Understanding Relativity: A Blind Man's Perspective on Time and Physics

[JesseM]

As far as Newtonian physics goes, won't the bus driver happily say he himself is accelerating and the man is standing still?

This fixation on claiming oneself to be at rest is more of an Einsteinian thing, isn't it? (By the way, you don't have to nominate yourself to be at rest, you just nominate a rest frame and do your calculations from there. I think.)

I've told you many times before that this is a misconception of yours--when solving problems in relativity you don't pick one frame to be "the rest frame", the phrase rest frame is only used in connection with particular objects, like "the rest frame of the rocket". You are free to do calculations in multiple frames, including ones where none of the objects you are analyzing are at rest; there is no "fixation" on claiming oneself to be at rest, although it is traditional in both Newtonian physics and relativity to use "my frame" as shorthand for "the frame where I am at rest".

 

[YellowTaxi]

I've told you many times before that this is a misconception of yours--when solving problems in relativity you don't pick one frame to be "the rest frame", the phrase rest frame is only used in connection with particular objects, like "the rest frame of the rocket". You are free to do calculations in multiple frames, including ones where none of the objects you are analyzing are at rest; there is no "fixation" on claiming oneself to be at rest, although it is traditional in both Newtonian physics and relativity to use "my frame" as shorthand for "the frame where I am at rest".

Well, I thought Einstein was interested in looking at things from the point of view of objects in free-fall. i.e. from a 'place' where the gravity field was undetectable and thus one which could easily be considered in effect a genuine inertial frame [a pure rest frame].
ie A specific frame in which the (possibly varying) acceleration through space is 'invisible' so long as you don't look out the window.

 

[Dale]

I can't follow you. Why should the inertial forces violate the third law?

What are you talking about? Newton’s laws are perfectly capable of taking care of all these examples.
At least for your example you have shown no violation or even a modification.

I echo the question that some others asked: why is the invocation of a fictitious force automatically a violation of any of Newton's 3 laws?

I would say that the same is true in the rocket frame. The spaceman observes some constant force acting on him and his collection of tennis balls, and uses Newton's law to predict - correctly - how they move when he tosses them.

IOW, Newton's laws seem to me to be independent of the nature or source of the forces; they simply say how massive bodies react in their presence.

OK, I see that I am going to have to justify my statements a little better. I apologize for thinking my points were obvious. I was obviously not writing as clearly as I thought I was.

The question I am addressing is: "How are Newton's laws violated/modified in an accelerated reference frame?" I will address only v<<c so that we don't need to worry about relativistic effects and we will stick with the rocket in deep-space scenario so that we don't need to worry about gravity, friction, air resistance or any other confounding factors.

First, let's start in the inertial frame where the rocket is initially at rest at the origin with the engines off. This will be the unprimed frame and all unprimed coordinates will reference this frame.

At time t=0 the rocket turns on generating a steady 2000 N of thrust in the x direction. The rocket masses 1000 kg. The thrust is the only force acting on the rocket so, by Newton's 2nd law, the acceleration is 2 m/s². In SI units, the equations of motion for the rocket are therefore:
r(t)=(t²,0,0)
vr(t)=(2t,0,0)
ar(t)=(2,0,0)
for t>0.

At t=1 a ball is released in the cockpit. After release there is no force acting on the ball, so by Newton's 1st law, the acceleration is 0. In SI units, the equations of motion for the ball are therefore:
b(t)=(2t+1,0,0)
vb(t)=(2,0,0)
ab(t)=(0,0,0)
for t>1.

So far we have considered Newton's 2nd and 1st law, but not the 3rd law. The only force considered thus far is the thrust. There is a thrust force which is acting on the rocket exhaust which is equal-and-opposite to the thrust force acting on the rocket. This force accelerates the exhaust and satisfies Newton's 3rd law.

Now we will consider the situation in the primed reference frame in which the rocket is at rest at the origin for t>0, i.e. this frame is accelerating at a rate of 2 m/s² in the positive x direction wrt the inertial frame. The transformation equations are:
t'=t
x'=x-t²
y'=y
z'=z

By transforming we obtain the equations of motion for the rocket in the primed frame:
r'(t')=(0,0,0)
vr'(t')=(0,0,0)
ar'(t')=(0,0,0)
t'>0.
Since there is a non-zero net force acting on the rocket and since the acceleration is zero f≠ma, a clear violation of Newton's 1st and 2nd laws.

Similarly, by transforming we obtain the equations of motion for the ball in the primed frame are:
b'(t')=(-t²+2t+1,0,0)
vb'(t')=(-2t+2,0,0)
ab'(t')=(-2,0,0)
t'>1.
Since there is no force acting on the ball and since the acceleration is non-zero f≠ma, a clear violation of Newton's 1st and 2nd laws.

However, we can easily "fix" the first and second laws simply by introducing an inertial force (-2M,0,0) where M is the mass of the body on which the force is acting. So, for the rocket, this inertial force exactly balances the thrust such that the net force is zero which explains the acceleration of zero. For the ball, this inertial force is the only force acting on it so there is a non-zero net force which explains the non-zero acceleration. Thus, with this modification Newton's 1st and 2nd laws are now satisfied.

Now, let's consider Newton's 3rd law in the accelerated reference frame. Again, the thrust force on the rocket is equal and opposite to the thrust force on the exhaust, forming a 3rd law pair. That leaves only the inertial forces. However, all of the inertial forces act in the same direction, so they do not form 3rd law pairs. The inertial forces violate the 3rd law because there are no equal and opposite inertial forces.

I hope this clearly explains my above statements, because I really would have a hard time being more explicit or concrete.

 

[JesseM]

I've told you many times before that this is a misconception of yours--when solving problems in relativity you don't pick one frame to be "the rest frame", the phrase rest frame is only used in connection with particular objects, like "the rest frame of the rocket". You are free to do calculations in multiple frames, including ones where none of the objects you are analyzing are at rest; there is no "fixation" on claiming oneself to be at rest, although it is traditional in both Newtonian physics and relativity to use "my frame" as shorthand for "the frame where I am at rest".

Well, I thought Einstein was interested in looking at things from the point of view of objects in free-fall. i.e. from a 'place' where the gravity field was undetectable and thus one which could easily be considered in effect a genuine inertial frame [a pure rest frame].
ie A specific frame in which the (possibly varying) acceleration through space is 'invisible' so long as you don't look out the window.

I was talking more about special relativity in the above comments, where one can analyze a situation from any inertial frame using the same equations for the laws of physics, and there is no convention that one of these inertial frames is selected as "the" rest frame as neopolitan seems to imagine. In general relativity it's true that an observer in free-fall will "locally" observe the laws of physics to work just like they do in an inertial frame from special relativity, where locally means that the observer can only look at the results of experiments in a very small region of space for a very small window of time, small enough so the effects of spacetime curvature are negligible. In GR you wouldn't say that this observer is "accelerating" in a local sense, in fact if a freefalling observer falls past an observer at a constant height above the ground--sitting on a platform, say--then in a local sense it is the observer on the platform who is accelerating, not the frefalling observer.

 

[YellowTaxi]

I was talking more about special relativity in the above comments, where one can analyze a situation from any inertial frame using the same equations for the laws of physics, and there is no convention that one of these inertial frames is selected as "the" rest frame as neopolitan seems to imagine. In general relativity it's true that an observer in free-fall will "locally" observe the laws of physics to work just like they do in an inertial frame from special relativity, where locally means that the observer can only look at the results of experiments in a very small region of space for a very small window of time, small enough so the effects of spacetime curvature are negligible. In GR you wouldn't say that this observer is "accelerating" in a local sense, in fact if a freefalling observer falls past an observer at a constant height above the ground--sitting on a platform, say--then in a local sense it is the observer on the platform who is accelerating, not the frefalling observer.

yep that's obvious.
but the guy that's actually falling will get hurt the most when the ground hits him

on a more serious note
But I didn't understand why you said 'only for a small region of time' for the falling guy.
because: If the g field varies (over time - or whatever), actually it doesn't matter to the falling guy. For him g is always 'invisible' whatever its actual value my seem to the guy on the platform.

 

[harryjoon]

What everyone seems to forget is that relativistic effects can only be observed across two reference frames. Lorentz transform equations relate measurements of two observers with a non-zero relative velocity. In order to maintain the universal validity of the laws of physics we need Lorentz transform between two inertial frames, and generally covariant transformation laws for non-inertial frames. This is indeed the very essence of the Special and the General relativity theories.

 

[belliott4488]

DaleSpam:

Thanks for taking the time to type all that up.

I'm afraid I still don't get it though. Why is it any more difficult to identify the "equal and opposite pairs" in this case than it is in the simple case of constant gravity = m*9.8 m/s^2 that we all worked with when we first learn Newton's laws? We were told that if a rock rests on a table, its weight exerts a force = mg on the table, and the table (thanks to molecular deformation, we later learned) exerts the same magnitude force upward. The same would happen in the rocket ship, right? The hardest one to grasp when we were kids was the equal and opposite force when we allowed that rock to fall toward Earth - but we accepted that the Earth moves upward ever so slightly - even though we certainly never saw any evidence to that effect.

[EDIT: I originally typed "as long as the spaceman holds the rocket" in the next paragraph ... oops. sorry if I confused anyone.]

Is it any harder to believe that the spaceship moves "up" (or "forward") when the rock is released in it? After all, as long as the spaceman holds the rock, the thrusters are accelerating the total mass of spaceship + spaceman + rock. When he releases the rock, the thrusters suddenly are accelerating only spaceship + spaceman (until the rock strikes the floor), so the spaceship will accelerate slightly faster forward.

Isn't this all the same thing? Mustn't it be, if we are to take the Equivalence Principle seriously? Newton derived his laws in the (approximately) constant gravitational field of the Earth before he ever put together the concept of gravity as the force the holds the planets in place as well as pull rocks to Earth. Can't our spaceman do exactly the same thing?

 

[JesseM]

But I didn't understand why you said 'only for a small region of time' for the falling guy.
because: If the g field varies (over time - or whatever), actually it doesn't matter to the falling guy. For him g is always 'invisible' whatever its actual value my seem to the guy on the platform.

I asked the same question on this thread a while ago, see dicerandom's answer in post #9--even in a very small room, if you had two test particles on opposite sides of the room they would begin to drift together over an extended period of time due to tidal forces, just like what happens if you have a large room and you observe what happens to test particles on opposite sides for a brief period of time--see the bottom illustration in this article.

 

[harryjoon]

There are a number of points that need clarifying within the above arguments. Einstein said in non-inertial frames Newton's law do not hold good - this is not the same as the statement "Newtons' laws are violated." They don't hold good because we cannot adhere to the classical definition of space-time in which we can measure a unit of length and time regardless of our position. In non-inertial frame this is no longer valid, different position in space-time may have different metrics, which cause a straight line to appear as a curved line etc. The Euclidean geometry does not hold and must be replaced with a non-Euclidean geometry. The notion of straight line which is used to define Newton's laws no longer has the same meaning.
As to your question, if I understand it correctly, no it is not any different. In fact it is the very point that Einstein used to explain his equivalence principle. That a man in an elevator or a rocket will be unable to distinguish his system from a gravitational system since he will observe an object at rest (in another frame) to be falling with an acceleration which is independent of the mass of the object. Since this peculiar behavior belongs to gravitational systems, hence he concluded that a gravitational system is no different to an accelerated system. A fact further supported by experiments of Eotvos who verified the equivalence of the gravitational mass with the inertial mass.

 

[belliott4488]

Thanks, harryjoon. One question in response: the point about Newton's laws not holding in a non-inertial frame is due to relativistic effects, is it not? In the limit of small velocities (and small accelerations or small gravitational fields), are Newton's laws not as good an approximation as they are in inertial frames?

 

[JesseM]

DaleSpam:

Thanks for taking the time to type all that up.

I'm afraid I still don't get it though. Why is it any more difficult to identify the "equal and opposite pairs" in this case than it is in the simple constant gravity = m*9.8 m/s^2 that we all worked with when we first learn Newton's laws? We were told that if a rock rests on a table, its weight exerts a force = mg on the table, and the table (thanks to molecular deformation, we later learned) exerts the same magnitude force upward. The same would happen in the rocket ship, right? The hardest one to grasp when we were kids was the equal and opposite force when we allowed that rock to fall toward Earth - but we accepted that the Earth moves upward ever so slightly - even though we certainly never saw any evidence to that effect.

Is it any harder to believe that the spaceship moves "up" (or "forward") when the rock is released in it? After all, as long as the spaceman holds the rocket, the thrusters are accelerating the total mass of spaceship + spaceman + rock. When he releases the rock, the thrusters suddenly are accelerating only spaceship + spaceman (until the rock strikes the floor), so the spaceship will accelerate slightly faster forward.

Isn't this all the same thing? Mustn't it be, if we are to take the Equivalence Principle seriously? Newton derived his laws in the (approximately) constant gravitational field of the Earth before he ever put together the concept of gravity as the force the holds the planets in place as well as pull rocks to Earth. Can't our spaceman do exactly the same thing?

Even if you can make correct predictions in a Newtonian accelerating frame by introducing "fictitious forces", doesn't the very fact that you have to include forces not present in inertial frames mean that the laws of physics have a different form in this frame? For example, in an inertial frame the forces on a test particle depend only on its distance from other objects which exert gravitational forces on it, but this is no longer true in a non-inertial frame with 'fictitious forces'. If you wrote down the equations of motion for a given system of particles, the equations couldn't have the same form in non-inertial frames as they do in inertial ones.

 

[Dale]

The hardest one to grasp when we were kids was the equal and opposite force when we allowed that rock to fall toward Earth - but we accepted that the Earth moves upward ever so slightly - even though we certainly never saw any evidence to that effect.

Is it any harder to believe that the spaceship moves "up" (or "forward") when the rock is released in it? After all, as long as the spaceman holds the rocket, the thrusters are accelerating the total mass of spaceship + spaceman + rock. When he releases the rock, the thrusters suddenly are accelerating only spaceship + spaceman (until the rock strikes the floor), so the spaceship will accelerate slightly faster forward.

Isn't this all the same thing? Mustn't it be, if we are to take the Equivalence Principle seriously? Newton derived his laws in the (approximately) constant gravitational field of the Earth before he ever put together the concept of gravity as the force the holds the planets in place as well as pull rocks to Earth. Can't our spaceman do exactly the same thing?

You are thinking about effects that are many orders of magnitude smaller than what we are talking about here. The inertial forces are relatively large forces (2000 N on the rocket and .2 N on a .1 kg ball), the gravitational force between them would be about 6.7E-9 N or even less because of the geometry. However, the important point is that all of the inertial forces point in the same direction! Because they are all in the same direction they cannot possibly form equal-and-opposite pairs that satisfy the 3rd law.

I really don't know what more I can do here. I have given a very clear and concrete example fully worked out. I have shown my reasoning step-by-step and backed up my claims with explicit demonstrations. Please do the math yourself. Work a few of these problems and see that the inertial forces do not follow Newton's 3rd law. If you cannot see that from what I have already presented then I am sure that further write-ups on my part will not be helpful. There is really no substitute for getting down into the details and going through the work yourself.

PS Please don't take this wrong. There is nothing wrong with needing to work a few examples to understand a concept, particularly in this case. Most people go through all of their college physics courses without ever working a problem in a non-inertial reference frame.

 

[belliott4488]

Even if you can make correct predictions in a Newtonian accelerating frame by introducing "fictitious forces", doesn't the very fact that you have to include forces not present in inertial frames mean that the laws of physics have a different form in this frame? For example, in an inertial frame the forces on a test particle depend only on its distance from other objects which exert gravitational forces on it, but this is no longer true in a non-inertial frame with 'fictitious forces'. If you wrote down the equations of motion for a given system of particles, the equations couldn't have the same form in non-inertial frames as they do in inertial ones.

Right, but that means Newton's laws are also "violated" on the surface of the Earth when we make the approximation that g is a constant, independent of height. I know that's only an approximation, but I wouldn't have called that a "violation" of Newton's laws; I would have said that it's an application of those laws that is as correct as the initial assumption, i.e. of a constant force.

My main point was really that I don't see how Newton's laws of motion say anything about the source or the nature of the forces. They simply say, "you give me a force, and I'll tell you how object will react to it." As far as that goes, they work equally well on the Earth's surface (which really is non-inertial, after all, not that anyone claims Newton's laws don't apply there) and in the accelerating rocketship/elevator.

 

[JesseM]

Right, but that means Newton's laws are also "violated" on the surface of the Earth when we make the approximation that g is a constant, independent of height. I know that's only an approximation, but I wouldn't have called that a "violation" of Newton's laws; I would have said that it's an application of those laws that is as correct as the initial assumption, i.e. of a constant force.

My main point was really that I don't see how Newton's laws of motion say anything about the source or the nature of the forces. They simply say, "you give me a force, and I'll tell you how object will react to it." As far as that goes, they work equally well on the Earth's surface (which really is non-inertial, after all, not that anyone claims Newton's laws don't apply there) and in the accelerating rocketship/elevator.

But Newton's laws of motion aren't the whole of what is meant by Newtonian physics--the three laws of motion are insufficient in themselves to calculate the dynamical behavior of objects given their initial conditions. You also need some set of force laws, such as the equation of Newtonian gravitation. For Galilei-invariant force laws like Newtonian gravity, the law will obey the same equations in every inertial frame, but the same equations will not give correct predictions in a non-inertial frame.

 

[belliott4488]

But Newton's laws of motion aren't the whole of what is meant by Newtonian physics--the three laws of motion are insufficient in themselves to calculate the dynamical behavior of objects given their initial conditions. You also need some set of force laws, such as the equation of Newtonian gravitation. For Galilei-invariant force laws like Newtonian gravity, the law will obey the same equations in every inertial frame, but the same equations will not give correct predictions in a non-inertial frame.

Oh, absolutely ... and to describe electromagnetic forces fully, you should have Maxwell's equations, too. I think the post that I initially responded to referred only to the laws of motion, however. Of course I agree that Newton's Law of gravity won't do you much good in an accelerating rocketship far from any gravitational field.

I think this is all kind of tangential to the original post. It asked about why Newton's laws are said to be violated in accelerating frames. If that referred to the law of gravity, then fine - there's a simple enough answer. If it refers to the laws of motion, however, then I think it's misleading to say that these laws are "violated." Yes, additional conditions, in the form of inertial forces, must be added, but then the laws still work.

 

[Dale]

If it refers to the laws of motion, however, then I think it's misleading to say that these laws are "violated." Yes, additional conditions, in the form of inertial forces, must be added, but then the laws still work.

No they dont!

 

[belliott4488]

No they dont!

Again I ask you, why not? You said earlier that the 3rd law doesn't work, but I responded (https://www.physicsforums.com/showpost.php?p=1671401&postcount=37") by pointing out that in fact it does, much as it does for the case of a constant gravitational field, to which this case is equivalent.

Do you in fact believe that Newton's 3rd law is violated in the approximation of a constant gravitational field that is typically invoked for motion on the Earth's surface? If so, then at least you're consistent, although rather unconventional, since that approximation is often used in the presentation of Newton's Laws for the first time in introductory Physics.

[Edit: I'm adding more, while I wait to see a response ...]
One more thing: Suppose I'm riding in my rocket ship/space elevator, and I'm doing all kinds of Physics - I'm shooting pool, I'm tossing balls through the air (it's a big elevator), I'm playing with pendula, I'm spinning gyroscopes - at what point will I find myself unable to explain or predict what I'm seeing simply by applying Newton's laws and the assumption of a constant force field, which I'll call, for lack of a better term, "gravity"? If you tell me the laws of motion break down, then I'll be able to tell that I'm in an accelerating frame and not in a constant gravitational field, and the Principle of Equivalence just got shot to Hell.

 

[Dale]

Again I ask you, why not? You said earlier that the 3rd law doesn't work, but I responded (https://www.physicsforums.com/showpost.php?p=1671401&postcount=37") by pointing out that in fact it does, much as it does for the case of a constant gravitational field, to which this case is equivalent.

Show your work and justify your claim that it is equivalent.

Do you in fact believe that Newton's 3rd law is violated in the approximation of a constant gravitational field that is typically invoked for motion on the Earth's surface? If so, then at least you're consistent, although rather unconventional, since that approximation is often used in the presentation of Newton's Laws for the first time in introductory Physics.

In Newtonian physics gravity is a real force with two bodies interacting. It satisfies the 3rd law, and I have no problem with the approximation to a uniform field.

The fact that the uniform approximation has the same form as the inertial force is irrelevant here. The two situations are not equivalent wrt the 3rd law because the gravitational force is due to the interaction with another body and the inertial force is not.

 

[Dale]

One more thing: Suppose I'm riding in my rocket ship/space elevator, and I'm doing all kinds of Physics - I'm shooting pool, I'm tossing balls through the air (it's a big elevator), I'm playing with pendula, I'm spinning gyroscopes - at what point will I find myself unable to explain or predict what I'm seeing simply by applying Newton's laws and the assumption of a constant force field, which I'll call, for lack of a better term, "gravity"? If you tell me the laws of motion break down, then I'll be able to tell that I'm in an accelerating frame and not in a constant gravitational field, and the Principle of Equivalence just got shot to Hell.

What body is causing this "gravity"? Or in other words, where do you put your reaction force to "gravity"?

 

[belliott4488]

DaleSpam:

As I said in an earlier response to JesseM, I completely agree that Newton's Law of Gravity does not apply in the accelerating frame - this is trivially true. My point is that the Laws of Motion say nothing about the source or nature of the force. They simply say, "give me a force and I'll tell you how objects will react to it."

As for your request that I "show my work" to show that a constant gravitational field is equivalent to a frame undergoing constant acceleration ... um, hasn't that been done - a lot? I know you know the Principle of Equivalence, which I keep mentioning just as a short-hand because I assume we all know the argument. Otherwise, Google it - you'll find a better explanation than I'll come up with off the top of my head, which is fine since it's not my argument.

As far as I can tell, our only real disagreement has to do with the reaction forces, or the "equal and opposite pairs," as we've called them. Did you disagree with my examples? So long as the objects are accelerated along with the frame, something is applying a force, which is what the inertial "reaction force" is reacting to. To the observer in the accelerating frame, he detects this mysterious force (the nature of which is immaterial to him, since he only needs to call it "F" in his equations), and sees that an equal and opposite force is needed to support it so that it doesn't fall to the floor.

When he drops the object, he feels - with his very sensitive feet - that the "ground" is accelerating upwards toward the object, as it must, since as far as he's concerned it feels the equal and opposite force. We, in our inertial frame outside the spaceship, attribute this to the increase in acceleration due to the slight decrease in accelerated mass, but the result is all the same.

I hate to keep appealing to authority, but doesn't Einstein's Principle of Equivalence make this kind of a non-issue? His whole point is that an observer in an accelerating frame cannot distinguish between his accelerating frame and a frame at rest in a constant gravitational field. If he could detect a violation of Newton's laws, then he could make the distinction. All of GR depends on this principle, so I don't see how you think it could be so wrong.

 

[Dale]

OK, belliott. I am willing to engage in a discussion about GR and the equivalence principle, but first I want you to address directly the following:

In my example I purposfully set up a scenario without gravitation so that there would be no confusion about if a given force was real or inertial. So, in the context of my example (i.e. flat spacetime, no gravity) I have performed an analysis demonstrating a clear violation of Newton's 3rd law in an accelerated reference frame. You are claiming that there is no such violation, so where did I make my error? Can you demonstrate that Newton's 3rd law is, in fact, satisfied in my scenario?

 

[belliott4488]

OK, belliott. I am willing to engage in a discussion about GR and the equivalence principle,

Nah, I don't think we need to go into GR - I haven't gotten the impression we have any disagreement there, unless you think it's all a crock, but you don't impress me as a crackpot , so I think we're fine just staying on-topic.

... but first I want you to address directly the following:

In my example I purposfully set up a scenario without gravitation so that there would be no confusion about if a given force was real or inertial. So, in the context of my example (i.e. flat spacetime, no gravity) I have performed an analysis demonstrating a clear violation of Newton's 3rd law in an accelerated reference frame. You are claiming that there is no such violation, so where did I make my error? Can you demonstrate that Newton's 3rd law is, in fact, satisfied in my scenario?

I agree this is the crux of our disagreement, but I don't see where you have demonstrated a "clear violation of Newton's 3rd law." You did show how the laws didn't work if the spaceman attempted to do Physics as if he were in an inertial frame, but then you showed how the 1st and 2nd laws did work if he invoked the fictitious inertial force. About the 3rd law, you simply said, "... all of the inertial forces act in the same direction, so they do not form 3rd law pairs. The inertial forces violate the 3rd law because there are no equal and opposite inertial forces."

This is what I've already responded to - I don't understand why you say this. First, the inertial forces are in the direction opposite to the acceleration, right? so if there were no other forces at work any objects under the influence of those forces would all be moving in that direction, which they aren't - at least, not until the spaceman drops the rock.

Let me go through it one more time, and tell me what it is that you object to (I've already been through this twice, but maybe you haven't been reading all the posts). If the spaceman stands in his rocket ship holding a rock, he feels that the rock exerts a "downward" force of ma on his hand. As long as he holds the rock motionless in his frame, he sees that he's exerting an equal and opposite force upward, ie. ma in the "upward" direction. We would describe it differently from outside, in our inertial coordinates, but you know how to do that and it doesn't matter for this discussion.

Now if he let's go of the rock and sees it fall to the floor, he will, of course, see it accelerate at a in the "downward" direction - again, we outside observers would say that it was moving inertially, but in his coordinates he would say it accelerates. If you now ask, "where's the force equal and opposite to the force he sees that causes this perceived acceleration?" the answer is essentially the same as it is when a student asks about the force equal and opposite to the gravitational force on a falling apple. In that case, the Earth is accelerated upward by a minuscule amount; in this case the rocket ship is accelerated "upward" when the spaceman releases the rock. In his frame, this is simply due to the equal and opposite force pulling "up" on the floor (which he can detect). In our frame, we see the additional acceleration on the ship because the rocket motor is now producing a greater acceleration of a' = F/M instead of the original acceleration a = F/(M+m), where F is the thrust of the rocket motor, m is the mass of the rocket, and M is the total remaining mass being accelerated by the motor.

Once the rock lands on the floor, the motor must now accelerate it as well (according to our inertial frame), so the acceleration we see returns to its original value, and the spaceman no longer feels the additional acceleration. (As an interesting side-note, his description would most likely be that he initially feels a downward force on his body, which he might call his "weight", due to the fictitious force, but when he drops the rock and feels the additional force, he would call that an inertial reaction force since he sees the floor (and himself) accelerating :"upward" in response to the equal and opposite force from the rock. In our inertial frame, we'd say it's all inertial reaction forces, just two different values for the two different acceleration magnitudes.)

That's not super rigorous, but do you really want me to go through the algebra? The reason I keep referring to the Equivalence Principle is just that I'm assuming you've seen this before and you know that the calculations in the accelerated frame are identical to the calculations in a constant gravitational field, so I'm not bothering to copy them here. I will, if you insist - or if you think there really is a subtlety there that I'm missing but will show up in the calculations.

I'll add one more thing, which is just to reinforce that I agree completely that the fictitious forces in non-inertial frames screw up the calculations. This is particularly evident in rotating frames, where we can't even attribute the centrifugal or Coriolis force to an apparent gravitational field. Nonetheless, we can write down expressions for what these forces look like in the coordinates of the rotating frame - in all their ugliness - and then do Physics using Newton's laws. This is a pain, but as far as I know, it works - do you disagree?

Again, my whole point boils down to this: whatever the origin of the forces, if you can write down expressions for them, then you can use Newton's laws to describe the resulting motion in the coordinates of the accelerated frame. I don't think you've actually denied this, except to assert that there aren't any equal and opposite forces. Can you show where this is true and you would be unable to describe the motion in the accelerated frame?

If you still think so, then how about we try the simple case of the spaceman tossing the rock in the air and recording its trajectory (just one dimension is fine, of course)? I predict that his application of Newton's laws will correctly describe the motion he observes. I'll gladly go through it in detail if you like, but I'd ask that you do the same so that we can see where our calculations diverge.

 

[Dale]

I agree this is the crux of our disagreement, but I don't see where you have demonstrated a "clear violation of Newton's 3rd law." You did show how the laws didn't work if the spaceman attempted to do Physics as if he were in an inertial frame, but then you showed how the 1st and 2nd laws did work if he invoked the fictitious inertial force. About the 3rd law, you simply said, "... all of the inertial forces act in the same direction, so they do not form 3rd law pairs. The inertial forces violate the 3rd law because there are no equal and opposite inertial forces."

This is what I've already responded to - I don't understand why you say this.

In SI units the inertial force on the rocket is (-2000,0,0) and the inertial force on the ball is (-0.2,0,0). I didn't explicitly mention the inertial force on the exhaust, but if the exhaust masses 1 kg then the inertial force on the exhaust is (-2,0,0). These are all of the inertial forces that exist in the example, they are all of different magnitudes and in the same direction so none of them are equal nor opposite to each other, so they cannot form 3rd law pairs. There are no missing forces and no missing bodies so the 3rd law is violated in the accelerated frame.

First, the inertial forces are in the direction opposite to the acceleration, right?

Yes.

so if there were no other forces at work any objects under the influence of those forces would all be moving in that direction, which they aren't - at least, not until the spaceman drops the rock.

I don't understand your comment here.

Let me go through it one more time, and tell me what it is that you object to (I've already been through this twice, but maybe you haven't been reading all the posts). If the spaceman stands in his rocket ship holding a rock, he feels that the rock exerts a "downward" force of ma on his hand. As long as he holds the rock motionless in his frame, he sees that he's exerting an equal and opposite force upward, ie. ma in the "upward" direction. We would describe it differently from outside, in our inertial coordinates, but you know how to do that and it doesn't matter for this discussion.

Now if he let's go of the rock and sees it fall to the floor, he will, of course, see it accelerate at a in the "downward" direction - again, we outside observers would say that it was moving inertially, but in his coordinates he would say it accelerates.

Yes.

If you now ask, "where's the force equal and opposite to the force he sees that causes this perceived acceleration?" ... in this case the rocket ship is accelerated "upward" when the spaceman releases the rock. In his frame, this is simply due to the equal and opposite force pulling "up" on the floor (which he can detect). In our frame, we see the additional acceleration on the ship because the rocket motor is now producing a greater acceleration of a' = F/M instead of the original acceleration a = F/(M+m), where F is the thrust of the rocket motor, m is the mass of the rocket, and M is the total remaining mass being accelerated by the motor.

This is where you need to run the numbers. I understand what you are saying here, but since you have not run the numbers you don't realize that you are wrong in your conclusions: the resulting forces in this slightly more detailed scenario will still not satisfy Newton's 3rd law. Do you want to do the math or do you want me to do it?

That's not super rigorous, but do you really want me to go through the algebra?

Yes. I honestly believe that is the best way to understand non-inertial frames and inertial forces.

The reason I keep referring to the Equivalence Principle is just that I'm assuming you've seen this before and you know that the calculations in the accelerated frame are identical to the calculations in a constant gravitational field, so I'm not bothering to copy them here. I will, if you insist - or if you think there really is a subtlety there that I'm missing but will show up in the calculations.

I will be glad to discuss GR later, but before we can have a productive conversation about the equivalence principle and curved spacetime we need to come to a mutual understanding about non-inertial reference frames and ficticious forces in flat space.

I'll add one more thing, which is just to reinforce that I agree completely that the fictitious forces in non-inertial frames screw up the calculations. This is particularly evident in rotating frames, where we can't even attribute the centrifugal or Coriolis force to an apparent gravitational field. Nonetheless, we can write down expressions for what these forces look like in the coordinates of the rotating frame - in all their ugliness - and then do Physics using Newton's laws. This is a pain, but as far as I know, it works - do you disagree?

Newton's 1st and 2nd laws I agree. Newton's 3rd law I disagree. In fact, you can make an arbitrarily accelerating reference frame with arbitrarily ugly inertial forces and you can use those inertial forces to correctly predict the outcome of traditional (non-relativistic) physics experiments in the non-inertial frame. But those inertial forces will always violate the 3rd law.

Again, my whole point boils down to this: whatever the origin of the forces, if you can write down expressions for them, then you can use Newton's laws to describe the resulting motion in the coordinates of the accelerated frame. I don't think you've actually denied this, except to assert that there aren't any equal and opposite forces. Can you show where this is true and you would be unable to describe the motion in the accelerated frame?

I never made such a claim. I have been very consistent with my statements.

In a non-inertial reference frame objects experiencing no real forces accelerate. This is a violation of Newton's 1st and 2nd law, however it is a "repairable" violation in the sense that we can introduce inertial forces. These inertial forces modify the normal laws but can be used to correctly predict the motion of objects in the non-inertial frame, but they do not satisfy Newton's 3rd law. This last is a "non-repairable" violation.

If you still think so, then how about we try the simple case of the spaceman tossing the rock in the air and recording its trajectory (just one dimension is fine, of course)? I predict that his application of Newton's laws will correctly describe the motion he observes. I'll gladly go through it in detail if you like, but I'd ask that you do the same so that we can see where our calculations diverge.

Sounds good. I will do that, and I predict that his application of Newton's first two laws, including the inertial forces, will correctly describe the motion he observes and that the inertial forces will not satisfy the 3rd law.

So that we can be consistent let's use the following:
mass of the ball = .1 kg
mass of the rocket (including ball) = 1000 kg
mass of the exhaust at t=1 s (if needed) = 1 kg
thrust of the engine = 2000 N
transformation equation between inertial and non-inertial frame: x' = x-t²
ball released gently at t=t'=1s

If I missed something just specify it explicitly.

 

[belliott4488]

In SI units the inertial force on the rocket is (-2000,0,0) and the inertial force on the ball is (-0.2,0,0). I didn't explicitly mention the inertial force on the exhaust, but if the exhaust masses 1 kg then the inertial force on the exhaust is (-2,0,0). These are all of the inertial forces that exist in the example, they are all of different magnitudes and in the same direction so none of them are equal nor opposite to each other, so they cannot form 3rd law pairs. There are no missing forces and no missing bodies so the 3rd law is violated in the accelerated frame.

Sorry, I do think you're missing some forces here, specifically the "real" forces that cause the objects to accelerate and hence to give rise to the inertial reaction forces in the first place. In fact, I believe inertial reaction forces are often defined as the equal and opposite forces applied by massive objects back on whatever is applying the forces on them. Here are the specifics for your case:

1. The ball: In the accelerating frame the ball is subject to an inertial force of -2 N, as you've said. It has this because it is being accelerated along with the rocket (relative to the inertial frame), which means it is motionless in the accelerated frame. So, if it is subect to this force in this frame, why isn't it moving? Simply because the spaceman is holding it in his hand, exerting an equal and opposite force of +2 N. Better yet, he can place it on a spring scale, which will show the "weight" of the ball as 2 N, which of course means that the spring is exerting a +2 N force "upward" on the ball.

2. The rocket: In the accelerating frame the spaceman notices that the ball has a "weight" (2 N), he has a weight (his mass X the acceleration), and he quickly deduces that the rocket itself must have weight, so something must be holding it up, since it is motionless in his frame. Looking out his port hole, he sees the rocket plume and realizes that the rocket motor is supplying thrust in the "upward" direction, which must be exactly equal and opposite to the "weight" of the rocket+spaceman+ball. In other words, the rocket is "hovering" in his frame due to the thrust force, which is exactly equal and opposite to the inertial force on the rocket. We know that this must be, because the thrust of the motor is exactly what is needed to accelerate this same mass in the inertial system to the acceleration that produces the "weight" in the accelerated system.

If you now ask, "where's the force equal and opposite to the force he sees that causes this perceived acceleration?" the answer is essentially the same as it is when a student asks about the force equal and opposite to the gravitational force on a falling apple. In that case, the Earth is accelerated upward by a minuscule amount; in this case the rocket ship is accelerated "upward" when the spaceman releases the rock. In his frame, this is simply due to the equal and opposite force pulling "up" on the floor (which he can detect). In our frame, we see the additional acceleration on the ship because the rocket motor is now producing a greater acceleration of a' = F/M instead of the original acceleration a = F/(M+m), where F is the thrust of the rocket motor, m is the mass of the rocket, and M is the total remaining mass being accelerated by the motor.

This is where you need to run the numbers. I understand what you are saying here, but since you have not run the numbers you don't realize that you are wrong in your conclusions: the resulting forces in this slightly more detailed scenario will still not satisfy Newton's 3rd law. Do you want to do the math or do you want me to do it?

Okay, let's do this more carefully, then. First, let's look at it in inertial coordinates, where we know we have no disagreement. Before the spaceman releases the ball, the rocket thrust is accelerating the entire system of rocket + ball, so the acceleration is a_1=F/(M+m), where, as before, F is the thrust of the motor, m is the mass of the ball (I think it was a rock last time ...), and M is the remaining accelerated mass. After he releases the ball, the thrust is applied only to the remaining mass, so the acceleration is now a_2 = F/M. After the ball hits the floor and comes to rest, the acceleration will return to its initial value.

What is the difference in the acceleration: it's simply
\delta a = a_2 - a_1 =\frac{F}{M} - \frac{F}{(M+m)} = \frac{F}{(M+m)} * \frac{m}{M}

But since the original acceleration is a_1=\frac{F}{(M+m)} we have,
\delta a = a_1 * \frac{m}{M}

Now, what does the spaceman see in the accelerated frame? He feels the ball with its "weight" of ma, and when he releases it, he sees it accelerate "downward" with an acceleration of a. He also feels the floor accelerate upward, however, due to the equal and opposite force exerted by the ball on the floor. What acceleration does he detect? Well, the force is given by ma, so that is the force "upward" on the floor. That force must accelerate the entire rocket (minus ball), so it produces an acceleration
a = \frac{F}{M} = a*\frac{m}{M}
which is the same acceleration we saw from the inertial point of view. Therefore by assuming a force equal and opposite to the inertial force he felt on the ball, he predicts exactly the same resulting acceleration as we did in the inertial frame.

I didn't explicitly plug any specific numbers in there, but since I have the same result for both frames, plugging in arbitrary numbers isn't going to prove anything.

For some reason the forum produces a database error if I post my whole response, so I'll leave this one here, and then post the rest in a subsequent response.

 

[belliott4488]

This is the remainder of my response, which the Forum wouldn't let me post in one piece:

Sounds good. I will do that, and I predict that his application of Newton's first two laws, including the inertial forces, will correctly describe the motion he observes and that the inertial forces will not satisfy the 3rd law.

So that we can be consistent let's use the following:
mass of the ball = .1 kg
mass of the rocket (including ball) = 1000 kg
mass of the exhaust at t=1 s (if needed) = 1 kg
thrust of the engine = 2000 N
transformation equation between inertial and non-inertial frame: x' = x-t²
ball released gently at t=t'=1s

If I missed something just specify it explicitly.

Okay, let's go through this. I'm not sure I really see the point, since we're just applying the 1st and 2nd laws, which you've already agreed work fine in the accelerated system, but since if you believe the error in my reasoning will be revealed, then on we go ...

Let's define some variables so that the equations aren't just a mash-up of numbers:

mass of ball: m_b = 0.1 \,\text{kg}
mass of rocket (including ball and spaceman): m_r = 1000.0 \,\text{kg}
thrust of rocket motor: T=2000.0 \,\text{N}

We're using primed coordinates, velocities, and accelerations for the accelerating system, unprimed for the inertial system.
I'm going to override your specification that the ball is released at t=1 s (gently - that's good! :-) we don't need any extraneous forces, here!), since there's nothing important happening 1 s before then to start our clocks. I'll just say that t=0 at the moment of release and avoid unnecessary extra terms that look like (t - 1s); if you've gone ahead with your definition, then there will be a lot of "t-1" factors that we can easily correct for.

I'll also say that the origins of both coordinate systems are aligned with the point of release of the ball at t=0, so that point is x = x' = 0 as well.

Then we have initial velocities, all at time t = 0:
rocket initial velocity in accel. stytem: v&#039;_{r0} = 0\,\text{km/sec}
rocket initial velocity in inertial system: v_{r0} = 10.0 \,\text{km/sec}
ball initial velocity in accel. system: v&#039;_{b0} = 0.1 \,\text{km/sec}
ball initial velocity in inertial systme: v_{b0} = v_{r0} + v&#039;_{b0} = 10.1 \,\text{km/sec}

For the acceleration of the rocket in the inertial frame we have:
a = F/m_r = 2 \,\text{km/sec}^2

Next we'll work out the trajectory of the ball in the inertial frame, where we are in agreement about everything.
The position of the rocket (i.e. the origin of the accel. frame) in the inertial frame is
x_r(t) = v_r t + \frac{1}{2}at^2
where v_r(t) = v_{r0}+at

The ball moves inertially once it's released, so its position in the inertial frame is simply
x_b(t) = v_{b0} t

So, what is the distance from the point of release in the rocket (i.e. the spaceman's hand) to the ball at time t? It will be:
h(t) = x_b - x_r = v_{b0} t - v_r t - \frac{1}{2}at^2 = (v_{b0}-v_r)t - \frac{1}{2}at^2

Now let's go to the accelerated frame. The ball is subject to the inertial force in the "downward" direction, so its postion will be:
x&#039;_b(t) = v&#039;_b t - \frac{1}{2}at^2

But we note that since the ball's (constant) velocity in the inertial frame is just
v_b = v_{b0} = v_r + v&#039;_b
we get
v&#039;_b = v_b - v_r
so
x&#039;_b(t) = (v_{b0}-v_r)t - \frac{1}{2}at^2 = h(t)

In other words, by invoking Newton's Laws in his accelerated frame, the spaceman predicts exactly the same answer that we do in our inertial frame.

 

[Dale]

Hi belliott, here is my work. I have attached free-body diagrams for all four parts of this problem. All forces are indicated on the diagrams and the accelerations are worked out for the rocket and the ball. Note in particular that I have highlighted the blue 3rd law pair for the thrust and the green third law pair for the contact force. Note also that none of the inertial forces in black are part of a 3rd law pair, i.e. they do not satisfy the 3rd law.

Below are the resulting equations of motion where capital X indicates the position of the rocket, lower-case x indicates the position of the ball, subscript 1 indicates when the ball is held, subscript 2 indicates when the ball is released, unprimed is the inertial coordinate system, primed is the accelerated coordinate system with transformation equation between the two x' = x - t² and inertial forces of -2m on every mass in the accelerated frame.

The equations of motion for the inertial frame while the ball is held (upper left) are:
\begin{array}{l}<br /> X_1=t^2 \\<br /> x_1=t^2<br /> \end{array}

The equations of motion for the accelerated frame while the ball is held (lower left) are:
\begin{array}{l}<br /> X_1&#039;=0=X_1-t^2 \\<br /> x_1&#039;=0=x_1-t^2<br /> \end{array}As you can see, these equations transform correctly, indicating agreement between the frames. As you can also see, the inertial forces were required in order to achieve this agreement.

Evaluating the above expressions at t=1 for the initial conditions of the next part gives:
\begin{array}{l}<br /> X_1(1)=x_1(1)=1 \\<br /> V_1(1)=v_1(1)=2<br /> \end{array}

\begin{array}{l}<br /> X_1&#039;(1)=x_1&#039;(1)=0 \\<br /> V_1&#039;(1)=v_1&#039;(1)=0<br /> \end{array}

The equations of motion for the inertial frame while the ball is dropped (upper right) are:
\begin{array}{l}<br /> X_2=1.0001 t^2-0.0002 t+0.0001 \\<br /> x_2=2. t-1.<br /> \end{array}

The equations of motion for the accelerated frame while the ball is dropped (lower right) are:\begin{array}{l}<br /> X_2&#039;=0.0001 t^2-0.0002 t+0.0001=X_2-t^2 \\<br /> x_2&#039;=-t^2+2. t-1.=x_2-t^2<br /> \end{array}As you can see, these equations again transform correctly, indicating agreement between the frames. As you can also see, the inertial forces were again required in order to achieve this agreement.

Bottom line: in the accelerating frame the inertial forces do not follow Newton's 3rd law, as shown in the free-body diagrams, but are required in order to satisfy Newton's 1st and 2nd laws and to obtain the correct equations for motion. In order to satisfy Newton's 1st and 2nd laws the 3rd law is violated. I didn't look at your posts in detail, but from my cursory reading nothing even appeared to address the 3rd law nor rebut my claims that it is violated.

 

[Dale]

Sorry, I do think you're missing some forces here, ... The ball: ... the spaceman is holding it in his hand, exerting an equal and opposite force of +2 N.

You are correct, I neglected that force (+0.2 N contact force on the ball) in the initial description. I have included it and it's 3rd law pair (-0.2 N contact force on the rocket/spaceman) in my analysis above.

Okay, let's do this more carefully, then. First, let's look at it in inertial coordinates, where we know we have no disagreement. Before the spaceman releases the ball, the rocket thrust is accelerating the entire system of rocket + ball, so the acceleration is a_1=F/(M+m), where, as before, F is the thrust of the motor, m is the mass of the ball (I think it was a rock last time ...), and M is the remaining accelerated mass. After he releases the ball, the thrust is applied only to the remaining mass, so the acceleration is now a_2 = F/M. After the ball hits the floor and comes to rest, the acceleration will return to its initial value.

What is the difference in the acceleration: it's simply
\delta a = a_2 - a_1 =\frac{F}{M} - \frac{F}{(M+m)} = \frac{F}{(M+m)} * \frac{m}{M}

But since the original acceleration is a_1=\frac{F}{(M+m)} we have,
\delta a = a_1 * \frac{m}{M}

Yes.

Now, what does the spaceman see in the accelerated frame? He feels the ball with its "weight" of ma, and when he releases it, he sees it accelerate "downward" with an acceleration of a. He also feels the floor accelerate upward, however, due to the equal and opposite force exerted by the ball on the floor. What acceleration does he detect? Well, the force is given by ma, so that is the force "upward" on the floor. That force must accelerate the entire rocket (minus ball), so it produces an acceleration
a = \frac{F}{M} = a*\frac{m}{M}
which is the same acceleration we saw from the inertial point of view. Therefore by assuming a force equal and opposite to the inertial force he felt on the ball, he predicts exactly the same resulting acceleration as we did in the inertial frame.

OK, there are several errors in this analysis. First and most importantly, you left out the thrust force (2000 N) acting on the rocket. Second, you left out the inertial force of the rocket (-1999.8 N). These two forces are sufficient to account for the acceleration and obtain agreement between the two frames. By arbitrarily adding another 0.2 N force on the rocket you wind up with disagreement between the frames since the delta acceleration in the non-inertial frame would be twice the delta in the inertial frame.

 

[Dale]

This is the remainder of my response, which the Forum wouldn't let me post in one piece:

I think there must be a limit on the number of LaTeX objects in a post because I was getting database errors replying too.

Okay, let's go through this. I'm not sure I really see the point, since we're just applying the 1st and 2nd laws, which you've already agreed work fine in the accelerated system, but since if you believe the error in my reasoning will be revealed, then on we go ...

I have consistently agreed that, by including the inertial forces, the 1st and 2nd laws work. I have consistently stated that the inertial forces required for the 1st and 2nd law violate the 3rd law.

Let's define some variables so that the equations aren't just a mash-up of numbers:
...

We're using primed coordinates, velocities, and accelerations for the accelerating system, unprimed for the inertial system.
I'm going to override your specification that the ball is released at t=1 s ... I'll just say that t=0 at the moment of release ...

I'll also say that the origins of both coordinate systems are aligned with the point of release of the ball at t=0, so that point is x = x' = 0 as well.

Then we have initial velocities, all at time t = 0:
...

For the acceleration of the rocket in the inertial frame we have:
a = F/m_r = 2 \,\text{km/sec}^2

This is incorrect. It should be
a = F/(m_r - m_b) = 2.0002 \,\text{m/s}^2
for t>0

Next we'll work out the trajectory of the ball in the inertial frame, where we are in agreement about everything.
The position of the rocket (i.e. the origin of the accel. frame) in the inertial frame is
x_r(t) = v_r t + \frac{1}{2}at^2
where v_r(t) = v_{r0}+at

I think you mean x_r(t) = v_{r0} t + \frac{1}{2}at^2. I corrected the following equations too.

The ball moves inertially once it's released, so its position in the inertial frame is simply
x_b(t) = v_{b0} t

So, what is the distance from the point of release in the rocket (i.e. the spaceman's hand) to the ball at time t? It will be:
h(t) = x_b - x_r = v_{b0} t - v_{r0} t - \frac{1}{2}at^2 = (v_{b0}-v_{r0})t - \frac{1}{2}at^2

OK

Now let's go to the accelerated frame. The ball is subject to the inertial force in the "downward" direction, so its postion will be:
x&#039;_b(t) = v&#039;_{b0} t - \frac{1}{2}at^2

But we note that since the ball's (constant) velocity in the inertial frame is just
v_{b0} = v_{r0} + v&#039;_{b0}
we get
v&#039;_{b0} = v_{b0} - v_{r0}
so
x&#039;_b(t) = (v_{b0}-v_{r0})t - \frac{1}{2}at^2 = h(t)

In other words, by invoking Newton's Laws in his accelerated frame, the spaceman predicts exactly the same answer that we do in our inertial frame.

This is fine, your frame is accelerating at a different rate than mine, but everything works out OK.

However, you never addressed the 3rd law question here. Where is the 3rd law pair for the -0.20002 N "inertial force in the 'downward' direction" on the ball? You have an inertial force on the rocket of -2000 N which exactly balances the thrust of 2000 N, but that force is neither equal nor opposite. Again, from your own work, it is clear that in order to get Newton's 1st and 2nd laws to work in the accelerated frame you must postulate inertial forces which violate the 3rd law.

 

[newTonn]

This is fine, your frame is accelerating at a different rate than mine, but everything works out OK.

However, you never addressed the 3rd law question here. Where is the 3rd law pair for the -0.20002 N "inertial force in the 'downward' direction" on the ball? You have an inertial force on the rocket of -2000 N which exactly balances the thrust of 2000 N, but that force is neither equal nor opposite. Again, from your own work, it is clear that in order to get Newton's 1st and 2nd laws to work in the accelerated frame you must postulate inertial forces which violate the 3rd law.

Let us analys the problem logically.I think there is no need to eloborate the problem with putting y,z co-ordinate(we can take x co-ordinate which is the only relevant one in this example) .
Initally the relative velocity will be zero and hence there will be no relative change in position between the ball and spacecraft floor because ball is physically connected to the spacecraft by the person who holds it.
A thrust,applied to the spacecraft will accelerate both spacecraft and ball.F= (M+m)a will be the equation of motion.But again the relative velocity and relative motion will be zero because both the bodies are accelerated at same rate.
At the instance the ball is released,it will have a velocity which will be the velocity of spacecraft and ball at that instance as calculated from the above equation.the ball will continue it's state of motion with that uniform velocity.The position of ball after 't' seconds will be decided by this velocity.
Once the ball is released,the equation of motion will be F=Ma'.the position of spacecraft floor after 't' seconds can be calculated from a'.
In both these None of the laws are violated.
Now,if we don't have a clear picture as above.ie. we don't have a third reference which is absolute with respect to both object,still we can calculate the position of ball after 't' seconds by introducing a fictious force.
Since we know this is a fictious force,it is not necessary for it to have an equal and opposite pair.
Still if it is so necessary,i can say there is a fictious equal and opposite pair for it ,by using the same logic.

 

[Dale]

Let us analys the problem logically.I think there is no need to eloborate the problem with putting y,z co-ordinate(we can take x co-ordinate which is the only relevant one in this example) .
Initally the relative velocity will be zero and hence there will be no relative change in position between the ball and spacecraft floor because ball is physically connected to the spacecraft by the person who holds it.
A thrust,applied to the spacecraft will accelerate both spacecraft and ball.F= (M+m)a will be the equation of motion.But again the relative velocity and relative motion will be zero because both the bodies are accelerated at same rate.
At the instance the ball is released,it will have a velocity which will be the velocity of spacecraft and ball at that instance as calculated from the above equation.the ball will continue it's state of motion with that uniform velocity.The position of ball after 't' seconds will be decided by this velocity.
Once the ball is released,the equation of motion will be F=Ma'.the position of spacecraft floor after 't' seconds can be calculated from a'.
In both these None of the laws are violated.

Correct, this is all done in the inertial reference frame.

Now,if we don't have a clear picture as above.ie. we don't have a third reference which is absolute with respect to both object,still we can calculate the position of ball after 't' seconds by introducing a fictious force.
Since we know this is a fictious force,it is not necessary for it to have an equal and opposite pair.

If it is not necessary for a force to have an equal and opposite pair then Newton's 3rd law is not a law.

Still if it is so necessary,i can say there is a fictious equal and opposite pair for it ,by using the same logic.

No, you cannot, as I have demonstrated twice now. Inertial forces violate Newton's 3rd law. You are welcome to try to demonstrate otherwise, but handwaving like this is insufficient.