Re: Question on Spacetime dilation.

On Dec 17, 6:57 am, "Ken S. Tucker" <dynam...@xxxxxxxxxxxx> wrote:
A clock (in K) moving at 0.8c (relative to K') is
dilated 0.6 by t' = t*sqrt(1 - v^2/c^2),  so that
t'=(0.6)*t.

In GR that is generalized to be,

ds^2 =g_uv dx^u dx^v    ,  {u,v=0,1,2,3},

and then by association equatable to

= dx_u dx^u ,

= dx_0 dx^0 + dx_i dx^i   , {i=1,2,3} ,   Eq.(1).

I expect I should then obtain,

dt' = ds = (0.6) dt,                              Eq.(2).

What differential coefficients should be subbed
into Eq.(1) to yield Eq.(2)?

TIA
Regards
Ken S. Tucker

c^2dt'^2 = c^2dt^2 -dx^2 -dy^2 -dz^2
= c^2dt^2 -ds^2
= c^2dt^2(1 - v^2/c^2)
so
dt' = (1/g)dt, where g = 1/sqrt(1 - v^2/c^2)

I say, of course, that these are time dilation
equations only, and that the first equation has
been misinterpreted as a metric equation for
100 years now! You can write it in the form
using g_uv, and it works mathematically, with
g_uvs being 1 and -1, but all this doesn't really
mean anything of any significance physically.

Alen
.