Discrete Surface and Volume Integrals

[inflector]

In another forum, I have been challenged to prove mathematically that a certain idea which consists of fields of discrete elements will satisfy http://en.wikipedia.org/wiki/Divergence_theorem" .

The fields are not expressible in terms of a differentiable function but rather consist of discrete elements which have exact positions and values at those positions and no value at other positions either through the surface nor inside the volume. So the function is not smooth or continuous.

I know how to show that the sum of the vectors for my discrete fields inside the volume is equal to the sum of the vectors through the surface, but how do I prove that:

1) The surface integral for a discrete field is the same as the sum of the discrete vectors through the surface? Is there some rule or definition which shows this?

2) The volume integral for a discrete field is the same as the sum of the discrete vectors inside the volume? Is there some rule or definition which shows this?

So can I just assume that everyone knows that exact sums of the discrete elements in a volume and through a surface are both somewhat equivalent to their respective integrals?

They are not strictly speaking the same it seems to me but only analogous. The trouble is that a point standing alone doesn't have any volume or area. How, after all, does one determine the area under the curve for a value that exists only at (1,1,1) or (1.5, 3, 7) but no other points?

 

[HallsofIvy]

In another forum, I have been challenged to prove mathematically that a certain idea which consists of fields of discrete elements will satisfy http://en.wikipedia.org/wiki/Divergence_theorem" .

The fields are not expressible in terms of a differentiable function but rather consist of discrete elements which have exact positions and values at those positions and no value at other positions either through the surface nor inside the volume. So the function is not smooth or continuous.

I know how to show that the sum of the vectors for my discrete fields inside the volume is equal to the sum of the vectors through the surface, but how do I prove that:

1) The surface integral for a discrete field is the same as the sum of the discrete vectors through the surface? Is there some rule or definition which shows this?

2) The volume integral for a discrete field is the same as the sum of the discrete vectors inside the volume? Is there some rule or definition which shows this?

There is no standard definiton for integration on a discrete field. To devise your own, I suggest you try to set it up so that
\frac{\int_A f(x)dx}{|A|}
where A is some set of values of x, and |A| is some measure of the size of A, is equal to the average value of f(x) on that set. Whether or not the makes the integral the sum depends on exactly how you measure the set A. If |A| is just the number of values of x in A (which requires that A be finite) then, yes, the integral is the sum.

So can I just assume that everyone knows that exact sums of the discrete elements in a volume and through a surface are both somewhat equivalent to their respective integrals?

They are not strictly speaking the same it seems to me but only analogous. The trouble is that a point standing alone doesn't have any volume or area. How, after all, does one determine the area under the curve for a value that exists only at (1,1,1) or (1.5, 3, 7) but no other points?

What do you mean by "exists only at (1,1,1) or (1.5, 3, 7)"? I would have thought 1.5, 3, 7 were values of x but then (1,1,1) makes no sense.

 

[inflector]

There is no standard definiton for integration on a discrete field. To devise your own, I suggest you try to set it up so that
\frac{\int_A f(x)dx}{|A|}
where A is some set of values of x, and |A| is some measure of the size of A, is equal to the average value of f(x) on that set. Whether or not the makes the integral the sum depends on exactly how you measure the set A. If |A| is just the number of values of x in A (which requires that A be finite) then, yes, the integral is the sum.

I can specify in my proof that the set is finite. I'm showing a model of quantum gravity that is defined via a 3 dimensional tree structure consisting of points which can be enumerated. The points are connected by line segments which can also be enumerated. So the "field" values only exist at discrete points and the "field" is not a field in the traditional sense. But I am trying to adapt the spirit of Gauss's law to show that the concept obeys that spirit.

Gauss's law equates the volume integral of the divergence inside a closed surface with the surface integral of the flux through that closed surface. Now naively for a volume V, it seems like:

\iiint\limits_V \nabla\cdot F(p)

should be the sum of the divergence of the points in the volume V which are also contained in the set A where

F(p)

is a continuously differentiable function which defines the field value of the location p.

However, I see one big problem with meeting the continuously differentiable requirement of the divergence itself because a discontinuous field is not continuously differentiable. Therefore one cannot define the divergence for the function F:

F = Ui + Vj + Wk

which is equal to

\nabla\cdot F = \frac{\delta U}{\delta x} + \frac{\delta V}{\delta y} + \frac{\delta W}{\delta z}

because this requires that the function F is continuously differentiable in all three dimensions. This is not true for the field I am considering.

Nevertheless, the sum of F(p) for the set of points where F(p) is nonzero does represent the spirit of Gauss's Theorem in the sense that it defines the sources and sinks in the volume and the flux through the surface which is what Gauss's theorem is supposed to show. Namely that they are equal.

An analogous problem exists for the flux through the surface since in this quantum gravity approach the flux is defined by the line segments with pass through the surface. I can also enumerate and sum these line segments and their values at the surface which seems analogous to performing a surface integral of the flux, but I cannot perform an actual surface integral.

I'm just having trouble figuring out how to best explain this in a mathematical proof.

What do you mean by "exists only at (1,1,1) or (1.5, 3, 7)"? I would have thought 1.5, 3, 7 were values of x but then (1,1,1) makes no sense.

Sorry, I should have been more precise. The field's values are defined by a point defined in 3 dimensions. So my example was simply the three coordinates for two separate points.

In cartesian coordinates, my (1,1,1) is x=1, y=1, z=1 with a nonzero field value, and the second point, (1.5, 3, 7), was x=1.5, y=3, z=7 with a nonzero field value. Note, at x<>1, y=1, z=1 the field value is zero, likewise at x<>1.5, y=3, z=7, the field value is zero. This is what makes the field discontinuous and therefore not continuously differentiable.