Theory of spatial point processes has a notion of "compIete spatial randomness", which basically means that there is no interaction between points whatsoever. In your case it would be equivalent to add the points one after the other to the cylinder, and each new point is placed randomly in the same way as all other points, independently of what is there already. If this means that every location in the cylinder is chosen with the same probability, you get a homogeneous completely random distribution. It is also isotropic. Amazingly, this distribution still remains completely random and isotropic(!) if you stretch the cylinder.

However, the tiniest bit of interaction between the points will change the picture. Such interaction could mean that points cannot come closer than a given distance, which is the case if the points represent particles that cannot overlap. There are tons of mathematical models for random arrangement of points, and most of them are isotropic. If you started with an isotropic spatial distribution of particles that is not _completely_ random, then stretching actually will affect the distribution, and it will become anisotropic. Fry (1979, Random point distributions and strain measurement in rocks. Tectonophysics 60, 89–105 ) proposed a convenient plot to detect this anisotropy, and if you use R, you can find this as function fryplot in the spatstat package.

Here is a paper where several methods were compared to infer on the strain, the particles were air bubbles in glacial ice:

Redenbach, Särkkä, Freitag and Schladitz (2009) Anisotropy analysis of pressed point processes. Advances in Statistical Analysis, 93, 237-261.

Ian, I believe the following points could guide you in answering your questions

1. In this case, the operational definition of "random distribution"

2. Based on present literature, the factors that affect randomness (or make random non-random)

3. Based on present literature, measure of degree of randomness or non-randomness.

I think an understanding or knowledge about these three points (beyond your case) will help your case. Ed

What does "random" mean to you? In the most comprising way - to my knowledge - "random" simply means "lack of knowledge". Thus, if you have a random distribution of of points it simply means that you do not have the means and/or the knowledge to predict their positions with certainty. However, "random" does not mean that we do not know anything. We must have some assumptions, some expectations we can formulate. This means, we can say, for instance, that we can give some relative degrees of expectation where the points might be. If - just as example, the cylinder is huge (kilometers in height, say a thought column of the earth's atmosphere), we know that due to gravity the particle density must decrease with altitude. Physics tells us that the decrease in density is exponential, so we can use and exponential function to formulate out expectation that a particular small volume contains any particle. If we do not have such additional knowledge we have no reason to more likely expect a particle in any part of the cylinder than in any other part, leading to a simple constant function to formulate our relative expectations (i.e. we think of a uniform distribution).

As long as the cylinder is not squeezed into a single point, the distribution of the particles in the cyclinder will thus remain random. From an external standpoint, squeezing the cylinder changes the directional contributions, but relative to the cylinder the contributions are invariant. This is a purely geomotrical issue that does not affect the fact of a "random distribution" of the particles. A comparison might be possible, however, through the entropies of the distributions that might change with the shape of the cylinder.

Arun, what is a rationale behind calling the "isotropic" projection of a distribution "random" but not an "anisotropic" projection?

What would you say about the "randomness" of the particles if you were not told whether the cylinder was already transformed?

Thank you for the explanations. I agree that the shape of the distribution relative to an external inertial system changes. I do not understand what the central limit theorem has to do with all this and why you call the initial distribution "isotropic".

Theory of spatial point processes has a notion of "compIete spatial randomness", which basically means that there is no interaction between points whatsoever. In your case it would be equivalent to add the points one after the other to the cylinder, and each new point is placed randomly in the same way as all other points, independently of what is there already. If this means that every location in the cylinder is chosen with the same probability, you get a homogeneous completely random distribution. It is also isotropic. Amazingly, this distribution still remains completely random and isotropic(!) if you stretch the cylinder.

However, the tiniest bit of interaction between the points will change the picture. Such interaction could mean that points cannot come closer than a given distance, which is the case if the points represent particles that cannot overlap. There are tons of mathematical models for random arrangement of points, and most of them are isotropic. If you started with an isotropic spatial distribution of particles that is not _completely_ random, then stretching actually will affect the distribution, and it will become anisotropic. Fry (1979, Random point distributions and strain measurement in rocks. Tectonophysics 60, 89–105 ) proposed a convenient plot to detect this anisotropy, and if you use R, you can find this as function fryplot in the spatstat package.

Here is a paper where several methods were compared to infer on the strain, the particles were air bubbles in glacial ice:

Redenbach, Särkkä, Freitag and Schladitz (2009) Anisotropy analysis of pressed point processes. Advances in Statistical Analysis, 93, 237-261.

I think that degree of randomness it exactly the same. If you have some shape S and transform it by reversible transformation F then points which are randomly distributed in S have the images randomly distributed in F(S). If the used distribution depends from some geometrical characteristics of S then distribution in F(S) depends from transformed characteristics. If the degree of randomness changes after transformation we obtain the transformation which, for example, decrease randomness. In this case the inverse transformation increase randomness. It will be very strange effect.

You can estimate the distribution parameters bu set of points in original and transformed areas and compare it. The difference will depends from F only.

Could you refer to the concept of (representative volume element) RVE in non local continua, it might help cleared the answer.

.

if we assume a simple transformation of a [0, 1]x[0, 1] domain to [0, 2]x[0, 1/2]

x -> 2x

y -> y/2

it seems to me that a Kolmogorov complexity argument shows that if the initial distribution is random (in the Kolmogorov sense), the resulting distribution is random too

as Ute points out above, this of course ignores any kind of interactions that might happen between points or between points and the "substrate" (be it contact, short-range or long range interaction) ; we all know that if such interactions are taken into account, phase transitions may occur as the conditions change (pressure for instance)

.

Initially question asks about degree of randomness transformed area if the original area contains randomly distributed points. As Fabrice Clerot note, the Kolmogorov complexity theory state that measure of ranfomness of transformed deistribution is the same as original distribution.

I am not agree, that for transformed are can changes pressure because pressure is a function of the particles velocity, which have to be recalculated.

I think that general problem is exhausted. If the question include some specific options we can discuss it after the options will cpesified.

Thank you all for your answers to my question. It seems there are degrees of randomness depending on interactions between the particles and substrate?

I will provide some background to my question.

If a multi grain piece of metal is strained at low temperature the grains will be elongated. This elongation can be determined by quantitative metallography to give an independent measure of intra-granular strain. If we regard triple points as reference points could we deduce strain from the distribution of the triple points. A method that comes to mind is to select a triple point at random, then determine the angle of the nearest neighbouring triple point.

Repeat for many triple points selected at random, and plot the distribution of nearest triple point against angle from tensile axis. Would this distribution be random (circle) or non random (ellipse)?

The object of the exercise was to measure inter-granular strain (= total strain - intra-granular strain).

At higher temperatures the grain boundaries will migrate to reduce their surface energy.

This will change the anisotropy of grains and move the triple points, so these can no longer be used as markers.

Most metals contain inclusion particles. Could a change in their distribution after straining be used to measure intra-granular strain? Particles in grain boundaries would be ignored as these could have been swept up by grain boundary migration.

Ute is right. If the random distribution is modelised by an homogeneous Poisson point process, after streching, one can show that it is is still an homogeneous Poisson process. the intensity is just changed.

For example take a square A, The construction of the process of intensity a (mean number of points by area unit) consists in sampling a Poisson variable of parameter

a.area(A) to obtain the number N of points in A, then to draw N points independently with an uniform distribution in the square. This is done by sampling the x and the y coordinates of each points with two indeopendent uniform distribution. If you streched the result to a double square rectangle,, the number of points is unchanged and the uniform distribution stays an uniform distribution, so that the resulting process is an homogeneous Poisson process but as the area doubled the intensity has to be divided by 2.

If it is possibe to add a possible,further comment :stretchìng the original area, the general distribution of the points will remain a random distribution, but tne numerosity of the ponits per unit area will change ; that is to say, assuming the area as a volume of a body and the points to solid parts of the body ,the distributon of its particles will renainn the same but its density will change.

This recent paper is somehow related to your question:

Martin Fink, Jan-Henrik Haunert, Joachim Spoerhase, Alexander Wolff, "Selecting the Aspect Ratio of a Scatter Plot Based on Its Delaunay Triangulation", IEEE TRANSACTIONS ON VISUALIZATION AND COMPUTER GRAPHICS, VOL. 19, NO. 12, DECEMBER 2013.

Hi Giancario,

If we stretch a piece of metal there will a slight reduction in density for elastic deformation (characterised by a Poison's ratio of less than 0.5. However the density remains substantially constant for plastic deformation. Consequently I would expect the number of small sparse particles per unit volume to remain unchanged. My hypothesis is that relative to a randomly selected particle, neighbouring particles in line with the tensile axis will move away and those perpendicular to the tensile axis will move towards the reference particle. Unfortunately we cannot use the same reference particles before and after deformation. The answers suggest that any detectable deformation around specific particles is masked by the random nature of the original distribution of particles?

Hi Michael,

Thanks for the reference. I will look it up.

The answer to your question can be found by using elementary statistical definitions.

Firstly, your two versions of the problem (stretch a rectangle and stretch a cylinder are mathematically identical since an (infinitely thin) cylinder surface has an Euler curvature of zero and can be "unwrapped" to a rectangle without introducing any distortion.

Now assume that, before the stretching, the rectangular sheet has longitudinal (length) and lateral (circumference) dimensions of Lx and Ly, respectively. By assumption, the surface is dotted with points according to an uncorrelated twodimensional distribution of density

f(x,y)=g(x) h(y) = 1/(Lx Ly) with g(x)=1/Lx, h(y)=1/Ly

Now lets stretch the cylinder (the rectangular sheet) such that its new length and circumference are given by

Lx'=a Lx, Ly' = b Ly

where, generally, the relative increase of the surface area a*b>1 corresponding to a Poisson ratio of less than 0.5. If this elongation is applied uniformly, a point (x,y) on the surface will transform to the point

(x', y')=(ax, by).

Note that the x and y coordinates are transformed independently of each other. The new one-dimensional distribution functions can be obtained directly by the conservation law of probabilities,

g(x) dx = g(x(x')) dx/dx' dx' equiv g'(x') dx'

and analogously for h(y) dy. With x(x')=x'/a and y(y')=y'/b this gives

g'(x')=g(x'/a) / a, h'(y')=h(y'/b) / b,

and the new two-dimensional distribution

f'(x',y')= g'(x') h'(y') = g(x'/a) h(y'/b) / (a b)

this is valid for any distribution functions g(x) and h(y). For the uniform distributions g(x)=1/Lx and h(y)=1/Ly, we obtain

f'(x',y')=1/(Lx Ly a b) = f(x,y)/(a b)= 1/(Lx' Ly')

This means, f'(x',y') does not depend on x' and y'. so we get again a two-dimensional independent uniform distribution! The only thing which has changed is the density which is now smaller by a factor 1/(a b) corresponding to an increase of the area by a b. For an area-conserving stretch (a*b=1, Poisson ratio=0.5), we have f'(x',y')=f(x,y).

Since a two-dimensional distribution is completely and uniquely specified by its density f(x,y), this means, the points remain completely isotropically and uniformly distributed, whether a*b=1 or not. Thus, the stretching can *not* be deduced from the distribution of the points.

Hi Ian,

your hypothesis about the neighbouring particles is right. And, as you write, you cannot find out which particles were neighbours before the deformation. In the case that the points form a Poisson process (which we call "without interaction"), the distribution of neighbours around a point stays the same after deformation, even though the neighbours are not necessarily the same points as before. But there is some hope in your case. Reckon that triple points are the points that belong to the boundary of 3 grains? Such points are quite likely not distributed like a Poisson point process, and you should find some detectable anisotropy after deformation.

Looking at the direction to the nearest neighbour only is a good start, in particular if the pattern before transformation is more regular than a Poisson pattern. It might be even more informative to look at the number of points inside a sector of a circle of given radius, but let the sector move around (plotted against the radius, and with a fixed direction of the sector, you obtain the so called directional K-function). This also works if the original point pattern is more clustered than a Poisson pattern. The methods in Michael's reference look also promising.

Hi Michael,

Interesting reference! Voronoi / Delaunay tessellations are a great tool for analysis of point patterns, also popular among "point process theorists".

Hi Fabrice, and Evgeny

I meant interaction between points and points only, but interaction between points and substrate could also be interesting to look at. Interesting approach to interpret "degree of randomness" in the sense of information content :-)

Hi Martin,

The cylinder I had in mind was solid. For constant volume Poisson's ratio will need to be 0.5.

Stretching the 2D plane is purely hypothetical. For constant area the "Poisson's ratio" would need to be 1.0. To emulate this in a real material the latter would need to keep its thickness and volume constant as it is stretched.

Dear Jan, we have to agree on what we are discussing. My reply was concerning a theoretical geometrical cylinder of volume V : this cylinder does not have any particular physic properties as elasticity or resistance to the straining from which the persistence of the random distribution and the change of points density. But you and other researchers are evaluating the concerns of a material cylinder : I am not an expert in this field and I cannot participate to your discussion ; I have only to think that in such case could be suitable to define the particular material by which the cylinder is made, because I think that the physical properties of cylinders made by different plastic or iron cylinder could differently reply to the straining with different changings of particles position.

Ian, for as far as I understand what you want to do with it, if you make a 2D polar plot of the nearest triple point in a (distance, angle) type of way, an initial 'random' distribution, represented by a circle, would be transformed to an 'elongated' distribution, represented by an ellipse in your polar plot (I think this is clear if you think of the perfect grain with a hexagonal shape that gets compressed and or extended in a certain direction). This would however not be a perfect ellipse, as it will be distorted due to the elastic and or plastic anisotropy in your material (any single crystal material like a 'perfect' metal grain is anisotropic), except if you would be unlucky enough to have tested your material in a pseudo-isotropic direction. Now, a major concern that I have is how you are going to distinct between local (individual grains) and global (polycrystalline material) deformation in your measurement. The experiment you propose is going to 'assemble' data from a large number of different grains, that all have different crystal orientations. I don't seem how your technique will be able to unravel inter-grain strain from intra-grain strain without going over some sort of OIM measurement and crystal elasticity/plasticity analysis.

OK, undoubtedly this is all very interesting ,but this discussion is clearly going out any possible interest of Nephrology and I would add out of any field of appliance to Medicine : so I leave definitely this discussion and any alert of RG concerning the same problem will be neglected.

Just to add a small point to the discussion, since I am working on these things quite a while: Martin is completely right, stretching a uniform distibution results in a uniform distribution.

If, however you strech nonuniformly, you will deform the distribution, e.g., in the case of nonlinear materials. If you stretch the middle less than the outer regions, the distribution gets more "gaussian" (in that there is a peak in the middle which decays towards the outer regions. That's essentially how you produce numerically random distributions from uniform ones (one of the ways, to be more accurate(.

A function of a random variable is a random variable. In terms of RVs I think you mean is Y = a*X a RV if X is. Yes!!

Thanks for the answers. I would like to clarify a couple of points.

The initial random distribution of points is stretched in one direction and contracted in the other two directions so that the number of points per unit volume remains constant. Most metals contain small oxide or sulphide inclusions which approximate to points

Regarding grain shape analysis, this is most easily achieved by counting longitudinal and transverse grain boundaries per unit length.

intra-granular strain = (Nt/Nl)^2/3 - 1 where Nt and Nl are the number of grain boundaries per m in the transverse and longitudinal directions respectively.

See the following references in J. Institute of Metals:

W. A. Rachinger, 1952-53, V81, p33

J.H. Hensler and R.C. Gifkins, 1963-64, V92, p340

C.M. Sellars, 1964-65, V93, p365

The problem with this method is that any migration of the grain boundaries will cause errors.

Don´t you believe that depends on your definition of randomness?

A random variable under transformation should remain random with the exception of a transformation which is an imitation of Maxwell's demon. However, measurements appear to be random but may not be random.

Every sample space that you do not have exact prediction on the observations is random. What you mean by not random? If you mean unequal probabilities (densities) then we have randomness without equal probabilities. Jeffreys would look for transformations that produce uniform distributions.

I'm pointing to a real measurement space (i.e. in a lab, I measure x) in which the randomness is a measure of our inability to discern. The underlying attribute may look random due to inadequate definition of the model by which I interpret the measurements.

However, my first comment that a random variable remains random in a transformation that does not embed a demon remains valid. Yes, transformations that modify the distribution still produce random variables. It is that teh transformation from a random variable to a non-random varialbe would require an insertion of ntelligent transformation. Nothing grand; just a "random" observation of the communication stream in this list.

If you are reasoning in terms of finite number of random points, like specks of impurities in a molten metal cylinder, consider the reverse problem: can you define a transformation where , knowing the approximate location of these random specks by knowing their distribution function , can you define a transformation whereby this distribution is "approximately" not so random , down to a certain cell radius, say. For example you might want this transformation to make the density of impurities inside any cube 1 micron wide, homogeneous. If the initial random distribution is sufficiently well known and invariant , I think by invoking Radon-Nykodim or the likes you can define a transformation made of local adhoc transformations that can sort of recast the random into almost non random.

A one-to-one mapping of random points results in points that are still random.. The contracting along one dimension and expanding along another clearly meets this criterion; in fact, is linear. The mapping need not be linear but a linear mapping will preserve uniformity.

Assuming the deformation is uniform, the answer is yes  they are distributed randomly.  While the points' absolute positions will have changed, their *relative* positions have not: relative to each other, and relative to the coordinate system inside the volume/plane.

More generally, we can ask if any arbitrary deformation could induce the particle locations to become non-random.  I'm not sure about this.  If the answer is yes, it would surely be a strange deformation designed only for that purpose.

Hi. -  Interesting.  -   Being distributed randomly, and a uniform distribution, are two different things. If you had a gas in a confined space, it would probably be better to think of that as a uniform distribution, rather than random. All of the particles would seek to distribute themselves uniformly. But whether you have points uniformly or randomly distributed, increasing or decreasing the space they occupy sounds like changing a variance, but not the distributional form.  BTW, if random about a mean, points would only be approximately random, as it sounds like you have boundaries at which the distribution is truncated. ??

Ian - I see you have this: 

"The initial random distribution of points is stretched in one direction and contracted in the other two directions so that the number of points per unit volume remains constant. Most metals contain small oxide or sulphide inclusions which approximate to points."

Then consider this: If you have a uniformly applied contraction or stretching in one dimension for a randomly distributed set of points - however that set of points was randomly distributed in the first place - then the distribution remains of that distributional form, with changed variance.  I think that the same applies in each dimension.  I think that the constant-points-per-volume note is just additional information, with regard to your question.  

Thanks -  Jim   

Oops. Ah, maybe the unit volume part is associated with the last part of your question: 

"Can the change be measured to deduce strain?" 

And as for the change in variance for the distribution, you still have to consider the finite number of points that are actually there, over which you are summing.