# Spatial homogeneity and the functional form of two-point functions

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• Frank Castle
In summary: Yes that is correct. An affine space is a slight generalisation of the notion of a vector space. In a vector space the points themselves are vectors and so are their sums and differences. In an affine space, the points are not vectors, and the sum of points is not defined (nor is the scalar product of a point) but their differences are defined and are equal to members of the related vector space.
Frank Castle
Consider a two-point function $$f(\mathbf{r}_{1},\mathbf{r}_{2})$$ If one requires homogeneity, then this implies that for a constant vector ##\mathbf{a}## we must have $$f(\mathbf{r}_{1},\mathbf{r}_{2})=f(\mathbf{r}_{1}+\mathbf{a},\mathbf{r}_{2}+\mathbf{a})$$ How does one show that if this is true then the function must depend only on the difference between the two points ##\mathbf{r}_{1}## and ##\mathbf{r}_{2}## and not on each of them independently? i.e. $$f(\mathbf{r}_{1},\mathbf{r}_{2})\equiv f(\mathbf{r}_{1}-\mathbf{r}_{2})$$ I can kind of see why this is the case (since ##\mathbf{r}_{1}-\mathbf{r}_{2}=(\mathbf{r}_{1}+\mathbf{a})-(\mathbf{r}_{2}+\mathbf{a})## and hence ##f(\mathbf{r}_{1},\mathbf{r}_{2})=f(\mathbf{r}_{1}+\mathbf{a},\mathbf{r}_{2}+\mathbf{a})## is satisfied if ##f(\mathbf{r}_{1},\mathbf{r}_{2})\equiv f(\mathbf{r}_{1}-\mathbf{r}_{2})##), but is this an if and only if statement and can it be proven that if ##f(\mathbf{r}_{1},\mathbf{r}_{2})## is homogeneous, then ##f(\mathbf{r}_{1},\mathbf{r}_{2})\equiv f(\mathbf{r}_{1}-\mathbf{r}_{2})##.

Apologies if this is a really stupid question, but it's been bugging me and I'm hoping that someone can help me out.

Frank Castle said:
can it be proven that if ##f(\mathbf{r}_{1},\mathbf{r}_{2})## is homogeneous, then ##f(\mathbf{r}_{1},\mathbf{r}_{2})\equiv f(\mathbf{r}_{1}-\mathbf{r}_{2})##.
It seems that what you want to prove is that, given an affine space ##A## over vector space ##V##, if ##f:A\times A\to R## is homogeneous then there exists a function ##g:V\to\mathbb R## such that, for all ##x,y\in A## we have ##f(x,y)=g(\vec{(x-y)})##.

To prove this, pick any base point ##z_0\in A##. Then define ##g## by ##g(\vec v)=f(z_0+\vec v,z_0)##.

Now we test whether it has the required property. Take any two ##x,y\in A##. Then

$$g(\vec{(x-y)})\equiv f(z_0+\vec{(x-y)},z_0)=f(z_0+\vec{(x-y)}+\vec{(y-z_0)},z_0+\vec{(y-z_0)}) =f(z_0+\vec{(x-z_0)},y)=f(x,y)$$
as required (the second step uses the homogeneity of ##f##).

andrewkirk said:
To prove this, pick any base point z0∈Az_0\in A. Then define gg by g(⃗v)=f(z0+⃗v,z0)g(\vec v)=f(z_0+\vec v,z_0).

Now we test whether it has the required property. Take any two x,y∈Ax,y\in A. Then

g(→(x−y))≡f(z0+→(x−y),z0)=f(z0+→(x−y)+→(y−z0),z0+→(y−z0))=f(z0+→(x−z0),y)=f(x,y)​
g(\vec{(x-y)})\equiv f(z_0+\vec{(x-y)},z_0)=f(z_0+\vec{(x-y)}+\vec{(y-z_0)},z_0+\vec{(y-z_0)}) =f(z_0+\vec{(x-z_0)},y)=f(x,y)
as required (the second step uses the homogeneity of ff).

I'm not quite sure I understand the second step - isn't it required the the shift is by a constant vector?

Also, is the form ##f(x,y)=f(x-y)## a necessary or sufficient condition (or both) to satisfy spatial homogeneity? I've often seen it written as spatial homogeneity requires that any function of two positions ##\mathbf{r}_{1}## and ##\mathbf{r}_{2}## is dependent only on the difference between the these two positions and not each of them independently. They then proceed express this mathematically as ##f(\mathbf{r}_{1},\mathbf{r}_{2})=f(\mathbf{r}_{1}-\mathbf{r}_{2})##, is this just an abuse of notation or a definition of the function?

Frank Castle said:
I'm not quite sure I understand the second step - isn't it required the the shift is by a constant vector?
The condition, properly expressed, is that for all ##x,y\in A## and for all ##\vec v\in V##, we have ##f(x,y)=f(x+\vec v,y+\vec v)##. If the word 'constant' is used, it is referring to the fact that the vector added to ##x## is the same as the vector added to ##y##.

The condition I gave in post 2 is necessary and sufficient, ie it is logically equivalent to this definition of homogeneity.

Also, is the form ##f(x,y)=f(x-y)## ... They then proceed express this mathematically as ##f(\mathbf{r}_{1},\mathbf{r}_{2})=f(\mathbf{r}_{1}-\mathbf{r}_{2})##, is this just an abuse of notation or a definition of the function?
It's a horrendous abuse of notation that's pretty well guaranteed to sow confusion, as it has done here. I would have no confidence in any text that abuses notation in such a way.

The word-based definition in italics that you gave is OK. It's the attempt to render it in mathematical notation that is wrong.

andrewkirk said:
The condition, properly expressed, is that for all x,y∈Ax,y∈Ax,y\in A and for all ⃗v∈Vv→∈V\vec v\in V, we have f(x,y)=f(x+⃗v,y+⃗v)f(x,y)=f(x+v→,y+v→)f(x,y)=f(x+\vec v,y+\vec v). If the word 'constant' is used, it is referring to the fact that the vector added to xxx is the same as the vector added to yyy.

The condition I gave in post 2 is necessary and sufficient, ie it is logically equivalent to this definition of homogeneity.

Ah ok, I think I get it now. Thanks for your help!

Just to check on one point. Is the difference between two points in the affine space ##A## equal to a vector in ##V##, i.e. ##\vec{(x-y)}\in V##, with ##x,y\in A##? Then, when you wrote ##\vec{(x-y)}+\vec{(y-z_{0})}=\vec{(x-z_{0})}## is valid because (if what I've put above is correct) the vector ##\vec{(x-y)}## is just the difference between two points in ##A##?

Last edited:
Yes that is correct. An affine space is a slight generalisation of the notion of a vector space. In a vector space the points themselves are vectors and so are their sums and differences. In an affine space, the points are not vectors, and the sum of points is not defined (nor is the scalar product of a point) but their differences are defined and are equal to members of the related vector space.

andrewkirk said:
Yes that is correct

OK thanks.

One last thing. In the case of a one-point function, i.e. ##f(\mathbf{x})##, would it be correct to say that spatial homogeneity implies that its value is independent of position, i.e. it is everywhere equal to a constant? Mathematically, $$f(\mathbf{x})=f(\mathbf{x}+\mathbf{a})\iff f(\mathbf{x})=\text{constant}$$ that is, a one-point function is spatially homogeneous if and only if it is equal to a constant for all values of ##\mathbf{x}##, i.e. it is in fact independent of ##\mathbf{x}##?!

Frank Castle said:
In the case of a one-point function, i.e. ##f(\mathbf{x})##, would it be correct to say that spatial homogeneity implies that its value is independent of position, i.e. it is everywhere equal to a constant?
Yes.

andrewkirk said:
Yes.

Is the correct reasoning for this that homogeneity is the requirement that the given quantity is translation invariant and therefore it cannot be dependent on position, if it is described by a two point function, since the only way that it's value can be translation invariant for all values of ##\mathbf{x}## is it it's value is actually independent of position ##\mathbf{x}## (sorry to reiterate this point, but just want to check that I've understood the reasoning behind it correctly?!)
Additionally, if a given quantity is a function of two points, ##\mathbf{x},\mathbf{y}##, then translation invariance requires that it is actually only a function of the difference of these points as this is the only possible way to ensure that its value is independent of any particular point? Would this be a correct argument?

## 1. What is spatial homogeneity in relation to two-point functions?

Spatial homogeneity refers to the property of a system where the statistical properties of its components remain unchanged regardless of their spatial location. In the context of two-point functions, this means that the correlation between two points in space remains constant regardless of the distance between them.

## 2. How is spatial homogeneity tested in two-point functions?

Spatial homogeneity can be tested in two-point functions by measuring the correlation between points at different distances and comparing them. If the correlation remains constant, then the system is considered to be spatially homogeneous.

## 3. What is the functional form of a two-point function?

The functional form of a two-point function describes the mathematical relationship between two points in a system. It is typically represented by a function that takes into account the distance between the two points and any other relevant variables.

## 4. How is the functional form of a two-point function determined?

The functional form of a two-point function is determined through statistical analysis and modeling. This involves collecting data on the system and using mathematical techniques to determine the best fit function that describes the relationship between the points.

## 5. Why is understanding the spatial homogeneity and functional form of two-point functions important?

Understanding the spatial homogeneity and functional form of two-point functions is important because it provides insight into the underlying processes and dynamics of a system. It can also help in making predictions and understanding how changes in one point in the system may affect other points, leading to more efficient and accurate analysis and modeling.

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