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

1. Jan 31, 2013

### 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 FT 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!

2. Feb 1, 2013

### 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.

3. Feb 2, 2013

### 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 realise 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!

4. Feb 4, 2013

### 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.

5. Feb 4, 2013

### 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:
$$\sum_i m_i = m_c$$

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:
$$\int \mathbf{a}_c(\mathbf{x})\sum_i m_i(\mathbf{x}) \delta (\mathbf{x}-\mathbf{x}_i)d^3\mathbf{x} = \mathbf{F}_c$$

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.

6. Feb 5, 2013

### Jano L.

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.

7. Feb 6, 2013

### 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.

8. Feb 6, 2013

### 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.

9. Feb 7, 2013

### James MC

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?

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.

10. Feb 7, 2013

### Jano L.

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.

11. Feb 9, 2013

### James MC

Re: Distribution (Dirac&standard) formulations of f=ma... how do they

Apologies for the delay...

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:
$$\mathbf{a}_c\sum_i m_i = \mathbf{F}_c$$
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:
$$\sum_i m_i = m_c$$

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

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.

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.

12. Feb 10, 2013

### Jano L.

Re: Distribution (Dirac&standard) formulations of f=ma... how do they

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.

13. Feb 11, 2013

### James MC

Re: Distribution (Dirac&standard) formulations of f=ma... how do they

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.
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.

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.

14. Feb 12, 2013

### Jano L.

Re: Distribution (Dirac&standard) formulations of f=ma... how do they

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?

15. Feb 13, 2013

### James MC

Re: Distribution (Dirac&standard) formulations of f=ma... how do they

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.

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 realise 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.

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.

16. Feb 13, 2013

### Jano L.

Re: Distribution (Dirac&standard) formulations of f=ma... how do they

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 $\mathbf x$.

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.

17. Feb 14, 2013

### James MC

Re: Distribution (Dirac&standard) formulations of f=ma... how do they

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?

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 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.

18. Feb 14, 2013

### Jano L.

Re: Distribution (Dirac&standard) formulations of f=ma... how do they

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.

19. Feb 15, 2013

### James MC

Re: Distribution (Dirac&standard) formulations of f=ma... how do they

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

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.

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.

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?

20. Feb 15, 2013

### Khashishi

Re: Distribution (Dirac&standard) formulations of f=ma... how do they

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.

Last edited: Feb 15, 2013