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easy
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Joined: July 14th, 2002, 3:00 am

### Newton's shell theorem with random density

Does anyone have references for Newton's shell theorem for the case where the mass density is randomly distributed and the point of interest is inside the shell? I am interested in how the statistical moments appear in the gravitational field.

Alan
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### Re: Newton's shell theorem with random density

Is the putative random mass distribution still rotationally invariant?

easy
Topic Author
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Joined: July 14th, 2002, 3:00 am

### Re: Newton's shell theorem with random density

No, there's no symmetry assumed. But the mass density distribution can be considered to be independent of position.

Alan
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### Re: Newton's shell theorem with random density

I don't understand what 'independent of position' means. Can you give some examples?

Paul
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### Re: Newton's shell theorem with random density

It's just a blob. (Or several blobs?) Same density.

Alan
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### Re: Newton's shell theorem with random density

All too vague for me. How does it all relate to a "shell"? Somebody please use the LaTex or post a clarifying picture.

trackstar
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Joined: January 1st, 1970, 12:00 am

### Re: Newton's shell theorem with random density

All too vague for me. How does it all relate to a "shell"? Somebody please use the LaTex or post a clarifying picture.
Newton's Shell Theorem via math.ksu.edu in pdf form

Equations and pictures there, Your Majesty. : D

Paul
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Joined: July 20th, 2001, 3:28 pm

### Re: Newton's shell theorem with random density

I assume it's a question about gravitational fields. From outside, a spherical shell behaves like a point mass. From inside, there is zero gravity (from the shell).

For different shapes you get different fields. Can those fields be represented by something simpler? I don't know what the word 'statistical' is doing in there so that makes me doubt my understanding.

trackstar
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Joined: January 1st, 1970, 12:00 am

### Re: Newton's shell theorem with random density

"A solid, spherically symmetric body can be modeled as an infinite number of concentric, infinitesimally thin spherical shells. If one of these shells can be treated as a point mass, then a system of shells (i.e. the sphere) can also be treated as a point mass. Consider one such shell (the diagram shows a cross-section).." and so on.

Alan
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### Re: Newton's shell theorem with random density

A "random mass distribution" needs to be defined by the OP.

For example, does easy mean "N random point masses uniformly distributed within a sphere of radius R"?
Or, something else?

trackstar
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Joined: January 1st, 1970, 12:00 am

### Re: Newton's shell theorem with random density

Replacing previous diagram with one that will stay put:
NSTdiagram.png

katastrofa
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### Re: Newton's shell theorem with random density

I don't understand what 'independent of position' means. Can you give some examples?
The two nice words I constantly used in my physics times were isotropic and (vs) homogeneous.

easy
Topic Author
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Joined: July 14th, 2002, 3:00 am

### Re: Newton's shell theorem with random density

I don't understand what 'independent of position' means. Can you give some examples?
Hi Alan, by independent of positon I mean that the statistical moments, mean, variance, etc of the probability density function describing the mass density (apologies for the double use of density, but it should be clear) don't depend on the position on the shell. I'm not assuming that they are point masses inside a sphere. The standard result uses symmetry to calculate the acceleration due to gravity, starting with a diagram similar to trackstar's post, except the point where "m" is, is inside the shell, eg Figure 2 here https://www.math.ksu.edu/~dbski/writings/shell.pdf. If the mass is not evenly distributed then I can't use symmetry to give us a 1-D integral for the acceleration.
In reply to Paul, if the mass density is constant on the shell, then the gravitational field is zero in the shell. If the mass density wasn't constant, but could be characterised by a probability density function, and I'll assume all moments exists for the point of the argument, then it seems plausible that there's a fluctuation of the gravitational field inside the shell, but the expected value would be zero. is there a dipole or quadrupole term due to the uneven distribution of mass on the shell.
It seems like a plausible problem, but the discussions, lecture notes and references I can find assume that the density of mass on the shell is constant.

Alan
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### Re: Newton's shell theorem with random density

So, just to clarify, and tell me if I am wrong on any point:
- there is mass only on the shell (surface of the sphere).
- the mass is randomly distributed according to a probability density, call it $p(\Omega)$, where $\Omega = (\theta, \phi)$ are coordinates on the shell.
- the observation point is inside.

If this is correct so far, AND you expect the expected value of gravitational field at the observation point to be zero, then I am guessing you want $p(\Omega)$ to be a uniform probability density? So, N point particles of equal mass distributed on the surface of the sphere, where the location of each particle is an IID draw from the uniform distribution on the surface?

Otherwise, if  $p(\Omega)$ is not uniform, but arbitrary, the mean force (a vector) and the mean field will not generally be zero. For example, take the case where $p(\Omega)$ is concentrated at the north pole.

Finally, your statement about moments is confusing because moments are an integration over the surface. So moments are scalars independent of position --  no matter what the probability density.

Alan
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### Re: Newton's shell theorem with random density

Sorry, to clarify, -- I shouldn't have said "moments are scalars". The "statistical" moments here are moments of a two-dimensional density $p(\theta,\phi)$. So, the means form a two-dimensional list. There is a variance-covariance matrix (2 x 2), etc. The "statistical moments .. don't depend upon the position" still doesn't make sense to me for the reason I gave above -- moments are integrations over positions. For example in 1D, it is like saying the mean $\int x \, p(x) dx$ doesn't depend upon position. Well, how could it?