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

• I
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.

## Answers and Replies

andrewkirk
Homework Helper
Gold Member
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##).

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?

andrewkirk
Homework Helper
Gold Member
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.

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:
andrewkirk
Homework Helper
Gold Member
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.

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}##?!

andrewkirk