Endy0816 Posted February 17, 2016 Posted February 17, 2016 Just curious what the time dilation and length contraction equations would look like in a system based on: v = c - vNormal for finding velocity. I was thinking along the lines of how we convert to Kelvins. Wondering how things would look if we did something similar with velocity.
pzkpfw Posted February 17, 2016 Posted February 17, 2016 But what's VNormal?That implies there's some absolute velocity that other velocities can be compared to. There is no such thing.
timo Posted February 17, 2016 Posted February 17, 2016 (edited) [math] \gamma = \left( 1 - \frac{v_{\rm normal}^2}{c^2} \right)^{-1/2} = \left( 1 - \frac{(c - v_{\rm new})^2}{c^2} \right)^{-1/2} = \left( 1 - \frac{c^2 -2 c v_{\rm new} + v_{\rm new}^2}{c^2} \right)^{-1/2} = \left( 2 \frac{v_{\rm new}}{c} - \frac{ v_{\rm new}^2}{c^2} \right)^{-1/2}[/math] There is no real reason to do the conversion akin to Celsius->Kelvin, since the zero of our "normal" velocity measurement already correspond to "absolute zero". A more common conversion is to set the maximum to 1 (100%), i.e. [math]v_{\rm new} = v_{\rm normal}/c[/math]. Then, [math] \gamma = \left( 1 - v_{\rm new}^2 \right)^{-1/2}[/math], which in fact makes the time dilatation factor look simpler. Edited February 17, 2016 by timo 1
Endy0816 Posted February 17, 2016 Author Posted February 17, 2016 (edited) I meant normal only in how we normally measure. Start at zero and count up. This would reverse things. Start you at the value of c and decrease as you went faster. Think that is what I am looking for Timo. I'll take a look at it. Many thanks. Edited February 17, 2016 by Endy0816
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