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

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1. Apr 27, 2016

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.

2. Apr 29, 2016

andrewkirk

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 $f$).

3. Apr 29, 2016

Frank Castle

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?

4. Apr 29, 2016

andrewkirk

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.

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.

5. Apr 30, 2016

Frank Castle

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: Apr 30, 2016
6. Apr 30, 2016

andrewkirk

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.

7. May 1, 2016

Frank Castle

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 $\mathbf{x}$?!

8. May 1, 2016

Yes.

9. May 2, 2016

Frank Castle

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?

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