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Point masses versus infintesimal densities, is there a difference?

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James MC
#1
Jan27-13, 07:29 PM
P: 175
Hello,

Is a (classical) physical object that is correctly described as a point mass (i.e. as a mass value M at a point x) a physically different object from a (classical) physical object that is correctly described as a mass density distribution distributed/defined over an infintesimal region?

If the answer is no, then are the point mass description and the infintesimal mass density distribution description just two ways of saying the same thing? In other words, if I am told that there exists a point mass is it trivial to immediately infer that there exists a mass density distributed over an infintesimal region?

If the answer is yes, what exactly is the difference, and is the idea of a 'dirac delta function' (which I don't yet understand) an attempt to apply formalism originally intended for densities, to point masses?

Thanks!
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phinds
#2
Jan27-13, 07:36 PM
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"point" implies dimensionless, that is zero in x,y,z dimensions. "Infinitesimal" is not zero, so I don't see how they could be the same.
James MC
#3
Jan27-13, 08:03 PM
P: 175
I think "point" doesn't quite imply "dimensionless" given that the classical point masses I have in mind are partly defined in terms of their co-ordinates in three dimensional space. Though I take it you mean that while a classical point mass is not physically extended over any of the x,y,z dimensions that it exists in, an infintesimal mass density distribution IS extended over these dimensions, it's just that the region over which it is extended is "infinitely small"?

This means that there is a real physical difference between being extended over an infinitely small region whose centre is x,y,z and being located at the point x,y,z. I find it a little hard to wrap my head around what that difference is supposed to be, but perhaps the important point is just that an infinitely small region, whatever that is, is physically "bigger" than the point at the centre of that region.

BruceW
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Jan27-13, 08:04 PM
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Point masses versus infintesimal densities, is there a difference?

discrete and continuous are two different things. And they should be treated separately. Generally, physical phenomena will depend on whether we are talking about continuous or discrete stuff. (e.g. real numbers versus integers).

But luckily, there are some physical theories where discrete and continuous 'formalism' give highly analogous results. For example, in classical electrodynamics, the force on a charged point particle (discrete), or the force per volume on a continuous distribution of charge (continuous). In this case, the equations are very similar. And we can even think of the discrete case being approximately the same as the continuous case, when the charge density is very great in a very small volume. (This is essentially the idea of the Dirac-Delta).
Chestermiller
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Jan27-13, 08:07 PM
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Quote Quote by James MC View Post
Hello,

Is a (classical) physical object that is correctly described as a point mass (i.e. as a mass value M at a point x) a physically different object from a (classical) physical object that is correctly described as a mass density distribution distributed/defined over an infintesimal region?

If the answer is no, then are the point mass description and the infintesimal mass density distribution description just two ways of saying the same thing? In other words, if I am told that there exists a point mass is it trivial to immediately infer that there exists a mass density distributed over an infintesimal region?

If the answer is yes, what exactly is the difference, and is the idea of a 'dirac delta function' (which I don't yet understand) an attempt to apply formalism originally intended for densities, to point masses?

Thanks!
Yes. A point mass can be envisioned as having a limiting density distribution. Usually, the density distribution for a point mass is described using the Dirac delta function δ(x-x0), where x0 is the location of the mass, and where the delta function is zero everywhere except at x = x0, and the integral of the delta function from - ∞ to +∞ is equal to unity. The density distribution of a point mass using this formalism is represented by Mδ(x-x0). Of course, this is only the 1D version. An analogous expression can be written for 3D.
elfmotat
#6
Jan27-13, 08:14 PM
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I would define the mass density of a point particle to be:

[tex]\rho(x)=m\delta (x-x_0)[/tex]

where [itex]x_0[/itex] is the position of the particle and [itex]\delta (x)[/itex] is the Dirac delta function.

The DDF is just a function (well, technically it isn't really a function) that is defined to be zero everywhere except at the origin, where [itex]\delta (0)=\infty [/itex]. Its integral is defined such that:

[tex]\int_a^b \delta(x)dx=1[/tex]

as long as a<0 and b>0 (as long as the integral "covers" the part where delta shoots to infinity).


Now with these definitions and since mass is the volume integral of mass density, the mass contained in a region around [itex]x_0[/itex] is therefore:

[tex]M=\int \rho (x)dx=m\int \delta (x-x_0)dx=m[/tex]

This definition of mass density for a point particle can be easily generalized to three dimensions as follows:

[tex]\rho (x,y,z)=m\delta (x-x_0)\delta (y-y_0)\delta (z-z_0)[/tex]

where we define [itex]0*\delta (x)=0[/itex] for all x (even x=0). As you can see the density is zero everywhere except at the point (x0,y0,z0) where it shoots to infinity. The volume integral over a region containing the point (x0,y0,z0) is therefore:

[tex]M=\iiint \rho (x,y,z)dxdydz=m[/tex]
DrewD
#7
Jan27-13, 08:51 PM
P: 446
I don't think this is really a physics question. Mathematically, Phinds is absolutely correct. A true point particle would be a discrete closed set in the three dimensional world and it would be zero dimensional. If you define it as a dirac delta, it is the limit of a sequence of functions (distributions) and integrating this over all space is non zero. But in reality, neither is really correct, so you should choose whichever method suits the question. You see this all the time. Sigma notation when it is suitable and integration when that makes more sense.

So, I guess, interchangeable a far as physics goes, but certainly different.
James MC
#8
Jan27-13, 09:03 PM
P: 175
Thanks, these are extremely useful responses.

Whether they are interchangeable regarding the physics is important to me. The reason I ask is that my predicament is this: I have formulated a proof in the context of Newtonian mechanics, which makes use of mass density distributions. I have (rightfully) been told that "for this proof to have significance you really need to show that it works in a point particle setting too" and when seeking advice from a physicist he said "this might be possible if you use Dirac delta functions".

Here is a fragment of the proof. I consider Newton's gravitational field equation for N mass densities, where the value of the gravitational field g(x) is determined by N mass densities (indexed by i):

g(x) = G [itex]\sum[\int\frac{ρ_{i}(ζ)(x-ζ)}{|x-ζ|^{3}}[/itex]] dζ

Here, we are integrating over a whole volume of space and ζ is a dummy term for integration. The important part of the proof shows that because the integrals are all over the same interval, we can swap the sum and the integral to get:

g(x) = G [itex]\int[\frac{(x-ζ)}{|x-ζ|^{3}}\sumρ_{i}(ζ)][/itex]dζ

I hope that makes sense. (Part of what it's supposed to show is that Mass densities are additive, or more precisely, that it makes no difference to g(x) whether we have many densities or a single density that is there sum.)

So my ultimate question is whether this "sum/integral swap" can be shown to hold for point particles--or at least, for dirac delta functions? Can one simply replace the densities in these equations with delta functions? Perhaps more importantly, can dirac delta functions add together to compose delta functions with spikes in more than one position?
elfmotat
#9
Jan27-13, 09:39 PM
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Quote Quote by James MC View Post
Thanks, these are extremely useful responses.

Whether they are interchangeable regarding the physics is important to me. The reason I ask is that my predicament is this: I have formulated a proof in the context of Newtonian mechanics, which makes use of mass density distributions. I have (rightfully) been told that "for this proof to have significance you really need to show that it works in a point particle setting too" and when seeking advice from a physicist he said "this might be possible if you use Dirac delta functions".

Here is a fragment of the proof. I consider Newton's gravitational field equation for N mass densities, where the value of the gravitational field g(x) is determined by N mass densities (indexed by i):

g(x) = G [itex]\sum[\int\frac{ρ_{i}(ζ)(x-ζ)}{|x-ζ|^{3}}[/itex]] dζ

Here, we are integrating over a whole volume of space and ζ is a dummy term for integration. The important part of the proof shows that because the integrals are all over the same interval, we can swap the sum and the integral to get:

g(x) = G [itex]\int[\frac{(x-ζ)}{|x-ζ|^{3}}\sumρ_{i}(ζ)][/itex]dζ

I hope that makes sense. (Part of what it's supposed to show is that Mass densities are additive, or more precisely, that it makes no difference to g(x) whether we have many densities or a single density that is there sum.)

So my ultimate question is whether this "sum/integral swap" can be shown to hold for point particles--or at least, for dirac delta functions? Can one simply replace the densities in these equations with delta functions? Perhaps more importantly, can dirac delta functions add together to compose delta functions with spikes in more than one position?
I believe what you're looking for is the following identity:

[tex]\int \sum _i f(x) \delta (x-x_i)dx=\sum_i f(x_i)[/tex]

which hold for any function f(x) which is continuous in the domain of integration. Using [itex]\rho_i (\zeta)=m_i \delta (\zeta -x_i)[/itex], you get:

[tex]g(x)=G \int \sum_i \frac{x-\zeta}{|x-\zeta |^3}~m_i~ \delta (\zeta -x_i)d \zeta=G \sum_i m_i \frac{x-x_i}{|x-x_i |^3}[/tex]

And this is exactly what you'd expect - i.e. it's the rule for summing multiple point particles' field contributions.
elfmotat
#10
Jan27-13, 10:05 PM
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As an aside, the proof for the identity I provided is relatively simple. Consider the following integral:

[tex]\int f(x) \delta (x-x_0)dx[/tex]

Now, since [itex]\delta[/itex] is zero everywhere except at x0, the entire integrand is zero everywhere except at that point. This means that the only contribution of f(x) to the integral happens at x0, and it can therefore be written as:

[tex]\int f(x)\delta (x-x_0)dx=\int f(x_0)\delta (x-x_0)dx=f(x_0) \int \delta (x-x_0)dx=f(x_0)[/tex]

If we're considering the addition of multiple delta functions ("spikes" as you called them), we simply write:

[tex]\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[/tex]
[tex]=f(x_1)+f(x_2)+~.~.~.~+f(x_n)=\sum_i f(x_i)[/tex]
BruceW
#11
Jan28-13, 05:16 AM
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Quote Quote by James MC View Post
Whether they are interchangeable regarding the physics is important to me. The reason I ask is that my predicament is this: I have formulated a proof in the context of Newtonian mechanics, which makes use of mass density distributions.
In physics generally, they are not interchangeable. But in the context of Newtonian mechanics, using mass density distributions, they are interchangeable, as you and elfmotat have shown. ("They" being the continuous and discrete formalisms).


Quote Quote by James MC
Here, we are integrating over a whole volume of space and ζ is a dummy term for integration. The important part of the proof shows that because the integrals are all over the same interval, we can swap the sum and the integral to get:
Yes, it is fine to do this because integration acts linearly, so you can always take the summation inside the integral.
James MC
#12
Jan28-13, 06:22 AM
P: 175
Thanks elfmotat, this is very helpful. I can see now how to formulate the proof in terms of delta functions. It's slightly different to how you've formulated it, however, and I'm not sure if the identity you state (which I think is sometimes called the sifting, or selecting property) can help me. Though it would be great if it could.

Newton's gravitation equation in Dirac delta form is:

[tex]g(x)=G \sum_i [\int \frac{x-\zeta}{|x-\zeta |^3}~m_i~ \delta (\zeta -x_i)d \zeta][/tex]

And here's where my "sum/integral swap" differs from yours... the sum only takes the delta functions in its scope, not the fraction:

[tex]g(x)=G \int [ \frac{x-\zeta}{|x-\zeta |^3} \sum_i ~m_i~\delta (\zeta -x_i) ]d \zeta[/tex]

The point of this is that I've recovered the form of the single point mass law (or at least, the single delta function law). This is what enables me to infer from the form of the law that mass is additive. Though I've not quite deduced mass additivity, I've only deduced delta function additivity. In other words, if [itex]\rho[/itex]_c(x) is the mass of the composite of the point masses indexed by i, then I've shown that:

[tex]\rho_c(x) = \sum_i ~m_i~\delta (\zeta -x_i)[/tex]

I'm not sure if it's possible to derive Mc=SUM(Mi) from this. One could appeal to the identity you mention, though I think one would first have to motivate the claim that the mass of a complex delta function is determined through integration, which might be begging the question? Any thoughts on this would be most welcome.


BruceW, thanks for your comments. As I understand it, they are not interchangeable in special relativity partly because there, mass is not additive.
BruceW
#13
Jan28-13, 06:29 AM
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Quote Quote by James MC View Post
BruceW, thanks for your comments. As I understand it, they are not interchangeable in special relativity partly because there, mass is not additive.
Really? I thought rest mass was additive. (as long as the objects have no relative motion). Or maybe, it is the relative motion that you are speaking about?
James MC
#14
Jan29-13, 03:33 AM
P: 175
In SR mass (what you call 'rest mass') is not additive. Mass additivity states that the masses of composites are the sum of their atomic component parts. All composites, whether their parts are moving relative to each or not, are within the scope of this principle. So you only need to find one example of a composite whose mass is not additive in order to refute the principle.
Studiot
#15
Jan29-13, 05:33 AM
P: 5,462
A few thoughts:

Going back to your original question, no they are not the same, even in physics.

A point mass is an idealisation that can have no angular motion or moment.
It allows centre of mass constructs.

It has precisely 6 degrees of freedom in the generalised coordinates system.

Any distribution in space, however small, must offer a moment/angular motion construct.
It must possess three extra degrees of rotational freedom.

On the subject of discrete v continuous. One of the greatest discoveries of classical physics/ applied maths was that the solution to continuous equations can have discrete solutions.

go well
BruceW
#16
Jan29-13, 05:42 AM
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(In response to James MC 'rest masses not additive' post): true, true. But if we assume that any two points very close to each other must have the same velocity, then there is no problem, because locally rest masses will add up, and so there will be no difference between 'point masses' and 'infinitesimal densities'.

In other words, I am pretty sure you can still make use of the Dirac-Delta function to represent the density of a point particle in SR. (Not going into quantum mechanics).

Edit: this is also relevant to Studiot's post. I think that as long as we assume that in the limit of small distance between different parts of the fluid, we get a vanishingly small difference in velocities. And so this would disallow angular momentum of a very small distribution in space. Does this sound right? I have not really thought about it before. P.S. I totally agree that small densities and point masses are not the same, even in physics.
Studiot
#17
Jan29-13, 05:58 AM
P: 5,462
we get a vanishingly small difference in velocities
Just because something is multiplied by an effectively zero quantity does not mean that it does not exist or that the degree of freedom does not exist.
It simply means that the value is zero.
BruceW
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Jan29-13, 06:01 PM
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I'm going to go against what I said in my last post. In SR, rest mass is not additive in the discrete or continuous formalisms. In the discrete case, if we have several point masses, (with generally different velocities), then their rest mass is definitely not additive. And in the continuous case, if we are integrating over the density of some fluid, then over some region, the fluid may be at rest, so we are calculating its rest mass in this region. But generally, the fluid will have a non-uniform velocity field, so generally we will be calculating the relativistic mass, not the rest mass.

And the thing I said about vanishingly small difference in velocities: I don't think I was quite right. If we consider a continuous distribution, and imagine making its volume very small, then different parts of it can still have different velocities. So any angular motion of that tiny distribution will be very small, but as (I think) Studiot was saying, it will still be some non-zero value, which therefore means there is an associated degree of freedom.

It is only when we take some special limit, where we assume that the relative velocities literally go to zero, that the angular motion disappears. And this is the discrete (i.e. point-like) formalism (as Studiot was saying). So it is only when we pass over to the discrete formalism, that the degrees of freedom associated with the physical angular momentum of the distribution disappear. I think that how this limit is taken in quantum mechanics has something to do with renormalisation theory. I don't know, as I haven't learned that kind of stuff. But it seems pretty interesting.

Edit: sorry to start talking about quantum mechanics at the end there, we had been talking about classical physics. quantum is a different kettle of fish, or some similar idiom.


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