In the book physical foundations of cosmology, it says that Hubble's Law is unique and a problem seems to be a hint of proving that.
In order for a general expansion law,v=f(r,t), to be the same for all observers, the function f must satisfy the relation $$f(\bf{r_{CA}}−\bf{r_{BA}},t) = f(\bf{r_{CA}},t)−f(\bf{r_{BA}},t),$$ where ABC are three points in space. Show that the only solution of this equation is given by the Hubble law.
With a little help from a Taylor approximation, I can convince myself that $f$ should be a linear function without a constant. But it seems to me that this is not good enough for a proof. How can one prove it in a more mathematical way? Thanks!
I spent a lot of time reading about Hubble's Law only to find that in a math context this problem isn't too hard to explain.
This function says that the change in $r$ corresponds to a constant change in $f$. So, let $x=\frac{\partial f}{\partial r}+\frac{\partial f}{\partial t}\frac{dt}{dr}$. Since we assume that $t$ is a constantly changing function, it's derivative with respect to $r$ is 0. this means that $x=\frac{\partial f}{\partial r}$. We can integrate this function with respect to $r$ \begin{gather} \int \frac{\partial f}{\partial r}\,dr=\int x\,dr = xr + c \end{gather}
I am unsure of as to why $c$ would be set to 0. There's probably an argument in there somewhere but I'm physics'd out for the day.