Den 01.10.2023 06:51, skrev Richard Hertz:
>> On 9/30/23 7:52 PM, Laurence Clark Crossen wrote:
>>> The crux of the argument is that Einstein's time synchronization
>>> method is inconsistent with the LT as given by Crothers:
I assume the claim is:
According to the Lorentz transform, the coordinate clocks
in two inertial frames in relative motion can't be synchronous
in both their respective frames.
>
> x' = γ (x - vt)
> t' = γ (t - xv/c²)
>
> Replacing x by x' in t' gives
>
> t′ = (1/γ)t − x′v/c²
No, t' = f(t,x') gives:
t' = (γt + γ²x′v/c²)/(1 - γ²v²/c²)
Your error is that you have set x = γ (x' - vt')
and then the denominator would be (1 + γ²v²/c²) = γ².
But x = γ (x' + vt') and (1 - γ²v²/c²) ≠ γ²
>
> For every value of x' (where clocks are placed), this equation clearly shows that those clocks are out of sync
Your conclusion is correct, though, t'≠ t unless x'= x = 0
> in the same inertial frame:
Don't be ridiculous!
Let's call the frames K'(t'x') and K(t,x).
t'≠ t means that the temporal coordinates of the event
are different in K' and K.
That doesn't mean that the coordinate clocks i K
not are in sync with each other, and that the coordinate
clocks i K' not are in sync with each other,
>
> Being t = 0, v = 1 m/s, c = 1 and x' being increased in 1 m steps,
You can't mix natural and SI units, so with v = 1 m/s and x' = N m,
c = 299792458 m/s, γ ≈ 1+5.56E-18
>
> t = 0, t'(x') = − x′v/c²
The coordinates of event N in in K' = ( tₙ', N m)
the coordinates of event n in K = (0 s,xₙ)
The LT:
xₙ = γ(N + v⋅tₙ)m ≈ (1+5.56E-18)⋅(N+v⋅0)m = (1+5.56E-18)⋅N m
tₙ' = γ(0 - vxₙ/c^2) = -γvxₙ/c^2
tₙ' = -(1+5.56E-18)⋅1⋅(1+5.56E-18)⋅N⋅1.11E-17) s ≈ - N⋅1.11E-17 s
>
> t = 0, t'(0) = 0 -------> Clock C0 at origin x' = 0
> t = 0, t'(1) = -1 -------> Clock C1 at x' = 1 m
> t = 0, t'(2) = -2 -------> Clock C2 at x' = 2 m
> ....
> t = 0, t'(N) = -N -------> Clock CN at x' = N m
Nonsense.
The coordinates of event N are:
in K : tₙ = 0 s, xₙ = (1+5.56E-18)⋅N m
in K': tₙ'≈ -N⋅1.11E-17 s, xₙ'≈ N m
Approximated to 15 significant digits we get:
When tₙ = 0.00000000000000, tₙ'= 0.00000000000000
This shouldn't surprise you.
When v = 1m/s the result will to a very good approximation be equal
to the results of Galilean relativity, which is that all clocks
always show the same. t'= t, remember?
But you didn't think of that, and got a result which was
1E17 times too high! :-D
HOWEVER;
If the coordinates in K of event N are t = 0 s, x = γN m
then the coordinates in K' of event N are t'=γ²⋅N⋅v/c² s, x' = N m
So why is your statement below absolute ridiculous?
>
> All the N+1 clocks, equally spaced in the moving frame, are OUT OF SYNC (as perceived by the relativist at x = 0 when t = 0).
Because the N+1 different t' are the temporal coordinates of
N+1 different events. It is NOT N+1 coordinate clocks which,
simultaneous in K', show different values.
The N+1 events are simultaneous in K,
but they are NOT simultaneous in K'.
When the coordinate clock at x' = N m show t'=γ²⋅N⋅v/c² s,
then, simultaneously in K', all the other coordinate clocks
show the same.
This:
https://paulba.no/pdf/Mutual_time_dilation.pdf
PROVES that according to the Lorentz transform, the clocks
in both K' and K can be synchronous in their respective frames.
If you claim otherwise, you will have to show that my math is wrong
in the reference above.
--
Paul
https://paulba.no/