mcastillo356's latest activity

Thanks to everyone who voted for me. I must confess I feel more like a leg-end than a legend, but there you go.

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

My special thanks to Fresh42, Mark44, Bvu, Perok, SammyS, Haruspex, Orodruin, Chestermiller, WWGD, Martinbn, and the other person from...

$$\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...

An integral is an antiderivative. The only thing you need to check is that: $$mc^2\frac d {du} \bigg...

You can also derive ##d \gamma = \gamma^3 (\vec \beta \cdot d \vec \beta)## fairly easily and use that to do a change of variables...

Good afternoon, This is not Paul, but rather his son Robert. I am very sad to write here that my dad passed away on Saturday...

I think this is the integral you are attempting to compute. If so, I'd do it by inspection. For any integrand with a ##1/f^n(x)## it's...

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...

It's the end of the year and once again I wish to thank the mentors and admins for their stedfast commitment to the PF cause and for...

Note that you can avoid the differentiation, by expressing ##v## as a function of ##p##: $$p = \gamma mv \implies v = \frac p {\sqrt{m^2...

Understood inmediatly. Thanks Forum.

If you take the derivative of that, m just "gets out" because it is constant: $$dp =m \, \, d...

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...

But the mass does not dissapear, you have it right there in your expression for K. c is the speed of light, it is a constant.