Consider a(adsbygoogle = window.adsbygoogle || []).push({}); two-pointfunction $$f(\mathbf{r}_{1},\mathbf{r}_{2})$$ 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 thedifferencebetween 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 anif and only ifstatement 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.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

Tags:

Have something to add?

Draft saved
Draft deleted

**Physics Forums | Science Articles, Homework Help, Discussion**