Distribution (Dirac&standard) formulations of f=ma how do they go again?

[James MC]

Hi, I'm wondering how to formulate Newton's second law in terms of Dirac delta functions and standard mass density distributions.

We define the Dirac delta function of a point mass with mass m located at point x₀ as follows:
$$\rho(x)=m\delta (x-x_0)$$
If our point mass is represented by a Dirac delta function, our acceleration will need to be too. So we need an acceleration distribution for our point mass that assigns an instantaneous acceleration a to x₀:
$$a(x)=a\delta (x-x_0)$$
If that is correct, then is the correct Dirac delta function formulation of f=ma:
$$F_i = [\int m\delta (x-x_i)dx] [\int a\delta (x-x_i)dx]$$
...for the force of particle i? And, for the total force F_T for N particles, is it:
$$F_T = \sum_i [[\int m\delta (x-x_i)dx] [\int a\delta (x-x_i)dx]]$$
Okay, so if that's all good, is this the equation for standard mass density distributions (not for point particles, but for mass densities distributed over tiny regions):
$$F_T = \sum_i [\int [\rho(x) a(x)]dx]$$
Where ρ(x) is now defined at dM/dV and a(x) as da/dV. Notice that the Dirac delta formulation has TWO integrals while the mass density formualtion has ONE integral. I think that's right because you cannot multiply delta functions because, pre-integration, that would be multiplying infinities? Whereas with the densities, you are instead multiplying finite values and so you CAN multiply BEFORE the integration?

Or am I just completely confused here? Any advice would be most welcome!

 

[Jano L.]

Hello James,

using delta distributions is not very convincing way to find that expression. You can instead proceed directly from the definition of force

$$
\mathbf F = \frac{d\mathbf p}{dt}
$$

for the fluid in question. Enclose the whole body in surface S of volume V so that nothing comes in or out. Then momentum of the fluid is

$$
\mathbf p = \int_V \rho \mathbf v~ dV
$$

and the experienced force

$$
\mathbf F = \frac{d}{dt} \int_V \rho \mathbf v ~dV.
$$

If the surface does not move, we can move in with the time derivative into the integrand:

$$
\mathbf F = \int_V \partial_t \rho \,\mathbf v + \rho\mathbf a~dV.
$$

In case the fluid is incompressible (volume is constant), the first term vanishes and we end up with the formula you guessed above. The only thing is that there is no summation over $i$, as we already introduced integration over continuous functions.

 

[James MC]

Hi Jano, thanks for your response.

I don't understand what you mean when you say "using delta distributions is not a very convincing way to find that expression". If "that" refers to my final expression, I was not trying to derive that expression. I actually just want to know if the equations themselves are mathematically consistent. And I need to stick with delta functions for my purposes.

Another way to put my question is this: which is the correct formulation of Newton's second law in delta function terms:

$$F_i = [\int m\delta (x-x_i)dx] [\int a\delta (x-x_i)dx]$$
Or
$$F_i = \int [m\delta (x-x_i) a\delta (x-x_i)]dx$$

I would very much like to know the answer to this (I'm hoping it's the latter, but suspect due to the weird infinities, you can't multiply DDFs, so it's the former?).

I realize it's a non-standard formulation, but I don't see how the momentum formulation can help me. It's hard to say why without giving you some of the background of my project, which is somewhat complicated and so may discourage readers from responding, when really, I just need someone competent with DDFs to point out the correct formulation. Hope you can help!

 

[Jano L.]

The first expression can be used, but the second is, as you rightly suspect, problematic.

Delta distributions sometimes appear in a product, if they are concentrated at different points of real line. But two equal delta distributions are not meant to be multiplied together - it is not cleat what such object means.

It would help if you could describe at least telegraphically what is it you want to achieve with these delta distributions.

 

[James MC]

Textbook proofs of mass additivity go something like this: Where $F_T$ is the total force on a number of point masses indexed by i (due to an external source, which I do not specify, so as to simplify the equations), we begin with:
$$\sum_i m_i\mathbf{a}_i = \mathbf{F}_T$$
It is then assumed that the particles indexed by i compose a composite body and that the force on the composite is equivalent to the total force on the parts ($F_T=F_c$). A physical situation is then assumed in which the accelerations of the particles indexed by i are identical, which allows the following transformation:
$$\mathbf{a}_i\sum_i m_i = \mathbf{F}_c$$
It is then assumed that the acceleration of the composite $\mathbf{a}_c$ just is the acceleration of its parts so that:
$$\mathbf{a}_c\sum_i m_i = \mathbf{F}_c$$
Which is taken to show that given the force and acceleration of the composite, the composite's mass must be additive, that is:
$$\sum_i m_i = m_c$$
Hence mass is additive.

I am just trying to formulate a version of this proof that does not assume that the parts have identical accelerations. Because I wish to stay in the context of point particles, it seems Dirac delta functions may be the only hope.

So far, the generalised proof looks like this:

Fundamental many-particle law:
$$\sum_i \int m_i(\mathbf{x}) \mathbf{a}_i(\mathbf{x})\delta (\mathbf{x}-\mathbf{x}_i)d^3\mathbf{x} = \mathbf{F}_T$$
Recover the form of the single particle law (this involves deriving $\mathbf{a}_c(\mathbf{x})$, the acceleration distribution of the composite, from the acceleration distributions of the parts, which I'll just assume here):
$$\int \mathbf{a}_c(\mathbf{x})\sum_i m_i(\mathbf{x}) \delta (\mathbf{x}-\mathbf{x}_i)d^3\mathbf{x} = \mathbf{F}_T$$
Assume that the force on the composite is the force on the parts:
$$\int \mathbf{a}_c(\mathbf{x})\sum_i m_i(\mathbf{x}) \delta (\mathbf{x}-\mathbf{x}_i)d^3\mathbf{x} = \mathbf{F}_c$$
Infer inertial mass additivity.

I hope that makes sense. (In the recent past I have given a slightly more extensive discussion of the project here: https://www.physicsforums.com/showpost.php?p=4257563&postcount=15). Any thoughts would be most welcome.

 

[Jano L.]

I am just trying to formulate a version of this proof that does not assume that the parts have identical accelerations.

That makes a lot of sense. However, you do not need delta distributions for that.

You can repeat the procedure above by using the acceleration of the center of mass.

Total force is equal to sum of forces experienced by all particles:
$$
\mathbf F_T = \sum_i m_i \mathbf a_i.
$$

Acceleration is a second time derivative of the radius vector:

$$
\mathbf a_i = \frac{d^2}{dt^2 }\mathbf r_i,
$$

so

$$
\mathbf F_T = \sum_i m_i\frac{d^2}{dt^2 }\mathbf r_i.
$$

Since the masses $m_i$ are constant in time, the differential operator can be pulled in front of the whole expression:

$$
\mathbf F_T = \frac{d^2}{dt^2 } \sum_i m_i \mathbf r_i,
$$

which can be written as

$$
\mathbf F_T = \sum_i m_i \frac{d^2}{dt^2 } \frac{\sum_i m_i \mathbf r_i}{\sum_i m_i}.
$$

or

$$
\mathbf F_T = \sum_i m_i \frac{d^2}{dt^2 } \frac{\sum_i m_i \mathbf r_i}{\sum_i m_i}.
$$

So, the composite body moves in such a way that its center of mass

$$
\mathbf R = \frac{\sum_i m_i \mathbf r_i}{\sum_i m_i}~~(*)
$$

moves as if it was particle with mass $\sum_i m_i$.

Thus in this line of thought, the additivity of mass is a consequence of the choice of (*) for the definition of the center of mass. Other choices come to mind, but they are more complicated and is not clear whether they offer some advantage.

 

[James MC]

Thanks Jano, that's an interesting suggestion. If I understand you, your strategy falls under solution 1, whereas mine falls under solution 2:

Solution 1: Find a non-arbitrary, non ad hoc way to define the acceleration of a composite as a single point acceleration. Then, once one knows the force of the composite one can calculate the composite's mass using the standard equation for Newton's second law.

Solution 2: Find a non-arbitrary, non ad hoc way of reformulating the standard equation for Newton's second law, so that it applies to composites that have distributed locations, and distributed accelerations. Determine composite force then solve for composite mass.

Here is the concern I have with all instances of Solution 1 generally: I don't see how one can find a non-arbitrary way to relate (yet alone identify) the acceleration of a composite with the acceleration of the composite's centre of mass, without presupposing mass additivity. For that reason, I don't think one can appeal to centres of mass if the goal is to explain mass additivity.

I'll try to apply this to the particular instance of Solution 1 you've provided.

I take it we agree that the goal is start with our Newtonian fundamentals:
$$\sum_i m_i\mathbf{a}_i = \mathbf{F}_T$$
And to prove something like the following:
$$\mathbf{a}_c\sum_i m_i = \mathbf{F}_c$$
Because then if we know the composite force and the composite acceleration, then we can solve for composite mass, and thereby explain composite mass.

We can derive $F_T=F_c$ easily enough: it is a priori that the force on a composite is identical to the total force on its parts.
So the question is: how can we derive the acceleration of the composite? That is, how can we introduce $a_c$ in a way that does not presupoose what we are trying to explain?

Your suggestion, as I understand it, was: treat $a_c$ as whatever weighted average of component accelerations it needs to be in order to get $F_c$ assuming that $m_c$ = $\sum_im_i$. You're absolutely right that the acceleration of the centre of mass is what we get. But this presupposes mass additivity.

The alterntive route (Solution 2) is to derive a value for $a_c$ directly, without presupposing mass additivity. Here, one can simply treat the acceleration of a composite as what it is physically: a distribution of accelerations. In that case, we can describe the acceleration of the composite with a Dirac Delta function. This is why I think solution 2 can solve the problem and explain mass additivity.

Let me know your thoughts.

 

[Jano L.]

Your idea is not clear to me, because of you have distribution of accelerations, you have also distribution of mass and there is no one mass of the composite.In the above, I defined

$$
\mathbf R = \frac{\sum_i m_i \mathbf r_i}{\sum_i m_i},
$$

without supposing that the mass is additive. True, the definition is probably motivated by the additivity, but does not require it. If the definition is adopted, the additivity follows as a nice consequence of that definition.

Because there may be other definitions of the "center" of the body, I think that the additivity is not necessarily true in mechanics. For example, if we chose

$$
\mathbf R = \frac{\sum_i m_i \mathbf r_i}{\sqrt{\sum_i m_i^2}},
$$
to be the center, then the mass of the composite is

$$
M = {\sqrt{\sum_i m_i^2}}.
$$

This is strange, but it works as well.

However, it seems to unnecessarily cumbersome to calculate mass in this way - so we adopt the first definition.

 

[James MC]

Your idea is not clear to me, because if you have distribution of accelerations, you have also distribution of mass and there is no one mass of the composite.

This is a fair challenge. My transformations of the DDF formulation of F=MA, entail mass distribution additivity. That is, they show that $\sum_i m_i\delta (x-x_i)dx$ is the mass distribution of the composite. I take it your worry is that mass distribution additivity ≠ mass additivity. However, to derive mass additivity from mass distribution additivity, I only need to integrate $\sum_i m_i\delta (x-x_i)dx$. That is another upshot of using DDF's here: the value of a quantity represented by a DDF is by definition, its integral. This means that to derive mass additivity, I only need to appeal to the following equalities:
$$\int \sum_i f(x) \delta (x-x_i)dx = \int [ f(x)\delta (x-x_1)+f(x)\delta (x-x_2)+~.~.~.~+ f(x)\delta (x-x_n)]dx$$
$$=f(x_1)+f(x_2)+~.~.~.~+f(x_n)=\sum_i f(x_i)$$
Hence, $\int \sum_i m_i\delta (x-x_i)dx = \sum_i m_i$ Hence mass additivity.

Does that clarify the idea?

Because there may be other definitions of the "center" of the body, I think that the additivity is not necessarily true in mechanics.

I don't see how one could defend the position that additivity is not true in Newtonian mechanics. Newtonian mechanics is not some inapplicable abstract mathematical formalism, in which it is open to us to make such stipulations. Newtonian mechanics is a theory of the physical world, and answers to experiment. Mass additivity is empirically confirmed within Newtonian mechanic's domain of validity (e.g. slow moving macroscopic bodies) whereas this equation:
$$
M = {\sqrt{\sum_i m_i^2}}.
$$
is ruled out by experiment.

One can strengthen this argument by appeal to a more realistic case in which mass additivity does not hold: relativity theory. In relativity theory, there is a very specific formulation for the centre of mass of a composite:
$$
\mathbf R = \frac{\sum_i m_iγ_i \mathbf r_i}{\sum_i m_iγ_i},
$$
This is the correct formulation in virtue of the non-additivity of mass (or the additivity of $m_iγ_i$), which is an empirically verifiable phenomenon, not a stipulation.

Furthermore, an argument for why additivity is in fact true in Newtonian mechanics, is that I have proved it! Although, whether you're convinced by that argument, will depend, I guess, on whether you think I've successfully responded to your challenge from above.

 

[Jano L.]

Fundamental many-particle law:

∑i∫mi(x)ai(x)δ(x−xi)d3x=FT

Recover the form of the single particle law (this involves deriving ac(x), the acceleration distribution of the composite, from the acceleration distributions of the parts, which I'll just assume here):
∫ac(x)∑imi(x)δ(x−xi)d3x=FT

In the above, you seem to think that when

$$
\sum_i m_i \mathbf a_i
$$

is rewritten into


$$
\int \rho(\mathbf x) \mathbf a(\mathbf x) dV
$$

with $\rho(\mathbf x) = \sum_i m_i \delta(\mathbf x-\mathbf x_i)$ and some new convenient function $\mathbf a(\mathbf x)$, the additivity of mass is proven for point particles.

I do not see why do you think such thing. The mere rewrite of the expression does not prove anything. There is no reason to introduce mass distribution function and call

$$
\int \rho(\mathbf x) dV
$$

total mass unless one is sure/assumes that the mass is additive.

I think that the additivity of inertial mass can proven by considering the equation of motion as I did above. In relativity, the equation is different and thus we do not have such additivity.

I think the standard choice of center of mass is the best one, and the additivity follows as a consequence of this assumption, the equation of motion and the superposition principle.

But if somebody finds some other center $\mathbf C$ useful, he may use it, and for him the inertial mass $M$ in the equation

$$
M\ddot{\mathbf C} = \mathbf F
$$

may not be additive. This would be very cumbersome to use for description of experiments, because the new center would much differently than usually expected, but this cannot rule it out, as it is mathematically as correct as the standard choice.

So perhaps the better and more interesting question is, why do we choose $\frac{\sum_i m_i \mathbf r_i}{\sum_i m_i}$ to define the motion of the body, and not some other point?

And I think the answer is additivity of mass. I other words, the additivity is the basic motivation and the theory is so developed that it conforms, but it is also true that it is kind of luck, since in relativity, this is no longer possible - potential energy changes inertial mass of the composite body in a way that is not additive.

 

[James MC]

Apologies for the delay...

In the above, you seem to think that when
$$
\sum_i m_i \mathbf a_i
$$
is rewritten into
$$
\int \rho(\mathbf x) \mathbf a(\mathbf x) dV
$$
with $\rho(\mathbf x) = \sum_i m_i \delta(\mathbf x-\mathbf x_i)$ and some new convenient function $\mathbf a(\mathbf x)$, the additivity of mass is proven for point particles.
I do not see why do you think such thing. The mere rewrite of the expression does not prove anything. There is no reason to introduce mass distribution function and call
$$
\int \rho(\mathbf x) dV
$$
total mass unless one is sure/assumes that the mass is additive.

I don't believe I am assuming mass additivity. I will go through my proof step by step to try to show why. Hopefully that will make it easy for you to pick out the exact step that is concerning you.

Start with the textbook explanation of mass additivity that I mentioned a while back, that I thought you and I agreed was valid yet incomplete. (I take us to be disagreeing on how to complete it - Solution 1 (centre of mass) or Solution 2 (distributions)).

We begin with a simple physical situation involving Newtonian particles, and we state that the law applying to them is Newton's many particle law:
$$\sum_i m_i\mathbf{a}_i = \mathbf{F}_T$$
Now, I'll take the textbook proof step by step.

Step 1: Infer the existence of composite C from the existence of the particles. (Presumably this is trivial).
Step 2: Infer the force on C via the following principle: If the superposition principle tells us that the force on some particles that compose C, is F, then the force on C is F. (Again, this is presumably trivial.) We may now infer:
$$\sum_i m_i\mathbf{a}_i = \mathbf{F}_C$$
Step 3: Stipulate that the accelerations of the particles indexed by i are identical, which allows the following transformation:
$$\mathbf{a}_i\sum_i m_i = \mathbf{F}_C$$
Step 4: The crucial acceleration step: The acceleration of the composite $\mathbf{a}_c$ just is the acceleration of its parts so that:
$$\mathbf{a}_c\sum_i m_i = \mathbf{F}_c$$
Step 5: Infer mass additivity:
Mass is defined as that property of objects responsible for their resistance to changes in motion given applied forces. The equation in step 4 tells us that the property responsible for the composite's disposition to resist acceleration given applied forces is the sum of the masses of its parts. Hence we can derive:
$$\sum_i m_i = m_c$$
Hence mass is additive.

It strikes me that if you think this proof is valid and non-question-begging (for the limited situation it deals with) then you should also think that my proof is valid and non-question-begging (for the general situation); this is because my proof is structurally identical:

We begin with a physical situation involving Newtonian particles, and we state that the law applying to them is Newton's many particle law:
$$\sum_i m_i\mathbf{a}_i = \mathbf{F}_T$$
Now, independently of the issue of mass additivity, one can straightforwardly derive a distribution form of this law:
$$\sum_i[\int \rho(\mathbf x)_i \mathbf a(\mathbf x)_i dV] = \mathbf{F}_T$$
(Note that it might be easier to view these as standard distributions, rather than deltas, so as to avoid (for now) technical complications concerning distribution multiplication.) This does not presuppose mass additivity. If one grasps distribution formalism and is asked to formulate the equivalent of the second law + superposition principle, one infers this, and one does so without any thought of the properties of composites.

Now comes my structurally identical proof:

Step 1: Infer the existence of composite C from the existence of the particles. (Presumably this is trivial).
Step 2: Infer the force on C via the following principle: If the superposition principle tells us that the force on some particles that compose C, is F, then the force on C is F. (Again, this is presumably trivial.) We may now infer:
$$\sum_i[\int \rho(\mathbf x)_i \mathbf a(\mathbf x)_i dV] = \mathbf{F}_C$$
Step 3: Here, we do not stipulate that the accelerations of the particles indexed by i are identical, yet we still want an analogous transformation:
$$\int [[\sum_i \mathbf a(\mathbf x)_i][\sum_i\rho(\mathbf x)_i]] dV = \mathbf{F}_C$$

Step 4: The crucial acceleration step: The textbook explanation justifies this step by just saying "acceleration of the composite just is the acceleration of the parts". My justification is along the same lines, but I think a little more sophisticated: By definition, acceleration is the second time derivative of position. The composite is positioned where its parts are positioned - a set of positions (trivial). Therefore, the composite acceleration is a set of time derivatives. In distribution formalism, we may then deduce:
$$\mathbf a(\mathbf x)_C = \sum_i \mathbf a(\mathbf x)_i$$
Which yields the desired result, in which we recover the (distribution) form of the single particle law:
$$\int [[\mathbf a(\mathbf x)_C][\sum_i\rho(\mathbf x)_i]] dV = \mathbf{F}_C$$
Step 5: Infer mass additivity:
Mass is defined as that property of objects responsible for its resistance to changes in motion given applied forces. The equation in step 4 tells us that the property responsible for the composite's disposition to resist acceleration given applied forces is the sum of the mass distributions of its parts. Hence we can derive:
$$\mathbf \rho(\mathbf x)_C = \sum_i \rho(\mathbf x)_i$$
Hence mass distributions are additive. We can then derive mass additivity, from mass distribution additivity, as discussed in my previous post.

I do not think any step in either proof presupposes mass additivity, though if you think one or more steps are problematic, I would be very interested to know why.

I think that the additivity of inertial mass can proven by considering the equation of motion as I did above. In relativity, the equation is different and thus we do not have such additivity.

I think the standard choice of center of mass is the best one, and the additivity follows as a consequence of this assumption, the equation of motion and the superposition principle.

I completely agree. If you allow as a premise that the composite's position is the centre of mass, then the proof goes through. My point is that a proof containing an unmotivated premise is not an explanation. And to motivate that premise you require what you're trying to explain (mass additivity). Hence, the proof is not an explanation of mass additivity.

I think there are more powerful objections to this approach, but the above is sufficient for my purpose, which is to motivate the need for my explanation. (These objections appeal to the fact that nothing, let alone the composite, is positioned at the COM.)

But if somebody finds some other center $\mathbf C$ useful, he may use it, and for him the inertial mass $M$ in the equation

$$
M\ddot{\mathbf C} = \mathbf F
$$

may not be additive. This would be very cumbersome to use for description of experiments, because the new center would much differently than usually expected, but this cannot rule it out, as it is mathematically as correct as the standard choice.

While it is mathematically consistent it is still incorrect because it is inconsistent with, and refuted by, experiment. One can show $M = {\sqrt{\sum_i m_i^2}}.$ is false, by measuring an object's disposition to resist acceleration given a force, and then cutting the object in half, and determining their dispositions to resist acceleration given the same force.

So perhaps the better and more interesting question is, why do we choose $\frac{\sum_i m_i \mathbf r_i}{\sum_i m_i}$ to define the motion of the body, and not some other point?

And I think the answer is additivity of mass. In other words, the additivity is the basic motivation and the theory is so developed that it conforms, but it is also true that it is kind of luck, since in relativity, this is no longer possible - potential energy changes inertial mass of the composite body in a way that is not additive.

I think that the case of relativity strengthens my argument that the person who chooses a different centre, and hence a different composite mass, advocates a false theory. The relationship between the mass of the composite and the mass of the parts, is not a matter of convenient stipulation of a mass centre. Rather, it is a matter of experimental physics. And it was an extrodinary discovery about nature herself, that a box of hot gas offers more resistance to acceleration than the same box after it is cooled. The mass-composite/mass-parts relationship is a consequence of the fundamental laws alone and relativity shows that when you change the laws (and nothing else), you change the observed relationship. I intend to prove the relationship from the laws, starting with the Newtonian case.

My apologies for the length of the post, I hope it does not deter you from responding. I doubt that I've convinced you of my view just yet, and if so, I would be quite interested to learn why.

 

[Jano L.]

The problem is already beginning to appear in the expression

$$
\sum_i[\int \rho(\mathbf x)_i \mathbf a(\mathbf x)_i dV] = \mathbf{F}_T.
$$

How do you define $\rho(\mathbf x)_i$ and $\mathbf a(\mathbf x)_i$ ?

If we used DD in both quantities, then the integral is invalid expression, as two same DD cannot be multiplied.

If we used ordinary continuous functions, then it would be incorrect, as the term $\partial_t\rho\, \mathbf a$ would be missing.

We can try to introduce square-root DD (bit non-standard)

$$
\sqrt{\delta(\mathbf x-\mathbf r_i)}
$$

and use one for

$$
\rho_i(\mathbf x) = m_i \sqrt{\delta(\mathbf x- \mathbf r_i)}
$$

and one for acceleration

$$
\mathbf a_i(\mathbf x) = \mathbf a_i \sqrt{\delta(\mathbf x- \mathbf r_i)},
$$

so the first integral above gives $\sum_i m_i \mathbf a_i$ correctly, but then the integral of rho is wrong:

$$
\int \rho_i dV = 0.
$$

So I do not see how your expressions can make sense.

 

[James MC]

The problem is already beginning to appear in the expression

$$
\sum_i[\int \rho(\mathbf x)_i \mathbf a(\mathbf x)_i dV] = \mathbf{F}_T.
$$

How do you define $\rho(\mathbf x)_i$ and $\mathbf a(\mathbf x)_i$ ?

If we used DD in both quantities, then the integral is invalid expression, as two same DD cannot be multiplied.

This is certainly correct, and is what I see as the main outstanding problem with my argument. I have three completed proofs already (I think!), gravitation and inertia proofs in the continuous context, and a gravitation proof in the DD context, but I haven't been able to prove inertial mass additivity in the DD context for precisely this reason. I make a suggestion at the bottom though.

If we used ordinary continuous functions, then it would be incorrect, as the term $\partial_t\rho\, \mathbf a$ would be missing.

I don't really understand what you mean because I don't see what $\partial_t\rho\, \mathbf a$ would add to the equation.

Perhaps this points to the fact that I haven't been fully explicit about my simplified scenario. I'm working with a fixed volume of space, and my particles (arbitrarily small continuous mass density distributions) are confined to that volume, and do not leave that volume. Furthermore, my particles are frozen in time, so to speak. Obviously, they are accelerating, but I am taking a snap shot of them at a time, and then applying my equation to them, to determine total force, at that time. Given these simplifications, I don't see what a partial time derivative of both the mass and acceleration densities would add. Although, I'm also not entirely sure what work it would do if I removed these simplifications. Can you please say more to clarify what you have in mind?

In the continuous case, my definitions are as follows:
$\rho = M/V$ for an object whose mass is evenly distributed over V.
$\rho(\mathbf x) = ΔM(x)/ΔV(x)$ for non-uniform distributions.
$\rho(\mathbf x) = dM/dV$ taking the limit, where our volumes are infinitesimal.
My acceleration distributions are defined analogously, where instantaneous accelerations replace the masses...
$\mathbf a = \mathbf a/V$ for an object with a uniformly distributed instantaneous acceleration.
$\mathbf a(\mathbf x) = Δ\mathbf a(x)/ΔV(x)$ for non-uniform distributions.
$\mathbf a(\mathbf x) = d\mathbf a/dV$ taking the limit, where our volumes are infinitesimal.

We can try to introduce square-root DD (bit non-standard)

$$
\sqrt{\delta(\mathbf x-\mathbf r_i)}
$$

and use one for

$$
\rho_i(\mathbf x) = m_i \sqrt{\delta(\mathbf x- \mathbf r_i)}
$$

and one for acceleration

$$
\mathbf a_i(\mathbf x) = \mathbf a_i \sqrt{\delta(\mathbf x- \mathbf r_i)},
$$

so the first integral above gives $\sum_i m_i \mathbf a_i$ correctly, but then the integral of rho is wrong:

$$
\int \rho_i dV = 0.
$$

So I do not see how your expressions can make sense.

I'm inclined to think that because the three other proofs I mentioned work (I think!), then it must be that the DD verison of the inertial mass additivity proof works too, it's just a matter of getting the technicalities right.
Now, while you're correct in saying that this expression is invalid:
$$\sum_i \int m_i(\mathbf{x})\delta (\mathbf{x}-\mathbf{x}_i) \mathbf{a}_i(\mathbf{x})\delta (\mathbf{x}-\mathbf{x}_i)d^3\mathbf{x} = \mathbf{F}_T$$
It's not so clear to me that this is invalid:
$$\sum_i \int m_i(\mathbf{x}) \mathbf{a}_i(\mathbf{x})\delta (\mathbf{x}-\mathbf{x}_i)d^3\mathbf{x} = \mathbf{F}_T$$
In other words, the latter equation does not multiply DD's (the invalid bit); there is only one DD, and it assigns the product of m and $\mathbf{a}$ to $x_i$. So the m and a terms need to be defined as functions on the delta function and not delta functions themselves? I suspect this latter equation is the correct DD formulation of Newton's second law. Running my proof with it is another story.

 

[Jano L.]

Obviously, they are accelerating, but I am taking a snap shot of them at a time, and then applying my equation to them, to determine total force, at that time. Given these simplifications, I don't see what a partial time derivative of both the mass and acceleration densities would add.

Your definitions are not clear enough and however I try to interpret them, they are invalid. There is no sense in introducing distribution of acceleration by
$$
\mathbf a(\mathbf x) = \mathbf a / \Delta V.
$$

Acceleration is not distribution, but an ordinary function.


I think you would better read first chapters of some book on hydrodynamics, for example Landau and Lifgarbagez. Then you will see how the theory can be built consistently, when the additivity of mass is assumed right from the beginning.

In continuum mechanics, the total force acting on the volume of fluid contained within a fixed boundary is

$$
\mathbf F = \frac{d}{dt} \int_{V}\bigg( \rho \mathbf v \bigg)\,dV.
$$

Moving in with the derivative and differentiating the product, you get

$$
\int_V \partial_t \rho\, \mathbf v + \rho \mathbf a\, dV.
$$

The last integral in your last post is OK, provided it is meant for point-like particles and $m_i(\mathbf x),\mathbf a_i(\mathbf x)$ are ordinary functions of $\mathbf x$. How do you define them? Your previous post did not explain that. Is $\mathbf a_i(\mathbf x)$ nonzero also in places where the particle $i$ is not present?

 

[James MC]

Your definitions are not clear enough and however I try to interpret them, they are invalid. There is no sense in introducing distribution of acceleration by
$$
\mathbf a(\mathbf x) = \mathbf a / \Delta V.
$$

Acceleration is not distribution, but an ordinary function.

Acceleration distributions are certainly non-standard, although it appears they come up every so often in places (e.g. here.) But you're right that those definitions are insufficient without some background.

I'm not sure how to best formally define the acceleration distributions, but I'm basically thinking of them as something like functions from infinitesimal regions to instantaneous accelerations; or sets of ordered pairs.

Assume a physical situation in which we have two infintesimal mass densities located at distinct infinitesimal regions x₁ and x₂, at t₁. We can assign acceleration distributions to each particle at t1. One of these distributions will assign x₁ some (potentially) non-zero instantaneous acceleration value, zero elsewhere. The other distribution assigns x₂ some (potentially) non-zero instantaneous acceleration value, zero elsewhere. One can also define a composite acceleration distribution, which assigns (potentially) non-zero instantaneous accelerations to two infinitesimal regions, zero at every other infinitesimal region.

Then I think one just needs to define the integral accordingly - something like a step function, I take it, would be best suited.

Perhaps I'm missing something, but that idea seems reasonably clear to me, I'm just sure what the best mathematical vocabulary for expressing it precisely would be.

I think you would better read first chapters of some book on hydrodynamics, for example Landau and Lifgarbagez. Then you will see how the theory can be built consistently, when the additivity of mass is assumed right from the beginning.

In continuum mechanics, the total force acting on the volume of fluid contained within a fixed boundary is

$$
\mathbf F = \frac{d}{dt} \int_{V}\bigg( \rho \mathbf v \bigg)\,dV.
$$

Moving in with the derivative and differentiating the product, you get

$$
\int_V \partial_t \rho\, \mathbf v + \rho \mathbf a\, dV.
$$

Is v the velocity of the centre of mass in these equations? I take it it's not a velocity distribution?

I've gone over Truesdell's "A First Course in Rational Continuum Mechanics", but I didn't find it to be of much help. I think that's because I'm not trying to do continuum mechanics, and I'm not describing fluids. I'm describing fixed infintitesimal masses separated by relatively large regions of space. And I'm doing it in a way that should seem quite odd to someone who doesn't realize that I've just been trying to get this mass additivity proof to work.

It's important to the proof that one specifies the superposition principle separately form the integral, so as to indicate that we are describing (to begin with) many distinct particles. We then try to recover the single object law by moving the sum into the equation so that it attaches to the masses.

The last integral in your last post is OK, provided it is meant for point-like particles and $m_i(\mathbf x),\mathbf a_i(\mathbf x)$ are ordinary functions of $\mathbf x$. How do you define them? Your previous post did not explain that. Is $\mathbf a_i(\mathbf x)$ nonzero also in places where the particle $i$ is not present?

Yes, I use the DD equations when describing point particles, and the quasi-continuum-mechanics formalism when describing particles that are infinitesimal mass densities.

At first I was thinking of defining the Dirac delta function of a point mass with mass m located at point x₀ as follows:
$$\rho(x)=m\delta (x-x_0)$$
The acceleration distribution for the point mass that assigns an instantaneous acceleration a to x₀ is then:
$$\mathbf a(x)=\mathbf a\delta (x-x_0)$$
It then looks like if you multiply $\rho(x)$ with $\mathbf a(x)$ then you're multiplying two DD's, which is bad, but I think if you specify only one DD in the actual equation, as I did in the equation you thought valid, then this means that the DD is just assigning the product of two values to points in space, which avoids the problem.
To answer your question, yes because the subscript "i" serves to just pick out the right value, i.e., the acceleration of the particle located at i. Every point gets assigned that value and then the DD puts every point to zero except point i.

I hope I'm making some sense! Sorry that I can't be more mathematically rigorous about this, I've had a pretty non-standard physics education.

 

[Jano L.]

Is v the velocity of the centre of mass in these equations? I take it it's not a velocity distribution?

All the quanties $\rho(\mathbf x),\mathbf v(\mathbf x), \mathbf a(\mathbf x)$ in
$$
\int_V \partial_t \rho\, \mathbf v + \rho \mathbf a\, dV.
$$
are functions of position in space $\mathbf x$.

Every point gets assigned that value and then the DD puts every point to zero except point i.

I hope I'm making some sense!

Yes, except that i-th delta function can sometimes choose non-i-th acceleration and that will spoil the result. The step when you change sum of products into a product of two sums requires that no cross terms contribute. This is possible if you define

$$
\mathbf a_i(\mathbf x) = \mathbf a_i D(\mathbf x - \mathbf r_i),
$$

$$
\mathbf \rho_i(\mathbf x) = m_i \delta(\mathbf x - \mathbf r_i),
$$

where $D(\mathbf x - \mathbf r_i)$ is a characteristic function of some small ball, i.e. function that is equal to 1 if $|\mathbf x - \mathbf r_i| < a$ and equal to zero otherwise.

Then, if the mutual distances of the particles are greater than $a$, cross terms are zero and the manipulation works.



However, the resulting object

$$
\mathbf a_C = \sum_i \mathbf a_i D(\mathbf x-\mathbf r_i)
$$

is quite artificial, because it depends on the choice of $a$. But it is possible that particles can come closer than $a$ for any choice of $a$ and then the manipulation fails.

Disregarding that, even it the manipulation works for given situation, I do not think that success in re-expressing $F = ma$ by

$$
\int \rho_C \mathbf a_C dV
$$

is a derivation of mass additivity. The expression is not in the form $\mathbf F= m_C \mathbf a_C$, which is necessary to find composite mass $m_C$. Then I do not see any convincing reason to adopt the idea that $\mathbf a_c$ is or represents object's acceleration and $\rho_C$ its mass distribution; they are just strange new objects defined in the process.

 

[James MC]

All the quanties $\rho(\mathbf x),\mathbf v(\mathbf x), \mathbf a(\mathbf x)$ in
$$
\int_V \partial_t \rho\, \mathbf v + \rho \mathbf a\, dV.
$$
are functions of position in space $\mathbf x$.

I see. Then the use of an acceleration distribution (or acceleration function - perhaps I'm confusing these) within the integrals was not the problem. In that case, is it possible to run the argument with the equation you've provided? Something like this:

Step 1: Introduce the single density force law
$$
\mathbf F_i = \frac{d}{dt} \int_{V}\bigg( \rho_i \mathbf v_i \bigg)\,dV.
$$
Step 2: Introduce the many density total force law
$$
\mathbf F_T = \sum_i \frac{d}{dt} \int_{V}\bigg( \rho_i \mathbf v_i \bigg)\,dV.
$$
Step 3: Swap the sum and the integral
$$
\mathbf F_T = \frac{d}{dt} \int_{V}\bigg(\sum_i \rho_i \mathbf v_i \bigg)\,dV.
$$
Step 4: Choose some velocity function $v_?$ so that the following holds:
$$
\mathbf F_T = \frac{d}{dt} \int_{V}\bigg(\mathbf v_? \sum_i \rho_i \bigg)\,dV.
$$
Step 5: Argue that $F_T = F_C$ and $v_? = v_C$ so as to recover the form of the single density law applied to the composite C:
$$
\mathbf F_C = \frac{d}{dt} \int_{V}\bigg(\mathbf v_C \sum_i \rho_i \bigg)\,dV.
$$
Step 6: Infer mass density additivity

Are steps 2-4 valid?

Disregarding that, even it the manipulation works for given situation, I do not think that success in re-expressing $F = ma$ by

$$
\int \rho_C \mathbf a_C dV
$$

is a derivation of mass additivity. The expression is not in the form $\mathbf F= m_C \mathbf a_C$, which is necessary to find composite mass $m_C$. Then I do not see any convincing reason to adopt the idea that $\mathbf a_c$ is or represents object's acceleration and $\rho_C$ its mass distribution; they are just strange new objects defined in the process.

I think this might be the same concern that you raised in post 8? There, you said: "Your idea is not clear to me, because if you have distribution of accelerations, you have also distribution of mass and there is no one mass of the composite."

Is the concern that even if the steps of the proof work, I haven't proven mass additivity because I have only proven mass distribution additivity and mass ≠ mass distribution? If so, I think this is a fair concern, but I tried to answer it, at least for the DD case, in post 9. There, the claim was roughly that it is true by definition that the value of the quantity described by a Dirac delta function is its integral. So if I've inferred, for the composite, a DD whose integral is in fact the composite's mass (i.e., is additive) then I've proved mass additivity.

Yes, except that i-th delta function can sometimes choose non-i-th acceleration and that will spoil the result. The step when you change sum of products into a product of two sums requires that no cross terms contribute. This is possible if you define

$$
\mathbf a_i(\mathbf x) = \mathbf a_i D(\mathbf x - \mathbf r_i),
$$

$$
\mathbf \rho_i(\mathbf x) = m_i \delta(\mathbf x - \mathbf r_i),
$$

where $D(\mathbf x - \mathbf r_i)$ is a characteristic function of some small ball, i.e. function that is equal to 1 if $|\mathbf x - \mathbf r_i| < a$ and equal to zero otherwise.

Then, if the mutual distances of the particles are greater than $a$, cross terms are zero and the manipulation works.



However, the resulting object

$$
\mathbf a_C = \sum_i \mathbf a_i D(\mathbf x-\mathbf r_i)
$$

is quite artificial, because it depends on the choice of $a$. But it is possible that particles can come closer than $a$ for any choice of $a$ and then the manipulation fails.

Yes you're right that cross contamination of terms is certainly a problem in the most natural ways of fleshing out the DD inertial proof. I'm pretty sure this doesn't happen in the corresponding proof of gravitational mass additivity in DD form:
$$\mathbf{g}(\mathbf{x})_T=G \sum_i\int \frac{\mathbf{x}-\mathbf{r}}{|\mathbf{x}-\mathbf{r}|^3}~m_i~\delta(\mathbf{r}-\mathbf{x}_i)~\mathrm{d}^3\mathbf{r}$$
=
$$\mathbf{g}(\mathbf{x})_C=G\int \frac{\mathbf{x}-\mathbf{r}}{|\mathbf{x}-\mathbf{r}|^3}\sum_i ~m_i~\delta(\mathbf{r}-\mathbf{x}_i)~\mathrm{d}^3\mathbf{r}$$
I have wondered why an equation that takes the product of mass distribution and position, should be so different from an equation that takes the product of mass distribution with acceleration. I find it hard to believe that the proof is possible for the former, but not the latter!

I have some thoughts about how to remove cross contamination for the inertial DD proof, but they're in their infancy. It might help to first get your thoughts on the above issues, and by then my thoughts on cross contamination will be clearer in my head.

 

[Jano L.]

Is the concern that even if the steps of the proof work, I haven't proven mass additivity because I have only proven mass distribution additivity and mass ≠ mass distribution?

No, I think the problem is that your procedure does not even prove additivity of mass distributions in physical sense.

There are two different notions of distribution causing mess here.


1. distribution in the mathematical sense, as a linear functional acting on some neat function. These behave according to

$$
\sum_i \int D_i(x) f(x) dx = \int [ \sum_i D_i(x) ] f(x) dx,
$$

so one can say mathematical distributions are additive. Your procedure just uses and exhibits this property of delta distribution.

2. distribution in the physical sense, also called density, as an estimate of quantity of some stuff within some volume. In the case of mass, to define density, first you have to know mass M of some volume element V and then calculate

$$
\rho = \frac{M}{V}.
$$

You cannot circumvent this and redefine density as $\rho = \sum_i \rho_i$. That is just using the fact that you want to prove, that the mass density can be described by linear functionals which are additive by definition.

Other way to see this: try to repeat your procedure in the case of relativistic gas, $m_i$ being rest masses of the particles. You will end up with the same formula suggesting additivity, but in fact the rest mass of the gas is not sum of individual rest masses of the particles. The reason is that linear functionals are additive always, while rest mass is not. In order to restore the use of linear functionals in relativity, you have to use inertial mass (also called relativistic mass).

The point is, you have to know first what is additive to decide what to describe by linear functionals.

In Newtonian mechanics, Newton knew that mass is additive and therefore introduced density. It makes no sense to do it the opposite way.

 

[James MC]

No, I think the problem is that your procedure does not even prove additivity of mass distributions in physical sense.

There are two different notions of distribution causing mess here.


1. distribution in the mathematical sense, as a linear functional acting on some neat function. These behave according to

$$
\sum_i \int D_i(x) f(x) dx = \int [ \sum_i D_i(x) ] f(x) dx,
$$

so one can say mathematical distributions are additive. Your procedure just uses and exhibits this property of delta distribution.

Yes, one step of the overall proof uses this - what I below call "step 3".

2. distribution in the physical sense, also called density, as an estimate of quantity of some stuff within some volume. In the case of mass, to define density, first you have to know mass M of some volume element V and then calculate

$$
\rho = \frac{M}{V}.
$$

You cannot circumvent this and redefine density as $\rho = \sum_i \rho_i$. That is just using the fact that you want to prove, that the mass density can be described by linear functionals which are additive by definition.

But at no point do I redefine density as $\rho = \sum_i \rho_i$ I prove it using the below five step proof.

The additivity of linear functionals is insufficient to get my result. What's important are (i) the composition of forces law and (ii) the fact that the law treats masses and forces as linearly proportional. Both are entirely contingent features of the nomic structure of Newtonian worlds, and they can be appealed to to prove mass additivity. In particular, the composition of forces entails the additivity of forces and the linearity of the law entails that if forces are additive then so are masses.

Other way to see this: try to repeat your procedure in the case of relativistic gas, $m_i$ being rest masses of the particles. You will end up with the same formula suggesting additivity, but in fact the rest mass of the gas is not sum of individual rest masses of the particles. The reason is that linear functionals are additive always, while rest mass is not. In order to restore the use of linear functionals in relativity, you have to use inertial mass (also called relativistic mass).

I do not agree that you will end up with the same formula. One reason that one does not end up with the same formula is that the composition of forces does not hold in either special relativity or general relativity. So there is simply no sum that one can swap into the equation!

I suspect that the procedure would fail in relativity due to the non-linearity of the equations too - the introduction of gamma cubed into the inertial law, though I'm not totally sure.

The point is, you have to know first what is additive to decide what to describe by linear functionals.

In Newtonian mechanics, Newton knew that mass is additive and therefore introduced density. It makes no sense to do it the opposite way.

I agree with the first sentence, but not the second. Regarding the first: we have to know that forces are additive before we can describe them by linear functionals or integrals. And we do know this in advance, thanks to the compositions of forces law. The composition of forces law tells us that forces are additive. This means that to calculate forces, we can integrate over mass/position products (for gravitational forces) or mass/acceleration products (for inertial forces).

Thus, it is the additivity of forces that motivates the introduction of integrals, not the additivity of masses.

Since I've worked out both continuous and discrete proofs of gravitational mass additivity, I'll use one to spell out what I'm saying in more detail. I'll use the continuous case rather than DDs. I understand you as objecting to step 2:

Proof of gravitational continuous mass density additivity

Step 1: Introduce Newton's many particle law of gravitation
$$\mathbf{g}(\mathbf{x})=G\sum_i \frac{\mathbf{x}-\mathbf{x}_i}{|\mathbf{x}-\mathbf{x}_i|^3}~m_i~$$
Note that if we assume that the positions of the particles indexed by i are identical, we can show that it's as if there is just one particle whose mass is additive:
$$\mathbf{g}(\mathbf{x})=G \frac{\mathbf{x}-\mathbf{x}_1}{|\mathbf{x}-\mathbf{x}_1|^3}\sum_i~m_i~$$

Step 2: Derive the distribution formulation:
$$\mathbf{g}(\mathbf{x})=G \sum_i\int \frac{\mathbf{x}-\mathbf{r}}{|\mathbf{x}-\mathbf{r}|^3}~\rho_i(\mathbf{x}) ~\mathrm{d}^3\mathbf{r}$$
Key point: we know that we are allowed to integrate mass/position contributions to derive force because (i) integrating is just adding and (ii) the composition principle licences this.

Step 3: Algebraic transformation that recovers form of single density law:
$$\mathbf{g}(\mathbf{x})=G \int \frac{\mathbf{x}-\mathbf{r}}{|\mathbf{x}-\mathbf{r}|^3}~\sum_i\rho_i(\mathbf{x}) ~\mathrm{d}^3\mathbf{r}$$

Step 4: Total grav potential = composite grav potential & position of parts = position of composite:
$$\mathbf{g}_C(\mathbf{x})=G \int \frac{\mathbf{x}-\mathbf{r}}{|\mathbf{x}-\mathbf{r}|^3}~\sum_i\rho_i(\mathbf{x}) ~\mathrm{d}^3\mathbf{r}$$

Step 5: Infer gravitational mass distribution additivity.


Okay, that's the continuous gravitation proof. Now I would like to respond in more detail to your claim that this wrongly suggests that mass is additive in all cases because linear functionals (integrals and densities) are invoked. I said above, the procedure requires (i) the composition/additivity of forces and (ii) the linearity of the law. I give an example of when (i) breaks down and an example of when (ii) breaks down:


Example when (i) breaks down:

Step 1: Introduce law that replaces sum with product:
$$\mathbf{g}(\mathbf{x})=G\prod_i \frac{\mathbf{x}-\mathbf{x}_i}{|\mathbf{x}-\mathbf{x}_i|^3}~m_i~$$
Step 2: Derive the distribution formulation:
$$\mathbf{g}(\mathbf{x})=G \prod_i\int \frac{\mathbf{x}-\mathbf{r}}{|\mathbf{x}-\mathbf{r}|^3}~\rho_i(\mathbf{x}) ~\mathrm{d}^3\mathbf{r}$$
Step 3: Algebraic transformation that recovers form of single density law:
$$\mathbf{g}(\mathbf{x})=G \int \frac{\mathbf{x}-\mathbf{r}}{|\mathbf{x}-\mathbf{r}|^3}~\prod_i\rho_i(\mathbf{x}) ~\mathrm{d}^3\mathbf{r}$$
This fails - transformation not valid.


Example when (ii) breaks down:

Step 1: Introduce law requiring that we square the mass/mass-density:
$$\mathbf{g}(\mathbf{x})=G\sum_i \frac{\mathbf{x}-\mathbf{x}_i}{|\mathbf{x}-\mathbf{x}_i|^3}~m_i^2~$$
Step 2: Derive the distribution formulation:
$$\mathbf{g}(\mathbf{x})=G \sum_i\int \frac{\mathbf{x}-\mathbf{r}}{|\mathbf{x}-\mathbf{r}|^3}~[\rho_i(\mathbf{x})]^2 ~\mathrm{d}^3\mathbf{r}$$
Step 3: Algebraic transformation that recovers form of single density law:
$$\mathbf{g}(\mathbf{x})=G \int \frac{\mathbf{x}-\mathbf{r}}{|\mathbf{x}-\mathbf{r}|^3}~\sum_i[\rho_i(\mathbf{x})]^2 ~\mathrm{d}^3\mathbf{r}$$
Step 4: Total grav potential = composite grav potential & position of parts = position of composite:
$$\mathbf{g}_C(\mathbf{x})=G \int \frac{\mathbf{x}-\mathbf{r}}{|\mathbf{x}-\mathbf{r}|^3}~\sum_i[\rho_i(\mathbf{x})]^2 ~\mathrm{d}^3\mathbf{r}$$
Step 5: Infer gravitational mass distribution additivity.

This fails because in general: $[\sum_i\rho_i(\mathbf{x})]^2$ ≠ $\sum_i[\rho_i(\mathbf{x})]^2$

Therefore, my result does not depend on an arbitrary introduction of linear functionals. Instead, it depends on (i) the composition principle, and (ii) the linearily of the law. Either one of these or both could break down, despite the use of linear functions (integrals). When they break down, mass additivity breaks down.

What do you think?

 

[Khashishi]

It sounds like you are using the wrong approach here, and you should instead use a Boltzmann equation to calculate what you are doing. Instead of treating the acceleration as an object or a field, the velocity is included as a dimension of the distribution function. This way, the mass and momentum are integrated together into the same object.

The Boltzmann equation tells you how to evolve a continuous distribution of particles or mass.
$\frac{\partial f(x,p,t)}{\partial t}+\frac{p}{m}\cdot \nabla f(x,p,t)+F\cdot \frac{\partial f}{\partial p}=C$

If you need to track individual particles, you _can_ treat acceleration as an object, but then it should not be parametrized by the space coordinate, but rather a label of a specific particle. For example:
m(n), x(n), v(n), a(n)
this would give the equations of motion for each particle labeled by n. You can make n a continuous variable if you want. In this case, each particle is a dirac delta function in n, not in x.

 

[Jano L.]

But at no point do I redefine density as ρ=∑i ρi. I prove it using the below five step proof.

I do not agree. As I already said, I do not think that when you achieve the expression

$$
\mathbf F = \int \rho_C \mathbf a_c dV
$$

with $\rho_C =\sum_i \rho_i$ this leads to conclusion that $\rho_C$ is mass density distribution of the body. As I wrote, mass density distribution can only be introduced when we already know that mass is additive. If you do not agree, how else would you know that the function $\rho_C$ gives the mass of the body according to

$$
M = \int \rho_C dV ?
$$


You are right that your procedure will fail if the mass occurs with some other power than first. Also, you are right that superposition principle is necessary for your procedure.

I do not agree that you will end up with the same formula. One reason that one does not end up with the same formula is that the composition of forces does not hold in either special relativity or general relativity. So there is simply no sum that one can swap into the equation!

Then try it for special relativistic mechanics of particles - the composition of ordinary forces holds true as in Newtonian mechanics, every step of your procedure will pass.

Which page do you refer in the paper by Kholmetskii? I am quite sure that forces are additive in special theory relativity.

 

[James MC]

I do not agree. As I already said, I do not think that when you achieve the expression

$$
\mathbf F = \int \rho_C \mathbf a_c dV
$$

with $\rho_C =\sum_i \rho_i$ this leads to conclusion that $\rho_C$ is mass density distribution of the body.

I think I understand. At a certain point in the procedure, we reach the following expression:
$$
\mathbf F_C = \int [\sum_i \rho_i] \mathbf a_c dV
$$
At this stage, I do not define $\rho_C =\sum_i \rho_i$. Rather, I infer it with a substantive argument. The argument is necessary for the textbook proof too (the proof that assumes all accelerations are identical). Thus, the textbook proof gets to this expression:
$$
\mathbf F_C = \sum_i m_i \mathbf a_c
$$
To infer mass additivity from this expression, one must reason as follows:

(1) The mass of an object is the property of that object that explains its resistance to motion given applied forces (definition of 'mass').
(2) The property of C that explains C's resistance to motion given applied forces is $\sum_i m_i$ (entailed by the equation).
(3) Therefore, C's mass is $\sum_i m_i$ (entailed by (1) and (2)).

This is how we derive $\mathbf F_C = m_C \mathbf a_c$

Now, the same kind of reasoning applies to the distribution case. Thus, if our procedure can yeild:
$$
\mathbf F_C = \int [\sum_i \rho_i] \mathbf a_c dV
$$
Then we can derive mass distribution additivity from it by reasoning as follows:

(1) The mass distribution of an object is the property of that object that explains its resistance to motion given applied forces (definition of 'mass distribution').
(2) The property of C that explains C's resistance to motion given applied forces is $\sum_i \rho_i$ (entailed by the equation).
(3) Therefore, C's mass distribution is $\sum_i \rho_i$ (entailed by (1) and (2)).

Does that help? Sorry, perhaps I should have made that assumption more explicit earlier.

As I wrote, mass density distribution can only be introduced when we already know that mass is additive.

But I responded: we can motivate the introduction of mass density distributions by appeal to the fact that integrating over mass*position or mass*acceleration products, is an extension of the composition law. Hence, mass density distribution can be introduced into the law when we already know that forces are additive. We don't need to know that masses are additive.

In fact, this was meant to be the upshot of appeal to a law which squares the density. As I mention below, running the procedure that way enables us to infer a different principle from mass additivity.

If you do not agree, how else would you know that the function $\rho_C$ gives the mass of the body according to

$$
M = \int \rho_C dV ?
$$

It is important to distinguish two possible objections to my argument. One objection states that I have failed to derive mass distribution additivity. Another objection states that I have failed to derive mass additivity (even assuming that I've derived distribution additivity). I thought you were making the former objection, but now it looks like your making the latter.

I believe I have responded t the former objection. The latter objection is a good one, but I think I can answer it: it is true by definition that the value of the quantity described by a density is its integral. If that is correct, then I can derive mass additivity. But this doesn't mean that introducing densities assumes mass additivity, because we can show that the procedure fails to derive mass additivity for laws that describe densities in different ways...

You are right that your procedure will fail if the mass occurs with some other power than first. Also, you are right that superposition principle is necessary for your procedure.

...but if you agree that the procedure will fail is the mass is squared, then how can you say that the procedure is defective in virtue of assuming or always yeilding mass additivity? In the case where we square the density, the procedure entails the result that we should expect:
$$
M_C = {\sqrt{\sum_i m_i^2}}.
$$

Which page do you refer in the paper by Kholmetskii? I am quite sure that forces are additive in special theory relativity.

The footnote on page 185: "we cannot simply sum up the forces applied to the system, for different inertial observers"
In fact, I thought this was one reason why we don't formulate relativity in terms of force, but in terms of energy instead?

Then try it for special relativistic mechanics of particles - the composition of ordinary forces holds true as in Newtonian mechanics, every step of your procedure will pass.

Well, according to wiki the relevant equation is:
$F = \gamma(v)^3ma_| + \gamma(v)ma_-$
If we assume the composition of forces (which I'm not sure we can) and introduce densities we get:
$F_T = \sum_i \int \gamma_i(v_i)^3\rho_ia_i + \gamma_i(v_i)\rho_ia_i dV$
..and it looks like the sum cannot just be swapped into the equation, because there are two mass distributions!?

I think we can just appeal to the standard relativistic equations to make the key point, which is that introducing densities alone does not gaurantee mass additivity. Because the equations are a bit different, the relevant proof looks a little different. We start with the fundamenal law:
$$
M = {\sqrt{\frac{E^2}{c^4} - \frac{p^2}{c^2}}}.
$$
Now, as I understand it, rather than appealing to a force composition law (which I thought was no longer well defined), STR instead appeals to composition laws for both energy and momentum? In that case, the mass of the composite is equivalent to:
$$
M_C = {\sqrt{\frac{\sum_iE^2_i}{c^4} - \frac{\sum_ip^2_i}{c^2}}}.
$$
And I take it that from the previous two equations, one can easily derive that $M_C$ $\neq$$\sum_im_i$
Now, surely, we can represent all of these quantities as densities, and this very same argument against mass additivity fails? Doesn't that demonstrate that expressing an equation using mass densities does not presuppose mass additivity?

 

[James MC]

Khashishi, thanks for your input. That's an interesting point about treating acceleration as an object with particle label parameters. But if the mass and momentum are integrated together into the same object, then it's hard to see how we can prove mass additivity as a consequence of the law. Don't we need to keep mass separate for that? See post 5 for an explanaton of what I'm trying to achieve here.

 

[Jano L.]

Rather, I infer it with a substantive argument. The argument is necessary for the textbook proof too (the proof that assumes all accelerations are identical).

The textbook derivation for the particles which have the same acceleration is correct both in non-relativistic and relativistic theory(provided particles do not interact), but this is very special situation.

Typical situation is that the particles have different motions. Then the textbook argument fails and has to be replaced by different argument, like the one I gave in the first posts.

But I responded: we can motivate the introduction of mass density distributions by appeal to the fact that integrating over mass*position or mass*acceleration products, is an extension of the composition law. Hence, mass density distribution can be introduced into the law when we already know that forces are additive. We don't need to know that masses are additive.

OK, but this is mathematical distribution, i.e. linear functional that gives you the correct equation of motion. It does not follow that this mathematical distribution is the mass density.


Mass density is a different thing, defined as the function that gives the mass in any region $\Delta V$ by

$$
\Delta M = \int_{\Delta V} \rho\, dV .
$$

This definition of mass density is motivated by the idea that mass is additive.

The latter objection is a good one, but I think I can answer it: it is true by definition that the value of the quantity described by a density is its integral.

Exactly, that is true for density. Your procedure only leads to linear functional $\rho_c$, but this is not density.

In special relativity for non-interacting particles, you can run your argument in this way:

$$
\sum_i \mathbf F_i = \sum_i m_i \mathbf w_i,
$$

where $\mathbf w_i = \frac{d}{dt}(\gamma_i \mathbf v_i)$. Now treat $\mathbf w_i$ as you treated $\mathbf a_i$ and everything is the same. The reason is that linear functionals will add to linear functional provided the mass of the particle appears in the first power, which happens in both theories. In relativity, you will derive

$$
\sum_i\mathbf F_i = \int \rho_C \mathbf w_C\, dV
$$

with $\rho_C = \sum_i m_i\delta(\mathbf x - \mathbf r_i)$.

But that's just gymnastics with linear functionals. It does not follow that the rest mass is given by

$$
\int \rho_C dV.
$$

In fact, the rest mass of the composite will be higher, due to chaotic motion of the particles.

The reason for this is that the rest mass of the body is determined by the whole equation of motion, including the time dependent term ($\mathbf a_i$,$\mathbf w_i$), not just by the fact that mass appears in the first power.

You have to include consideration of the time dependent terms into the argument.One way to do so is to introduce the center of the body, like I explained in the first posts.

 

[James MC]

The textbook derivation for the particles which have the same acceleration is correct both in non-relativistic and relativistic theory(provided particles do not interact), but this is very special situation.

Typical situation is that the particles have different motions.

How could the textbook derivation applied to SR be correct, given your view? All that the textbook explantion does is make every variable self-identical, apart from what falls within the scope of the sum post-transformation. Thus, we begin with:
$$
\mathbf F_T = \sum_i m_i \mathbf w_i,
$$
And then the textbook explanation would make the following transformation:
$$
\mathbf F_i = \mathbf w_1\sum_i m_i,
$$
Hence, mass additivity in relativity theory! So doesn't the textbook explantion fail in the relativistic context, and so, given your argument, shouldn't you reject the textbook explanation for classical mass additivity given that the procedure it uses wrongly shows that rest mass is additive?

More generally, shouldn't you say that the textbook explanation for classical mass additivity also fails because my distribution explanation fails? In particular, my explanation of mass distribution additivity simply makes use of (i) the composition of forces and (ii) the fact that the law treats masses and forces as linearly proportional. That's all that the textbook explanation makes use of in the simple scenario. If you reject that mass additivity follows from (i) and (ii) then it seems that you should reject the textbook argument outright.

Then the textbook argument fails and has to be replaced by different argument, like the one I gave in the first posts.

But I thought we agreed that the COM explantion also fails, because in order to define a COM you must presuppose a value for the mass of the composite? It seems that your view entails that composite mass is an emergent property, that is inexplicable in lower level microphysical terms.

OK, but this is mathematical distribution, i.e. linear functional that gives you the correct equation of motion. It does not follow that this mathematical distribution is the mass density.


Mass density is a different thing, defined as the function that gives the mass in any region $\Delta V$ by

$$
\Delta M = \int_{\Delta V} \rho\, dV .
$$

This definition of mass density is motivated by the idea that mass is additive.

The question of mass additivity is the question of whether the mass of the composite is the sum of the masses of the atomic parts. If the atomic parts are infintesimal densities, then determining the mass of an atomic part by integrating its density, does not presuppose mass additivity, because it doesn't involve adding masses (because the elementary mass density is already infinitesimal). Therefore, in the specific situation of a set of infintitesimal densities, describing them using a density law cannot presuppose mass additivity.

Then the question is, can we transform this law, in such a way that we can recover the single density law, applied to the force and acceleration distribution of the composite of those various infinitesimal densities? It turns out, we can. Does this presuppose mass additivity? I say no because there is absolutely no gaurantee that the composite density function that one ends up with, will be such that if you integrate it, you get additivity. Indeed, I've provided a case in which one doesn't get this result, e.g. when density is squared. So what gaurantees that the mass density yields an additive result depends on the form of the law. And so if by this method I can show that the transformation yields mass density additivty, then I've shown that mass additivity is a consequence of the form of the law.

Thus, introducing the law for elementary atomic particles cannot presuppose additivity because the masses of the objects that the law applies to can't be added; and the transformation doesn't assume additivity either given that the same transformation for a different law doesn't yield it.

Exactly, that is true for density. Your procedure only leads to linear functional $\rho_c$, but this is not density.

In special relativity for non-interacting particles, you can run your argument in this way:

$$
\sum_i \mathbf F_i = \sum_i m_i \mathbf w_i,
$$

where $\mathbf w_i = \frac{d}{dt}(\gamma_i \mathbf v_i)$. Now treat $\mathbf w_i$ as you treated $\mathbf a_i$ and everything is the same. The reason is that linear functionals will add to linear functional provided the mass of the particle appears in the first power, which happens in both theories. In relativity, you will derive

$$
\sum_i\mathbf F_i = \int \rho_C \mathbf w_C\, dV
$$

with $\rho_C = \sum_i m_i\delta(\mathbf x - \mathbf r_i)$.

But that's just gymnastics with linear functionals. It does not follow that the rest mass is given by

$$
\int \rho_C dV.
$$

Yes, this is certainly a troubling objection. Hence why I opened up another thread regarding force composition in SR. Since that debate is still going, let's just assume for argument's sake that forces are additive in SR. In that case, I have two responses (or perhaps, questions, as I'm very unsure of these responses)...

Firstly, the law you appeal to isn't the correct law. What you've done is taken the law for the special situation in which the velocity and the direction of force are perpendicular. So your equation only deals with a limited case in which it's clear to me that the relevant effects relating to mass-non-additivity become apparent. Doesn't your argument need to appeal to three dimensions, like mine? In that case, you'll need an extra term allowing for the force to parallel to the velocity, which will introduce a cubed expression into the equation. Can you still make your argument work then?

Secondly, the point of the transformation is this: you start with an expression for the force on many particles, and you want to know if you can transform it in such a way that you recover the form of the single particle law, which gives an expression for the inertia of the composite. So you need to break up w into gamma and acceleration, and when the sum is swapped in, it attaches to gamma and mass (not just mass) because those are the combined facts relevant to the composites disposition to resist acceleration given force. In that case the transformation gives exactly the right result, because the mass of the composite just is the sum of the mass/gamma products of the atomic parts.

In fact, the rest mass of the composite will be higher, due to chaotic motion of the particles.

Sometimes it can be lower e.g. deuterium nucleus.

The reason for this is that the rest mass of the body is determined by the whole equation of motion, including the time dependent term ($\mathbf a_i$,$\mathbf w_i$), not just by the fact that mass appears in the first power.

You have to include consideration of the time dependent terms into the argument.One way to do so is to introduce the center of the body, like I explained in the first posts.

The COM explanation, it seems to me, will fail for the same reason as in classical mechanics: you need to know the composite mass before you define the COM.

Here is an explanation that attempts to show that composite mass in SR is a consequence of the form of the microphysical SR laws:

We begin with:
$$
M = {\sqrt{\frac{E^2}{c^4} - \frac{p^2}{c^2}}}.
$$
Now, if force composition is true in SR, as a matter of fundamental law, then since momentum is just force accumulated over time, and since energy is just force accumulated over space, then we can infer the additivity of energy and momentum, and then reapply the law to determine composite mass:
$$
M_C = {\sqrt{\frac{\sum_iE^2_i}{c^4} - \frac{\sum_ip^2_i}{c^2}}}.
$$

Now, why was this proof so easy. Why, in SR can we derive composite masses so easily, yet in Newtonian mechanics we must fiddle around with COM's or distributions?

 

[Jano L.]

In the first part of your post, you wonder how can we derive mass addition in relativity - isn't it true that mass is not additive there? Yes, generally it isn't. But as I said, we (and the textbook) consider special situation, when all particles moves with the same velocity all the time. In that case, even in relativity, the rest mass is sum of rest masses.

In order to obtain non-additivity, the particles of the composite have to move in various directions, or have some mutual interactions.

The explanation based on center of mass is perhaps motivated by additivity, but it shows WHY we can assume this additivity, and what it exactly means, in terms of the motion of the center of mass.

If the atomic parts are infintesimal densities, then determining the mass of an atomic part by integrating its density, does not presuppose mass additivity, because it doesn't involve adding masses (because the elementary mass density is already infinitesimal).

Well, if we talk about density, we have to know what it is. The standard definition I gave above is based on the additivity of mass, because integrating is just a refined form of summation. In my view, additivity of mass is basic concept, while density is more of a technical device. I do not think one can infer the former from the latter.

The procedure I suggest for the relativistic case is alright, there is nothing restricting it to perpendicular force - why would you think that?

I do not understand how you arrived at sum of squares in the end of your post. I do not think that is correct. Rest mass can be obtained as the ratio rest energy/c^2, and the latter is for non-interacting particles given by $E = \sum_i \gamma_i m_i c^2$, in the frame where the composite is at rest, or its center is at rest.

 

[James MC]

In the first part of your post, you wonder how can we derive mass addition in relativity - isn't it true that mass is not additive there? Yes, generally it isn't. But as I said, we (and the textbook) consider special situation, when all particles moves with the same velocity all the time. In that case, even in relativity, the rest mass is sum of rest masses.
In order to obtain non-additivity, the particles of the composite have to move in various directions, or have some mutual interactions.

Yes okay I see what you mean. I think this adds to the strength of the objection to the classical textbook explantion of additivity, because it shows that the simple scenario appealed to fails to show that the form of the law entails additivity. It shows that the explanation needs to be extended to be adequate (perhaps by your centre of mass approach, or some other approach).

The explanation based on center of mass is perhaps motivated by additivity, but it shows WHY we can assume this additivity, and what it exactly means, in terms of the motion of the center of mass.

Perhaps we disagree here on what counts as an explanation. I am assuming that explanations can't be circular. You say it shows why we can assume mass additivity. Yet, mass additivity is itself the (or one) reason why we can assume that the COM is the position of the composite. Don't you think this reasoning is circular? It also doesn't seem to show what mass additivity means, given that the COM does have any mass, let alone the sum of the mass of the elementary parts (nothing is literally located at the COM, and no forces literally apply to it).

Here's a question I think will help: imagine you know nothing about the world, yet you have infinite reasoning powers. Now imagine you are informed of the Newtonian microphysical laws (f=ma etc.) and the kinematics, masses, and forces, of all elementary Newtonian particles. Do you think you could deduce mass additivity from this information alone, with certainty?

Well, if we talk about density, we have to know what it is. The standard definition I gave above is based on the additivity of mass, because integrating is just a refined form of summation. In my view, additivity of mass is basic concept, while density is more of a technical device. I do not think one can infer the former from the latter.

Yes, I now see what you mean when you say that I've only shown the additivity of mathematical distributions, and not the additivity of density distributions. You're saying that we cannot relate mathematical distributions to real physical things until we say what quantities they represent, and we cannot say what quantities we represent until we say "integrate to get the value of the quantity", but integrating is adding, and hence presupposes the additivity of the quantity being represented. Thanks.

The procedure I suggest for the relativistic case is alright, there is nothing restricting it to perpendicular force - why would you think that?

Because the correct equation must also account for the parallel force, in which case, gamma must be squared. However, I've tried to do the calculations, and it looks like it makes no difference. So if you are right about force composition in SR, then it seems you are right that my procedure wrongly shows mass to be additive in SR.

I do not understand how you arrived at sum of squares in the end of your post. I do not think that is correct. Rest mass can be obtained as the ratio rest energy/c^2, and the latter is for non-interacting particles given by $E = \sum_i \gamma_i m_i c^2$, in the frame where the composite is at rest, or its center is at rest.

I'm sorry, that was a typo, I forgot the brackets.

The fundamental microphysical SR law is:
$$
M = {\sqrt{\frac{E^2}{c^4} - \frac{p^2}{c^2}}}.
$$
If force composition is true in SR, as a matter of fundamental law, then since momentum is just force accumulated over time, and since energy is just force accumulated over space, then we can infer the additivity of energy and momentum. Then, we can reapply the law to the composite energy and the composite momentum to determine composite mass:
$$
M_C = {\sqrt{\frac{[\sum_iE_i]^2}{c^4} - \frac{[\sum_ip_i]^2}{c^2}}}.
$$
The question I would like to ask you is this: Why, in SR can we derive composite masses so easily, yet in Newtonian mechanics we must fiddle around with COM's (or distributions)?

These two equations are frame invariant. That is, they yield the same mass no matter the frame-variant energies and momentum we input. So I don't think this proof needs to rely on anything about centre of mass, or any particular frame; though if you think I'm wrong please let me know.

 

[Jano L.]

I do not think the explanation is circular. Whatever reason for introducing the standard center of mass, the additivity of mass can be derived based on its motion. One reason for COM is just that we want additivity of mass, but I think this is not the only one. Other reason is that the standard center of mass is the simplest definition which has useful properties - it stays near the body in all situations and it plays a special role in the description of motion of free rigid bodies.

Another point of view, probably the historic one (as Newton saw it) is to assume that density of body is the basic concept and the mass is derived as the integral of this density. Then again the additivity of mass is assumed just from the start, as an independent postulate of the Newtonian mechanics.

imagine you know nothing about the world, yet you have infinite reasoning powers. Now imagine you are informed of the Newtonian microphysical laws (f=ma etc.) and the kinematics, masses, and forces, of all elementary Newtonian particles. Do you think you could deduce mass additivity from this information alone, with certainty?

This is hard to imagine, but I think the answer is that pure mathematician with no empiric knowledge about motion and mass would finally arrive at COM and mass additivity, since they are quite simple and have interesting mathematical properties - the addition is one of the simplest thing mathematician can do.

Why, in SR can we derive composite masses so easily, yet in Newtonian mechanics we must fiddle around with COM's (or distributions)?

I do not think the formula you assume can be taken as a basis for derivation, because it is not immediately clear that it is valid for composite object. That should be inferred from the behaviour of the constituents, just like in the non-relativistic case. Then it is in fact more difficult in relativity than in Newtonian mechanics.

 

[James MC]

I do not think the explanation is circular. Whatever reason for introducing the standard center of mass, the additivity of mass can be derived based on its motion.

I agree. I just think the derivation is not an explanation given that a premise of the derivation cannot be motivated without assuming the explanandum.

One reason for COM is just that we want additivity of mass,

I suspect that is the only reason, which is why one cannot appeal to it to explain mass additivity. After all, the very idea of assuming that that composite is located at the COM is confused: the composite is located where its parts are located.

but I think this is not the only one. Other reason is that the standard center of mass is the simplest definition which has useful properties - it stays near the body in all situations and it plays a special role in the description of motion of free rigid bodies.

Simplicity and staying near the body don't help, I don't think, because there are points that are simpler, and that stay closer. For example, the average position when not weighted by masses. And one could only know that COM plays a special role in the description of composite bodies if one already knows that mass is additive. So again, this route won't work as an explanation, because it assumes the explanandum. I honestly don't think there is any way of explaining mass additivity in terms of COM. Have you ever seen a textbook, or academic paper, which attempts to explain mass additivity in terms of COM?

Another point of view, probably the historic one (as Newton saw it) is to assume that density of body is the basic concept and the mass is derived as the integral of this density. Then again the additivity of mass is assumed just from the start, as an independent postulate of the Newtonian mechanics.

Newton made the metaphysical assumption that mass is a measure of the amount of matter in a region. I think you can derive mass additivity from this metaphysical assumption, but I think the metaphysical assumption does not follow from the physics.

This is hard to imagine, but I think the answer is that pure mathematician with no empiric knowledge about motion and mass would finally arrive at COM and mass additivity, since they are quite simple and have interesting mathematical properties - the addition is one of the simplest thing mathematician can do.

I don't see why. If the pure mathematician is asked to (i) derive the truth about the composite mass and (ii) has infinite reasoning powers, then the pure mathematician will care little about what is "simple". Unless of course she is told in advance that what is true and what is simple coincide in the world under consideration; but that isn't a given.

Here is the conclusion I have come to: (inertial) mass is defined in terms of the "most natural" property responsible for resistance to changes in motion given applied forces. The acceleration for a non-rigid composite body is simply not well defined. And because mass is defined in terms of acceleration, mass is not well defined for non-rigid bodies either. In that case, there is no need to extend the textbook explanation - it is complete, as only rigid bodies can be said to have mass in the first place!

There is one thing that needs to be added though. Recall the textbook explanation:
$$
\mathbf F_T = \sum_i m_i \mathbf a_i,
$$
$$
\mathbf F_T = \mathbf a_1\sum_i m_i,
$$
$$
\mathbf F_C = \mathbf a_C\sum_i m_i,
$$
As you helped me see, this is inadequate due to the fact that such a simplified situation (identical accelerations) is appealed to. The simplified situation is unhelpful because it does not allow us to see what the crucial difference is between NM and SR:
$$
\mathbf F_T = \sum_i m_i \mathbf w_i,
$$
$$
\mathbf F_T = \mathbf w_1\sum_i m_i,
$$
$$
\mathbf F_C = \mathbf w_C\sum_i m_i,
$$
However, I now think the solution is to appeal to neither distributions (for reasons you patiently spelled out), nor to COM's (for the reasons I gave above). Rather, I think the solution requires that we appeal to a special kind of "rigid" body, which I will call an "a-rigid" body. An a-rigid body is a body whose parts have identical accelerations but distinct velocities. Now if our situation appeals to a-rigid bodies rather than rigid bodies, then we can save the textbook explanation from the objection, and distinguish NM from SR. We do that as follows:
$$
\mathbf F_T = \sum_i \gamma_i m_i \mathbf a_i,
$$
$$
\mathbf F_T = \mathbf a_1\sum_i \gamma_i m_i,
$$
$$
\mathbf F_C = \mathbf a_C\sum_i \gamma_i m_i,
$$

So the result we get is that the mass of the composite is equivalent to $\sum_i \gamma_i m_i$ except we add the fact that because we are after the "most natural" property that comes closest to playing the inertial role for the composite, we require a frame-invariant, and so we infer that the mass of the composite is equal to $\sum_i \gamma_i m_i$ as calculated in the composite's rest frame (i.e. where total momentum is zero).

I now think that the above constitutes the complete explanation of both Newtonian and relativistic composite mass (in terms of the microphysical laws). What do you think?

I do not think the formula you assume can be taken as a basis for derivation, because it is not immediately clear that it is valid for composite object. That should be inferred from the behaviour of the constituents, just like in the non-relativistic case. Then it is in fact more difficult in relativity than in Newtonian mechanics.

I'm almost certain that the formula is valid - it is appealed to in many texts, for example p651 in this little gem.
My objection to this explanation of composite mass in relativity theory is that the M that is derived for the composite is not obviously composite mass because it is not obviously connected to our concept of inertia. So we really need the force/acceleration laws, it seems, to properly derive composite mass, as above.

 

[Jano L.]

An a-rigid body is a body whose parts have identical accelerations but distinct velocities. Now if our situation appeals to a-rigid bodies rather than rigid bodies, then we can save the textbook explanation from the objection, and distinguish NM from SR. We do that as follows:
$$
\mathbf F_T = \sum_i \gamma_i m_i \mathbf a_i,
$$

This is not well. First, the formula you assume in invalid, because the gamma factor has to be differentiated too. The correct equation is

$$
\mathbf F = \sum_i m_i \frac{d}{dt}(\gamma_i \mathbf v_i).
$$

Second, the same acceleration of the parts is very special situation in both nonrel. and rel. mechanics, which almost never happens. You would derive the additivity only for such special situation.

So we really need the force/acceleration laws, it seems, to properly derive composite mass, as
above.

Yes, this was my point. I think that the derivation in rel. mechanics will be necessarily different and more difficult than in nonrel. mechanics, because in relativity, the interactions between the parts are not instantaneous anymore and this has consequence on the mass and motion of the composite.

I am afraid that you won't be able to find explanation which will be more satisfactory to you and correct, because physics is not so tight axiomatic system like mathematics. In my view, the additivity of mass in nonrel. physics is primarily a generalization of daily experience with weighing, and this later was vindicated in the form of motion of COM as implied by Newton's laws. In this way, I consider it very satisfactory and I would not attempt to derive it from anything more simpler. Because, what is more simple?

But if you think there is something to discover, keep trying, who knows, perhaps it is possible.

 

[James MC]

This is not well. First, the formula you assume in invalid, because the gamma factor has to be differentiated too. The correct equation is

$$
\mathbf F = \sum_i m_i \frac{d}{dt}(\gamma_i \mathbf v_i).
$$

Huh? But I see this all the time. Take the proof in the wikipedia article. Do you think there is something wrong with the wikipedia example? Surely I only need to separate things into the part parallel to the velocity and the part perpendicular to it. My example just used the perpendicular component for simplicity.

Second, the same acceleration of the parts is very special situation in both nonrel. and rel. mechanics, which almost never happens. You would derive the additivity only for such special situation.

I don't think that's a special situation because the acceleration of a composite is not well defined if its parts have distinct accelerations. And if an object's acceleration is not well defined, then neither can its mass be, because mass is defined in terms of it. Of course, any such composite may still have a mass in the following sense: it resists acceleration given applied forces when its acceleration is well defined. That means that for any composite, we can define its mass by considering counterfactual scenarios in which its components have identical accelerations, and then running my argument. So there is no composite that my argument does not apply to.

Yes, this was my point. I think that the derivation in rel. mechanics will be necessarily different and more difficult than in nonrel. mechanics, because in relativity, the interactions between the parts are not instantaneous anymore and this has consequence on the mass and motion of the composite.

I am afraid that you won't be able to find explanation which will be more satisfactory to you and correct, because physics is not so tight axiomatic system like mathematics. In my view, the additivity of mass in nonrel. physics is primarily a generalization of daily experience with weighing, and this later was vindicated in the form of motion of COM as implied by Newton's laws. In this way, I consider it very satisfactory and I would not attempt to derive it from anything more simpler. Because, what is more simple?

But if you think there is something to discover, keep trying, who knows, perhaps it is possible.

I guess I'll have to wait and see what you think of the wikipedia proof, that derives a form of the law that separates out gamma from acceleration.

 

[Jano L.]

I think you take very special situation:

1) accelerations are all the same
2) they are all in the direction of velocities

From this it follows that all particles have velocities in the same direction. This is just too special for composite body.

But if you insist on this situation, take the particular one in which all particles have the same velocity. Your formula gives inertial (relativistic) mass, while the original question is how the rest masses of the particles form the rest mass of the composite. In this case, the rest mass is just sum of rest masses

$$
M = \sum_i m_i.
$$

EDIT: Disregard, I misunderstood. I'll write the answer in while.

 

[Jano L.]

OK, you take this situation:

1) accelerations are all the same
2) they are all perpendicular to velocities

From this it follows that all velocities lie in a plane. This is the only case when the textbook procedure works in rel. mechanics, but you in fact derive inertial(relativistic) mass.

But the basic question is how the rest masses compose to form rest mass of the composite. If all velocities are the same, the rest mass is just sum of individual masses. Your formula gives something different.

 

[James MC]

This is a good point, but I think I in fact derive both the composite relativistic mass and the composite invariant mass.

Recall the reason why I rejected the energy/momentum based explanation of composite invariant mass. That's because the "M" that was derived, had no obvious connection to our concept of mass. I recall you agreed with me on this point.

The concept of mass is best characterised as "that property that explains the object's disposition to resist changes in motion given applied forces". According to the equation I appeal to, relativistic mass best fits this definition for elementary particles, because m*gamma is the proportionality constant between force and acceleration. And as you rightly note, I end up inferring the composite relativistic mass, because I show that the sum of the relativistic masses of the parts is the proportionality constant between composite force and composite acceleration.

However, when we are intentionally using theoretical vocabulary to pick out the most natural, fundamental properties of nature, we must add a clause to our definitions if our definitions are to capture our actual use of the vocabulary. There is therefore another notion of 'mass' that is defined as "the most natural/fundamental property that is involved in explaining resistance to changes in motion given applied forces". This definition does not pick out relativistic mass because relativistic mass is a frame variant (and hence non-natural/fundamental) property. For elementary particles, what therefore fits this definition is the property that plays the most substantive role in explaining the inertia of the particle, that is frame invariant. This is the invariant mass, which is equal to the relativistic mass when momentum = 0 (I think that's right). In that case, when we infer the relativistic mass of the composite, we simply infer its invariant mass by moving to the frame in which the composite has zero total momentum, and seeing what the real invariant mass is.

So the proof infers both composite relativistic mass and composite invariant mass (as it presumably should). I also think it illuminates why two notions of 'mass' arose in the first place.

 

[Jano L.]

In that case, when we infer the relativistic mass of the composite, we simply infer its invariant mass by moving to the frame in which the composite has zero total momentum, and seeing what the real invariant mass is.

How do you "see what the real invariant mass is"? Your procedure of finding inertial mass depends on the assumption that accelerations are perpendicular to velocities. If this is so in the original reference frame, it need not be so in the frame where total momentum is zero. Imagine swarm of electrons with equal velocities perpendicular to a uniform magnetic field. In this frame, forces and accelerations are perpendicular to velocities, but in the frame of the electrons, this is no longer true, since the speed of the particles change.

Try to think of a more general procedure, where velocities and accelerations can be arbitrary.