In a certain rest frame, we have a right cuboid box moving in the +x direction with velocity v. As measured in the rest frame the box has one axis aligned with the x axis and of length A and the other axes aligned with the Y and Z axes and of identical length B. At one point the center of the box coincides with the center of our coordinate system and a flash of light is emitted and bounces off a mirror at the leading edge of the box and returns to the center of the box. Also it hits a detector at the center of one of the faces parallel to the direction of the movement.
Question 1. Describe all events in the rest frame if the event where the center of the box coincides with the center of coordinate system is labeled as (T, 0, 0, 0).
Event O -- The light leaves the center of the cube
$$ O = (T, 0, 0, 0)$$
Event D -- the light hits the detector in the center of a face parallel to the direction of movement.The detector starts a distance of B/2 from the origin of the light and moves with speed v in a direction initially perpendicular to the spatial separation, so from the theorem of Pythagoras:
$$ (c \Delta t_D) = (v \Delta t_D)^2 + (\frac{B}{2})^2$$ or $$\Delta t_D = \frac{B}{2} \times \frac{1}{\sqrt{c^2 - v^2}}$$ so:
$$ D = \left(T + \frac{B}{2} \times \frac{1}{\sqrt{c^2 - v^2}} , \quad \frac{B}{2} \times \frac{v}{\sqrt{c^2 - v^2}}, \quad 0, \quad \frac{B}{2} \right)$$
(Because of the ambiguity of the specification, any of $$\tiny \left(T + \frac{B}{2} \times \frac{1}{\sqrt{c^2 - v^2}} , \; \frac{B}{2} \times \frac{v}{\sqrt{c^2 - v^2}}, \; 0, \; - \frac{B}{2} \right) , \quad \left(T + \frac{B}{2} \times \frac{1}{\sqrt{c^2 - v^2}} , \; \frac{B}{2} \times \frac{v}{\sqrt{c^2 - v^2}}, \; - \frac{B}{2} , \; 0 \right) , \quad \textrm{or} \left(T + \frac{B}{2} \times \frac{1}{\sqrt{c^2 - v^2}} , \; \frac{B}{2} \times \frac{v}{\sqrt{c^2 - v^2}}, \; + \frac{B}{2} , \; 0 \right)$$ are also acceptable.) Event M -- the light bounces off the mirror in the leading direction
Here, as a simple rate problem, the light catches up with the runaway mirror with a head-start of $$\frac{A}{2}$$. So $$\frac{A}{2} + v \Delta t_M = c \Delta t_M$$ or $$\Delta t_M = \frac{\frac{A}{2}}{c-v}$$ so:
$$M= \left(T + \frac{A}{2} \times \frac{1}{c - v} , \quad \frac{A}{2} \times \frac{c}{c - v}, \quad 0, \quad 0 \right)$$
Event R -- the bounced light returns to the center of the moving cubeThis happens a time in the future of B by the amount $$\Delta t_R = \frac{\frac{A}{2}}{c+v}$$ so:
$$R = \left(T + \frac{c A}{c^2 - v^2} , \quad \frac{c A v}{c^2 - v^2}, \quad 0, \quad 0 \right)$$
Motor Daddy's exact example is with T = 0, A = B = 1 light-second, $$v = \frac{\sqrt{69}}{13} c = \sqrt{\frac{69}{169}} c \approx 0.6390 c \approx 191.6 \, \textrm{Mm} / \textrm{s}$$. So:
$$ O = (0, 0, 0, 0)
D = \left( \frac{13}{20} \, \textrm{s} , \quad \frac{\sqrt{69}}{20} \, c \cdot \textrm{s}, \quad 0, \quad \frac{1}{2} \, c \cdot \textrm{s} \right) \approx \left( 0.6500 \, \textrm{s} , \quad 0.4153 \, c \cdot \textrm{s}, \quad 0, \quad 0.5000 \, c \cdot \textrm{s} \right) \approx \left( 194.9 \, \textrm{Mm} / c , \quad 124.5 \, \textrm{Mm}, \quad 0, \quad 149.9 \, \textrm{Mm} \right)
M = \left( \frac{ 13 (13+\sqrt{69}) }{200} \, \textrm{s} , \quad \frac{ 13 (13+\sqrt{69}) }{200} \, c \cdot \textrm{s}, \quad 0, \quad 0 \right) \approx \left( 1.385 \, \textrm{s} , \quad 1.385 \, c \cdot \textrm{s}, \quad 0, \quad 0 \right) \approx \left( \quad 415.2 \, \textrm{Mm} / c , \quad 415.2 \, \textrm{Mm}, \quad 0, \quad 0 \right)
R = \left(\frac{169}{100} \, \textrm{s} , \quad \frac{13 \sqrt{69}}{100} \, c \cdot \textrm{s}, \quad 0, \quad 0 \right) \approx \left( 1.690 \, \textrm{s} , \quad 1.080 \, c \cdot \textrm{s}, \quad 0, \quad 0 \right) \approx \left( 506.6 \, \textrm{Mm}/c , \quad 323.7 \, \textrm{Mm}, \quad 0, \quad 0 \right)
$$
D = \left( \frac{13}{20} \, \textrm{s} , \quad \frac{\sqrt{69}}{20} \, c \cdot \textrm{s}, \quad 0, \quad \frac{1}{2} \, c \cdot \textrm{s} \right) \approx \left( 0.6500 \, \textrm{s} , \quad 0.4153 \, c \cdot \textrm{s}, \quad 0, \quad 0.5000 \, c \cdot \textrm{s} \right) \approx \left( 194.9 \, \textrm{Mm} / c , \quad 124.5 \, \textrm{Mm}, \quad 0, \quad 149.9 \, \textrm{Mm} \right)
M = \left( \frac{ 13 (13+\sqrt{69}) }{200} \, \textrm{s} , \quad \frac{ 13 (13+\sqrt{69}) }{200} \, c \cdot \textrm{s}, \quad 0, \quad 0 \right) \approx \left( 1.385 \, \textrm{s} , \quad 1.385 \, c \cdot \textrm{s}, \quad 0, \quad 0 \right) \approx \left( \quad 415.2 \, \textrm{Mm} / c , \quad 415.2 \, \textrm{Mm}, \quad 0, \quad 0 \right)
R = \left(\frac{169}{100} \, \textrm{s} , \quad \frac{13 \sqrt{69}}{100} \, c \cdot \textrm{s}, \quad 0, \quad 0 \right) \approx \left( 1.690 \, \textrm{s} , \quad 1.080 \, c \cdot \textrm{s}, \quad 0, \quad 0 \right) \approx \left( 506.6 \, \textrm{Mm}/c , \quad 323.7 \, \textrm{Mm}, \quad 0, \quad 0 \right)
$$
But none of these calculations relate to Special Relativity which is the assertion that all inertial coordinate systems are valid descriptions of the same physics and that the speed of light is the same for all directions in all inertial coordinate systems. Motor Daddy also calculates $$v \Delta t_M = \frac{69+13 \sqrt{69}}{200} c \cdot \textrm{s} \approx 0.8849 \, c \cdot \textrm{s} \approx 265.3 \, \textrm{Mm}$$ as the x-coordinate of the center of the box simultaneous with event M and $$c \Delta t_R = \frac{ 13 (13-\sqrt{69}) }{200} \, c \cdot \textrm{s} \approx 0.3051 \, c \cdot \textrm{s} \approx 91.46 \, \textrm{Mm}$$ which is the difference between the position of event M and the position of event R. These calculations don't tie to any events and represent Motor Daddy's endorsement of the concept of absolute time and absolute space and complete rejection of Special Relativity.
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