# Isotropy/Homogeneity of Spacetime and Inertial Frame Equivalence

1. Jan 17, 2012

### Mindscrape

So this is problem 11.1 out of Jackson Electrodynamics:
Two equivalent intertial frames K and K' are such that K' moves in the positive x direction with speed v as seen from K. The spatial coordinate axes in K' are parallel to those in K and the two origins are coincident at times t=t'=0. (a) Show that the isotropy and homogeneity of space-time and the equivalence of different intertial frames (first postulate of relativity) require that the most general transformation between the space-time coordinates (x,y,z,t) and (x',y',z',t') is the linear transformation,
$$\begin{eqnarray}x'&=&f(v^2)x-vf(v^2)t \\ t'&=&g(v^2)t-vh(v^2)x \\ y'&=&y \\ z'&=&z \end{eqnarray}$$

My problem with Jackson so far, after one problem set, is that his problems can just be so vague. So the linear transformations of the boost parameter/rapidity, seem to me, are made under the assumption of homogeneity and isotropy of space time. If so, then the linear transformation of the rapidity gives the answer. But, I just don't know if this is really what Jackson has intended from the problem.

I'm also a bit confused on how the isotropy and homogeneity of space-time is not an equivalent way of saying that different inertial frames are equivalent. If a linear transformation gives me the same invariant, under the assumption of isotropy and homogeneity, then that's the same as saying my linear transformation (essentially a space-time rotation) implies that the event metric (the interval) in the rotated axes (the new inertial frame) must be equivalent. Is this circular logic though?

I'm pretty sure I can do parts b and c if I can just understand what this problem wants from me.

Last edited: Jan 17, 2012