What Do Newton's Laws Say When Carefully Analysed

In summary: N2 and N3.Is that really the historical context? Do you have any sources discussing that? I'm seeing a lot discussing the need for an explicit refution of Aristotelian physics, but then only... considering the matter closed, and moving on to N2 and N3.In summary, the conversation discusses the issue of Newton's Laws and their significance in understanding nature. While the first law is often considered a definition, it has been experimentally tested and forms the basis of Einstein's first postulate of relativity. The third law, while important, is not equivalent to conservation of momentum. Additionally, there is a suggestion to view force as a prescription for analyzing mechanical problems rather than a
  • #141
DrStupid said:
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?
 
  • Like
Likes Dale
Physics news on Phys.org
  • #142
vis_insita said:
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?
 
  • #143
vis_insita said:
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.
 
  • #144
Dale said:
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?
 
  • Like
Likes olgerm
  • #145
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.
 
  • #146
andresB said:
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.
 
  • #147
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.
 
  • #148
DrStupid said:
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?
 
  • #149
andresB said:
How do you know if the particle is non-interacting? The answer comes from N3.
N3 is silent on non-interacting bodies.

andresB said:
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.
 
Last edited:
  • #150
atyy said:
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.
 
  • Like
Likes Dale
  • #151
Dale said:
How can you apply N3 to a single non-interacting particle at all?

See #85 for an example.
 
  • #152
vis_insita said:
In reality lines and circles are defined as geometric objects with coordinate-independent properties.
Yes, and using curved coordinates to define a straight line doesn't make the straight line curved nor does it make the curved coordinate lines straight.

As I mentioned above, I don't think that Newton's original intentions are really important to modern science, but he didn't explicitly describe things in terms of coordinates or frames (both of which were developed later). So his statements could just as easily be taken as coordinate- and frame-independent geometric statements rather than frame- or coordinate-based statements. Those would then be formalized in modern geometric notation as you posted earlier. (Except for N3)
 
  • Like
Likes vanhees71
  • #153
vis_insita said:
The question I was answering is whether you have demonstrated in #85 that Newton's axioms are violated in a rotating frame. As I have shown, your conclusion becomes false if acceleration is defined as second derivative

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

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

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

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

In order to do that he need to know “precisely that acceleration”. Where does it come from? The satellite is at rest from his point of view. What does he need to do in order to know that it is actually accelerated and how?
 
  • #154
Dale said:
N3 is silent on non-interacting bodies.

No, it says that non-interacting bodies are free of forces.
 
  • #155
vis_insita said:
Not any more than the axioms of Euclidean geometry are violated in curvilinear coordinates. In geometry a line can be defined by ##\ddot x^i = 0##, but only if you are using a Cartesian or at least affine coordinates. In spherical coordinates the same condition can describe a circle. Of course, you could say everything depends on how you define "line", but that would be a very strange position to take in this regard. In reality lines and circles are defined as geometric objects with coordinate-independent properties. No one makes the difference between them a matter of definition, because that would utterly confuse every discussion of geometry.

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

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

Anyway, as you have given it, how does your preferred formulation distinguish between gravity being formulated as a force or not a force, ie. do test particles in gravity follow geodesic or non-geodesic paths?
 
  • #156
If energy and momentum are defined in all frames of references as same functions of velocities as in inertial frame of reference.
Definition is: Inertial frame of of reference is frame of reference where momentum and energy of any system(aka setup) is conserved.

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

I do not know which of these definitions of energy and momentum is more correct/standard.
 
Last edited:
  • #157
atyy said:
I don't agree with your insistence on your preferred interpretation, when both interpretations can be made physically and mathematically equivalent.

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

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

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

Loosely speaking one could say that fictitious forces are everything that can be included in the definition of ##\nabla##, while real forces can not and must appear on the right hand side of ##m\nabla_u u = F##. The reason will of course be, that those forces depend on instrinsic properties of the particle other than mass.
 
  • #158
olgerm said:
I do not know which of these definitions of energy and momentum is more correct/standard.
an someone tell me?
 
  • #159
vis_insita said:
In Newton-Cartan-Theory free falling particles follow geodesics, i.e. the same trajectories as do force-free particles according to my formulation of the 1st Axiom. Thus, the Equivalence Principle can be interpreted as "free fall = inertial motion." This means that spacetime curvature and gravity are the same thing. The only non-vanishing component of the Ricci tensor is related to mass density, via the field equation
$$R_{00} \propto \rho.$$
(Absolute space is still completely flat.) This eliminates the gravitational potential as well as its gradient, both of which are observer-dependent quantities and not uniquely definable as fields on spacetime.

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

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

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

In the Newton-Cartan system, a different definition of ##\nabla## is used, and there are no global inertial frames, and test particles follow geodesic paths in gravity.
 
  • #160
DrStupid said:
The example in #85 is based on the results of measurements using the different frames of reference. These measurements result in the coordinate acceleration and I just used them unchanged. You are right that the resulting violation of the laws of motion can be avoided using proper acceleration. But that is not the primary result of the measurement.

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

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

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

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

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

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

He only needs to know the force. From this he immediatley knows the proper acceleration from the 2nd law. Remember F=ma is valid independently of reference frames if a is interpreted as proper acceleration. I don't infer proper acceleration from coordinate acceleration alone, because I don't have to.
 
  • #161
atyy said:
I see, would it be right to say that there is freedom in the definition of ##\nabla##?

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

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

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

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

Under special circumstances, however, this is possible. If mass is confined to a bounded region, you could impose the boundary condition ##\Phi\to 0## if ##r\to\infty## on the potential to eliminate the ambiguity ##\Phi' = \Phi + \boldsymbol{b}\cdot \boldsymbol{r}## caused by the relative acceleration ##\boldsymbol{b}## of two observers. This singles out a preferred class of local inertial frames, which map onto each other by means of the usual Galilean transformations.
 
  • Informative
Likes atyy
  • #162
vis_insita said:
That's true, but in general there is no natural choice for global inertial frames, with respect to which you can define a gravitational force.

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

So we still need additional conditions to define a global inertial frame (and not just local inertial frames)? Could Newton's third law (or equivalently momentum conservation or spatial translation invariance) provide or help to provide the additional conditions?
 
  • #163
DrStupid said:
No, it says that non-interacting bodies are free of forces.
Not according to Calkin
 
  • #164
It's a bit strange to claim that non-interacting bodies are not free of forces. Are you claiming that there are forces which are not due to interactions (and here I don't consider inertial forces which are just reinterpretations of parts of the covariant time-derivative in non-inertial frames brought to the other side of the equation as forces).
 
  • #165
vanhees71 said:
It's a bit strange to claim that non-interacting bodies are not free of forces.
Nobody is claiming that. We all agree that non-interacting bodies are free of forces. The question is if it is a claim made in the third law or elsewhere.
 
  • #166
atyy said:
So we still need additional conditions to define a global inertial frame (and not just local inertial frames)? Could Newton's third law (or equivalently momentum conservation or spatial translation invariance) provide or help to provide the additional conditions?

I'm not sure how. Fundamentally the question is how to distinguish physically a curved spacetime from a flat one. As far as I understand it, the logic of Newton-Cartan-Theory is quite the same as in General Relativity: Equivalence principle -> free falling particles follow geodesics -> spacetime is curved -> there are no global inertial frames. The only step that has physical content is the first one, identifying test particle trajectories with geodesics. This must be false if there are supposed to be global inertial frames. But I don't see any relation to the Third Law or momentum conservation.
 
  • Like
Likes vanhees71
  • #167
The 3rd law is simply saying that two-body interactions are equal in magnitude and antiparallel, i.e., ##\vec{F}_{12}=-\vec{F}_{21}## and thus that (for this special case) of interactions momentum conservation holds. In standard Newtonian theory by assumption there exists a global inertial frame (and thus infinitely many related by Galilei transformations).
 
  • Like
Likes Dale
  • #168
vis_insita said:
You just have to measure the rotation of your frame of reference too. In practice this will be done by observing the apparent movement of other systems, e.g. a Foucault pendulum or a gyroscope.

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

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

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

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

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

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

Yes, but that just shifts the problem from the force to the acceleration. In my example in #85 I conclude that the forces (and therefore the accelerations) in the rotating frame are not real and you say that the accelerations (and therefore the forces) in the rotating frame are not real. That's pretty much the same.
 
  • #169
vanhees71 said:
The 3rd law is simply saying that two-body interactions are equal in magnitude and antiparallel, i.e., →F12=−→F21F→12=−F→21 and thus that (for this special case) of interactions momentum conservation holds.

Newton's original version of the 3rd law also says that every force results from such two-body interactions. The use of different versions of the laws of motion is a major problem in this thread. As we obviously can't agree about a single version, we will need to indicate what we refer to - for every single statement - if we don't want to keep on talking at cross-purposes.
 
  • #170
vis_insita said:
I'm not sure how. Fundamentally the question is how to distinguish physically a curved spacetime from a flat one. As far as I understand it, the logic of Newton-Cartan-Theory is quite the same as in General Relativity: Equivalence principle -> free falling particles follow geodesics -> spacetime is curved -> there are no global inertial frames. The only step that has physical content is the first one, identifying test particle trajectories with geodesics. This must be false if there are supposed to be global inertial frames. But I don't see any relation to the Third Law or momentum conservation.

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

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

If global conservation of momentum is absent in Newton-Cartan, then the third law or global momentum conservation could be one criterion that helps to differentiate thee Newton connection from the Newton-Cartan connection.
 
Last edited:
  • #171
DrStupid said:
Newton's original version of the 3rd law also says that every force results from such two-body interactions. The use of different versions of the laws of motion is a major problem in this thread. As we obviously can't agree about a single version, we will need to indicate what we refer to - for every single statement - if we don't want to keep on talking at cross-purposes.
May be, but that's not true (e.g., for the forces acting within nucleons, i.e., protons and neutrons in the non-relativistic approximation). Newtonian mechanics is more general than Newton's formulation of it ;-).
 
  • Like
Likes Dale
  • #172
DrStupid said:
And how do you know that these measurements show you the rotation of the frame of reference and not just forces acting on the pendulum or gyroscope? In my example in #85 I used Newton III to conclude that there are no such forces because there can't be corresponding counterforces (due to the lack of another body). That tells me that the "forces" are not real. How do you do that?

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

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

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

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

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

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

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

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

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

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

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

That's not exactly what I was saying. I was saying that ##\boldsymbol{a} = \frac{d\boldsymbol{ v}}{d t}## is the same in both frames (because it is independent of any frame). Therfore there is no violation of Newtons laws (formulated in terms of ##\boldsymbol{a}##) in any of the frames.
 
Last edited:
  • Like
Likes Dale and vanhees71
  • #173
vanhees71 said:
May be, but that's not true (e.g., for the forces acting within nucleons, i.e., protons and neutrons in the non-relativistic approximation). Newtonian mechanics is more general than Newton's formulation of it ;-).

How do the nucleons violate the third law?
 
  • Like
Likes DrStupid
  • #174
vis_insita said:
How do you know the Third Law is correct?

It's a definition.
 
  • Skeptical
Likes Dale
  • #175
atyy said:
I don't think I'm asking a "physical" question, since the two formalisms are identical (I think) in physical content. It's just that the connection is defined differently in each, such that gravity is a force in one formalism but not the other.

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

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

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

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

Similar threads

Replies
117
Views
8K
Replies
10
Views
1K
Replies
5
Views
1K
Replies
53
Views
3K
Replies
11
Views
1K
Replies
10
Views
4K
Replies
30
Views
4K
Replies
2
Views
1K
Replies
5
Views
5K
Back
Top