Clarification of "inertial mass" explanation

Question

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

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

---

Answer 1

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.

Answer 2

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

Answer 3

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 $\pmb{a_1}=-m_2 \pmb{a_2}$.

Answer 4

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,

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

Answer 5

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} $

Answer 6

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.