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Why the velocity $v$ is taken as a value and the definition of velocity not applied on a relativistic equations?


The equations of time dilation and length contractions as we know are
$$L = {L_0}{\left( {1 - {{{v^2}} \over {{c^2}}}} \right)^{{1 \over 2}}}$$ and
$$\Delta t = {{\Delta {t_0}} \over {{{\left( {1 - {{{v^2}} \over {{c^2}}}} \right)}^{{1 \over 2}}}}}$$
Here, ${\Delta {t_0}}$ is time interval between two events, and ${L_0}$ is length in rest frame. An observer moving with velocity $v$ measures the time interval $\Delta t$ and the length $L$.

The velocity $v$ that I am using has some value and for some reason I take it as a value and do not apply the definition
$$v = {{dx} \over {dt}} = \mathop {\lim }\limits_{\Delta t \to 0} {{\Delta x} \over {\Delta t}} = \mathop {\lim }\limits_{t \to {t_0}} {{x - {x_0}} \over {t - {t_0}}}$$
Velocity as I know is "change in displacement over change in time". Let's take length contraction equation, as the length contract there must be some variation for the end points of a rod (say) to move across a point, causing change in $v$. So, why I am not taking some relativistic equation of $v$ here in the equation. I feel, somehow there is some connection between length contraction, time dilation and this velocity. There must be something in this $v$ which is not factored in these equations.

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  • $\begingroup$ Would the fact that $v$ represents the velocity of the moving frame (relative to a frame considered stationary) help see why it's a constant value? $\endgroup$ – Kyle Kanos Jul 5 '19 at 14:22
  • $\begingroup$ @kyle the moving frame will have contraction, and the question remains the same. $\endgroup$ – Roopesh Singh Jul 5 '19 at 16:37
  • $\begingroup$ The frame doesn't have a contraction, but objects within the two frames will have different lengths based on the relative speed of the moving vs stationary frame. $\endgroup$ – Kyle Kanos Jul 5 '19 at 18:10
  • $\begingroup$ Even though a stationary observer will see an object as contracted, the front of that object will pass by him with the same velocity $v$ as the rear of that object. $\endgroup$ – Bill Watts Jul 5 '19 at 22:55
  • $\begingroup$ @KyleKanos Why not? Isn't frame all about space-time? $\endgroup$ – Roopesh Singh Jul 8 '19 at 13:17

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