Recent content by mcastillo356

How does my arithmetic end up with ##1089##?. I don't understand. Maybe missing something of the statement?. Attempt The statement imposes to arrange the digits so they must be always three.

If I have understood, your example ends up with ##999## Counterexample 1- ##495## 2- ##594## 3- ##594-495=99## 4- ##99## 5- ##99+99=198## ##999\neq{198}##

Last step is this way $$mc^2t^{-1/2}\bigg |_{1}^{1-\frac{u_f^2}{c^2}}=mc^2\left (\frac{1}{\sqrt{1-u_f^2/c^2}}-1\right )$$

$$\displaystyle\int_0^{u_f}m\Bigg (1-\cfrac{u^2}{c^2}\Bigg )^{-3/2}\,u\,du=m\displaystyle\int_0^{u_f}\Bigg (1-\cfrac{u^2}{c^2}\Bigg )^{-3/2}\,u\,du$$ $$1-\cfrac{u^2}{c^2}=t\Rightarrow{-\cfrac{2u}{c^2}}\,du=dt\Rightarrow{u\,du=-\cfrac{c^2}{2}\,dt}$$ $$m\Big (-\frac{c^2}{2}\Big...

As in classic mechanics, we will define kinetic energy as the work done by the net force in accelerating a particle from rest to some final velocity ##u_f##. Considering one dimension only, we have...

Understood inmediatly. Thanks Forum.

Let me think it over again. Your reasoning is perfect. I must be wrong somewhere, but still the change of variable in ##R-10## keeps me wondering.

Why in the change of variable from ##p## to ##v## vanishes ##m##. And which is the behavior of ##c##: is this last an invariant of the derivative?. These are my doubts.

Relativistic Momentum, Mass, and Energy Momentum and mass (...), the classic equations for conserving momentum and energy are not adequate for the analysis of high-speed collisions. (...) The momentum of a particle moving with velocity ##v## is given by...

Merry Christmas from Spain.

berkeman, pasmith, Mark44,.. So many

Thanks, @pervect ,@PeterDonis

Really true. But I won't try any self made explanation. Thank you, @PeterDonis.

https://en.wikipedia.org/wiki/Proper_reference_frame_(flat_spacetime)

The source is Wikipedia. Thanks indeed. I'm reading a pdf of the Uned, a Spanish reliable University.. Anyhow, thank you for the advice.