Velocity and Acceleration of a Classical SHO

Posted by doug
summary: 
The first term of the 4-acceleration is frozen at 0, an observations whose implications I do not understand.
description: 
The oscillator is in yellow, its first time derivative in blue, and second time derivative in red. I understand why the velocity has a fixed scalar value equal to 1, that is what gamma is for low velocities. Acceleration in special relativity must be handled with care. The math is easy: tack the derivative of a constant, and zero will result. The implications of that math are not clear to me.
command: 
q_graph -out sho-x_and_v_and_a -dir sho -box 10 -command 't_function -mqs 2 -mqs 1 -t_function polynomial -t_function one -x_function cos -x_function polynomial -y_function zero -y_function zero -z_function zero -z_function zero 1 2 1 2 1 3 0 0 0 1 0 0 -shift "-11 5 0 0" -pi 8 -n_steps 1600' -color yellow -command 't_function -mqs 2 -mqs 1 -t_function one -t_function one -x_function sin -x_function polynomial -y_function zero -y_function zero -z_function zero -z_function zero -shift "0 5 0 0" -pi 8 -n_steps 1600 -- 1 2 1 2 1 3 0 0 0 -2 0 0' -color blue -command 't_function -mqs 2 -mqs 1 -t_function zero -t_function zero -x_function cos -x_function polynomial -y_function zero -y_function zero -z_function zero -z_function zero -shift "0 5 0 0" -pi 8 -n_steps 1600 -- 1 2 1 2 1 3 0 0 0 -4 0 0' -color red
equation: 

(t, x, y, z) = (t, cos(2 t + 3), 0, 0)
(\gamma, \gamma \beta_x, \gamma \beta_y, \gamma \beta_z) = (1, -2 sin(2 t + 3, 0, 0)
(a_t, a_x, a_y, a_z) = (0, -4 cos(2 t + 3, 0, 0)



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