega}\times\boldsymbol{r}$$
You assumed $\dot{r}^i \equiv 0$, so $\boldsymbol{v}=0$.

Note that the velocity w.r.t both systems is just $\boldsymbol{v}=0$. It is not given by $\dot{\tilde r}^i\boldsymbol{\tilde e}_i$ in the rotating system, because that vector simply isn't equal to the time derivative
$$\boldsymbol{v} = \frac{\mathrm{d}\boldsymbol{r}}{\mathrm{d}t},$$
which I just defined as velocity. Although velocity is relative to the frame of reference, this relativity only produces a difference if both origins are in relative motion, which they are not in the situation you described. (If you think of it, it really makes sense that way. Relative velocity is just a vector in absolute space. It may depend on time, but why would it depend on an arbitrary frame, that someone sets up in space? A vector, as a geometric object, is supposed to be independent of basis.)

Now proceeding to acceleration, using $\boldsymbol{v}=0$, $\boldsymbol{\omega}=\text{const.}$, and -- obviously -- $\boldsymbol{r}\cdot\boldsymbol{\omega}=0$, we get
$$ \ddot{r}^i\boldsymbol{e}_i = \boldsymbol{a} = \ddot{\tilde r}^i \boldsymbol{\tilde e}_i + \dot{\tilde r}^i \boldsymbol{\omega}\times \boldsymbol{\tilde e}_i = \ddot{\tilde r}^i \boldsymbol{\tilde e}_i - \boldsymbol{\omega}\times(\boldsymbol{\omega}\times\boldsymbol{r}) = \ddot{\tilde r}^i\boldsymbol{\tilde e}_i + \omega^2 \boldsymbol{r}$$
which again vanishes since the left hand side vanishes by assumption. Now it is clear that what you call "acceleration" is just the term
$$\ddot{\tilde r}^i\boldsymbol{\tilde e}_i = -\omega^2 \boldsymbol{r}.$$
But again, this is not the time derivative of velocity relative to the rotating system, because it neglects the rotation of the axes, which change with time. It is no accident that textbooks introduce a special notation for this kind of operation, usually
$$\left(\frac{\mathrm{d}^2\boldsymbol{r}}{\mathrm{d}t^2}\right)_{S'}$$
because it needs to be distinguished from the covariant time derivative $\mathrm{d}/\mathrm{d}t$.

Now another point needs to be mentioned. Just as the relative velocities are equal, iff the origins are not in relative motion, so the accelerations -- calculated in the way above -- are equal only if the origins have constant relative velocity. Now, from what I just said it seems to follow that the application of Newton's axioms will be ambiguous if both origins are accelerating relative to each other. Because in that situation both systems seem to assign different accelerations to the same particle. But this is only apparent. Let's review the basic assumption that real forces, according either to the Third Law or to our general understanding of the interactions of that particular system, represent the cause for change of the particle's state of motion. Any causal effects on this state can only depend on the particle's state and properties, not on how arbitrarily definable "origins" are moving. This is why the acceleration in $m\boldsymbol{a} = \boldsymbol{F}$ must be covariantly defined, by

$$\nabla_{\boldsymbol{u}} \boldsymbol{u} = \boldsymbol{a},$$

which manifestly depends only on the particles motion through spacetime. This acceleration equals the second derivative of position $\frac{\mathrm{d}^2 \boldsymbol{r}}{\mathrm{d}t^2} = \nabla_{\boldsymbol{e}_0}\boldsymbol{v} = \nabla_{\boldsymbol{e}_0}(\boldsymbol{u} - \boldsymbol{e}_0) = \boldsymbol{a} - \nabla_{\boldsymbol{e}_0} \boldsymbol{e}_0$ only under the condition that $\nabla_{\boldsymbol{e}_0}\boldsymbol{e}_0 = 0$, i.e. if the origin itself is moving on a straight and uniform trajectory (geodesic). This should be an immediately intuitive result.

That a definition based on the (second order) change of position can only be correct in special situations is, I think, not surprising. "Change of position" is an ill-defined concept anyway in light of the Principle of Relativity. It is only defined relative to an object of reference, which can itself change its state of motion arbitrarily. What objects are suitable references for this purpose is completely explained within the theory by means of the analysis above. What is always well-defined, however, without any reference are deviations from geodesic (straight and uniform) motions in spacetime.

In elementary treatments this complication doesn't show up, because there is always one inertial system involved relative to which all accelerations can be unambiguously equated with the second derivative of position (which is therefore used as a universal definition of acceleration). This makes it less obvious how a fully covariant formulation looks like and it seems that one inertial frame is always required. That this is not so, is only apparent when the laws are formulated in Newtonian spacetime. In any case, as shown by the Newton-Cartan-Theory, it is possible to formulate Newton's laws in a fully covariant way, which means the claim that some of these laws must be violated in non-inertial reference frames seems untenable to me. Furthermore, this spurious violation cannot be a valid basis for the definition of inertial frames.

 

[vis_insita]

Yes, if the gravitational field is truly uniform over all space, then one cannot distinguish it from an accelerated frame. However, suppose we have an infinite plane mass that produces a uniform gravitational field over the half space. N3 can distinguish between a real uniform gravitational field and an accelerated frame, since N3 would predict that the real gravitational field is produced by matter.

Newton-Cartan-Theory also predicts that all gravity is produced by matter. In both formulations of the theory you can cancel a homogeneous gravitational field by constant acceleration. Where exactly do you see a difference here?

Also, I am not arguing that the Third Law is useless for everything. I am just saying that it is not needed to define inertial frames.

 

[vanhees71]

Indeed, the reason for the confusion when considering non-inertial (particularly rotating) frames is to mix up the components of vectors wrt. the rotating basis with the vectors themselves.

You can simply work with the vectors, and nothing particular happens, i.e., if $\vec{e}_j$ are (time-independent) Cartesian basis vectors wrt. the intertial frame and $\vec{e}_k'(t)$ the basis vectors wrt. the rotating frame you have for any vector (Einstein summation convention used)
$$\vec{V}(t)=V_j(t) \vec{e}_j=V_k'(t) \vec{e}_k'(t),$$
and then
$$\dot{\vec{V}}(t)=\dot{V}_j (t) \vec{e}_j =\dot{V}_k'(t) \vec{e}_k'(t) + V_k'(t) \dot{\vec{e}}_k'(t).$$
Now since $\vec{e}_k'(t)$ also Cartesian since they are just the rotated vectors of the inertial Cartesian basis vectors, you have
$$\dot{\vec{V}}(t)=[\dot{V}_k'(t) + \epsilon_{klm} \omega_l(t) V_{m}'(t)]\vec{e}_k',$$
i.e., there's an additional term from the time derivative of the rotating basis vectors. Defining the covariant time derivative for vector components (not vectors!) as
$$\mathrm{D}_t V_k'=\dot{V}_k' + \epsilon_{klm} \omega_l V_m',$$
The equation of motion reads
$$m \mathrm{D}_t^2 x_k'=F_k'.$$
There are only "true forces", no "fictitious" or "inertial" forces. Only if you insist to bring the corresponding terms of the 2nd covariant time derivative to the other side solving for $m \ddot{x}_k'$, you get these additional "forces".

If all forces are pair forces obeying Newton III, momentum conservation holds in any frame too. You just have in terms of the components wrt. the inertial frame
$$p_k'=m \mathrm{D}_t x_k'.$$

 

[Dale]

Also, I am not arguing that the Third Law is useless for everything. I am just saying that it is not needed to define inertial frames

Did you ever mention how you include the third law in your description of the force as a field?

 

[vis_insita]

Did you ever mention how you include the third law in your description of the force as a field?

I tried to give an answer here. The short version is that it is not possible. If you want to include the Third Law in a covariant way, you have to substitute force fields for other geometric objects depending on 2 (or n) simultaneous events, not just a single event. Thus they can't be seen as fields or even functions F(x, y, ...) of multiple events in spacetime. I don't think it would be impossible to introduce such a concept to Newtonian spacetime. But it is probably unnecessary to do so. I'm not aware that such objects are studied in the context of Newton-Cartan-Theory, which seems to be really only concerned with gravity. And gravity is described by a field equation (reinterpreting the Laplacian of the potential as Ricci-Curvature). But for me this only underlines that the Third Law is of minor importance to the theory. What do you think?

 

[Dale]

I tried to give an answer here. The short version is that it is not possible.

Ah, I don't know how I missed that, particularly since you tagged me even!

If you want to include the Third Law in a covariant way, you have to substitute force fields for other geometric objects depending on 2 (or n) simultaneous events, not just a single event. Thus they can't be seen as fields or even functions F(x, y, ...) of multiple events in spacetime. I don't think it would be impossible to introduce such a concept to Newtonian spacetime. But it is probably unnecessary to do so. I'm not aware that such objects are studied in the context of Newton-Cartan-Theory, which seems to be really only concerned with gravity. And gravity is described by a field equation (reinterpreting the Laplacian of the potential as Ricci-Curvature). But for me this only underlines that the Third Law is of minor importance to the theory. What do you think?

I don't think that I would say the 3rd law is of minor importance since it produces conservation of momentum. However, I guess that the proper approach would be to apply Noether's theorem to the covariant formulation (with the assumption of spatial homogeneity), and derive the corresponding restriction on F that leads to conservation of momentum. Then that restriction could be considered a generalization of Newton's 3rd law.

 

[DrStupid]

I think nothing in your (or Newton's) formulation of the axioms implies that switching to a rotating frame changes the particle's acceleration to $-\omega^2 r$.

Your argumentation is based on the assumption that Newton's laws of motion must be used with proper acceleration and not with coordinate acceleration. Is there any indication that Newton didn't refer to positions, velocities and accelerations that are measured within (and with respect to) a given frame of reference?

 

[DrStupid]

But how would you define inertial frame of reference WITH Newton's 3. law?

Inertial frames of reference always comply with the laws of motion.

That's it.

 

[olgerm]

OK, let's say you have your non-interacting particles. How do you define an inertial frame with Newton I but without Newton III?

I think you may have be even right all along. misunderstanding is due to vague definition of ineraction and force (are inertial forces forces, are inrtial interactions interactions).

Just definition: "inertial frame of reference is frame of reference in which, forceless particles are moving without acceleration"
is not fine, because according to $F=\frac{\partial^2 x}{\partial t^2}*m$ inertial force is force and according to following this bad definition all frames of references were inertial.

definition: "inertial frame of reference is frame of reference in which, non-interacting particles are moving without acceleration"
is fine only if inertial forces are not considered interactions. Maybe it is possible to state it without using Newton's 3. law.

 

[Dale]

Is there any indication that Newton didn't refer to positions, velocities and accelerations that are measured within (and with respect to) a given frame of reference?

I don't think that the concept of a reference frame had even been invented at that time. Certainly, he does not explicitly discuss distinctions between inertial frames, non-inertial frames, and covariant formulations.

This is partly why the seminal authors do not get the final say on their work, there is simply no way that they could foresee subsequent theoretical and experimental developments. His unavoidable ignorance on such topics makes his stance somewhat irrelevant to a modern dialog on these specific aspects of his theory.

This is physics, not history, and physics can be rewritten by each new generation of physicists (whereas history should not!).

 

[vis_insita]

I don't think that I would say the 3rd law is of minor importance since it produces conservation of momentum. However, I guess that the proper approach would be to apply Noether's theorem to the covariant formulation (with the assumption of spatial homogeneity), and derive the corresponding restriction on F that leads to conservation of momentum. Then that restriction could be considered a generalization of Newton's 3rd law.

Yes, but still you would have presuppose non-local interactions for this to work, wouldn't you? The only homogeneous force field that conserves momentum is F(x) = 0. Only if F depends on more than one particle location at the same time can there be less trivial solutions like F(x-y) etc.

I guess that non-locality is why I said "unimportant". The other two laws appear much more universal. They can be carried over to relativistic theories almost verbatim. (Of course the meaning of the terms change in accordance with space time structure.) But in relativity, momentum conservation cannot be formulated as a condition on forces anymore. But maybe what I really meant is not "unimportant", but rather "imposing too strong conditions on interactions".

 

[Dale]

Yes, but still you would have presuppose non-local interactions for this to work, wouldn't you?

I would believe so, with the caveat that I have not worked out the math to make sure. But without going through the math I cannot see any way that it could work out with purely local interactions. However, in Newtonian theory there is no a priori reason to forbid non-local interactions.

I guess that non-locality is why I said "unimportant". The other two laws appear much more universal. They can be carried over to relativistic theories almost verbatim. (Of course the meaning of the terms change in accordance with space time structure.) But in relativity, momentum conservation cannot be formulated as a condition on forces anymore. But maybe what I really meant is not "unimportant", but rather "imposing too strong conditions on interactions".

I like that better. I think you are correct here.

 

[vis_insita]

Your argumentation is based on the assumption that Newton's laws of motion must be used with proper acceleration and not with coordinate acceleration.

No, I only said that a covariant formulation of Newton's axioms is possible using proper acceleration.

But my critique of your argument wasn't based on proper acceleration at all. It was based on the assumption that relative velocity is a vector in absolute space and acceleration is its time derivative. Do you disagree?

 

[DrStupid]

It was based on the assumption that relative velocity is a vector in absolute space and acceleration is its time derivative. Do you disagree?

Yes, but that's not the point. The question is, how do you know, wheather a measured velocity is a vector in absolute space or not? Let's take a geosynchronous satellite. It seems to rermain at rest from the view of an observer on Earth. How do the observer know that the satellite is actually accelerated in the absolute space?

 

[atyy]

In any case, as shown by the Newton-Cartan-Theory, it is possible to formulate Newton's laws in a fully covariant way, which means the claim that some of these laws must be violated in non-inertial reference frames seems untenable to me.

It depends on how one defines Newton's laws and what one includes in the laws. In a non-covariant definition, the laws are violated in a non-inertial frame. Of course, one can also rewrite them such that they are not violated in a non-inertial frame.

 

[andresB]

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

N3 tells you that how to identify the behaviour of two interacting particles. Without N3, how to define non-interactivness without circular reasoning?

 

[andresB]

There have been several approaches to Newtons laws. For Marion and Thorton (classycal dynamics of particles and systems) the first two laws are just definitions and the real physical content is in the third law is the only one with real physical content, just the opposite of the approach of Lindsay and Margenau.

My take is: Newtons laws, unlike axioms in mathematics, can't be untangled from each other. Newtons formulation of mechanics isn't axiomatic, and the understanding of the laws comes from taking them all togheter at the same time and correctly applying them to solve problem and predict the behaviour of mechanical systems.

 

[Dale]

N3 tells you that how to identify the behaviour of two interacting particles. Without N3, how to define non-interactivness without circular reasoning?

How would you define non-interacting particles with N3? N3 deals with the behavior of pairs of interacting particles so it doesn't describe single non-interacting particles at all. How can you apply N3 to a single non-interacting particle at all?

If you follow it down carefully enough, any system of definitions is going to have some undefined primitives, like Euclid's points. A non-interacting body is similar. Just like Euclid could give a generic description of a point or draw an approximation to a point, we can give generic descriptions of non-interacting bodies (far away from other matter) or approximations to non-interacting bodies (interacting bodies with balanced forces).

So we start with the undefined primitive concepts of interacting and non-interacting bodies. Then N1 describes the behavior of non-interacting bodies and defines inertial frames in terms of them. Meanwhile N3 describes the behavior of pairs of interacting bodies and together with N2 describes their behavior with respect to the inertial frames defined by non-interacting bodies in N1.

 

[andresB]

You see the motion of a particle. The motion can be accelerated. How do you know if the particle is non-interacting? The answer comes from N3. If the acceleration comes from a "real" force then there will be some reaction force in some other particle in the universe, if there is no such reaction force then you are not in a inertial frame.

EDIT: Not saying the approach is unique, but it's valid.

 

[vis_insita]

Yes, but that's not the point. The question is, how do you know, wheather a measured velocity is a vector in absolute space or not?

On the contrary, that is precisely the point. 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. Hence, the alleged violation is not justified on the basis of Newton's axioms, but hinges entirely on a peculiar definition of acceleration.

Because your definition is not logically required by those axioms it needs to be stated explicitly as an additional assumption. This is all I wanted to point out.

Let's take a geosynchronous satellite. It seems to rermain at rest from the view of an observer on Earth. How do the observer know that the satellite is actually accelerated in the absolute space?

He simply argues that Earth's gravity is affecting the satellite in a way that causes precisely that acceleration,
$$m\boldsymbol{a} = \boldsymbol{F}.$$
Since he knows the difference between a vector and its components, he is not in danger of confusing his observation that $\ddot{r}^i = 0$ with $\boldsymbol{a} = 0$.

What is your own answer to your question?

 

[Dale]

How do you know if the particle is non-interacting? The answer comes from N3.

N3 is silent on non-interacting bodies.

My take is: Newtons laws, unlike axioms in mathematics, can't be untangled from each other. Newtons formulation of mechanics isn't axiomatic, and the understanding of the laws comes from taking them all togheter at the same time and correctly applying them to solve problem and predict the behaviour of mechanical systems.

I like that approach also. That does seem reasonable.

 

[vis_insita]

It depends on how one defines Newton's laws and what one includes in the laws. In a non-covariant definition, the laws are violated in a non-inertial frame.

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.