Three Experiments Challenging SRT
- Thread starter Masterov
- Start date
That is not a good summary of SRT. SRT also has a relativity of simultaneity effect which is necessary for invariance of the (one way) speed of light. The full relation between coordinates for a boost along the x axis is given by a Lorentz transformation:
$$
\begin{eqnarray}
t' &=& \gamma (t \,-\, \frac{v}{c^{2}} x)
x' &=& \gamma (x \,-\, vt)
y' &=& y
z' &=& z \,,
\end{eqnarray}
$$
\begin{eqnarray}
t' &=& \gamma (t \,-\, \frac{v}{c^{2}} x)
x' &=& \gamma (x \,-\, vt)
y' &=& y
z' &=& z \,,
\end{eqnarray}
$$
with $$\gamma \,=\, (1 \,-\, v^{2}/c^{2})^{-1/2}$$. SRT length contraction and time dilation can be derived as special cases of this transformation. Your summary is also misleading because if two events occur a distance $$\Delta x$$ from one another, the distance between them in a different frame is generally not $$\Delta x \sqrt{1-v^2/c^2}$$.
The Lorentz transformation leaves the speed of light invariant, e.g. $$x \,=\, ct \,\Rightarrow\, x' \,=\, ct'$$.
Multiplying the Lorentz transformation by a (possibly velocity-dependent) scaling factor still leaves the speed of light invariant. The problem is that the resulting set of transformations will generally not form a symmetry group.
Multiplying the top line $$t' = \gamma (t \,-\, \frac{v}{c^{2}} x)$$ by a scaling factor will still not make time absolute as it is in Newtonian physics, because scaling won't eliminate the $$-\, \frac{v}{c^{2}} x$$ term which represents the relativity of simultaneity effect.
1. Linear transformations in space-time can not affect the symmetry.
2. In MT attach Galilean transformations, and relativistic effects are visual.
They can, because not all linear transformations are symmetries of physics, and not all collections of linear transformations form a valid symmetry group.
Dilation (scaling of all coordinates by a fixed factor) is not a symmetry in physics. This is evidenced by the fact we see characteristic distance scales in nature. All hydrogen atoms are the same size, for instance. If physics had scaling symmetry, a hydrogen atom of any size should be possible.
What does this mean?
Coordinates do not really exist in nature. They are things we define for convenience. Often we use coordinates that count units like metres and seconds, but in principle we can use any coordinate systems we like.
So what do you mean when you call some coordinates "real" and others "visual"?
If you compose a Galilean transform (+v) with a Galilean transform (-v) you get a Galilean transform of (0)
If you compose a Lorentz transform (+v) with a Lorentz transform (-v) you get a Lorentz transform of (0)
If you compose a Generalized Galilean transform (+v) with a Generalized Galilean transform (-v) you get a Generalized Galilean transform of (0)
(See post #21 of "Tutorial: Relativity - what is a reference frame?")
If you compose a MT transform (+v) with a MT transform (-v) you do not get a MT transform of (0) -- In fact, if u and v are arbitrary non-zero velocities, then no MT transform (u) composed with MT transform (v) equals a MT transform of (0). Thus a MT transform cannot be interpreted as a change of coordinates describing the same physical universe.
--------------------
Length contraction in relativity is part-and-parcel of relativity of simultaneity. Let A & B be objects moving in parallel world lines separated only by a fixed amount in the x coordinate direction.
Then
$$\begin{pmatrix} \rule{0pt}{20ex} 0 & 0 & 0 & 0 \\ \rule{0pt}{20ex} -u_x & 1 & 0 & 0 \\ \rule{0pt}{20ex} -u_y & 0 & 1 & 0 \\ \rule{0pt}{20ex} -u_z & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \rule{0pt}{20ex} t_A \\ \rule{0pt}{20ex} x_A \\ \rule{0pt}{20ex} y_A \\ \rule{0pt}{20ex} z_A \end{pmatrix} = \begin{pmatrix} \rule{0pt}{20ex} 0 \\ \rule{0pt}{20ex} x_0 \\ \rule{0pt}{20ex} y_0 \\ \rule{0pt}{20ex} z_0 \end{pmatrix} $$
and
$$\begin{pmatrix} \rule{0pt}{20ex} 0 & 0 & 0 & 0 \\ \rule{0pt}{20ex} -u_x & 1 & 0 & 0 \\ \rule{0pt}{20ex} -u_y & 0 & 1 & 0 \\ \rule{0pt}{20ex} -u_z & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \rule{0pt}{20ex} t_B \\ \rule{0pt}{20ex} x_B \\ \rule{0pt}{20ex} y_B \\ \rule{0pt}{20ex} z_B \end{pmatrix} = \begin{pmatrix} \rule{0pt}{20ex} 0 \\ \rule{0pt}{20ex} x_0 - L \\ \rule{0pt}{20ex} y_0 \\ \rule{0pt}{20ex} z_0 \end{pmatrix} $$
Subtracting one from the other, we find:
$$\begin{pmatrix} \rule{0pt}{20ex} 0 & 0 & 0 & 0 \\ \rule{0pt}{20ex} -u_x & 1 & 0 & 0 \\ \rule{0pt}{20ex} -u_y & 0 & 1 & 0 \\ \rule{0pt}{20ex} -u_z & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \rule{0pt}{20ex} t_A - t_B \\ \rule{0pt}{20ex} x_A - x_B\\ \rule{0pt}{20ex} y_A - y_B\\ \rule{0pt}{20ex} z_A - z_B \end{pmatrix} = \begin{pmatrix} \rule{0pt}{20ex} 0 & 0 & 0 & 0 \\ \rule{0pt}{20ex} -u_x & 1 & 0 & 0 \\ \rule{0pt}{20ex} -u_y & 0 & 1 & 0 \\ \rule{0pt}{20ex} -u_z & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \rule{0pt}{20ex} \Delta t \\ \rule{0pt}{20ex} \Delta x \\ \rule{0pt}{20ex} \Delta y \\ \rule{0pt}{20ex} \Delta z \end{pmatrix} = \begin{pmatrix} \rule{0pt}{20ex} 0 \\ \rule{0pt}{20ex} L \\ \rule{0pt}{20ex} 0 \\ \rule{0pt}{20ex} 0 \end{pmatrix}$$
so $$\Delta x = L + u_x \Delta t$$ and $$\sqrt{(\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2} = \sqrt{(L + u_x \Delta t)^2 + (u_y^2 + u_z^2 ) (\Delta t)^2 } = \sqrt{L^2+ (u_x^2 + u_y^2 + u_z^2 ) (\Delta t)^2 - 2 L u_x \Delta t}$$.
The point that this emphasizes is that $$\Delta t = 0 \; \Rightarrow \; \Delta x = L \, \wedge \, \sqrt{(\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2} = \left| L \right|$$ or for parallel world-lines, what we call their separation in space at the same time is constant.
That's a simple consequence of working in Cartesian coordinates for time and space. Now lets look at the generalization of the Galilean transformation for a boost in the x direction.
$$ \begin{pmatrix} \rule{0pt}{20ex} \Delta t' \\ \rule{0pt}{20ex} \Delta x' \\ \rule{0pt}{20ex} \Delta y' \\ \rule{0pt}{20ex} \Delta z' \end{pmatrix} = \begin{pmatrix} \rule{0pt}{20ex} \frac{1}{\sqrt{1 - K v^2}} & \frac{K v}{\sqrt{1 - K v^2}} & 0 & 0 \\ \rule{0pt}{20ex} \frac{v}{\sqrt{1 - K v^2}} & \frac{1}{\sqrt{1 - K v^2}} & 0 & 0 \\ \rule{0pt}{20ex} 0 & 0 & 1 & 0 \\ \rule{0pt}{20ex} 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \rule{0pt}{20ex} \Delta t \\ \rule{0pt}{20ex} \Delta x \\ \rule{0pt}{20ex} \Delta y \\ \rule{0pt}{20ex} \Delta z \end{pmatrix}$$
Multiplying both sides by an inverse transformation we get
$$\begin{pmatrix} \rule{0pt}{20ex} \frac{1}{\sqrt{1 - K v^2}} & \frac{K v}{\sqrt{1 - K v^2}} & 0 & 0 \\ \rule{0pt}{20ex} \frac{v}{\sqrt{1 - K v^2}} & \frac{1}{\sqrt{1 - K v^2}} & 0 & 0 \\ \rule{0pt}{20ex} 0 & 0 & 1 & 0 \\ \rule{0pt}{20ex} 0 & 0 & 0 & 1 \end{pmatrix}^{-1} \begin{pmatrix} \rule{0pt}{20ex} \Delta t' \\ \rule{0pt}{20ex} \Delta x' \\ \rule{0pt}{20ex} \Delta y' \\ \rule{0pt}{20ex} \Delta z' \end{pmatrix} = \begin{pmatrix} \rule{0pt}{20ex} \frac{1}{\sqrt{1 - K v^2}} & \frac{-K v}{\sqrt{1 - K v^2}} & 0 & 0 \\ \rule{0pt}{20ex} \frac{-v}{\sqrt{1 - K v^2}} & \frac{1}{\sqrt{1 - K v^2}} & 0 & 0 \\ \rule{0pt}{20ex} 0 & 0 & 1 & 0 \\ \rule{0pt}{20ex} 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \rule{0pt}{20ex} \Delta t' \\ \rule{0pt}{20ex} \Delta x' \\ \rule{0pt}{20ex} \Delta y' \\ \rule{0pt}{20ex} \Delta z' \end{pmatrix} = \begin{pmatrix} \rule{0pt}{20ex} \Delta t \\ \rule{0pt}{20ex} \Delta x \\ \rule{0pt}{20ex} \Delta y \\ \rule{0pt}{20ex} \Delta z \end{pmatrix} $$
Substituting into our expression for coordinate differences between parallel world-lines we get:
$$\begin{pmatrix} \rule{0pt}{20ex} 0 & 0 & 0 & 0 \\ \rule{0pt}{20ex} -u_x & 1 & 0 & 0 \\ \rule{0pt}{20ex} -u_y & 0 & 1 & 0 \\ \rule{0pt}{20ex} -u_z & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \rule{0pt}{20ex} \frac{1}{\sqrt{1 - K v^2}} & \frac{-K v}{\sqrt{1 - K v^2}} & 0 & 0 \\ \rule{0pt}{20ex} \frac{-v}{\sqrt{1 - K v^2}} & \frac{1}{\sqrt{1 - K v^2}} & 0 & 0 \\ \rule{0pt}{20ex} 0 & 0 & 1 & 0 \\ \rule{0pt}{20ex} 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \rule{0pt}{20ex} \Delta t' \\ \rule{0pt}{20ex} \Delta x' \\ \rule{0pt}{20ex} \Delta y' \\ \rule{0pt}{20ex} \Delta z' \end{pmatrix} = \begin{pmatrix} \rule{0pt}{20ex} 0 & 0 & 0 & 0 \\ \rule{0pt}{20ex} \frac{-(u_x + v)}{\sqrt{1 - K v^2}} & \frac{1 + K u_x v}{\sqrt{1 - K v^2}} & 0 & 0 \\ \rule{0pt}{20ex} \frac{-u_y}{\sqrt{1 - K v^2}} & \frac{K u_y v}{\sqrt{1 - K v^2}} & 1 & 0 \\ \rule{0pt}{20ex} \frac{-u_z}{\sqrt{1 - K v^2}} & \frac{K u_z v}{\sqrt{1 - K v^2}} & 0 & 1 \end{pmatrix} \begin{pmatrix} \rule{0pt}{20ex} \Delta t' \\ \rule{0pt}{20ex} \Delta x' \\ \rule{0pt}{20ex} \Delta y' \\ \rule{0pt}{20ex} \Delta z' \end{pmatrix} = \begin{pmatrix} \rule{0pt}{20ex} 0 \\ \rule{0pt}{20ex} L \\ \rule{0pt}{20ex} 0 \\ \rule{0pt}{20ex} 0 \end{pmatrix}$$
and if we assume $$\Delta t' = 0$$, we have an equation which can be solved:
$$\begin{pmatrix} \rule{0pt}{20ex} \frac{1 + K u_x v}{\sqrt{1 - K v^2}} & 0 & 0 \\ \rule{0pt}{20ex} \frac{K u_y v}{\sqrt{1 - K v^2}} & 1 & 0 \\ \rule{0pt}{20ex} \frac{K u_z v}{\sqrt{1 - K v^2}} & 0 & 1 \end{pmatrix} \begin{pmatrix} \rule{0pt}{20ex} \Delta x' \\ \rule{0pt}{20ex} \Delta y' \\ \rule{0pt}{20ex} \Delta z' \end{pmatrix} = \begin{pmatrix} \rule{0pt}{20ex} L \\ \rule{0pt}{20ex} 0 \\ \rule{0pt}{20ex} 0 \end{pmatrix}$$
with solution:
$$ \begin{pmatrix} \rule{0pt}{20ex} \Delta x' \\ \rule{0pt}{20ex} \Delta y' \\ \rule{0pt}{20ex} \Delta z' \end{pmatrix} = \begin{pmatrix} \rule{0pt}{20ex} \frac{\sqrt{1 - Kv^2}}{1 + K u_x v} & 0 & 0 \\ \rule{0pt}{20ex} \frac{-K u_y v}{1 + K u_x v} & 1 & 0 \\ \rule{0 pt}{20ex} \frac{- K u_z v}{1 + K u_x v} & 0 & 1 \end{pmatrix} \begin{pmatrix} \rule{0pt}{20ex} L \\ \rule{0pt}{20ex} 0 \\ \rule{0pt}{20ex} 0 \end{pmatrix} = \begin{pmatrix} \rule{0pt}{20ex} \frac{\sqrt{1 - Kv^2}}{1 + K u_x v} L \\ \rule{0pt}{20ex} \frac{-K u_y v}{1 + K u_x v} L \\ \rule{0 pt}{20ex} \frac{- K u_z v}{1 + K u_x v} L \end{pmatrix} $$
In the case $$\Delta t' = 0 \, \wedge \, u_y = u_z = 0$$ we are permitted to simplify even further and write:
$$\Delta x' = \frac{\sqrt{1 - Kv^2}}{1 + K u_x v} L = L'$$
which has two important cases: $$\Delta t' = 0 \, \wedge \, u_y = u_z = 0 \, \wedge \, v = -u_x \; \Rightarrow \; L = \sqrt{1 - K v^2} L' \\ \Delta t' = 0 \, \wedge \, u_x = u_y = u_z = 0 \; \Rightarrow \; L' = \sqrt{1 - K v^2} L$$
In Special Relativity, $$K = c^{\tiny -2}$$, so the separation between the world-lines (L or L' depending on the relation between $$u_x$$ and v) is longer in the coordinate system where A &B don't move.
In Galilean Relativity, $$K = 0$$, so the separation between the world-lines is constant.
In Fan's pet theory, $$K = -c^{\tiny -2}$$, so the separation between the world-lines (L or L' depending on the relation between $$u_x$$ and v) is longer in the coordinate system where A & B move.
But Fan's experiment favors K > 0 over K = 0 or K < 0.
If you compose a Lorentz transform (+v) with a Lorentz transform (-v) you get a Lorentz transform of (0)
If you compose a Generalized Galilean transform (+v) with a Generalized Galilean transform (-v) you get a Generalized Galilean transform of (0)
(See post #21 of "Tutorial: Relativity - what is a reference frame?")
If you compose a MT transform (+v) with a MT transform (-v) you do not get a MT transform of (0) -- In fact, if u and v are arbitrary non-zero velocities, then no MT transform (u) composed with MT transform (v) equals a MT transform of (0). Thus a MT transform cannot be interpreted as a change of coordinates describing the same physical universe.
--------------------
Length contraction in relativity is part-and-parcel of relativity of simultaneity. Let A & B be objects moving in parallel world lines separated only by a fixed amount in the x coordinate direction.
Then
$$\begin{pmatrix} \rule{0pt}{20ex} 0 & 0 & 0 & 0 \\ \rule{0pt}{20ex} -u_x & 1 & 0 & 0 \\ \rule{0pt}{20ex} -u_y & 0 & 1 & 0 \\ \rule{0pt}{20ex} -u_z & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \rule{0pt}{20ex} t_A \\ \rule{0pt}{20ex} x_A \\ \rule{0pt}{20ex} y_A \\ \rule{0pt}{20ex} z_A \end{pmatrix} = \begin{pmatrix} \rule{0pt}{20ex} 0 \\ \rule{0pt}{20ex} x_0 \\ \rule{0pt}{20ex} y_0 \\ \rule{0pt}{20ex} z_0 \end{pmatrix} $$
and
$$\begin{pmatrix} \rule{0pt}{20ex} 0 & 0 & 0 & 0 \\ \rule{0pt}{20ex} -u_x & 1 & 0 & 0 \\ \rule{0pt}{20ex} -u_y & 0 & 1 & 0 \\ \rule{0pt}{20ex} -u_z & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \rule{0pt}{20ex} t_B \\ \rule{0pt}{20ex} x_B \\ \rule{0pt}{20ex} y_B \\ \rule{0pt}{20ex} z_B \end{pmatrix} = \begin{pmatrix} \rule{0pt}{20ex} 0 \\ \rule{0pt}{20ex} x_0 - L \\ \rule{0pt}{20ex} y_0 \\ \rule{0pt}{20ex} z_0 \end{pmatrix} $$
Subtracting one from the other, we find:
$$\begin{pmatrix} \rule{0pt}{20ex} 0 & 0 & 0 & 0 \\ \rule{0pt}{20ex} -u_x & 1 & 0 & 0 \\ \rule{0pt}{20ex} -u_y & 0 & 1 & 0 \\ \rule{0pt}{20ex} -u_z & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \rule{0pt}{20ex} t_A - t_B \\ \rule{0pt}{20ex} x_A - x_B\\ \rule{0pt}{20ex} y_A - y_B\\ \rule{0pt}{20ex} z_A - z_B \end{pmatrix} = \begin{pmatrix} \rule{0pt}{20ex} 0 & 0 & 0 & 0 \\ \rule{0pt}{20ex} -u_x & 1 & 0 & 0 \\ \rule{0pt}{20ex} -u_y & 0 & 1 & 0 \\ \rule{0pt}{20ex} -u_z & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \rule{0pt}{20ex} \Delta t \\ \rule{0pt}{20ex} \Delta x \\ \rule{0pt}{20ex} \Delta y \\ \rule{0pt}{20ex} \Delta z \end{pmatrix} = \begin{pmatrix} \rule{0pt}{20ex} 0 \\ \rule{0pt}{20ex} L \\ \rule{0pt}{20ex} 0 \\ \rule{0pt}{20ex} 0 \end{pmatrix}$$
so $$\Delta x = L + u_x \Delta t$$ and $$\sqrt{(\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2} = \sqrt{(L + u_x \Delta t)^2 + (u_y^2 + u_z^2 ) (\Delta t)^2 } = \sqrt{L^2+ (u_x^2 + u_y^2 + u_z^2 ) (\Delta t)^2 - 2 L u_x \Delta t}$$.
The point that this emphasizes is that $$\Delta t = 0 \; \Rightarrow \; \Delta x = L \, \wedge \, \sqrt{(\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2} = \left| L \right|$$ or for parallel world-lines, what we call their separation in space at the same time is constant.
That's a simple consequence of working in Cartesian coordinates for time and space. Now lets look at the generalization of the Galilean transformation for a boost in the x direction.
$$ \begin{pmatrix} \rule{0pt}{20ex} \Delta t' \\ \rule{0pt}{20ex} \Delta x' \\ \rule{0pt}{20ex} \Delta y' \\ \rule{0pt}{20ex} \Delta z' \end{pmatrix} = \begin{pmatrix} \rule{0pt}{20ex} \frac{1}{\sqrt{1 - K v^2}} & \frac{K v}{\sqrt{1 - K v^2}} & 0 & 0 \\ \rule{0pt}{20ex} \frac{v}{\sqrt{1 - K v^2}} & \frac{1}{\sqrt{1 - K v^2}} & 0 & 0 \\ \rule{0pt}{20ex} 0 & 0 & 1 & 0 \\ \rule{0pt}{20ex} 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \rule{0pt}{20ex} \Delta t \\ \rule{0pt}{20ex} \Delta x \\ \rule{0pt}{20ex} \Delta y \\ \rule{0pt}{20ex} \Delta z \end{pmatrix}$$
Multiplying both sides by an inverse transformation we get
$$\begin{pmatrix} \rule{0pt}{20ex} \frac{1}{\sqrt{1 - K v^2}} & \frac{K v}{\sqrt{1 - K v^2}} & 0 & 0 \\ \rule{0pt}{20ex} \frac{v}{\sqrt{1 - K v^2}} & \frac{1}{\sqrt{1 - K v^2}} & 0 & 0 \\ \rule{0pt}{20ex} 0 & 0 & 1 & 0 \\ \rule{0pt}{20ex} 0 & 0 & 0 & 1 \end{pmatrix}^{-1} \begin{pmatrix} \rule{0pt}{20ex} \Delta t' \\ \rule{0pt}{20ex} \Delta x' \\ \rule{0pt}{20ex} \Delta y' \\ \rule{0pt}{20ex} \Delta z' \end{pmatrix} = \begin{pmatrix} \rule{0pt}{20ex} \frac{1}{\sqrt{1 - K v^2}} & \frac{-K v}{\sqrt{1 - K v^2}} & 0 & 0 \\ \rule{0pt}{20ex} \frac{-v}{\sqrt{1 - K v^2}} & \frac{1}{\sqrt{1 - K v^2}} & 0 & 0 \\ \rule{0pt}{20ex} 0 & 0 & 1 & 0 \\ \rule{0pt}{20ex} 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \rule{0pt}{20ex} \Delta t' \\ \rule{0pt}{20ex} \Delta x' \\ \rule{0pt}{20ex} \Delta y' \\ \rule{0pt}{20ex} \Delta z' \end{pmatrix} = \begin{pmatrix} \rule{0pt}{20ex} \Delta t \\ \rule{0pt}{20ex} \Delta x \\ \rule{0pt}{20ex} \Delta y \\ \rule{0pt}{20ex} \Delta z \end{pmatrix} $$
Substituting into our expression for coordinate differences between parallel world-lines we get:
$$\begin{pmatrix} \rule{0pt}{20ex} 0 & 0 & 0 & 0 \\ \rule{0pt}{20ex} -u_x & 1 & 0 & 0 \\ \rule{0pt}{20ex} -u_y & 0 & 1 & 0 \\ \rule{0pt}{20ex} -u_z & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \rule{0pt}{20ex} \frac{1}{\sqrt{1 - K v^2}} & \frac{-K v}{\sqrt{1 - K v^2}} & 0 & 0 \\ \rule{0pt}{20ex} \frac{-v}{\sqrt{1 - K v^2}} & \frac{1}{\sqrt{1 - K v^2}} & 0 & 0 \\ \rule{0pt}{20ex} 0 & 0 & 1 & 0 \\ \rule{0pt}{20ex} 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \rule{0pt}{20ex} \Delta t' \\ \rule{0pt}{20ex} \Delta x' \\ \rule{0pt}{20ex} \Delta y' \\ \rule{0pt}{20ex} \Delta z' \end{pmatrix} = \begin{pmatrix} \rule{0pt}{20ex} 0 & 0 & 0 & 0 \\ \rule{0pt}{20ex} \frac{-(u_x + v)}{\sqrt{1 - K v^2}} & \frac{1 + K u_x v}{\sqrt{1 - K v^2}} & 0 & 0 \\ \rule{0pt}{20ex} \frac{-u_y}{\sqrt{1 - K v^2}} & \frac{K u_y v}{\sqrt{1 - K v^2}} & 1 & 0 \\ \rule{0pt}{20ex} \frac{-u_z}{\sqrt{1 - K v^2}} & \frac{K u_z v}{\sqrt{1 - K v^2}} & 0 & 1 \end{pmatrix} \begin{pmatrix} \rule{0pt}{20ex} \Delta t' \\ \rule{0pt}{20ex} \Delta x' \\ \rule{0pt}{20ex} \Delta y' \\ \rule{0pt}{20ex} \Delta z' \end{pmatrix} = \begin{pmatrix} \rule{0pt}{20ex} 0 \\ \rule{0pt}{20ex} L \\ \rule{0pt}{20ex} 0 \\ \rule{0pt}{20ex} 0 \end{pmatrix}$$
and if we assume $$\Delta t' = 0$$, we have an equation which can be solved:
$$\begin{pmatrix} \rule{0pt}{20ex} \frac{1 + K u_x v}{\sqrt{1 - K v^2}} & 0 & 0 \\ \rule{0pt}{20ex} \frac{K u_y v}{\sqrt{1 - K v^2}} & 1 & 0 \\ \rule{0pt}{20ex} \frac{K u_z v}{\sqrt{1 - K v^2}} & 0 & 1 \end{pmatrix} \begin{pmatrix} \rule{0pt}{20ex} \Delta x' \\ \rule{0pt}{20ex} \Delta y' \\ \rule{0pt}{20ex} \Delta z' \end{pmatrix} = \begin{pmatrix} \rule{0pt}{20ex} L \\ \rule{0pt}{20ex} 0 \\ \rule{0pt}{20ex} 0 \end{pmatrix}$$
with solution:
$$ \begin{pmatrix} \rule{0pt}{20ex} \Delta x' \\ \rule{0pt}{20ex} \Delta y' \\ \rule{0pt}{20ex} \Delta z' \end{pmatrix} = \begin{pmatrix} \rule{0pt}{20ex} \frac{\sqrt{1 - Kv^2}}{1 + K u_x v} & 0 & 0 \\ \rule{0pt}{20ex} \frac{-K u_y v}{1 + K u_x v} & 1 & 0 \\ \rule{0 pt}{20ex} \frac{- K u_z v}{1 + K u_x v} & 0 & 1 \end{pmatrix} \begin{pmatrix} \rule{0pt}{20ex} L \\ \rule{0pt}{20ex} 0 \\ \rule{0pt}{20ex} 0 \end{pmatrix} = \begin{pmatrix} \rule{0pt}{20ex} \frac{\sqrt{1 - Kv^2}}{1 + K u_x v} L \\ \rule{0pt}{20ex} \frac{-K u_y v}{1 + K u_x v} L \\ \rule{0 pt}{20ex} \frac{- K u_z v}{1 + K u_x v} L \end{pmatrix} $$
In the case $$\Delta t' = 0 \, \wedge \, u_y = u_z = 0$$ we are permitted to simplify even further and write:
$$\Delta x' = \frac{\sqrt{1 - Kv^2}}{1 + K u_x v} L = L'$$
which has two important cases: $$\Delta t' = 0 \, \wedge \, u_y = u_z = 0 \, \wedge \, v = -u_x \; \Rightarrow \; L = \sqrt{1 - K v^2} L' \\ \Delta t' = 0 \, \wedge \, u_x = u_y = u_z = 0 \; \Rightarrow \; L' = \sqrt{1 - K v^2} L$$
In Special Relativity, $$K = c^{\tiny -2}$$, so the separation between the world-lines (L or L' depending on the relation between $$u_x$$ and v) is longer in the coordinate system where A &B don't move.
In Galilean Relativity, $$K = 0$$, so the separation between the world-lines is constant.
In Fan's pet theory, $$K = -c^{\tiny -2}$$, so the separation between the world-lines (L or L' depending on the relation between $$u_x$$ and v) is longer in the coordinate system where A & B move.
But Fan's experiment favors K > 0 over K = 0 or K < 0.
If we assume that there is a slowing down of time, and one observer sees the slowing of time of the other, then that (the second) must see acceleration time of the first. But this is impossible, since in this case violated the equality of inertial reference frames.
LT says that both observers will see each other's time dilation.
This violates the principle of causality.
LT says that both observers will see each other's time dilation.
This violates the principle of causality.
Not true -- all physical phenomena at high altitude run faster than at low altitude.
These are your claims -- they are not based on facts or reasoned argument.
This is not true.
Because you don't understand the Lorentz coordinate transform or the related Galilean coordinate transform, so you don't understand the connection between length contraction and the relativity of simultaneity. Slices of space-time identified by $$\Delta t = 0$$ are distinct from slices identified by $$\Delta t' = 0$$.
Did you notice that if K=0, then the generalized Galilean transform I used in post #465 is just a Galilean transform with neither length contraction nor relativity of simultaneity.
Please demonstrate this claim with math. I believe it is true.
Please demonstrate this claim with math. I believe it is false.
Gravity is beyond the scope topic of which we are discussing.
My arguments (it's not only one):
If we assume that there is a slowing down of time, and one observer sees the slowing of time of the other, then that (the second) must see acceleration time of the first. But this is impossible, since in this case violated the equality of inertial reference frames.
LT says that both observers will see each other's time dilation.
This violates the principle of causality.
As will be true?
Show.
By the equality of inertial reference frames: both observers will see time dilation.
This violates the principle of causality.
This is a trivial consequence of the equality of inertial reference frames.
Mathematics would be required.
This is a logical consequence of the foregoing.
In the foregoing non violates the principle of causality.
The order of events of the foregoing (slow+rapid) not violates the principle of causality. (In this case violated the equality of inertial reference frames.)
Other order of events violates the principle of causality necessarily.
Two logically consistent sequence of events of one situation can not exist.
This reasoning only holds if clocks in the two reference frames being compared are synchronised the same way. In SRT, they are not. That is why the time dilation rates do not necessarily have to be inverses of one another. It makes a difference that the time dilation rates are being compared along different trajectories.
If you are interested in understanding this in more detail, I suggest an exercise for you. You know that the Lorentz transformation is
$$
\begin{eqnarray}
t' &=& \gamma (t \,-\, \frac{v}{c^{2}} x) \\
x' &=& \gamma (x \,-\, v t) \,.
\end{eqnarray}
$$
\begin{eqnarray}
t' &=& \gamma (t \,-\, \frac{v}{c^{2}} x) \\
x' &=& \gamma (x \,-\, v t) \,.
\end{eqnarray}
$$
Given the coordinates $$x$$ and $$t$$ of an arbitrary event, the transformation tells you the corresponding coordinates $$x'$$ and $$t'$$ in a different reference frame for the same event. SRT length contraction and time dilation can both be derived as special cases of this transformation.
Question: what is the inverse transformation?
$$
\begin{eqnarray}
t &=& ? \\
x &=& ? \,.
\end{eqnarray}
$$
\begin{eqnarray}
t &=& ? \\
x &=& ? \,.
\end{eqnarray}
$$
If you are given $$x'$$ and $$t'$$, what are the corresponding coordinates $$x$$ and $$t$$?
Is no need slow down time in order to reconcile the physics of Newton and the Michelson-Morly.
You can fabricate of stories about clock synchronization in order to deceive ourselves.
I do not need such tales.
I need not time dilation to wholly perceive the world without cut and deletion a logical no-correspondences.
You can fabricate of stories about clock synchronization in order to deceive ourselves.
I do not need such tales.
I need not time dilation to wholly perceive the world without cut and deletion a logical no-correspondences.
There is no deception. Your argument about inverse time dilation rates only works if different observers share the same concept of simultaneity. In SRT, they do not, and this effect has to be taken into account.
If you ignore an important part of relativity in order to criticise relativity, it is you who are deceiving yourself.
What "screen movies"?
Calling a theory "stupid" is not in itself a good argument against it.
Let there be a fixed screen, and let there be two mobile observer, which move in opposite directions.
And let each observer projected onto the screen his film.
Consider the question: what will be the sequence of frames in the presence of time dilation?
You force me to refute the verbiage that you are deceiving your own common sense. You do not believe in it already, but force to me, that I have refuted.
It's tedious and not interesting.
You would get a rather messy picture, since the screen has a certain size and light emitted from a moving source will not reach the entire screen all at the same time.
So what?