Density, Point Particles, and Poisson's equation

[JDoolin]

The definition for density

$$\rho\equiv \frac{m}{V}$$


is troublesome because it involves this arbitrary Volume. Let's say we were asked to find the density of our solar system. The value of the density would vary greatly depending on where we decide to draw the sphere. Should the sphere go from the sun out to the gas giants, or should we enclose the outermost orbit off the Oort clouds?

There is no problem with the definition of density, within the contexts in which it is usually applied, but I think there is clearly a problem with applying it where you have point-particles; like planets and sun on the scale of the entire solar system.

In this situation I'm thinking there must be some kind of way to treat the different point masses with a point-discontinuity in the density (Dirac Delta Function, perhaps). Then the density of the solar system is basically a flat map, with the density everywhere zero, except where there is a planet, and at those points, the density jumps.

Poisson's Equation, $$\nabla^2 \Phi = 4 \pi G \rho$$, has two very different interpretations depending on whether you think of density as dirac-delta functions around point masses, or if you think of density as the mass divided by the volume enclosed.

Now, in our regular routine, of course, if I'm asked for the density of a bowl, the density of water, the density of air, etc, I first note that the distribution of particles in one small region is relatively constant, and I make sure not to enclose a volume containing other substances.

But what does this really give us? The density we find is still only an average. Within that bowl are molecules, and within the molecule are tiny nuclei, and vast regions of empty space occasioned by electrons. And within those nuclei are protons and neutrons, etc.

If you pick any point in the universe; even in the structure of solid objects, the density is zero, with dirac delta functions of density in precise spots where the nuclei are.

Another problem with using $$\rho\equiv \frac{m}{V}$$
where V is an arbitrary but definitely finite value, density becomes a matter of action at a distance! If you have to to draw a sphere (of undetermined volume) around a region, and figure out whether the particles are in that sphere or not, then the density at any given point must depend on something that is going on far away fom the point.

You may think obviously no one would try to actually claim this, however:

In Misner/Thorne/Wheeler's Gravitation, Chapter 7, they give $$(7.2) \nabla^2 \Phi = 4 \pi G \rho$$ as part of Newton's formulation of gravity. (also known as Poisson's Equation). The author goes on to say "The field equation (7.2) is not Lorentz-invariant, since the appearance of a three-dimensional Laplacian operator instead of a four-dimensional d'Alembertian operator means that the potential [tex\Phi[/tex] responds instantaneously to changes in the density $$\rho$$ at arbitrarily large distances away."

To me, it appears that MTW are misusing the definition of density to make the claim that Poisson's Law is an action-at-a-distance law.

 

[quZz]

$m/V$ is an average density, it depends on which volume you choose.

In Poisson's equation $\rho$ is a point density $\rho(\textbf{x},t)$ that is a limit of average density when the volume tends to zero around point $\textbf{x}$ at time [tex]t[/itex]. For a system of point-like particles that are located in $\textbf{x}_a(t)$ you will have $\rho(\textbf{x},t) = \sum_a m_a \delta(\textbf{x}-\textbf{x}_a(t))$.

 

[JDoolin]

$m/V$ is an average density, it depends on which volume you choose.

In Poisson's equation $\rho$ is a point density $\rho(\textbf{x},t)$ that is a limit of average density when the volume tends to zero around point $\textbf{x}$ at time [tex]t[/itex]. For a system of point-like particles that are located in $\textbf{x}_a(t)$ you will have $\rho(\textbf{x},t) = \sum_a m_a \delta(\textbf{x}-\textbf{x}_a(t))$.

But what does that sum involve, precisely? Is it the sum over the nearest four particles? It still involves an arbitrary volume.

There are two different questions here; one asks for the density of a substance at a point in space. The other asks for the density of a point in space.

Now if you're calculating density of a particular homogeneous substance, then it makes good sense to use an arbitrary volume, or an arbitrary number of particles so long as all of the volume and particles belong to that substance.

However, your description "a limit of average density when the volume tends to zero around point $\textbf{x}$ at time [tex]t[/itex]. " will gve either zero, if no point particle is present, or infinity (=finite mass/zero volume), if there is a point particle present. There can be no sum over particles at finite distances, when you bring your volume down to zero.

 

[quZz]

But what does that sum involve, precisely? Is it the sum over the nearest four particles? It still involves an arbitrary volume.

It is the sum over all particles.

There are two different questions here; one asks for the density of a substance at a point in space. The other asks for the density of a point in space.

Density of a point particle is (as you said from the beginning) its' mass times Dirac delta function.
In macroscopic physics (e.g. fluid dynamics) you average this microscopic density over a small finite volume (small comparative to all volumes under consideration) but large enough to include a lot of molecules. So microscopic delta-functions vanish and you get a smooth function of coordinates - density of the substance.

 

[Nabeshin]

But what does that sum involve, precisely? Is it the sum over the nearest four particles? It still involves an arbitrary volume.

It's the sum over the particles enclosed within the volume you're solving the Poisson equation on! Only in this sense is it arbitrary. When you specify your boundary conditions, you limit the volume on which you solve the equation, which in turn tells you what "volume" to use for the density function.