DrGreg said:
The answer to your question is, in general, for "two-way absolute simultaneity",
t'=a t
for some a independent of t and x. I'll assume here that t is Einstein-synchronised in the reference frame R, and t' is a time coordinate in another frame I. However, if you want your time coordinate to coincide with proper time in the I-frame, a must be given by
a = \frac{1}{\gamma}
I should clarify what I said yesterday. To use the notation of post #4, the condition for "2-way" absolute simultaneity is
t_S(I) = a(I) t_S(R) ...(1)
where a(I) does not depend on
x or
t, but may vary from frame
I to another.
The Mansouri-Sexl condition that
a(I) = \frac{1}{\gamma} ...(2)
applies only in the special case when t_S(R) = t_E(R), i.e. when our family of coordinates uses Einstein sync in the reference frame R. This is true of the example I gave in post #4 (the Selleri or Tanglerhini coordinates) but not true for Leubner's "everyday" coords.
The condition for Leubner coords is
x_S(I) = x_E(I)
t_S(I) = t_E(I) - \frac{|x_E(I)|}{c}
if you define it literally as described, i.e. with the light signal traveling outward from the spatial origin to the secondary clock. However if you modify this slightly and agree always to synchronise "from left to right" (in the 1-D case), i.e. in the positive
x direction, i.e. whoever has the smallest
x-coord transmits the signal and the person with the largest
x-coord receives the signal, then you can use the modified equation
t_S(I) = t_E(I) - \frac{x_E(I)}{c}
which defines a "2-way" absolute simultaneity.
If you do the calculation with Lorentz tranforms you should get (if I haven't made a mistake)
t_S(I) = \gamma \left(1 + \frac{v}{c} \right) t_S(R)
for "left-to-right Leubner sync", confirming my equation (1) above.
If you wanted to come up with a 3-space-dimension version of Leubner-sync which was 2-way absolute, the only way I can think of doing it would be by a 3-stage process. For a given clock at (
x,
y,
z) introduce some intermediate clocks at (
x,0,0) and (
x,
y,0). Perform a 2-way sync along the
x-axis from (0,0,0) to (
x,0,0); then along the
y-axis to (
x,
y,0); then along the
z-axis to (
x,
y,
z). That's a rather awkward way to do it. The equation would be
t_S(I) = t_E(I) - \frac{x_E(I) + y_E(I) + z_E(I)}{c}
You can of course perform an "outward" sync direct from (0,0,0) to (
x,
y,
z), but that would only be "1-way" absolute simultaneity:
t_S(I) = t_E(I) - \frac{\sqrt{x_E(I)^2 + y_E(I)^2 + z_E(I)^2}}{c}
bernhard.rothenstein said:
Thank you. Is
t_S(I) = t_E(I) + \frac{v x_E(I)}{c^2}
a guess. Is there a relationship between that equation, apparent position and actual position?
I believe this formula for t_S(I) is the only one that is compatible with
t_S(R) = t_E(R) and (1) and (2). Of course, in 3-dimensions it becomes
t_S(I) = t_E(I) + \frac{\textbf{v} \cdot \textbf{x}_E(I)}{c^2}.
I just got this from the Mansouri-Sexl paper referred to via the references in my earlier post. I'm not quite sure what you mean by "apparent position" and "actual position" in this context.