High School Needed tips to understand Relativistic Energy in Especial Relativity

[mcastillo356]

TL;DR

I don't understand how Tipler/Mosca famous book accomplishes Relativistic Energy

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

$$K=\displaystyle\int_{u=0}^{u=u_f}F_{net}\,ds=\displaystyle\int_{u=0}^{u-u_f}\cfrac{dp}{dt}\,ds=\displaystyle\int_{u=0}^{u-u_f}u\,dp=\displaystyle\int_{u=0}^{u-u_f}u\,d\Bigg (\cfrac{mu}{\sqrt{1-(u^2/c^2)}}\Bigg )\qquad{39-21}$$

where we have used $u=ds/dt$. It is left as a problem for you to show that

$$d\Bigg (\cfrac{mu}{\sqrt{1-(u^2/c^2)}}\Bigg )=m\Bigg (1-\cfrac{u^2}{c^2}\Bigg )^{-3/2}\,u\,du$$

(I needed a quick browsing to understand it)

$$K=\displaystyle\int_{u=0}^{u=u_f}u\,d\Bigg (\cfrac{mu}{\sqrt{1-(u^2/c^2)}}\Bigg )=\displaystyle\int_0^{u_f}m\Bigg (1-\cfrac{u^2}{c^2}\Bigg )^{-3/2}\,u\,du$$

$$=mc^2\Bigg (\cfrac{1}{\sqrt{1-(u^2/c^2)}}-1\Bigg )$$

This last step is what I don't understand.

Attempt

$$(1-u^2/c^2)=t\Rightarrow{dt=-2u/c^2}$$

But it doesn't work: I can't get rid of $u$. I neither know how to deal with $c$

$$K=\cfrac{mc^2}{\sqrt{1-(u^2/c^2)}}-mc^2\qquad{39-22}$$

REST ENERGY

$$E_0=mc^2\qquad{39-23}$$

RELATIVISTIC ENERGY

$$E=K+mc^2=\cfrac{mc^2}{\sqrt{1-(u^2/c^2)}}\qquad{39-24}$$

 

[Ibix]

$$\displaystyle\int_0^{u_f}m\Bigg (1-\cfrac{u^2}{c^2}\Bigg )^{-3/2}\,u\,du$$

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 worth asking if $1/f^{n+1}(x)$ might be the solution and seeing if the rest of the integrand turns out to be $f'(x)$.