Lorentz and Voigt transforms are errors.
Look my post: [POST=2928447]Lorentz's error[/POST].
Your post is full of conceptual errors. The first one is on the first line:
Lorentz's relativism is based on the formula:$$x^2-(ct)^2=(x')^2-(ct')^2=0$$
No it's not. For
Lorentz transformations (which preserve the coordinate origin), and in one spatial dimension (since that's what you did) the defining property is just
$$(ct')^{2} \,-\, x'^{2} \,=\, (ct)^{2} \,-\, x^{2} \,, \quad (\ast)$$
i.e.
without the "= 0" at the end. If you extend to allow translations - as in
Poincaré transformations - then the relation is only true of coordinate intervals:
$$(c \Delta t')^{2} \,-\, (\Delta x')^{2} \,=\, (c \Delta t)^{2} \,-\, (\Delta x)^{2} \,,$$
For the
special case of a light pulse moving in the positive or negative
x direction, and passing through the origin at
t = 0, then
$$
\begin{eqnarray}
x &=& \pm ct \\
\Rightarrow\quad x^{2} &=& (ct)^{2} \\
\Rightarrow \quad x^{2} \,-\, (ct)^{2} &=& 0 \,,
\end{eqnarray}
$$
and applying (*) tells you $$x'^{2} \,-\, (ct')^{2} = 0$$, or $$x' \,=\, \pm ct'$$. That's just a statement of the invariance of
c: if something is moving at the speed of light in one inertial reference frame, it has to be moving at the speed of light in another inertial reference frame.
The key point is that this is a constraint on a coordinate transformation: (
x,
t) and (
x',
t') are the coordinates of
different inertial reference frames. You have clearly misunderstood this when you go on to ask your question:
I remind to you of a school-task about the two foot-passengers and the dog:
1. Two travelers go on the road with the same velocity ($$v$$) at a distance ($$L$$) from each other (one behind the other).
2. A dog runs between travelers (at velocity $$c$$).
QUESTION: how long time the dog ran ahead, and how long - ago.
Every schoolboy knows the answer to this ask: ldog's time will be different because:
1. When the dog runs back (to meet to lagging-traveler) - the dog's time will be the lesser: $$T_1=L/(c+v)$$.
2. When the dog runs forward (to rush to the advance-traveler) - the dog's time will be the greater: $$T_2=L/(c-v)$$.
Yes, this is a simple highschool kinematics problem
that has nothing to do with the definition of Lorentz transforms. You solve this problem in relativity exactly the same way you would have done it in highshool. You have (implicitly) asked the problem in just one reference frame - the one in which the dog is running at
c, the travellers are both moving with velocity
v, and they're a distance
L from one another. You only need a Lorentz transformation or Poincaré transformation if you've calculated when these events occur in the reference frame you set the problem up in, and you want to know how a different observer will view the same things. This:
$$x^2-(ct)^2=0$$
$$t_1=L/c$$
$$t_2=L/c$$
is a completely meaningless application of (*). Einstein, Lorentz, and Minkowski say no such thing for your problem. You've completely misunderstood what a Lorentz transformation is and when you actually need to apply one.
Now that you know what a Lorentz transformation is (or at least what it definitely is
not), would you care to reply to [POST=2957966]
post #63[/POST] in a meaningful way?
