Force/ gravity/ mass/ acceleration

[joigus]

21 minutes ago, Genady said:

Are you sure? I seem to remember (I think, from Shankar's lectures, but would need to look back to make sure), that force is defined first statically, like the following. Take a spring. Some force needs to be applied to stretch it say by 10 cm. Take two identical springs like that attached parallelly. A force needed to stretch both of them together by 10 cm is defined as double the first force. Etc. After force is defined in such a way, its use in dynamics is a law / equation.

Well, good point. It's a bit subtler than what I said. I do remember a similar operational definition to what you say in Mechanics by Keith R. Symon. But even for your operational definition of force, you need to set a unit of mass. So it's kind of circular. Mass helps you define force, while you need a standard of force to define mass. They're tied to each other, really.

Let me put it in my words: If you think it makes sense to fix a standard of force independent on anything it acts on, to that extent, you can define mass. If you think it makes sense to fix a standard of mass independent of the force that acts on it, then you can define force. It could be more complicated. It could be that there is no way to abstract the 'push or pull' that you exert on a body from the parameters that define the body.

Maybe it's something closer to what I called a formula (mathematicians use that distinction, I know). What it is not is an equation, unless you use the formula to plug in numbers and solve for the unknown, of course.

 

[MigL]

I'll side with Janus.
F=ma simply defines the relationship between two ( or possibly 3 ) variables.
Usually force is proportional to acceleration when mass is constant, but the other way F is proportional to m when a is constant, also works.

[Genady]

2 hours ago, joigus said:

you need a standard of force to define mass

I am not sure it is necessary. We only need to define that in any given conditions (i.e. under given albeit undefined forces) mass is inversely proportional to acceleration. We take a spring, again, stretch it by 10 cm, attach a "unit" of mass, release and measure the acceleration. Then we attach another body to the same spring stretched by the same amount and measure its acceleration. We get the mass of the second body. Etc.

Now we have independent definitions of mass and of force and can put them together without any circularity. We could've discovered that \(F \propto m^2a^2\) or \(F \propto e^{ma}\), ... 

PS. To establish that "mass is inversely proportional to acceleration under given conditions" we can do, e.g., the following. First, the same spring, body, stretch, release, measure acceleration procedure. Then, halve the body and define that its half has half the mass. Repeat the procedure and find that the acceleration doubles. ...

Edited February 15 by Genady

[joigus]

 

1 hour ago, Genady said:

I am not sure it is necessary. We only need to define that in any given conditions (i.e. under given albeit undefined forces) mass is inversely proportional to acceleration. We take a spring, again, stretch it by 10 cm, attach a "unit" of mass, release and measure the acceleration. Then we attach another body to the same spring stretched by the same amount and measure its acceleration. We get the mass of the second body. Etc.

(My emphasis.)

There it is. There's your standard of force if you want to make the definition operational. The fixed spring is your standard of force. It's actually inescapable that one needs the other, as F=ma involves both and neither F, nor m, is a primitive concept with a direct observational interpretation, like time or space have.

I see no a priori reason to rule out a more complicated mathematical dependence like, say,

F=(m+C₂m²+C₃m³+...)a

with m being the additive parameter representing the "amount of stuff", F being our standard spring, C₂, C₃, etc, very small coefficients under a wide range of dynamical conditions, and the other force laws that we know and love later accomodating this complicated dependence. I'd challenge anybody to provide a robust argument why it cannot be that way.

Newton's choice is very sound and very natural, and harmonises wonderfully with symmetries, known behaviours, etc, but I see no a priori reason why it should be that way, and not some other.

I don't essentially disagree with anything Janus has said, by the way.

[Genady]

1 hour ago, joigus said:

There it is. There's your standard of force if you want to make the definition operational. The fixed spring is your standard of force. It's actually inescapable that one needs the other, as F=ma involves both and neither F, nor m, is a primitive concept with a direct observational interpretation, like time or space have.

Sure, we need a standard of force, but I don't see how it makes F and m interdependent. I see that F and m are defined separately while F=ma sets their interdependence.

[joigus]

10 hours ago, Genady said:

Sure, we need a standard of force, but I don't see how it makes F and m interdependent. I see that F and m are defined separately while F=ma sets their interdependence.

They are not defined separately.  Not operationally at least. They are inextricably linked, and hidden assumptions operate between both concepts, as I will try to show. Consider,

F = ma = (2m)(a/2) = (3m)(a/3) = ... for different test masses m, 2m, 3m, etc we measure to good approximation impinged accelerations a, a/2, a/3, etc for a fixed spring of given elastic constant. This measures F. There is a hidden assumption here. Namely: masses are additive, and so I will be able to stick together identical test objects and assume they operate in Newton's law as twice the mass, three times the mass, etc. Something by no means obvious. In fact we know that to be false from relativity.

OTOH, for a fixed m and a fixed direction, we apply different sets of springs by hooking them together (in parallel, so that the spring constants are additive) and measure to good approximation that m = F/a = 2F/2a = 3F/3a =... Mind you, this also assumes something about connecting springs together. This measures m.

I don't think this is what one does to measure either mass or force, but it's introduced in books of mechanics from the 50's to the 60's or so. I would think methods based on displacement from an equilibrium position would be more accurate. But I'm not sure. Even in that case, hidden assumptions about mass and force are operating there that boil down to additivity, I'm sure.

In fact, the whole lot of Newton's mechanics can be put more simply in terms of F=ma plus a principle of additivity or external transitivity, if you like; and sub-divisibility or internal transitivity, if you like. Then it becomes just one law, instead of three, plus this principle that Newton's laws are to be applied to any level of integrating sub-parts or conglomerates of parts:

F=ma (first and only law)

1) System is free (F=0) => a=0 => v=constant (first law)

2) System is free F=0 but it's made of sub-parts 1 and 2. 

F=F₁₂+F₂₁ => F₂₁=-F₁₂ (third law)

And the principle (not so hidden, but explicitly stated throughout history) that certain "magical" frames of reference exist (inertial frames) where the conglomerate of all the parts can be looked upon as free, and ultimately all the dynamics can be analysed in terms of internal forces. Laboratory -> Earth -> Solar System -> etc.

The traditional expositions that there are 3 laws, inertial systems, inertia, and the whole shebang are very good to get started, but only obscure these very simple principles IMO, of F=ma (very, very strong assumption that we can isolate interactions into action on something, F, and reaction to that action, -ma, and ultimately bound to be false, as we know); and applicability both to sub-parts and super-parts.

[joigus]

17 hours ago, MigL said:

F=ma simply defines the relationship between two ( or possibly 3 ) variables.

Exactly (my highlight in bold red).

This is at no detriment to the use that @Janus and @Genady here in particular (and most physicists elsewhere as well) have given to Newton's third law as an equation. Any definition, identity, or formula can be postulated as an equation the moment one gives numerical values to any of the terms involved, or values in terms of further parameters. So, for example: sin²a+cos²a=1 is an identity. It says something obvious. A further substitution, eg sin²a=1/2 makes it an equation.

 

[MigL]

One thing I forgot to mention in this post

22 hours ago, MigL said:

In the case of M being the mass of the Earth, we realize that climbing to the top of the leaning Tower of Pisa to drop a bowling ball and a golf ball, and timing their time to fall, is not necessary.
They both fall at 9.8 m/s/s.

If you did go ahead and repeat Galileo's supposed experiment of dropping two unequal weights from the leaning tower of Pisa, you would find that they both fall at the same rate, and you would measure the acceleration of the two unequal masses at 9.8 m/s/s.
Further you would find that the a in F_i=m_ia is equivalent to GM_e/r², and, upon setting F_i=F_g ( or m_ia=GM_em_g/r^₂ ) that m_i and m_g are, in fact,  equal.
This is actually a deep, and surprising, result, as it shows inertial mass is equivalent to gravitational mass.
Even A Einstein pondered why this should be so, and may be what Joigus was referring to here

On 2/15/2024 at 6:16 AM, joigus said:

We should all be amazed that this little m disappears from the equation. It goes very deep in the nature of gravitation.

 

Edited February 16 by MigL

[Genady]

A gravitational acceleration being independent of a body's mass also follows from the Kepler's third law.

[joigus]

23 hours ago, joigus said:

have given to Newton's third law as an equation

I should have said Newton's 2nd law, obviously. I have this kind of dyslexic-like glitch that makes me do that very much like @studiot's problem with the typing.

23 hours ago, MigL said:

This is actually a deep, and surprising, result, as it shows inertial mass is equivalent to gravitational mass.
Even A Einstein pondered why this should be so, and may be what Joigus was referring to here

Yes, that's exactly what I meant. Now, if all forces of Nature were like that, I wouldn't find it surprising at all. After all, the word "surprise" has to do with contrast in comparison to previous experience, or inference from that. Electricity is not like that, nor any other interaction. Thereby the word "amazed".

23 hours ago, Genady said:

A gravitational acceleration being independent of a body's mass also follows from the Kepler's third law.

True. In fact Newton used Kepler's third law to guess his inverse-square law. Textbooks generally point out that the power law is implied. But the equivalence principle is too. The thing is, because the mass on the receiving end of the gravitational interaction (not the mass as a source) disappears from all the physics, it is almost inescapable that the distorsion that a source introduces around it can be described in some geometric way as a distorsion of space-time itself.

I think this is amazing even after one learns about GR.

[Genady]

3 minutes ago, joigus said:

I think this is amazing even after one learns about GR.

Another way to look at it is that mass of an affected body was introduced into the equation to make it fit into the force-based model. Then, it disappears when the model is not force-based anymore.

[joigus]

On 2/17/2024 at 3:12 PM, Genady said:

Another way to look at it is that mass of an affected body was introduced into the equation to make it fit into the force-based model. Then, it disappears when the model is not force-based anymore.

That's certainly what happens in GR. In QFT I think the process is much more painful. The theory is not force-based either, but we must start with mass being a parameter that discriminates between different types of fields (massless vs massive). But the physical mass (inertia) becomes more of a dynamical attribute that depends on the state and has to be calculated perturbatively. And there is no explanation for the spectrum of masses.

[Genady]

On 2/19/2024 at 6:43 AM, joigus said:

That's certainly what happens in GR. In QFT I think the process is much more painful. The theory is not force-based either, but we must start with mass being a parameter that discriminates between different types of fields (massless vs massive). But the physical mass (inertia) becomes more of a dynamical attribute that depends on the state and has to be calculated perturbatively. And there is no explanation for the spectrum of masses.

Are these QFT or rather SM issues?

[joigus]

27 minutes ago, Genady said:

Are these QFT or rather SM issues?

Ultimately it's a major SM issue, I think. But there are very general arguments in QFT in which Yang-Mills pretty much appears as the only interesting generalisation for gauge invariance. So what I mean I suppose is that from QFT to SM there's "just" (ahem) a choice of symmetry groups, generations, and mixing parameters. A very wise expert in QFT nothing fundamentally different from the general principles of QFT conveniently generalised. 

[Genady]

1 hour ago, joigus said:

Ultimately it's a major SM issue, I think. But there are very general arguments in QFT in which Yang-Mills pretty much appears as the only interesting generalisation for gauge invariance. So what I mean I suppose is that from QFT to SM there's "just" (ahem) a choice of symmetry groups, generations, and mixing parameters. A very wise expert in QFT nothing fundamentally different from the general principles of QFT conveniently generalised. 

QFT is a framework that fits models with massive as well as massless neutrinos, with one as well as five generations of particles, with photons as well as phonons, with vacuum as well as solid state, etc. I mean that all the missing questions are specific to the model, i.e., SM, and are in addition to QFT.

[joigus]

4 hours ago, Genady said:

QFT is a framework that fits models with massive as well as massless neutrinos, with one as well as five generations of particles, with photons as well as phonons, with vacuum as well as solid state, etc. I mean that all the missing questions are specific to the model, i.e., SM, and are in addition to QFT.

SM is but one particular QFT. Are you sure you're not splitting hairs here?

SM cannot be in addition to QFT the same way the statistical mechanics of an Ising magnet wouldn't be in addition to statistical mechanics. It's given by a particular choice of Hamiltonian within the general procedures of quantum statistical mechanics. SM is QFT under a particular choice of Lagrangian, including gauge groups, global gauge groups, and Higgs multiplets. 

Unless I overlooked an essential point you made, which is certainly possible, especially of late.

[Genady]

4 minutes ago, joigus said:

SM is but one particular QFT. Are you sure you're not splitting hairs here?

SM cannot be in addition to QFT the same way the statistical mechanics of an Ising magnet wouldn't be in addition to statistical mechanics. It's given by a particular choice of Hamiltonian within the general procedures of quantum statistical mechanics. SM is QFT under a particular choice of Lagrangian, including gauge groups, global gauge groups, and Higgs multiplets. 

Unless I overlooked an essential point you made, which is certainly possible, especially of late.

I'd refer to the following analogy.

QFT is like GR when SM is like LCDM.

GR can fit many different cosmological models, and the open questions are about actual contents and history of the universe, which are the aspects I refer to as being "in addition" to the GR framework. 

[Genady]

To clarify the analogy in my previous post. I don't mean that QFT is like GR, nor that SM is like LCDM. I mean that SM relates to QFT like LCDM relates to GR. 

[joigus]

9 minutes ago, Genady said:

To clarify the analogy in my previous post. I don't mean that QFT is like GR, nor that SM is like LCDM. I mean that SM relates to QFT like LCDM relates to GR. 

Yes, very much so. I'm re-reading what I said as well as your comment that motivated it. I was kinda losing track of what I was trying to say, and thinking 'why the hell did I mention QFT?' And (after re-reading) I see it's because you said,

On 2/17/2024 at 3:12 PM, Genady said:

[...] mass of an affected body was introduced into the equation to make it fit into the force-based model. Then, it disappears when the model is not force-based anymore.

The reason I mentioned QFT (or the SM as a particular case) is because I wanted to point out that sometimes, even though force and mass are not pillars of the theory, you still have to do a lot of work with this mass, so it's very far from disappearing from most considerations. But it's not like the theory is telling you what this parameter actually is or does.

[Genady]

16 minutes ago, joigus said:

Yes, very much so. I'm re-reading what I said as well as your comment that motivated it. I was kinda losing track of what I was trying to say, and thinking 'why the hell did I mention QFT?' And (after re-reading) I see it's because you said,

The reason I mentioned QFT (or the SM as a particular case) is because I wanted to point out that sometimes, even though force and mass are not pillars of the theory, you still have to do a lot of work with this mass, so it's very far from disappearing from most considerations. But it's not like the theory is telling you what this parameter actually is or does.

Yes, I certainly agree with this.