- #1

Frank Castle

- 580

- 23

*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.