What exactly is the reactive centrifugal force (split)

stevendaryl

In this particular case, we can certainly make the distinction between vectors, whose components are covariant under coordinate transformations, and terms that can be made to vanish by choosing a particular coordinate system. That's a mathematical, not a philosophical distinction, and it's an important distinction for working with the equations of motion.

DaleSpam

I actually think that this is an excellent example. I too think that the term "fictitious force" is unfortunate. Nonetheless, I recognize that it isn't up to me and that people who use it in the standard way are making correct statements from an authoritative definition. Similarly, you should realize that the definition of "physics", although you think it is unfortunate, it is not up to you and that people who use it in the standard way are making correct statements from an authoritative definition.

If I were having your argument with A.T. I would either reference a definition of physics that included the philosophical bits you like or I would challenge him to reference a definition of physics that excluded them. Many times terms have multiple authoritative definitions, but that doesn't mean that it is good practice to simply go about redefining them yourself, since nobody is likely to consider your definition authoritative.

stevendaryl

I disagree. There is nothing sacred about terminology, other than the fact that if one is using a technical term that is not standard, then there is time that must be spent explaining the terminology that could possibly be better spent on something else.

When it comes to technical terminology, such as "fictitious force" versus "inertial force" versus what I consider better "connection coefficients", different authors use different terminology. There is no absolute authority about what terms to use.

As for the definition of "physics" itself, I can't see any point whatsoever in worrying about it, or caring what the official definition is. A particular journal can decide to accept or reject a paper based on whatever criteria they choose; it doesn't need to be a universal definition of what counts as "physics".

Andrew Mason

This is not a good example because the gravitational force and normal force are not third law pairs. They are not equal and opposite, for one thing. The normal force is always a bit less than the gravitational force except at the poles.

The third law pair to the gravitational force of the earth on a body is necessarily the gravitational force of the body on the earth. Gravitational force supplies the centripetal acceleration required of the body on the surface of the earth. The body supplies the earth with its centripetal acceleration about the body-earth centre of mass (which is obviously extremely close to the earth's centre of mass).

I disagree. Your example illustrates why it can be difficult to apply Newton's third law. Where there is a body experiencing a change in motion there is always some other body experiencing an equal and opposite change in motion. Those are the third law pairs.

AM

A.T.

Who said something against making a distinction? I said that I prefer the terms "inertial forces" and "interaction forces". These are two distinct names.

Also note that I used the word "prefer", clearly indicating that it is just personal preference. Contrary to you, who is using terms like "falsehood", "crippled physics" etc. to denounce a naming convention which you don't like.

A.T.

What acceleration? Consider a non-rotating planet with you on it. Newtonian gravity. There is no acceleration. It is all static. But there is one interaction (electro-magnetic) between your feet and the ground, with a 3rd law force pair. And there is a second interaction (gravitational) between you and the planet, with a second 3rd law force pair. Two times action & reaction without any change in motion.

Nope, the third law pairs are individual forces, not accelerations.
http://en.wikipedia.org/wiki/Newton%27s_laws_of_motionYour problem is as usual: You think that the most trivial and primitive example (two bodies, one interaction) represents the general case. So you fail to apply the laws correctly to an even slightly more complex scenario (like the one above, with two interactions).

Andrew Mason

It depends on where the force acts. If one says that a force acts on the centre of mass of the body (ie. the space station), then the force is centripetal because it causes the space station and other astronaut (as the other body) to prescribe a centripetal acceleration towards the centre of rotation.

It is not wrong. It is just a way of looking at the third law that differs slightly, but not fundamentally, from your view.

The fact is that Newton refers to "forces" in his first and second laws. He does not use the word "force" in the third law.

Your view, I gather, is that his omission of the word "force" and his use of "action" and "reaction" were not deliberate. My view is that this was deliberate.

The third law can be viewed as the law of interacting bodies. If there is a change in motion of a body there must be another equal and opposite change in motion of some other interacting body or bodies. That, it seems to me, is the very essence of the third law. If there are forces but no actions (changes in motion), there is no need for the third law. The first law covers the situation.

AM

stevendaryl

Sure, but "connection coefficient" is what they are, mathematically, and has the advantage of having a precise definition saying how to compute them. Connection coefficients are needed whenever one needs to take a derivative of a vector or tensor field. Calling these terms "fictitious forces" doesn't make sense, because not all uses of derivatives have anything to do with forces. They really don't have anything to do with forces, in general.

DaleSpam

I never said they were. They act on the same body so they cannot be third law pairs. Each of those two forces (gravitational and normal) have their own third law reaction forces acting on the earth.

Fine. Consider the person to be standing on a turntable at the pole such that he is at rest in a Newtonian inertial frame. And consider the planet to be isolated far from any significant gravitational body.

Yes, this is correct. Note that the gravitational force (mg) is not equal to the change in momentum (ma=0). So your correct statement here contradicts your own incorrect statement elsewhere.

Not only is this not how Newtons third law works, it is not even true.

Consider three identical masses, each connected by identical springs of neglible mass to both of the others and starting at rest from a position where the springs are all equally stretched. Each object is experiencing a change in motion and there is no other body experiencing an equal and opposite change in motion.

A.T.

Look at the picture again. Try to figure out where the blue arrows start. That's where the astronaut applies the force to the space station.



...then one obviously has a serious case of cognitive dissonance.

Doesn't matter. In today's physics the 3rd law refers to forces. Every physics book will confirm this.

Andrew Mason

I agree. Those are the forces and the interactions. They all sum to 0. But you don't need the third law to analyse that. There is no "action" (ignoring the acceleration relative to the sun and other planets etc.) . Newton made the distinction between force and action. I didn't.


Wikipedia does not state Newton's the third law. It states one interpretation of Newton's third law but it does not use the terms "action" and "reaction" that Newton used. In my view, it implicitly combines the first, second and third laws of Newton so it makes it difficult to see how the third law is distinct from the first and second. If forces are balanced so there is no net force, the first law applies. You don't need the third law to determine that if a body's motion is not changing all the forces must sum to 0. You just need the first.

I am not disagreeing with you on the physics. All I am saying is that if there is no acceleration, you don't need the third law. The first law suffices.

An approach that says that people who take a valid but different view of things are wrong is not particularly helpful.

AM

stevendaryl

I agree with you that thinking of equal and opposite forces works better than thinking of equal and opposite changes of motion, but in the case you are talking about, the change in momentum of one of the masses is equal and opposite to the vectorial sum of the changes of momentum of the other two masses. That might be what AM meant.

A.T.

It is the interpretation of Newton's third that you will find in every modern physics book. It doesn't matter how you call this Law ("3rd law" or "generalized 3rd law"). All your hand waving cannot change the fact that:
- It is a law of physics.
- It states two equal but opposite forces.
- It applies to static cases too.
- It means that my astronauts exert a centrifugal reaction force on the station

stevendaryl

Hmm. Let's see if that's true. Suppose you have three objects 1, 2, 3. There are correspondingly 6 forces:

1. [itex]\stackrel{\rightarrow}{F_{12}}[/itex] = force of 1 on 2
2. [itex]\stackrel{\rightarrow}{F_{21}}[/itex] = force of 2 on 1
3. [itex]\stackrel{\rightarrow}{F_{13}}[/itex] = force of 1 on 3
4. [itex]\stackrel{\rightarrow}{F_{31}}[/itex] = force of 3 on 1
5. [itex]\stackrel{\rightarrow}{F_{23}}[/itex] = force of 2 on 3
6. [itex]\stackrel{\rightarrow}{F_{32}}[/itex] = force of 3 on 2

The fact that the situation is static means that the force on any object is zero. So we have:

1. [itex]\stackrel{\rightarrow}{F_{12}} + \stackrel{\rightarrow}{F_{32}} = 0[/itex]
2. [itex]\stackrel{\rightarrow}{F_{13}} + \stackrel{\rightarrow}{F_{23}} = 0[/itex]
3. [itex]\stackrel{\rightarrow}{F_{21}} + \stackrel{\rightarrow}{F_{31}} = 0[/itex]

That's three constraints. I don't see how you can derive that
[itex]\stackrel{\rightarrow}{F_{12}} + \stackrel{\rightarrow}{F_{21}} = 0[/itex]

DaleSpam

Andrew Mason, please provide a mainstream scientific reference for Newton's 3rd law being as you describe, equal and opposite net forces or equal and opposite changes in momentum.

DaleSpam

In fact, you cannot derive it. Here are a set of forces that satisfy Newton's 1st law but not his 3rd:

1. [itex]\stackrel{\rightarrow}{F_{12}}=2[/itex]
2. [itex]\stackrel{\rightarrow}{F_{21}}=1[/itex]
3. [itex]\stackrel{\rightarrow}{F_{13}}=3[/itex]
4. [itex]\stackrel{\rightarrow}{F_{31}}=-1[/itex]
5. [itex]\stackrel{\rightarrow}{F_{23}}=-3[/itex]
6. [itex]\stackrel{\rightarrow}{F_{32}}=-2[/itex]

Paul Andersen

Yes.
You can study the centrifugal and Coriolis forces here:
http://www.gethome.no/paulba/Spaceship.html

Hit the "slow motion" button.