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

[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 \vec{F} = m\vec{A} as a vector equation, then one way to explain "inertial forces" is as follows:

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

F^j = m \frac{d^2 x^j}{dt^2}

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

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}]

where the coefficients g^j, B^j_k and C^j_{kl} 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 j component of the acceleration, which is more complicated than just the "coordinate acceleration" a^j = \frac{d^2 x^j}{dt^2}. The idea of "inertial forces" is simply to rewrite this equation of motion with just the coordinate acceleration on the right-hand side:

F_{real}^j + F_{inertial}^j = m a^j

where 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}]

F_{inertial} 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: m g^j include the "g-forces" due to an accelerating frame, m \sum_k B^j_k \frac{dx^k}{dt} include the "Coriolis forces" due to using a rotating coordinate system, and m \sum_{kl} C^j_{kl} \frac{dx^k}{dt} \frac{dx^l}{dt}] 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 \vec{F} = m\vec{A} as a vector equation, then one way to explain "inertial forces" is as follows:

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

F^j = m \frac{d^2 x^j}{dt^2}

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

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}]

where the coefficients g^j, B^j_k and C^j_{kl} 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 j component of the acceleration, which is more complicated than just the "coordinate acceleration" a^j = \frac{d^2 x^j}{dt^2}. The idea of "inertial forces" is simply to rewrite this equation of motion with just the coordinate acceleration on the right-hand side:

F_{real}^j + F_{inertial}^j = m a^j

where 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}]

F_{inertial} 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: m g^j include the "g-forces" due to an accelerating frame, m \sum_k B^j_k \frac{dx^k}{dt} include the "Coriolis forces" due to using a rotating coordinate system, and m \sum_{kl} C^j_{kl} \frac{dx^k}{dt} \frac{dx^l}{dt}] 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
m g^j

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
m g^j

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 m g^j 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 m g^j 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]

F_{inertial} 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.