Improved Andrew Banks said:Let a point-like detector exist, stationary in coordinate system k, somewhere to the left of the $$\eta$$-axis and only capable of detecting light to its right (including light originating at the origin of coordinate system k).

Assuming everything stationary in system k moves to the right with velocity v (in the x-direction) in system K, assume the origins of system k and K correspond at their respective zero times. Thus

$$ t = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \left( \tau + \frac{v}{c^2} \xi \right) \\ x = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \left( \xi + v \tau \right) \\ y = \eta $$

Is it possible that for $$0 < v < c $$ the mirror could have such a large $$\eta$$ value that in the K frame a light flash from the time the origins were at the same position arrives at the detector from the left, preventing detection in one description of reality but not the other, supposedly equivalent one?

Andrew Banks correctly decides that the pulse from $$(\tau, \xi, \eta) = (0,0,0)$$ to $$(\tau_0, -\xi_0, +\eta_0)$$ would be seen in system K as a pulse from $$(t, x, y) = (0,0,0)$$ to $$(t_0, +x_0, +\eta_0)$$ whenever certain geometrical constraints are met, but ignores the question of what "to the right" means in system K.

First, what is the minimum value of v such that in system K the light pulse to the detector in purely in the $$+\eta$$ direction? That would mean $$x_0 = 0$$. Thus

$$v_0 = \frac{c \xi_0}{\sqrt{ \xi_0^2 + \eta_0^2}} < c $$

Then for any v such that $$v_0 < v < c$$ and assuming $$-\xi_0 < 0, \; \eta_0 > 0, \; \tau_0 = \frac{1}{c} \sqrt{\xi_0^2 + \eta_0^2} > 0$$ we have :

$$x_0 = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \left( -\xi_0 + v \tau_0 \right) = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \left( -\xi_0 + \frac{v}{c} \sqrt{\xi_0^2 + \eta_0^2} \right) { \Large \quad > \quad } \frac{1}{\sqrt{1-\frac{v_0^2}{c^2}}} \left( -\xi_0 + \frac{v_0}{c} \sqrt{\xi_0^2 + \eta_0^2} \right) = \frac{1}{\sqrt{1-\frac{v_0^2}{c^2}}} \left( -\xi_0 + \frac{\xi_0}{\sqrt{ \xi_0^2 + \eta_0^2}} \sqrt{\xi_0^2 + \eta_0^2} \right) = 0$$

But that, importantly, still doesn't answer if the light comes into the left or the right of the detector, which is answered by the sign of the cross product of the light ray movement and an extension of the detector (finite or infintesimal) in the $$\eta$$ direction.

Equivalently, we may look at the sign of the cross product of light striking two ends of the detector.

So we have a detector which runs from $$(-\xi_0, \eta_0)$$ to $$(-\xi_0, \eta_0 + \Delta)$$ in system k where the detector is stationary.

Thus we have events O, A and B where the light flash occurs and is received at the two points respectively. These same events have representations in the system K (with coordinates t, x and y).

$$\begin{array}{c|ccc} & O & A & B \\ \hline \\ \tau & 0 & \frac{1}{c} \sqrt{\xi_0^2 + \eta_0^2} & \frac{1}{c} \sqrt{\xi_0^2 + \eta_0^2 + 2 \Delta \eta_0 + \Delta^2} \\ \xi & 0 & -\xi_0 & -\xi_0 \\ \eta & 0 & \eta_0 & \eta_0 + \Delta

\\ \hline

\\ t & 0 & \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \left( \frac{1}{c} \sqrt{\xi_0^2 + \eta_0^2} - \frac{v}{c^2} \xi_0 \right) & \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \left( \frac{1}{c} \sqrt{\xi_0^2 + \eta_0^2 + 2 \Delta \eta_0 + \Delta^2} - \frac{v}{c^2} \xi_0 \right)

\\ x & 0 & \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \left( -\xi_0 + \frac{v}{c} \sqrt{\xi_0^2 + \eta_0^2} \right) & \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \left( -\xi_0 + \frac{v}{c} \sqrt{\xi_0^2 + \eta_0^2 + 2 \Delta \eta_0 + \Delta^2} \right)

\\ y & 0 & \eta_0 & \eta_0 + \Delta

\end{array}$$

So the cross product in system k is $$\xi_A \eta_B - \xi_B \eta_A = - \Delta \xi_0 < 0$$

And the cross product in system K is $$x_A y_B - x_B y_A = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \left( - \Delta \xi_0 + \frac{v}{c} \left( ( \eta_0 + \Delta) \sqrt{\xi_0^2 + \eta_0^2} - \eta_0 \sqrt{\xi_0^2 + (\eta_0 + \Delta)^2} \right) \right) < 0$$

This does not undergo a sign change at any speed $$0 < v < c$$ including $$v_0$$.

Thus because of the finite propagation speed of light, the expanding beam of light sweeps across the object in the same way, hitting the same face, a face which effectively has been rotated.

This interaction between ray-tracing beams of propagating light and the Lorentz transformation is known as Terrell rotation, not to be confused with Thomas precision which is another relationship between rotation and Lorentz transforms.

http://journals.cambridge.org/action/displayAbstract?fromPage=online&aid=2049244

http://prola.aps.org/abstract/PR/v116/i4/p1041_1

http://math.ucr.edu/home/baez/physics/Relativity/SR/penrose.html

So not only is Andrew Banks completely wrong about Einstein's approach in 1905, but the effect he ignores was demonstrated as a general geometric effect in 1959.

*Any*legitimate scientific venue for publication of matters related to the math of special relativity would have realized this along with many other flaws of this paper.