Re: Question about light clock and derivation of time dilation

I think we agree Daryl...

Daryl McCullough wrote:
> Ken S. Tucker says...
>
> >Daryl McCullough wrote:
>
> >> >Daryl McCullough wrote:
> >>
> >> >> and (2) there
> >> >> is no need (and no point) to using a non-diagonal
> >> >> metric tensor.
> >> >
> >> >Actually there is, it's aberration,
> >> >
> >> >g_i0 = - g_ij dx^j/dt.
>
> >> No. In Minkowsky coordinates, g_00 = -1, g_i0 = 0, g_ii = +1.
> >> Aberration has nothing to do with off-diagonal elements of the metric.
> >
> >Daryl solve U_i=0.
>
> You speak in non-sequiters, Ken. Aberration has
> nothing to do with off-diagonal elements of the metric.
>
> As far as solving U_i=0, what does that have to do
> with anything? It's impossible to solve without
> specifying a metric, but if the metric is diagonal
> g_00 = -1, g_11 = g_22 = g_33 = +1, then a
> solution is the spacetime path x^u(s) given by
>
> x^0 = s
> x^j = 0
>
> (assuming that you mean U^u = dx^u/ds)
>
> --
> Daryl McCullough
> Ithaca, NY

Using the light clock, the time axis "t"
is the path of the light-ray as it's used,
and that's fine.

In the rest frame vectors

t.x=0 , g_01=0

The same clock in the moving frame has

t'.x' = -v/c == g'_01 = -dx'/dt'

which is due to aberation.

Generally use the metric

g_00 = g_11 = 1, g_01=-v/c

then

ds^2 = (cdt)^2 - dx^2 .

That's consistent with

U_i = dx_i/ds =0 , (i=1,2,3).

That's ok too.

The problem with the light clock itself is a
need to understand g_0i, which most can't.

Regards
Ken S. Tucker

.

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