where clock \(C\) is exactly between clock \(A\) and Clock \(B\) and at noon both clocks flash in the directions of \(C\). If \(A\) and \(B\) are synchronized, then \(C\) receives both flashes of light simultaneously. A forth observer travelling to the right with velocity \(v\), will not receive the pulse simultaneously and the argument leads to Lorentz transform.
If our focus is on the clock being EXACTLY synchronized, then the directions of the light flashes are SPECIFICALLY in the \(C\) directions. \(X\) in the first place, does not first receive a pulse from clock \(B\) and then a pulse from clock \(A\). \(X\), while traveling misses both pulses aimed at \(C\). \(X\) travelling at velocity \(v\), do not see the light AT ALL. Neither does the position \(x\), where the moving clock \(X\) is instantaneously at, in space, while traveling at velocity, \(v\), at noon, receive any light pulse from \(A\) and \(B\).
If \(X\) happens to be at \(C\) at noon, it will not receive any pulses as it passes \(C\) because the pulses arrive after \(X\) has passed by. If \(X\) happens to be at \(C\), a little after noon, when the pulses arrive, \(X\) will receive both pulses. It cannot receive one pulse after another and conclude that the clocks \(A\) and \(B\) are not synchronized. \(X\) receives both pulses or none, and can only affirms that the clocks are synchronized and but cannot deduce that the clocks are not synchronized.
If X does receive a pulse, it is one aimed specifically at it and is necessarily of a different path length; ray \(a\ne b\), unless \(X\) is at \(C\) when the pulses arrive.
If we are to specify time exactly then light as a vector, its direction must also be specified exactly.
Otherwise, we rely on the conservation of energy, that in all frames of reference, the total energy contained within those frames, in the system, is the same.
where a frame with a moving origin at velocity, \(v\) and an \(x\) axis moving at one unit per unit time with respect to the origin is transformed back to a frame with a stationary origin and an \(x\) axis moving at one unit per unit time. Yes, the \(x\) axis is moving at one unit per unit time with respect to the origin. And the origin, a point, is orthogonal to all axes. The origin of the moving frame moving at \(v\),
\(x^{'}=x-vt\)
at the same time, one unit along \(x^{'}\) is actually longer than one unit along \(x\) in the rest frame,
\(x^{'}cos(\theta)=x^{'}.\cfrac{\sqrt{1-v^2}}{1}=x\)
So, in the rest frame units along \(x\),
\(x^{'}=\cfrac{x}{\sqrt{1-v^2}}\)
thus,
\(x^{'}=\cfrac{x-vt}{\sqrt{1-v^2}}\)
we move the origin in time by \(vt\) and scaled the units along \(x\) axis by \(\cfrac{1}{\sqrt{1-v^2}}\) to obtain the equivalent of a moving frame with respect to the rest frame.
And the rest of Lorentz transform between rest frame and moving frames follows. It is conservation of energy because we are dealing with velocity squared terms.
