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I have not well understood the picture of geogebra regarding the angle of time (t') that is inclined compared to (t) of 26.57°angle . In the picture we see that the velocity is setted at 0.5c, for which i belivied that if with c = 1 is represented by a 45° inclined line, and now we want draw a speed of 0.5c, all we have to do is divide 45° by 2 and we obtain an inclinatio of 22.5°... Somewere i have read that to have the correct inclinatio we have to do this calculation : new angle = arctan(0.5/1) = 26.57° This is the correct solution but i have not understood why...enter image description here

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Coordinates of the original frame as functions of coordinates of moving frame are given by inverse lorentz transform: $$ ct=\gamma\left(ct'+\beta x'\right) $$ $$ x=\gamma\left(x'+\beta ct'\right) $$ Now imagine point on the $ct´$ axis, lets say $(ct´,x´)=(1,0)$. This point will have coordinates in the original frame $(ct,x)=(\gamma,\gamma\beta)$ from which the slope is given by: $$ \tan\alpha=\frac{\gamma\beta}{\gamma}=\beta $$

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