What Do Newton's Laws Say When Carefully Analysed

[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.