Is Newton's third law of motion consistent with GTR?

[stevendaryl]

Reversing the causality does seem to be one of the major sources of confusion. That needs to be considered when switching from an inertial frame to an accelerated one. Let’s say Object A is being accelerated by surface B in an inertial frame. Call the force that surface B exerts on object A forceBA, and the “reactive” force that object A exerts on Surface B forceAB. If these two 3rd law forces are to be considered “real” forces, then they are also real when switching from an inertial frame to an accelerated one. It seems to me that arguments against this are purely semantic in nature.

In the accelerated frame, the force that object A exerts on surface B is due to its inertia, but instead of calling it an inertial force, let us call it forceAB. Surface B exerts a “reactive” force back on object A, forceBA. These two forces are Newton’s 3rd law pair in the accelerated frame, and match identically with their counterparts in the inertial frame.

You are confused about what an "inertial force" is. The force of A on B is not an inertial force. The force of B on A is not an inertial force. Those are both real forces. The inertial force is an additional force on A in the opposite direction from the force from B. If you are in an accelerating rocket, the inertial force is the apparent force pushing you against the floor of the rocket. This force has no 3rd law counterpart. There is also a real force, the floor pushing up against you. That does have a 3rd law counterpart, namely the force that you exert on the floor.

So in the case of a rocket accelerating upward: A is you, standing on the rocket floor. B is the floor of the rocket.

The forces on you (as computed in the noninertial frame of the rocket) are:

1. forceBA: The force of the floor pushing up against you.
2. gForce: The inertial force that apparently pushes you against the floor.

The first force is a real force, and has a 3rd law counterpart, namely forceAB, the force of you pushing downward on the floor. The second force is "fictitious" because it has no 3rd law counterpart. Think about it: Since gForce is exerted on you, and is directed downward, the 3rd law counterpart, if there were one, would be a force exerted BY you in an UPWARD direction. There is no such force. You are not exerting an upward force on anything.

 

[ShayanJ]

When you look at gravity as a local theory, yes, you can do this. If you want to make it a global theory a la classical Newtonian gravity, this is no longer possible. You can make the gravitational field zero at a given point, but not globally, and globally there is a 3rd law pair to each gravitational force.

Yes, you're right; I forget the important word "local".

Its not about the word local, because also in GR you can only do that locally. The point is that with the interpretation of gravity as a force, you can't say in one frame masses attract each other and in another frame they don't!

 

[A.T.]

As far as I know Newton Cartan makes all of the same experimental predictions as standard Newtonian physics. So I would consider it to be a reformulation of Newtonian physics, not a new theory (despite the treatment of gravity as a fictitious force).

Yes, when I say that Newtonian gravity is a real/interaction force, I mean Newton's formulation.

 

[A.T.]

Reversing the causality does seem to be one of the major sources of confusion.

Your confusion comes from introducing a physically irrelevant notion of causality.

and the “reactive” force that object A exerts on Surface B forceAB... Surface B exerts a “reactive” force back on object A

The labels "active" and "reactive" for 3rd Law forces are arbitrary and physically irrelevant.

the force that object A exerts on surface B is due to its inertia

That "due to its inertia" bit is another arbitrary causation assignment. A and B repel each other via EM forces, because their atoms are too close to each other. That's all there is to say about the "cause" of both these forces, if you feel the need to say anything at all about it.

 

[haushofer]

Its not about the word local, because also in GR you can only do that locally. The point is that with the interpretation of gravity as a force, you can't say in one frame masses attract each other and in another frame they don't!

Mmmm, I have to think about this. I don't see the problem here; the equation F=ma is only a tensor equation for Galilean transformations. So comparing with my earlier example of the rotating disc, I'd say that the phrase "masses attract each-other" has the same status as "the engine makes the disc rotate". So what is the crucial difference, according to you?

 

[ShayanJ]

Mmmm, I have to think about this. I don't see the problem here; the equation F=ma is only a tensor equation for Galilean transformations. So comparing with my earlier example of the rotating disc, I'd say that the phrase "masses attract each-other" has the same status as "the engine makes the disc rotate". So what is the crucial difference, according to you?

I wasn't referring to that post. I agree with you there. What I said was directed towards the posts I quoted.
My point is that GR and Newtonian gravity, in the point that you can get rid of gravity only locally, are like each other, as is obvious from the fact that you can have something like Newton-Cartan theory. So there is nothing physical in Newtonian gravity that says gravity has to be a real force! Its only that people thought it is.

 

[MikeGomez]

The force of A on B is not an inertial force. The force of B on A is not an inertial force. Those are both real forces.

Correct.

The inertial force is an additional force on A in the opposite direction from the force from B.

The situation is already described completely by using real forces. No “additional” forces are needed. How is that useful?

If you are in an accelerating rocket, the inertial force is the apparent force pushing you against the floor of the rocket.

The force pushing you against the floor can either be described as Newton’s 3rd law pair (the so called “reaction”) force in response to the floor pushing up against you, or as your own inertia as per Newton’s 1st and the EP.

This force has no 3rd law counterpart.

It is not correctly established as a force of any kind in the first place, so a better way to say it would be that it doesn't exist at all, and looking for a 3rd law pair is pointless.

There is also a real force, the floor pushing up against you. That does have a 3rd law counterpart, namely the force that you exert on the floor.

So in the case of a rocket accelerating upward: A is you, standing on the rocket floor. B is the floor of the rocket.

The forces on you (as computed in the noninertial frame of the rocket) are:
1. forceBA: The force of the floor pushing up against you.

Correct.

2. gForce: The inertial force that apparently pushes you against the floor.

“Apparently” yes, but actually, no. Einstein said that all motion is to be considered relative. Newton said that every force has an equal and oppositely directed reaction force. Instead of trying to reformulate Newtonian mechanics or relativity, take a closer look at what you are doing. You have a simple sign flip problem.

Think about it: Since gForce is exerted on you, and is directed downward, the 3rd law counterpart, if there were one, would be a force exerted BY you in an UPWARD direction.

An excellent indication that your force is pointing in the wrong direction.

 

[MikeGomez]

The labels "active" and "reactive" for 3rd Law forces are arbitrary and physically irrelevant.

The labels are arbitrary, and that is why I put quotes around them. It wasn't meant to be as significant as that.

On the other hand, the fact that opposing force pairs (whatever we call them) always exist as per Newton’s 3rd is far from physically irrelevant.

That "due to its inertia" bit is another arbitrary causation assignment. A and B repel each other via EM forces, because their atoms are too close to each other. That's all there is to say about the "cause" of both these forces, if you feel the need to say anything at all about it.

I didn’t bring it up. Haushofer did in post#34.

What was meant by “causality” by Haushofer is the problem of identifying the source and direction of the force, and contrary to being irrelevant, it was spot on.

 

[Orodruin]

The force pushing you against the floor can either be described as Newton’s 3rd law pair (the so called “reaction”) force in response to the floor pushing up against you, or as your own inertia as per Newton’s 1st and the EP.

This is just wrong. The 3rd law pair of the floor force on you is your force on the floor.

 

[MikeGomez]

This is just wrong. The 3rd law pair of the floor force on you is your force on the floor.

Sorry if the phrasing wasn’t clear. The forceAB that object 'A' exerts on the floor can be viewed either as a so called “reaction” force to the forceBA that the floor 'B' exerts on the object in the inertial frame, or as the object’s inertia as per Newton’s 1st in the accelerated frame.

 

[A.T.]

The situation is already described completely by using real forces. No “additional” forces are needed. How is that useful?

Inertial forces are introduced to make Newtons 2nd Law applicable to non-inertial frames.

The force pushing you against the floor can either be described as Newton’s 3rd law pair (the so called “reaction”) force in response to the floor pushing up against you,

No. A force "pushing you" is never a 3rd law partner of another force "pushing up against you". The two forces in 3rd law always act on two different bodies.

 

[A.T.]

Sorry if the phrasing wasn’t clear. The forceAB that object 'A' exerts on the floor can be viewed either as a so called “reaction” force to the forceBA that the floor 'B' exerts on the object in the inertial frame,

Yes.

or as the object’s inertia as per Newton’s 1st in the accelerated frame.

No. forceAB & forceBA are 3rd law partners in every frame. In the accelerated frame forceBA is still there, and is still the 3rd law partner of forceAB.

 

[MikeGomez]

No. forceAB & forceBA are 3rd law partners in every frame. In the accelerated frame forceBA is still there, and is still the 3rd law partner of forceAB.

Phrasing is corrected in post #45.

No. forceAB & forceBA are 3rd law partners in every frame. In the accelerated frame forceBA is still there, and is still the 3rd law partner of forceAB.

Why do you say “no”? I didn’t say differently.

Inertial forces are introduced to make Newtons 2nd Law applicable to non-inertial frames.

Herein lies the problem. In the accelerated frame, the application of Newton’s 2nd would correctly be calculated from the inertia (in the sense of Newton’s 1st) of object ‘A’.

 

[stevendaryl]

The situation is already described completely by using real forces. No “additional” forces are needed. How is that useful?

I think you're confused about what is meant by "inertial forces".

If you're standing on the floor of an accelerating rocket, you have an upward force pressing against your feet. But your (coordinate) acceleration is zero. (You aren't moving, relative to the rocket). So if by "acceleration" you mean "coordinate acceleration", then [itex]F = ma[/itex] does not work in a noninertial coordinate system.

There are two approaches to recovering [/itex]F = ma[/itex]: One is to introduce fictitious, or inertial, forces. Then your situation would be described as follows:

1. The floor exerts an upward force on your feet.
2. There is a downward inertial force on your whole body.
3. The net force is zero. (Which explains why your acceleration is zero).

The second approach, which is better, in my opinion, since it generalizes to General Relativity, is to distinguish between "coordinate acceleration" and "proper acceleration". Then you can say that proper acceleration is always due to real forces (not inertial forces), and real forces always have a 3rd law counterpart (well, actually, if the force is due to a field, then the 3rd law must be generalized: if the field is imparting momentum to you, then you are imparting momentum to the field, as well).

 

[stevendaryl]

The force pushing you against the floor can either be described as Newton’s 3rd law pair (the so called “reaction”) force in response to the floor pushing up against you...

No, it cannot! The floor is pushing up on YOU. The 3rd law counterpart is a force pushing DOWN on the FLOOR. It's not the force pushing down on YOU.

In the noninertial coordinate system of the rocket, there are a number of forces involved:

1. The floor pushing up on you.
2. The (apparent) g-force pushing down on you.
3. Your feet pushing down on the floor.

The 3rd law counterpart to Force 1 is Force 3. Force 2 has no 3rd law counterpart. You can't say that Force 2 is the 3rd law counterpart to Force 1 because they are both forces on YOU. The 3rd law counterpart to Force 1 is Force 3, a force on the FLOOR, not a force on you.

Think about the case with Newtonian gravity. You're standing on a floor. The floor is pushing up on you. The Earth is pulling down on you. Then we have the following forces:

1. The floor pushing up on you.
2. The Earth's gravity pulling down on you.
3. Your feet pushing down on the floor.
4. Your gravity pulling up on the Earth.

In this case, it is obvious that the 3rd law counterpart to Force 2 is Force 4, not Force 3. It's not a force acting on the floor, but a force acting on the Earth. But in the case of an accelerating rocket, there is no Force 4.

 

[stevendaryl]

Sorry if the phrasing wasn’t clear. The forceAB that object 'A' exerts on the floor can be viewed either as a so called “reaction” force to the forceBA that the floor 'B' exerts on the object in the inertial frame, or as the object’s inertia as per Newton’s 1st in the accelerated frame.

It's not a matter of phrasing. It's a matter of your being confused. There are two forces on object A: an apparent (fictitious, inertial) downward force, and a real upward force (forceBA as you call it). The upward force has a 3rd law counterpart (forceAB). The downward force does not.

 

[A.T.]

Why do you say “no”?

I quoted the part of your post, which is wrong. You conflate the contact force forceAB with the "objects inertia" in the accelerated frame.

 

[MikeGomez]

In the inertial frame, accelerating the object changes it's momentum.
In the accelerated frame, we cancel out only the object's velocity, not it's change in momentum.
If we don't take this into account, then a mysterious non-3rd law force appears.

Is that correct?

 

[A.T.]

In the inertial frame, accelerating the object changes it's momentum.

Coordinate acceleration changes momentum in every frame (if mass is constant).

 

[stevendaryl]

In the inertial frame, accelerating the object changes it's momentum.
In the accelerated frame, we cancel out only the object's velocity, not it's change in momentum.
If we don't take this into account, then a mysterious non-3rd law force appears.

Is that correct?

That might be a way to say it. But it's actually clearer when you look at the mathematics. If you view [itex]\vec{F} = m\vec{A}[/itex] as a vector equation, then one way to explain "inertial forces" is as follows:

If we are using inertial, Cartesian coordinates [itex]x^j[/itex], then this equation looks like:

[itex]F^j = m \frac{d^2 x^j}{dt^2}[/itex]

If you switch to noninertial, curvilinear coordinates, the form of the vector [itex]\vec{A}[/itex] becomes more complicated:

[itex]F^j = m [\frac{d^2 x^j}{dt^2} + g^j + \sum_k B^j_k \frac{dx^k}{dt} + \sum_{kl} C^j_{kl} \frac{dx^k}{dt} \frac{dx^l}{dt}][/itex]

where the coefficients [itex]g^j[/itex], [itex]B^j_k[/itex] and [itex]C^j_{kl}[/itex] depend on the coordinate system (and in general, these coefficients are not constants).

In this equation, the left-hand side is the net real force on the object, and it has a 3rd law counterpart. The right-hand side is just the [itex]j[/itex] component of the acceleration, which is more complicated than just the "coordinate acceleration" [itex]a^j = \frac{d^2 x^j}{dt^2}[/itex]. The idea of "inertial forces" is simply to rewrite this equation of motion with just the coordinate acceleration on the right-hand side:

[itex]F_{real}^j + F_{inertial}^j = m a^j[/itex]

where [itex]F_{inertial} = - m [ g^j + \sum_k B^j_k \frac{dx^k}{dt} + \sum_{kl} C^j_{kl} \frac{dx^k}{dt} \frac{dx^l}{dt}][/itex]

[itex]F_{inertial}[/itex] is not actually a force, but is just the extra terms in the acceleration that arise from using noninertial, curvilinear coordinates.The extra terms are given special names in special coordinate systems: [itex]m g^j[/itex] include the "g-forces" due to an accelerating frame, [itex]m \sum_k B^j_k \frac{dx^k}{dt} [/itex] include the "Coriolis forces" due to using a rotating coordinate system, and [itex]m \sum_{kl} C^j_{kl} \frac{dx^k}{dt} \frac{dx^l}{dt}][/itex] include the "Centrifugal forces" due to using polar coordinates. These "inertial forces" do not have a 3rd law counterpart.

 

[MikeGomez]

That might be a way to say it. But it's actually clearer when you look at the mathematics. If you view [itex]\vec{F} = m\vec{A}[/itex] as a vector equation, then one way to explain "inertial forces" is as follows:

If we are using inertial, Cartesian coordinates [itex]x^j[/itex], then this equation looks like:

[itex]F^j = m \frac{d^2 x^j}{dt^2}[/itex]

If you switch to noninertial, curvilinear coordinates, the form of the vector [itex]\vec{A}[/itex] becomes more complicated:

[itex]F^j = m [\frac{d^2 x^j}{dt^2} + g^j + \sum_k B^j_k \frac{dx^k}{dt} + \sum_{kl} C^j_{kl} \frac{dx^k}{dt} \frac{dx^l}{dt}][/itex]

where the coefficients [itex]g^j[/itex], [itex]B^j_k[/itex] and [itex]C^j_{kl}[/itex] depend on the coordinate system (and in general, these coefficients are not constants).

In this equation, the left-hand side is the net real force on the object, and it has a 3rd law counterpart. The right-hand side is just the [itex]j[/itex] component of the acceleration, which is more complicated than just the "coordinate acceleration" [itex]a^j = \frac{d^2 x^j}{dt^2}[/itex]. The idea of "inertial forces" is simply to rewrite this equation of motion with just the coordinate acceleration on the right-hand side:

[itex]F_{real}^j + F_{inertial}^j = m a^j[/itex]

where [itex]F_{inertial} = - m [ g^j + \sum_k B^j_k \frac{dx^k}{dt} + \sum_{kl} C^j_{kl} \frac{dx^k}{dt} \frac{dx^l}{dt}][/itex]

[itex]F_{inertial}[/itex] is not actually a force, but is just the extra terms in the acceleration that arise from using noninertial, curvilinear coordinates.The extra terms are given special names in special coordinate systems: [itex]m g^j[/itex] include the "g-forces" due to an accelerating frame, [itex]m \sum_k B^j_k \frac{dx^k}{dt} [/itex] include the "Coriolis forces" due to using a rotating coordinate system, and [itex]m \sum_{kl} C^j_{kl} \frac{dx^k}{dt} \frac{dx^l}{dt}][/itex] include the "Centrifugal forces" due to using polar coordinates. These "inertial forces" do not have a 3rd law counterpart.

Thanks A.T. and stevenDaryl.

Sorry for the dumb question.
What is the difference in magnitude and direction of
[itex]m g^j[/itex]

and

3.Your feet pushing down on the floor.

 

[stevendaryl]

Thanks A.T. and stevenDaryl.

Sorry for the dumb question.
What is the difference in magnitude and direction of
[itex]m g^j[/itex]

and

3.Your feet pushing down on the floor.

If you are stationary in the accelerated coordinate system, then the two will be equal in magnitude and direction. If the floor is spongy (so that you sink into the floor a little ways--and every floor is spongy to a certain extent), then the two will be slightly different. In the extreme case in which there is no floor, then of course [itex]m g^j[/itex] will be nonzero, but your force on the floor will be zero. So they aren't the same on all cases.

 

[A.T.]

If you are stationary in the accelerated coordinate system, then the two will be equal in magnitude and direction. If the floor is spongy (so that you sink into the floor a little ways--and every floor is spongy to a certain extent), then the two will be slightly different. In the extreme case in which there is no floor, then of course [itex]m g^j[/itex] will be nonzero, but your force on the floor will be zero. So they aren't the same on all cases.

I would add, that just because two forces have the same magnitude and direction, it doesn't make them the same force. In this case they don't even act on the same object or at the same point of application.

 

[MikeGomez]

[itex]F_{inertial}[/itex] is not actually a force, but is just the extra terms in the acceleration that arise from using noninertial, curvilinear coordinates.

So the answer the OP is yes, GTR is fully consistent with Newton’s 3rd. Real forces in GTR have 3rd law counter parts. Inertial forces in GTR aren’t real forces, and aren’t required to have 3rd law counter parts. The name ‘inertial forces’ can mislead us into thinking that they required 3rd law counter parts, but that’s terminology, not a physical equality to real forces. They might have been called something else to begin with, like “inertial terms” or something.

 

[stevendaryl]

So the answer the OP is yes, GTR is fully consistent with Newton’s 3rd. Real forces in GTR have 3rd law counter parts. Inertial forces in GTR aren’t real forces, and aren’t required to have 3rd law counter parts. The name ‘inertial forces’ can mislead us into thinking that they required 3rd law counter parts, but that’s terminology, not a physical equality to real forces. They might have been called something else to begin with, like “inertial terms” or something.

Yes, that's the way I like to think of it. Newton's third law works fine in noninertial frames, as long as you don't consider inertial forces to be forces.