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I am currently studying Classical Mechanics, 5th edition, by Kibble and Berkshire. Chapter 1.3 The concepts of Mass and Force says the following:

Clearly, we can compare the inertial masses of two bodies by subjecting them to equal forces and comparing their accelerations, but this does not help unless we have some way of knowing that the forces are equal. However there is one case in which we do know this, because of Newton's third law. If we isolate the two bodies from all other matter, and compare their mutually induced accelerations, then according to (1.1) and (1.3),

$$m_1 \mathbf{a}_1 = -m_2 \mathbf{a}_2, \tag{1.7}$$

so that the accelerations are oppositely directed, and inversely proportional to the masses. If we allow two small bodies to collide, then during the collision the effects of more remote bodies are generally negligible in comparison with their effect on each other, and we may treat them approximately as an isolated system. (Such collisions will be discussed in detail in Chapters 2 and 7.) The mass ratio can then be determined from measurements of their velocities before and after the collision, by using (1.7) or its immediate consequence, the law of conservation of momentum,

$$m_1 \mathbf{v}_1 + m_2 \mathbf{v}_2 = \text{constant}. \tag{1.8}$$

If we wish to separate the definition of mass from the physical content of equation (1.7), we may adopt as a fundamental axiom the following:

In an isolated two-body system, the accelerations always satisfy the relation $\mathbf{a}_1 = -k_{21} \mathbf{a}_2$, where the scalar $k_{21}$ is, for two given bodies, a constant independent of their positions, velocities and internal states.

If we choose the first body to be a standard body, and conventionally assign it unit mass (say $m_1 = 1 \ \text{kg}$), then we may define the mass of the second to be $k_{21}$ in units of this standard mass (here $m_2 = k_{21} \ \text{kg}$).

Note that for consistency, we must have $k_{12} = 1/k_{21}$. We must also assume of course that if we compare the masses of three bodies in this way, we obtain consistent results:

For any three bodies, the constants $k_{ij}$ satisfy $k_{31} = k_{32} k_{21}.$

It then follows that for any two bodies, $k_{32}$ is the mass ratio: $k_{32} = m_3/m_2$.

I am having difficulty understand the following section:

Note that for consistency, we must have $k_{12} = 1/k_{21}$. We must also assume of course that if we compare the masses of three bodies in this way, we obtain consistent results:

For any three bodies, the constants $k_{ij}$ satisfy $k_{31} = k_{32} k_{21}.$

It then follows that for any two bodies, $k_{32}$ is the mass ratio: $k_{32} = m_3/m_2$.

Precisely what of the content that came before this implies that we should "note that for consistency, we must have $k_{12} = 1/k_{21}$"? Furthermore, exactly what is the justification for "for any three bodies, the constants $k_{ij}$ satisfy $k_{31} = k_{32} k_{21}$"? And how exactly does "it then follows that for any two bodies, $k_{32}$ is the mass ratio: $k_{32} = m_3/m_2$"?

I would greatly appreciate it if people would please take the time to clarify what the authors are trying to explain here.


Edit

I am disappointed with the quality of answers this question has received. Despite being a seemingly simple question on classical mechanics, there have been a number of incorrect answers, and all of the other answers to date have been unclear and/or lack explanation.

So far, the only thing that I have been able to understand for myself is $k_{12} = 1/k_{21}$:


We have adopted as axiom that

$$\mathbf{a}_1 = -k_{21} \mathbf{a}_2 \\ \Rightarrow \mathbf{a}_2 = - \dfrac{1}{k_{21}}\mathbf{a}_1$$

We have also defined that $m_1 = 1 \ \text{kg}$ and $m_2 = k_{21} \ \text{kg}$.

This means that we also have that

$$-k_{12} \mathbf{a}_1 = \mathbf{a}_2$$

by symmetry of the indices. Therefore, we have that

$$k_{12} = \dfrac{1}{k_{21}}$$


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  • $\begingroup$ So you still need the reason for the mass ratio and the relation between the 3 k's right? $\endgroup$ Jun 21, 2020 at 6:12

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Oh dear. This really seems to be an example of people making a simple question complicated, probably for no better reason that they do not understand it themselves. My advice would be to ignore the textbook and go back to the mathematical meaning of Newton's laws. I really do not want to go into analysis of everything I see wrong in the original treatment, but I would only suggest, get a better text book.

Imv, the best way to understand Newton's laws is that they are an expression of a fundamental principle, conservation of momentum. This principle is actually provable for particle interactions in relativistic quantum mechanics, and it is equivalent to Newton's third law, together with Newton's second law treated as the definition of an active force. It is perfectly possible to express everything said in your text book in a very simple manner by thinking only of conservation of momentum. Then there is no need to enter into the complications which they introduce.

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  • $\begingroup$ Thanks for the answer, Charles. Given that the answers so far have been mostly unclear and lacking in detail, I did indeed suspect that people themselves didn't fully understand what the author was saying. With that said, Classical Mechanics by Kibble and Berkshire comes highly rated and recommended, as shown by the Amazon reviews. Do you have suggestions for a superior textbook? $\endgroup$ Jun 20, 2020 at 21:40
  • $\begingroup$ I was fortunate enough to have had a brilliant mechanics teacher when I was at school, which meant I never needed to refer to a textbook. It is possible to derive everything directly from Newton's laws, and really there is no better way to understand it. By being able to derive things from first principles, one is not dependent on what one learns from others, but I am afraid that is not much help to you. $\endgroup$ Jun 20, 2020 at 22:07
  • $\begingroup$ @ThePointer I am slightly late, but I think it is rather harsh to claim that I "didn't fully understand". I too was baffled at the author's approach, but I thought I would take the time to try and unpack their reasoning. I completely agree with everything Charles Francis says, but I don't believe it clarifies the given explanation. That said, I too will upvote this answer since it is nonetheless my preferred stance also. $\endgroup$
    – 13509
    Aug 7, 2020 at 16:23
  • $\begingroup$ @JamesWirth I understand less about this than either of you, so I'm just taking Charles's word for it. As I said in my comment to your answer, it seems to me that, out of all of the answers that were posted, yours was the best, which is why I awarded you the initial bounty. $\endgroup$ Aug 7, 2020 at 16:29
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We consider three bodies. It is postulated that if we take any two of them, and let them interact as an isolated system, then their accelerations are related by a scalar multiple. To be more specific, in the experiment with the $i^{\text{th}}$ and $j^{\text{th}}$ particle, we write

$$\vec{a}_{i}^{(ij)} = -k_{ji}\vec{a}_{j}^{(ij)}$$

Note that in each experiment, the acceleration of the same body will be different unless the other two masses are equal (i.e. $\vec{a}_{1}^{(12)}$ generally won't equal $\vec{a}_{1}^{(13)}$). Now the scalar multiple $k_{ji}$ that we can measure in each experiment serves as an indicator of the relative masses of the bodies. You may also rearrange the above to

$$-\frac{1}{k_{ji}}\vec{a}_{i}^{(ij)} = \vec{a}_{j}^{(ij)}$$

and by noticing that this must take exactly the same form due to symmetry under interchanging the indices, we obtain $k_{ji} = \frac{1}{k_{ij}}$.

Now consider the measure of the ratio of the masses of bodies $2$ and $3$, $k_{23}$. We can either measure this directly, by performing a single experiment with bodies $2$ and $3$, or we can perform two experiments with bodies $3$ & $1$ and $2$ & $1$, with body $1$ as a common reference. You will notice that the ratio of $k_{21}$ and $k_{31}$ must then also equal $k_{23}$, since $k_{21}$ and $k_{31}$ are measures of the masses of bodies $2$ and $3$ respectively w.r.t. the same reference. This is what prompts $\frac{k_{21}}{k_{31}} = k_{23}$.

So we finally deduce $k_{ij} = \frac{m_i}{m_j}$, with which you can easily verify the relations we came up with above. Namely

$$k_{21} = \frac{m_2}{m_1}= \frac{m_2}{m_3} \frac{m_3}{m_1} = k_{23}k_{31}$$ and $$k_{12} = \frac{m_1}{m_2} = \left(\frac{m_2}{m_1} \right)^{-1} = \frac{1}{k_{21}}$$

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    $\begingroup$ @ThePointer No problem. At the end of the day, this is really just a motivation for Newton's third law. You can of course "work backwards" to find out what the author has defined $k_{ij}$ to be. So I wouldn't worry too much about this part of the chapter, it's not so important in the grand scheme of things! $\endgroup$
    – 13509
    Jun 7, 2020 at 23:00
  • $\begingroup$ Why the downvote? Is there an error? $\endgroup$ Jun 8, 2020 at 13:26
  • $\begingroup$ @ThePointer I don't know why there is a downvote. Everything I said is pretty standard. $\endgroup$
    – 13509
    Jun 8, 2020 at 13:28
  • $\begingroup$ Is it possible for an interacting three-body system to know the accelerations (and thus the masses) in the first place? (+1, by the way!) $\endgroup$ Jun 9, 2020 at 9:44
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    $\begingroup$ @descheleschilder I'm not actually sure if such a method would work if we considered an interacting 3-body system. There we could write $m_1 \vec{a}_1 + m_2 \vec{a}_2 + m_3 \vec{a}_3 = \vec{0}$ but I think we would need a force law of some sort to determine anything further. I imagine it would get quite messy ;) $\endgroup$
    – 13509
    Jun 9, 2020 at 12:50
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In the quote from your book it is written:

If we wish to separate the definition of mass from the physical content of equation (1.7), we may adopt as a fundamental axiom the following: In an isolated two-body system, the accelerations always satisfy the relation $\pmb{a_1}=−k_{21}\pmb{a_2}$, where the scalar $k_{21}$ is, for two given bodies, a constant independent of their positions, velocities, and internal states.

Now, why would we separate the definition of mass from the physical content of an equation (or introduce the law of conservation of linear momentum)? These actions only obscure the real physics when introducing this silly $ f_{ij}$ symbols (while mass is a constant independent of the mass's positions, velocities, and internal states just as well), and I advise you not to take it too seriously. It makes things just more complicated (as your question bears witness to). But then again, maybe the $f_{ij}$ tensor is handy for later use, though I doubt it.

Just imagine three bodies (or point-particles) with masses $m_1$, $m_2$, and $m_3$. The only force we consider (as far as I understand it) is the gravitational force in 3d (in which case we measure the gravitational mass instead of the inertial mass, which up to now are considered equal; there is only a difference in interpretation, but that aside). These forces lay on the lines connecting the particles (a triangle) because otherwise, we would have to deal with an insolvable three-body problem (in general, apart from some particular cases). The $\frac{1}{r^2}$ dependence of the force is retained though.

Now for the magnitudes of the forces we can write (the forces are all attractive so the only thing that counts are their magnitudes):

$$m_1 a_1=m_2 a_2$$ $$m_1 a_3=m_3 a_4$$ $$m_2 a_5=m_3 a_6$$

which implies:

$$\frac {m_2}{m_1}=\frac {a_1}{a_2}$$ $$\frac {m_3}{m_1}=\frac {a_3}{a_4}$$ $$\frac {m_3}{m_2}=\frac {a_5}{a_6}$$

All ratios are constant (the gravitational force is time-independent, though this is not necessary). This means:

$$\frac{m_3}{m_1}=\frac {m_3}{m_2}\frac {m_2}{m_1}, $$ or $$k_{31}=k_{32}k_{21}$$

Of course, you can give any of the three bodies mass $m_3$ (and $m_1$ or $m_2$), so this holds for any two bodies.

In the same way, we could write a relation between the magnitudes of the accelerations:

$$\frac {a_3}{a_4}=\frac {a_5}{a_6}\frac{a_1}{a_2}$$

Returning to the $k_{ij}$'s, it's obvious that $k_{12}=\frac{m_1}{m_2}=\frac{1}{\frac{m_2}{m_1}}=\frac{1}{k_{21}}$. So because $m_1=1$, $m_2=k_{21} kg$, $k_{32}=\frac{k_{31}}{k_{21}}=\frac{{m_3}{m_1}}{{m_2}{m_1}}=\frac{m_3}{m_2}$, and from $\pmb{a_1}=−k_{21}\pmb{a_2}$ it follows $\pmb{a_1}=-m_2 \pmb{a_2}$.

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  • $\begingroup$ I think it would be best to delete the first two paragraphs. Paragraph 1 sounds like incoherent rambling. Paragraph 2 sounds irrelevant to my questions (where did $\frac{1}{r^2}$ come from?). $\endgroup$ Jun 14, 2020 at 21:16
  • $\begingroup$ The point of my question was to understand the authors' explanation. But it is not clear to me that your answer attempts to clarify the authors' explanation. Rather, it seems that you use your own explanation, and then attempt to somehow tie it back to the authors' explanation in some convoluted way. $\endgroup$ Jun 14, 2020 at 21:19
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The author has chosen to define $k_{ij}$ as $m_1/m_2$.

this implies $k_{12}=\frac{m_1}{m_2}$

and

$k_{21}=m_2/m_1$

Hence, $k_{12}=1/k_{21}$

Similarly, for 3 bodies,

$k_{32}=\frac{m_3}{m_2}=\frac{m_3}{m_1}.\frac{m_1}{m_2}=k_{31}.k_{12}$

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Well I am assuming you still want to know

i) How $k_{32}=m_3/m_2$:

By Newton's third law,

$m_2a_2$ = $-m_3a_3$

Also $a_2$ = $-k_{32}a_3$

If you divide these equations,

You get $k_{32}$=$m_3$/$m_2$

ii)$k_{31}=k_{32}k_{21}$

You can follow from the result in (i) that

$k_{32}k_{21}$=$(m_3/m_2)(m_2/m_1)$

Simplifying you get

$k_{32}k_{21}$=$m_3/m_1$

So

$k_{32}k_{21}=k_{31} $

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  • $\begingroup$ Division does not make sense for vectors; what you're looking for in (1) is substitution. $\endgroup$ Jun 21, 2020 at 21:35
  • $\begingroup$ Well i had taken the magnitudes into account as the direction vectors cancel out if I take a minus sign.However substitution works too. $\endgroup$ Jun 22, 2020 at 6:40
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The constants $k_{12}$ and $k_{21}$ are nothing but $m_1/m_2$ and $m_2/m_1$ respectively.So $k_{12} = 1/k_{21}$ and no energy is lost in the collision. $k_{31} = m_3/m_1$ and $k_{32} = m_3/m_2$. Therefore $m_3/m_1 = (m_3/m_2)(m_2/m_1)$. So the statements are nothing except maths.

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    $\begingroup$ I don't understand. It seems that you're claiming that $k_{12} = m_1/m_2$ and $k_{21} = m_2/m_3$, but then we would have that $k_{12} = 1/k_{21} = m_3/m_2$. Your answer is missing context, so it's very unclear to me. $\endgroup$ Jun 7, 2020 at 16:42
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    $\begingroup$ @ThePointer This first sentence of this answer is not correct. $k_{ij} = \frac{m_i}{m_j}$, and as such $k_{12} = \frac{m_1}{m_2} = \left( \frac{m_2}{m_1} \right)^{-1} = \frac{1}{k_{21}}$. It is likely a typo? $\endgroup$
    – 13509
    Jun 7, 2020 at 22:10
  • $\begingroup$ @JamesWirth I'm not sure. As you can see from my previous comment, this answer didn't seem correct to me either. $\endgroup$ Jun 7, 2020 at 22:26
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    $\begingroup$ @ThePointer I meant to say it is definitely a typo. I gave some more details in my answer. $\endgroup$
    – 13509
    Jun 7, 2020 at 22:41
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    $\begingroup$ Im so sorry everyone It was a typo.I wrote M3 instead of m1 $\endgroup$ Jun 8, 2020 at 8:18

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